1st Year Chemistry (Nasrat Katozai Nawaz Usb)

CONTENTS Chapter-1:

Stoichiometry

1

Chapter-2:

Atomic Structure

18

Chapter-3:

Theories of Covalent Bond and Shape of Molecules

36

Chapter-4:

Gases

50

Chapter-5:

States of Matter Liquid

65

Chapter-6:

States of matter III

76

Chapter-7: Chapter-7:

Chemical Equilibrium

83

Chapter-8:

Acids, Basis and Salts

98

Chapter-9:

Chemical Kinetics

108

Chapter-10: Chapter-10: Solution and Collides

120

Chapter-11: Thermochemistry Thermochemist ry

142

Chapter-12: Electrochemistry

158

Nasrat Ullah Katozai (Chemistry)

CHAPTER-1:

Master Coaching Academy

STOICHIOMETRY

i. Stoichiometry is the combination of two Greek words. ii. Stoikhein means elements iii. Metron means measurement. So, The measurement of elements from balance chemical equation is called stoichiometry. Definition:

That branch of chemistry in which we study the quantitative relationship between reactants and products in a balance chemical equation is called stoichiometry. Explanation:

It deals with the calculations involved in the interconversion of matter during any physical ph ysical or chemical change. It deals that how much matter is required for a specific amount of another matter. If the amount of reactants are known then th en the amount of products can be determined from stoichiometry. For example:

Consider the formation of water from oxygen and Hydrogen. Unbalance reaction:

⟶

H2 + O2 H2O In order to determine d etermine that the above equation is stoichiometric or not consider the following table. Conclusion: This is balanced because Reactants and products are equal.

Element

Reactant

Products

So this reaction is not balance because reactants and products are not equal.

H

2

2

Note:

O

2

1

It is not stoichiometry because it does not obey law of conservation of mass.

Total

4

3

Reactants

Products

Balance reaction:

Now consider the balance balance chemical reaction. reaction. 2H2 + O2 2H2O

H

4

4

Conclusion:

O

2

2

Total

6

6

⟶

Since it is stoichiometric equation because it obey law of conservation o f mass.

Elements

Conditions for stoichiometry:

There are two conditions; i. Number of atoms must be balanced. balanced. ii. Number of charges must must be balanced. Stoichiometric amount:

The amount of reactants & products Obtained from balance chemical equation. The amount of reactants and products can be expressed in four terms; i. Mole relation ii. Mass relation iii. Volume relation iv. Particle relation Relative atomic mass:

1. 2. 3. 4. 5.

It is also called unified atomic mass unit. It is statistical term It is dimensionless physical quantity It has no unit. It symbol is Ar.

Definition:

The mass of one atom of any element compared with the mass of one twelfth of one atom of C 12 is called relative atomic mass. Explanation:

1.

One gram atom of any element is the relative atomic mass of that element expressed in gram.

Example:

i. ii.

The relative atomic mass of hydrogen is 1 amu.

i.

The unit used to express relative atomic mass is called atomic mass unit (amu).

The relative atomic mass of “S” is 32 amu.

Note:

1



The term relative is used because atoms are very small i .e. in the range of o f 10 -24 g to 10-22 g. And there is not a single balance in the u niverse which can measure such small tiny particles. Therefore they are measured by means of comparison with some standard which is C 12.

ii. iii. iv.

Measuring Standard

The measuring standard with which tiny sensitive particles are compared is called C 12.Carbon C12is taken as a standard because of the following reasons. 1) 2) 3) 4)

Stability Can enter into many chemical reactions. It’s isotopes are stable.

Easily available.

Note: it atomic mass is not infraction.

There is not a single carbon atom in the universe which has atomic mass 12.01 i.e. in friction but 12.01 is average atomic mass. Mathematically:

Ar = (Atomic mass of first isotopes isoto pes x relative abundance) + (Atomic mass of second isotopes relative abundance) 100 Calculate the relative atomic mass of “ Chlorine” when Cl35is 75% and Cl37 is 25% Cl35 = 75%

Q-52:

1 amu is equall to 1.66×10 -24 g then 1.0g will be equal to……….?

Cl37 = 25%

As we know 1.66×10-24g = 1 amu.

   +   

1 g = χ.

Ar=

By crossmultiplication χ x 1.66×10-24 g = 1g× 1amu

 .×  × 10 χ= .   0.602×10

Atomic Mass Unit:

`

χ=

The unit used to express relative atomic as is called atomic mass unit.

+24

24

Representation:

χ = 6.02×10

It is represented by amu.

23

χ = 6.02×1023 amu

Definition:

“It is

112

part of the mass of one atom atom of C-12”

Mathematically:

Note:

1amu = 1.66x10 -24 g OR 1 amu = 1.66×10 -27kg.

I amu = mass of one atom of C12 12 Mole:

1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16)

Similarly

The term mole is a Latin word. The term mole is derived from “molus” Molus means “pile” or heaps.

The term mole was used by Wilhelm Ostwald in 1900 for the first time. The symbol of mole is “mol”.

Mole is a quantity as well as number. Mole is the SI unit of amount of substances. It is the quantity of substances in gram which contain 6.02×1023 particles. MOLE DAY: 23 October is celebrated as mole day every year since 190 0, because at that day mole was discovered. 1 mole = 6.02×1023 Actually 6.02 is time because mole day start from 6:02 am and end at 6:02 pm. 10: it is the month of October. 23: date. This style is American. Mole is also called Gram atom. Note: i. One mole contain 6.02×10 23 particles irrespective of Mole is a counting unit just like the size and shape. 1 dozen = 12 similar things ii. If a substance is small or large number of particle 1 century = 100 similar things will be same. 1 gross = 144 similar things 1 ream = 500 similar things 1 mol = 6.02 × 10 23 similar particles

Definition of Mole:

2



Atomic mass, molecular mass, formula mass or ionic mass expressed in gram is called mole. OR It is the amount of substances which contain as many elementary entities as in 12.0 g or 0.012kg of carbon 12. Representation:

It is represented by “x” Mathematically:

It can be written as: n = = in gram and M = molecular mass of substance.

      

For Example:

1) 2) 3)

Atomic mass of sodium is 23 amu so 23 g of sodium is equal to one mole Molecular mass of H2O is 18 amu so 18g of H 2O is equal to one mole. Formula mass of NaCl is 58.5 amu. So 58.5g of NaCl is equal to one mol.

Avogadro’s Number

“One mole of any substance contain a definite n umber of particles, i.e. (atoms, ions, molecules or formula units) which is equal to 6.02×10 at STP this definite number is called Avogadro’s Number” 23

Representation:

It is represented by N A.

Example- 1.1

How many moles are there in 60g of sodium hydroxide (NaOH)?

Explanation:

1. 2.

One mole of any substance contains 6.02×10 23 similar particles. Avogadro’s number was discovered by Italian scientist. Amedoe Avogadro.

Mathematically:

Solution:

It can be solved in the following steps Step-1

It can be written as; First form n = Number of particles NA Second form= If number of particles are to be find then we will use the following: Number of particles = n × NA× No of atoms in formula Third form =

Given data Mass in gram=|m|=60g Required Data: Mol=|n|=? Solution:

As we know: that the molar mass of NaOH can be determined as; as; Na+O+H Na=23 =23+16+1 O=16 H=1 =40g/mol

If we determine the mass of one particle, then we will use the following formula: Mass of one particle = Molar mass NA For example: Formula: 23

Now using the formula formula n =m/M n= = 1.5 mol

1 mole of Na = 23 g of Na = 6.02×10 atoms of Na

 /

1 mole of O = 16 g of O = 6.02×1023 atoms of oxygen. 1 mole of H2O = 18g of H2O = 6.02×1023 molecules of H2O

Conclusion: Since 60g of NaOH contain 1.5 mol of

NaOH.

1 mole of NaCl = 58.5 g of NaCl = 6.02×10 23 formula units of NaCl. 1 mole of AgNO3 = 170g of AgNO3 = 6.02×1023 of formula units of AgNO3.

Example:

Mole Calculations:

Solution:

Mole can be calculated in the following steps

What is the mass of 0.5 moles of calcium carbonate (CaCO 3)? It can be solved in the following steps. 1.

Give Data:

2.

Required Data:

No of moles=n=0.5mol moles=n=0.5mol

Step-1

In this step, determine the molar mass of a substance.

Mass in gram=m=? Solution:

Step-2.

Now write down the formula formula of the substance. For Example:

Sulphuric acid H2SO4 Step-3.

Now write positive sign between between each elements elements i.e H2 + S + O 4 Step-4.

Now place multiplication between between elements and and digits (H×2) + S+ (O×4)

In the first step we will determined the molar mass of CaCO 3. Ca+C+O3 as ca=40 40+12+(16x3) c = 12 40+12+48 o = 16 100g/mol Formula:

As we know n=m/M nxM=m Putting values of (n) and “M”

m=nxM m=0.5x100 mass=50g Conclusion:

Since 50g mass is present in 0.5 mol of CaCO 3.

3



Step-5.

Now write the atomic atomic mass of each element. element. (1×2) + 32 + (16×4) 2 + 32 + 64 M = 98amu Molar mass = 98 g/mol

H=1 S = 32 O = 16

Step-6

Now write the mol formula. n = mass/molar mass Example 1.3:

Note:

The above equation is used for the calculation of mass and mole.

In a certain experiment 8.50 x 10 25 molecules of water were used, calculate the number of moles of water. Given Data:

The mole and chemical equations:

No of H2O molecule = 8.50 x 10 25

Chemical Equation:

Required:

The shorthand representation of a chemical change in term of symbol of elements and formula of compounds of a substance sub stance in a chemical reaction is called chemical equation.

Solution:

n=? As we know that number of moles= Numer f H

Note:

Any chemical change is called chemical reaction.

Conclusion:

Reactant:

i. ii. iii. iv.

Those substances which react with each other are called reactants. They are also called reagent. They are present before a chemical reaction occur. They are present at the left hand side. For example: 2H2 + O2 2H2O Reactants are H 2 and O2

→

Products

i. ii. iii.

Since n= 1.41 x 10 2mol 8.50 x 1025 molecules of H 2O contain 1.41 x 102mol of H2O. Example 1.4:

How many formula unit are present in 125g of hydrated copper Sulphate(CuSo 4.5H2O)? Given data:

m=125g Required Data:

Those substances which are newly formed after a chemical reaction are called products. They are present at the right hand h and side. They are formed after a chemical reaction.

Representation: H2O is a product in above reaction.

The chemical reaction is represented by an arrow directing from reactants towards products. (i)

Formula units? Solution:

Now calculating molar mass of CuCo4.5H2O. M = Cu+S+(Ox4)+5x (H 2O) M = Cu+S+(16x4)+(5xH Cu+S+(16x4)+(5xH2)+ (5xO)

irreversible

Reactants

M = 64+32+64+(5x2)+ 64+32+64+(5x2)+ (5x16) M = 64+32+64+10+80 64+32+64+10+80 M = 250g/mol

products (ii)

reversible

Reactants

Formula:

products

As we have that n = m/M

For example:

i. ii.

n=

When magnesium react with oxygen it form, white magnesium oxide. It can be represented in the form of o f equation as

⟶

2Mg + O2

2MgO

Conclusion:

Since 125g of hydrated Copper Sulphate contain 0.5mol. Now formula units can be derived as.

Iron react with Sulphur if form iron oxide. Note:

 /

n = 0.5mol

Similarly when:

Fe + S



 . x   n= .    = 1.41 x 102mol

numer  .   umer   n= .   

⟶

n=

FeS

Rearranging

Balance chemical chemical equation is used in stoichiometry. Reason:

1023

Number of formula units= n x 6.02 x

Putting Values

A balance chemical equation is used in stoichiometry because it o bey. i. ii.

Law of Conservation of mass. Law of definite proportion.

Note:

Number of formula formula units=0.5 x 6.02 x 1023 Number of formula units = 3.011 x 1023 Conclusion:

Total mass of reactants and products are equal.

Since there are 3.011 x 10 23 formula units of CuSO4.5H2O in 125g of the salt.

Example:

i.

Practically, During experiment we cannot use 56g of Fe because 56g is much more greater quantity, therefore, we simplify it; 56:32 as 7:4

4


ii.


But 7g of Fe and 4g of Sulphur is also greater and in laboratory work we cannot use such a large quantity so we will take 0.1 mol of Fe & S: So Fe + S FeS 0.1 mol 0.1 mol 0.1 mol n= 0.1 x 88 nxM=m nxM=m 8.8g 0.1 x 56 = 5.6 0.1 x 32 = m m=5.6g m = 3.2g m = 8.8g 5.6g +3.2g=8.8g

⟶

 

Conclusion:

Now if 5g of (S) react react with 5.6g of Fe then only 3.2g of Sulphur with react 5.6g of Fe and the excess excess will remain remain unreacted 5 – 3.2 3.2 = 1.8g of S Stoichiometric Calculation

There are four types i. ii. iii. iv. 1.

Mole mole calculation. Mass mass calculation. Mole mass calculation. Calculations involving gases (mole volume calculation) Mole mole calculation: In this type calculation the number of mole of one substance is given and other should be calculated for example 10 mol of H2O required ……mol of O 2.

→

2H2 + O2 2H2O 1mol of O2 gives



2mol of H2O

10mol of H2O

By cross multiplication



10 × 1 = × 2

= ×  = m m× m = m 

2.



= 5 mol

Mass mass calculation: In this type of calculation the mass of one substance is given & mass of other substance should be

determined? Steps: It contains the following steps. i. Convert mass into mole. ii. Mole ratio is determination. iii. Convert mole again into mass. For example:

5g of H2 can produce…..g of H 2O 1. Convert mass into mole

n=

 

 n

=

n= 2.5 mol 2.

 

→

2H2 + O2 2H2O

2 mol

2 mol

(by cross multiplication) 3.



× 2 = 2×2.5 = x = 2.5g

Converting mole in mass.

m = n×M = 2.5×18 = 50.4g c. Mass mole calculation: The mass should be given & moles should be determined. For example:

2g of O2 can form ….. mole of H2O. First convert mass into mol. n =

→ 

2H2O + O2 1 mol 0.06 mol

2H2O

   = 0.06 mol  

2 mol

5



×12×0.06

x = 0.12mol

mole volume calculation: In this calculation volume is given & mole should be determined what volume occupied by 0.4 mole of

d.

CO2at STP? So

 .  

 while molar molar    volume = 22.4 As n =

 × 22.4 0.4 × 22.22.4 8.961 

V=



V=

So V= 

Percentage composition:

The mass of an element present in 100gm of a compound is called percentage composition. OR The percentage by mass of an element in a compound is called percentage composition. Mathematically:

It can be written as. First form:

          x 100 Second form:      umer    x 100 Percentage of an element =    Third form:       ×Numer   x 100 Percentage of element =     ×    Percentage of an element =

Unit:

It has no unit. Determination of Percentage Composition:

Percentage composition can be determined in the following steps: Step-1:

In this step we will identify atoms or elements in a compound. Whose percentage composition composition is to be determined. Step-2:

Now determine the molar molar mass of the compound.

MCQS (ETEA)

Pretest:

chapter-1 Page-40 MCQS-21 Which one of the following is the percentage of Ca in CaCO3? C2=40, C=12 and O=16 (a)10

(b) 20

(c) 40

(d) 80

Step-3:

Now write the mass of the element above and the total mass of the compound compoun d below and multiplying it by 100. %= 

  ×100

Solution:

It can be solved in the following steps. Step-1:

ii.

What is the percentage composition of each element in a. b.

For example:

Calculate the percentage composition of each element in sulphuric acid. [H2SO4]

i.

Example 1.12:

In this step we will write the formula of the compound. ie. H2SO4 Identifying elements and its total mass. H2 = 2g S = 32g O = 64g

Benzene C6H6 Glucose C6H12O6

Solution:

a.

Writing formula of Benzene C6H6 Molar mass C6+H6 (6x12) + (1x6) 72+6

M = 78g/mol a. Percentage determination: % of C =

% of C = 92.31% % of H =

Step-2

In this step calculating molar mass.

 x 100 

 

x 100

% of H = 7.69%

Molar mass of H2SO4= 98g/mol M = 98g/mol Step-3:

Now calculating calculating the percentage of each each element by using formula. 1.

Percentage of H =

   ×          x 100

b.

Molar mass of Glucose: C 6H12O6 Identify elements C = 6 x 12 = 72g H = 1 x 12 = 12g O = 16 x 6 = 96g

Molar mass:

M = C6+H12+O6 = (Cx6)+(Hx12)+(Ox6) (Cx6)+(Hx12)+(Ox6)

6



×   ×100  % of H =  x 100 % of H = 2.04%    "" Percentage of “S” =       x 100  % of S =  x 100 % of H=

2.

% of S = 32.653%

3.

% of 0 = 100 – (% (% of H + % of S) = 100 – (2.04 (2.04 + 32.6) = 65.3%

Note:

 

Sum of the individual percentages of all the elements must be equal to 100. Percentage composition is quantitative in nature.

Limiting reagent

Excess reagent

Definition:Those reactants which consume earlier in a chemical

Definition: Those substances which does not consume earlier is

reaction is called limiting reagent.

called excess reagent.

Stoichiometric amount: They are less than stoichiometric

Stoichiometric amount: They are greater than stoichiometric

amount.

amount.

reaction. Consumption: They consume first in a chemical reaction.

Consumption: They does not consume first in a chemical.

Controlling of reaction: when the limiting reagent finishes it

Controlling: It is present in larger quantity so it not finish earlier, so

control the reaction and hence the reactions stop.

it cannot controlled the chemical reaction.

Quantity of product: When limiting reactant is used up, no more

Quantity of product: It can not controlled the quantity of the

product can be formed, since since it controlled the quantity of product.

products.

Note: Excess reagent: Those reactants which is present in greater quantity than required is called excess reagent. Note:

In a balance chemical equation none of them will be the limiting reagent. STEPS FOR IDENTIFICATION OF LIMITING REAGENT:

Following steps are involved in the identification of limiting reagent. a. Convert mass into mole: Convert mass of reactants into mole. Less mole indicate liming reagent. b. balance chemical equation: Using balance chemical equation for the calculation of required product. c. Least amount of product: Those reactants which gives least number of mole of the required product will be called Limiting reagent. 1. Comprision: Now compare the moles of reactants with the help of balanced chemical equation. This also give information about a bout limiting

reagent. Example 1.13:

Carbonic acid H 2CO3 can be made according to the reaction. Reaction:

CO2 + H2O

→

H2 CO3

If 120 g of CO 2 is d issolved In 80g of water then a) b) c)

Identify the limiting reagent Calculate the maximum amount product Excess reagent?

a. Limiting Reagent:

First we will convert mass of both reactants into moles.

  n= 

1.

Moles of CO2 = n=

2.

n= 2.727 mol Mass of H 2O = 80g Molar mass =

18 

Moles of H2O = n = 4.44 moles 3. Moles of H2CO3: for H2O: Now calculate calculate the number of moles moles of H 2CO3 from balance chemical equation. H2O + CO2 H2CO3. 1 mol of H2O 1 mol of carbonic acid 4.44 mol By cross multiplication



→ 

b. Maximum Amount of Product (In Gram):

As we know that the amount of product is controlled by limiting reagent .  Therefore, the maximum amount of product is formed is 2.727 mol. Molar mass of H 2CO3 = H2CO3 H2+CO3 (H×2) + C+(0×3) (1×2) + C+(16×3) 2+c+48 2+12+48 M = 628/mol. ?? x = 2.727 Mol So, for mass in grams n = M × n = m m = 169.07 g

 

c. Excess Reagent: Reagent:

From balance chemical of equation CO2 + H2O H2CO3 2.727 2.727. 1 mol = 1 mol So excess reagent : 4.44 _ 2.727 = 1.717 mol. But they require mass so : m = x×molar mass of H 2O. m = 1.717×18 m = 31g

→ → →

7

Nasrat Ullah Katozai (Chemistry) x × 1 = 1 × 4.44 x = 4.44 ofH2CO3 mol of of H2CO3 = 4.44 mol

Now moles of H2CO3 for CO2:


→

(1)

Again writing stoichiometric equation. H2O + CO2 H2CO3 1 mol = 1 mol 3.727 = x By cross multiplication x×1 = 2.727×1 x= 2.728 mole of H2CO3 2

→

→

Conclusion:

Since 120g of CO 2 can produce less amount of products ie (H 2CO3) which is 2.727 mole so it is limiting reagent and H 2O is excess reagent.

YIELD:

The amount of products obtained from a balance chemical equation is called yield. Types of yield:

There are three types of yield. 1. Theritical yield 2. Actual yield. 3. Percent yield. 1)

Absolute Yield:

When the amount of product are expressed in gram is called molar yield. Molar yield: When products are expressed in term of mol.

Theoretical yield:

1. 2. 3. 4. 5.

It is also called imaginary yield. It is also called maximum yield. It is also called ideal yield. It is also called calculated yield. It is also called Hypothetical yield.

Explanation:

i. ii. iii. iv. v.

It has no limiting reagent. It has no side r  n It gives 100% products. It obey law of conservation of mass. No loss of product

Example:

If 2mol of H2 react with one mole of oxygen, then it will give ------ mol of pro ducts So we will calculate from balance chemical equation. 2H2 + O2 2H2O 2mol 2 mol. 2 mol =

→ →   ×   ×  

= x = (2) mol

Conclusion:

Since 2mol of H2 gives 2mol of H2O, with no lose is theoretical yield. 2)

Actual yield:

1. 2. 3. 4.

It is also called practical yield. It is also called experimental yield. It is also called non ideal yield. It is called minimum yield.

Note:

Actual yield is always less then theoretical yield. Reason:

Actual yield is always less than theatrical yield, because of the following reasons. 1. Unwanted reaction 2. Unsuitable condition 3. Reversibility 4. Mechanical loss 5. Random error 6. Personal error 7. Systematic error. 1.

Unwanted reaction:

Those reactions which occur without the intervention of th e with of the chemist. Types of unwanted reaction:

i. ii. iii.

Side reaction Parallel reaction Chain reaction

Percent yield:

8



The ratio between actual yield and theoretical yield is called percent yield. Mathematically:

It can be written as: % yield =

    × 100

Explanation:

1. Percent yield ranges from 1-100% 2. Chemist always try to increase percent yield. 3. If percent yield is high, then it means that “more product has been formed” 4. It percent yield is 90% then the reaction is considered as excellent. 5. No unit. Note: The efficiency of a chemical reaction is expressed in term of percent yield. 1. If percent yield is 100% reaction is quantitative 2. If percent yield is =90%reaction will be excellent. 3. If percent yield is = 80% very good 4. If percent yield is = 70% good 5. If percent yield is =50% fair 6. If percent yield is = 40% poor.

Note: If the amount of ideal yield become

equal to theoretical yield then the percent yield is equal to 100%.

Example:

Heating 24.8g of copper carbonate in a crucible produced only 13.9g of copper oxide. What is the percentage yield of copper oxide? The reaction is;

CuC → Cu  

Solution:

+

C 

As we know that the actual yield is 13.9g Step.1:

The theoretical yield is calculated from balance chemical equation.

→

CuCO3 CuO + CO2 1mol

1mol 1mol

64 + 12+48

64+16

124g

80g

124g of (CuCO3) = 80g 24.8g

= x

By cross multiplication x×124 = 24.8×80

 = . × 80  

Now:

x= 16g theoretical yield.

Percent yield = Putting values

% yield =

    × 100

. × 100 .

Percent yield 86.87% Exercise: 1)

The branch of chemistry which deals with the calculations based on balanced chemical equations is called ___________?

a) Environment chemistry b) Physical chemistry c) Stoichiometry d) Industrial chemistry 2.

The mass of an atom (element) compared with the mass on atom of c-12 is called __________ of that element.

1)

a) One mole b) Gram atomic mass c) Atomic number d) Relative atomic mass Which of the following is not true for a mole? 9


2) 3)

4)

5) 6) 7)

8)

9) 10)


a) It is a counting unit b) It is the gram atomic o r gram molecular mass of a substance. c) It contains 6.023x 1023 particles d) It contains different number of particles of different substances What is the mass (g) of 5 moles of H2O (water)? a) 90g b) 36g c) 18g d) 100g The number of molecules in 22g of CO2 is _________ ? a) 6.023x1023 b) 3.011 x1023 c) 6.023x1021 d) 6.023 x1022 Which of the following conditions of temperature and pressure are the standard conditions (STP)? a) 0 C, 1atm b) 273k, 1atm c) 273k, 760mmtHg i) A only ii) b only iii) a & b only iv) all of these The molar volume of SO2 gasat STP is _______? a) 64dm3 b) 24dm3 c) 22.4dm3 d) 100cm3 Percentage of calcium in CaCO 3 is ___________? a) 12% b) 100% c) 48% d) 40% Given the equation : CO2(g) 2CO(g) Which of the following equivalences is not correct for this reaction a) 1 mot CO2 2mol CO b) 44 g CO2 56 g CO c) 44 g CO2 28 g CO d) 44 g CO2 12 g CO Theoretical yield is always less than actual yield because: a) Some product is lost in the experiment b) Reversible reaction may occur c) Error are made in weighing the reactants or the products d) The given statement is not correct Actual yield will reach the ideal (theoretical) value if the % yield of the reaction is _______? _______ ? a) 50% b) 90% c) 100% d) 10% The largest number of molecules are present in _______? a) 44g of CO2 b) 98 g of H2SO4 c) 36 g of H2O d) 180 g of C6H12O6

°

→

≅≅≅ ≅

i

c

ii

d

iii

d

iv

a

v

b

vi

a

vii

c

viii

d

ix

c

x

d

xi

c

xii

c

Answer the Following Questions Briefly:

1)

What is gram atom? Why the concept of gram atom is useful in chemistry?

1.

Gram Atom:

“Atomic mass of an element expressed in grams is called its gram atom” 2.

For example;

3.

1 gram atom of C = 12g 1 gram atom of Na = 23g Note: It is also called gram atomic mass. 1 gram atom of Cl = 35.5g 1 gram atom of Ca = 40g

4.

Uses:

Gram atom is generally used for the th e atomic form of an element. For example 1 g atom of O = 16g 1 g atom of Cl = 35.5g Gram atom atom can also be defined as, as, the mass of Avogadro number of atoms of that element element in grams grams

5.

Note:

6.

Concept of Gram atom in chemistry:

The concept of gram atom is used u sed in chemistry because Atoms are very small particles and can not be measured even by using a high sensitive balance. But, the mass gram atom if (6.02 (6.0 2 x 1023atom) can be measured easily. Take a scientific digital balance and measure 23g of sodium metal by means of it.23 g of Na= 1 gram atom of Na. 2.

The mass of 5 moles of an element x in 60 g. calculate the molar mass of this element. Name the element?

Formula:

10


Number of moles = Give Data:


  

n= 5 moles m= 60g Required = M = ? Solution:

  M =  = 12g M=

Molar mass of x = 12

Conclusion: It shows that the element is carbon having a relative atomic mass of 12.

3.

Explain why balance chemical equations are used in stoichiometric problem?

1.

Mole Relation:We use balance chemical equation because it gives the following information

In this type of relation the amount of product and reactants can be expressed in term of mole. For examples:

Consider the following reaction. 2H2 + O2 2H2O 2 mol 1 mol 2 mol Note:

In the above equation 2 mole H2 react with one mol of oxygen to from 2 mol of H2O.

Note:

A balance chemical equation shows the Number of moles of reactants and p roducts. 2.

Number of particles:

It gives us information about the Number of particles of reactant and products. 2H2

23

+

2 (6.02 × 10 ) 3.

O2

6.02 ×10

⟶

2H2O 2 (6.02 × 10 23)

23

Volume:

It gives us information about the volumes of reactants and products. 2H2

+

2 (22.4) 4.

O2

2H2O

22.4

2 (22.4)

Mass:

It gives us information about the mass of reactant and products ie 2H2 + O2 2H2O

4g

32g

⟶

36g of water

Definition:

“The reactant which forms least amount of the product is called limiting reagent”. Actual Yield:

“In actual practice, we often get less amount of the product than the theoretical amount. This is called actual yield” Theoretical Yield:

“The amount of the product calculated on the basis of balanced chemical equation is called theortical yield of the reaction”



Actual yield is less than theoretical yield due to following reasons:

i. Reversibility

The reaction may be reversible, therefore reaction do not proceed to completion. ii. Unwanted reaction:

It is possible that some of the reaction form a b yproduct due to side reaction / parallel reaction or chain reaction. iii. Mechanical loss:

Mechanical loss is possible during transferring the product. It is called human error. iv. Unfavorable condition:

The formation of product may be affected by the disturbance in the conditions cond itions like temperature and pressure of the experiment 3(a) what is formula mass of a compound? What are the steps involved in calculating the formula mass of a compound. Explain with an example. Formula Mass:

11



“The sum of atomic masses of all the atoms or ions present in a formula unit is called its formula mass” Explanations:

Ionic compounds are represented by formula units instead of molecules. As ionic compound is formed by the aggregation of positive and negative ions to form a crystalline compound. They is why formula mass is used instead of molecular mass for ionic compounds. NaCl is the formula unit of sodium chloride and its formula mass = 23+35.5 = 58.5g 58.5g Steps to calculate the formula mass of a co mpound: First write the formula unit of the ionic compound. For example, magnesium bromide is an ionic compound. Its formula unit will be MgBr 2. It means each magnesium ion Mg++ is surrounded by two bromide ions, 2Br.

̅

But formula mass of the compound will be the sum of ionic masses of Mg++ and 2B ions ions expressed in grans, ++

Mg = 24 2Br – = 80 2Br – = 2 × 80 = 160 Formula Mass!

MgBr 2 = 24 + 160 = 184g (b)

i)

Calculate formula masses of the following compounds: (i) HNO3

(ii) C6H12O6

(v) Al2 O3

(vi) K 2 Cr 2 O7

(iii) C3 H8

(iv) C2 H5 OH

HNO3 atomic mass of H = 1 Atomic mass of N = 14 Atomic mass of O = 3 × 16 = 48 Formula mass of HNO3 = 1 + 14 + 48 = 63g

ii)

C6 H12 O6 6(12) + 12(1) + 6 (16) = 72 + 12 + 96 = 180 Formula mass of C6 H12 O6 = 180

iii)

C3 H8 3 (12) + 8 x 1 = 36 + 8 = 44 Formula mass of C2 H5 OH = 46g

iv)

C2 H5 OH 2(12) + 5(1) + 16(1) + 1 = 24 + 5 1 + 1 = 46 Formula mass of C2 H5 OH = 44g

v)

Al2 O3 2(27) + 3(16) = 54 + 48 = 102 Formula mass of Al2 O3 = 102g

vi)

K 2 Cr 2 O7 2(39) + 2 (52) + 7(16) = 78 + 104 + 112 = 294 Formula mass of K 2 Cr 2 O7 = 294g

4.

Define and explain mole and Avogadro’s nu mber with examples?

Mole:

a)

“Actually mole is a latin word which means a huge mass”.

Mole is the unit of amount of substance. Mole is a number as well as a quantity. It is represe nted by ‘n’. As, 2 similar things = 1 pair 12 similar things = 1 Dozen 100 similar things = 1 century 144 similar things = 1 gross Similarly 6.023 × 1023 similar things = 1 mole Definition-1:

“A group of 6.023 × 1023 similar things (ions, atoms, mo lecules) is called mole”. Definition-2

“The atomic masses, molecular masses, formula masses or ionic masses of a substance expressed (taken) in grams is called mole”. Examples:

12



1 mole of hydrogen atoms = 1.008 grams 1 mole of carbon atoms = 12.01 grams 1 mole of H2O molecules = 18.016 1 8.016 grams Avogadro’s Number:

“The number of particles present in one mole of a substance is called avogadro’s number”. It’s value is 6.023 × 1023. This constant number present in one mole of a substance was experimentally determined by an Italian

scientist Ameado Avogadro. So, it is named after the n ame of that scientist. It is represented by NA. For example:

1 mole of sodium = 23 grams Na = 6.023 × 1023 Na atoms 1 mole of water = 18g of H2O = 6.023 × 1023 H2O molecules. 1 mole of AgNO3 = 170g AgNO3 = 6.023×AgNO3 f. units b)

Given the Equation:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + Heat How can this equation be read in terms of particles, moles and masses? Equation:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) i)

Number of Particles:

1 molecule of CH 4 reacts with 2 molecules of O 2 to form 1 molecule of CO 2 and 2 molecules of H2O. ii)

Number of Moles:

1 mol CH4(g) + 2 mo O2(g) 1 mol CO2(g) + 2 mol H2O(g) Masses:

16gCH4(g) + 32gO2(g) 44gCO2(g) + 36g H2O(g) iii)

Number of Particles;

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) 6.02 × 1023 + 2 × 6.02 × 10 23 6.02 × 1023 + 2 × 6.02 × 10 23 Molecules Molecules Molecules Molecules 5.

What do you mean by percentage composition of a compound?

a)

How the %age of an element is calculated in a compound.

Percentage Composition:

“The percent by mass of an element in a compound is called its percentage composition”. Explanation:

Once the element present in a compound are identified, and the molar mass (or formula units) for the compound are d etermined, it is straight forward to calculate the percentage composition. Formula:

      × 100     OR,     ×.    × 100 Percentage of an element =     Percentage of an element =

Sum of the individual percentages of all the elements must be equal to 100. Calculate the % composition of each of the th e following compounds. (Given atomic weights of the elements).

(i) MgSO4

(ii) C3 H6 O

(iii) KMnO4

(v) NoAl (SO4)2

(vi) CaCO3

(vii) CH4

(iv) C6 H6

Ans.

In case case of ionic compounds, the term formula mass mass is used while while in case of molecular molecular compounds, compounds, molar mass is used.

(i)

MgSO4 Formula mass of MgSO4 = 24 + 64 = 120g

 × = 20%   × %age of S =  = 26.6%  × %age of S =  = 53.3%

%age of Mg =

13


(ii)


C3 H6 O Molar mass of C3 H6 O = 3(12) + 6(1) + 1(16) = 36 + 8 + 16 = 60g

 × = 60%   × %age of H =  = 13.33%  × %age of O =  = 26.66%

%age of C =

(iii)

KMnO4

Molar mass of KMnO4= 34 + 55+ 64 = 158g

 × = 24.68%   × %age of Mn =  = 34.81%  × %age of O =  = 40.50% %age of K =

(iv)

C6 H6

Formula mass of C6 H6= 72 + 6 = 78g

 × = 92.3%   × %age of H =  = 7.7%

%age of C =

(v)

MaAl (SO4)2

Formula mass of NaAl (SO4)2 = 23 + 27+ 3 (96) = 50 + 288 = 338g

 × = 6.80%   × %age of Al =  = 7.6%  × %age of S =  = 28.40%  × %age of O =  = 56.80% %age of Na =

(vi)

CaCO3

Formula mass of CaCO3 = 40 + 12 + 48 = 100g

 × = 40%   × %age of C =  = 12%  × %age of O =  = 48%

%age of Ca =

(vii)

CH4

Molar mass of CH4 = 12 + 4 = 16g

 × = 75%   × %age of H =  = 25%

%age of C =

6.(a)

Differentiate between a “Limiting reagent” and “a reagent in excess”. How will you identify the limiting reagent in a chemical reaction?

See page (9) Table 6.

How does a limiting reagent control the amount of the product formed? Give an example.

Ans.

When the reagents reagents are not taken in stoichiometric stoichiometric ratio i.e. one reagent reagent is taken in large amount than the required required one and the other in small amount, then the limiting reagent will be finished earlier and hence further products formation will be stopped. So, a limiting reagent controls the amount of products formation due to its earlier consumption.

Example:

Many reactions taking place in our surroundings involve limiting reagents e.g., burning of carbon and natural gas. C + O2 CO2 CH4 + 2O2 CO2 + 2H2O In these reactions, oxygen is in excess and carbon and CH 4 are limiting reactants. The use of limiting reactant has the following advantages. 1.

They make the reaction faster and faster.

2.

They make the reaction 100% complete.

7.

A technician weight 40g of sodium chloride. How many moles of formula units are in the sample?

14



Solution:

Moles of formula units of NaCl=? Given mass of NaCl = 40g Formula mass of NaCl = 58.5g Formula units of NaCl =

     . = 0.68 moles.

8.

Calculate the mass in grams of (a) 7.75 moles of Al 2O3 (b) 15 moles of H 2SO4 (c) 1.0 × 10 25 molecules of H2O.

a)

Mass =? Al2O3 = 7.75 mol Molar mass Al 2O3 = 54 + 48 = 102g

 

n = or m = n × M Mass = No. of moles × molar mass = 7.75 mol × 102g/mol = 790.5g 790.5 g b)

Massing of H2SO4 =? Molar mass of H2SO4 = 2 + 32 + 64 = 98g/mol Given moles of H2SO4 = 1.5 mol Mass = moles × molar mass = 1.5 mol ×18g/mol = 147g

c)

Massing of H2O=? Avogadro’s No. NA = 6.02 × 1023/ mol = 18g/mol Given number of H2O molecules = 1.0 × 1025 Mass of 6.02 × 1023 molecules = 18gH2O Mass of 1.0 × 1025 molecules = = 2.99 × 102g

i)

 ×. ×   . ×

= 29gH2O

How many moles are present in each of the following samples. (a) 30g of MgS

(b) 75g of Ca

(d) 40dm3 of O2 gas at STP

(c) 8.85Rg of CO2

(e) 7.5 × 1020 molecules of C6H5

Solution:

a)

Moles of Mgs = ? Formula mass of MgS = 24 + 32 = 56g/ mol Given mass of MgS = 30g

 = 0.53 mol /

Moles of MgS = b)

Moles of Ca= ?

Molar mass of Ca = 40g/mol Given mass of Ca = 75g Moles of Ca = c)

    / = 1.875 mol

Moles of CO2= ?

Molar mass of CO2 = 44g/mol Given mass of CO2 = 8.85kg = 8.85 × 1000 = 8850g Moles of CO2 = d)

 / = 201.1 mol

Moles of C6 H6= ? NA = 6.02 × 1023

Given,number of molecules of C6H6 = 7.5 × 1020 Moles of C6 H6 = 10.

.   . ×   = 1.24 x10  . ×

– 3

= 0.00124 mol

Calculate the mass % of a metal in a compound that is formed by 0.233g of metal combining with 0.354g of oxygen.

Solution:

Metal + Oxygen  Metal oxide 0.233 0.354 0.233 + 0.35 = 0.587g

15


% age=?

   × 100 = .× 100 39.69%     .

% age of metal = 11.


Given the equation: 2H2(g) + O2(g)

2H2O(g)

a.

How many moles of water will be obtained by burning 5.6 moles of O2 in an excess of H2?

b.

How many moles of O2 would be needed to react 58.5g of H 2 to form water?

c.

How many grams of H2 would be needed to form 120g of H2O?

Solution:

Equation: 2H2(g) + O2(g) 2H2O(g)

a)

excess 5.6 moles moles=? 1 mole of O2 forms = 2 mole of H 2O 5.6 mole of O2 forms = 2 × 5.6 = 11.2 11 .2 mol of H2O 2H2(g) + O2(g) 2H2O(g)

b)

58.5g

Moles=?

4g

1 mol

4gH2needs = 1 mol O2 58.5g H2 needs =

  ×. = 14.62 mol O 

2

2H2(g) + O2(g) 2H2O(g) g=? 120g

c)

4g

36g

36g H2O needs = 4gH2 120g H2O needs 12.

 × = 13.33gH 

2

Given the equation N2(g) + 3H2(g) an excess of H2?

2NH3(g) at STP. How many moles of NH 3 would be formed if 6.3dm3 of N2 gas react with

Solution:

N2(g) + 3H2(g) 2NH2(g) 6.3 dm3 Excess Moles=? 1 mol 2 mol 3 22.4dm 2mol 22.4dm3 N2 gives = 2 mol NH3 6.3dm3 N2 gives = 13.

  ×. .

= 0.56mol NH3

Calculate the mass of Mg metal required to consume 2560kg of CO2 in the reaction. 2MG(s) + CO2(S)

2MgO(S) + C(S)

Solution:

Reaction: 2Mg(s) + CO2(S) 2MgO(S) + C(S) Mass = ? 2560g 48g 44g 44g CO2 needs = 48g Mg 2560g CO2 needs = 14.

 × = 2792.72g Mg 

When steam is passed through red hot carbon, a mixture of H2 and CO gas, called water gas, is formed. H2O(g) + C(S)

a)

CO(g) + H2(g)

Which is the limiting reagent if 24.5g of carbon is mixed with 1.89 moles of water vapours?

Solution:

H2O(g) + C(S) CO(g) + H2(g) 1.89mol 24.5g Limiting reagent=? Converting 24.5g of C into mole =

   .     = 2.04 mol

Now, calculate calculate the amount of water gas gas (product) formed from 1.8 mol H2O and then form 2.04 2.0 4 mol C separately. i)

1 mole H2O forms = 1 mol of water gas (CO+H 2) 16


ii)


1 mol C forms = 1 mol of water gas (CO + H 2)

The amount of product formed from 1.89 mol H2O is 1.89 mol of CO + H2 product which is less than 2.04 mol, that is i s formed from 24.5g of C. The former, 1.89 mol of H2O in the limiting reagent. b)

Calculate the amount amount (in grams) of the excess reagent reagent left unreacted; The product formed from 1.89 mol H2O is less than that of 2.04 mol C. It means 1.89 mol H2O is limiting reagent while 2.04 mol C is in excess. Excess amount of C = 2.04 – 1.89 1.89 = 0.15 mol OR, 0.15 × 12 = 1.8g of C will remain uncreated. un created.

15.

Calculate the percent yield if 6.53g of hydrogen gas is produced when 5 moles of zinc is consumed in the reaction: Zn(S) + 2Hcl(aq)

ZnCl2(g) + H2

Solution:

Zn(S) + 2HCl(aq) ZnCl2(g) + H2 5 mol

6.53g

% yield = ?

From balanced chemical equation; 1 mole Zn gives = 1 mol H2 1 mole Zn gives = 2g H2 5 mol Zn gives = 2 × 5 = 10g of H2 It means, theoretical yield = 10gH 2 But, actual yield = 6.53g of H2 % yield = = 16.

    × 100

.× 100 

= 65.3%

The percentage yield of the following reaction is 85% 2Al(S) + 3Cl2(g)

2AlCl3(S)

How many grams of AlCl 3 will be obtained from 100g of aluminum metal. Solution:

2Al(S) + 3Cl2(g) 2AlCl3(S) Actual yield of AlCl 3 = 85% 2Al = 2 × 27 = 54

 g AlCl  100g Al gives =  × 100 = 44.4g (Theoretical yield) 1g Al gives =

3

AlCl3 = 27 + 106.5 = 133.5

But actual yield = 85%

2AlCl3 = 2 × 133.5 = 267

Therefore, actual yield obtained = 44.4 × 85%

54g Al give = 267g AlCl3

=

. × = 420g 

17



CHAPTER-2: CHAPTER-2:

ATOMIC STRUCTURE

Introduction:

In this chapter we will study about Important Point, about structure of Atom

1.

Concept of atom was given by Democritus.

2.

Atom is Greek word “atomous”.

3.

John Dalton presented Dalton atomic theory

4.

Atomous mean indivisible.

5.

Electrons revolves anticlockwise around the nucleus.

6.

Atom contain 3-fundamental particles i. Proton ii. Neutron iii. Electron

7.

Among these 3-particles electrons are fundamental

8.

Because electron electron is not no t composed from any sub atomic particle

9.

Proton is not fundamental because it is composed from uud i.e. up, up & down quarks

10.

Neutron is not fundamental because because it is composed from from udd. i.e. up, down down quarks.

11.

The moment of an electron in an atom is out word.

12. 13.

The symbol of electron is . 19 Charge on an electron is -1.602 × 10 – 19 col.

14.

31 Mass of electron is 9.11 × 10 – 31 Kg.

15.

Relativistic rest mass of electron is 0.511 ev/c 2

16.

Electron was discovered by J.J Thamson.

17.

Chemical properties of atoms depends on electrons.

18.

Electron is 1836 times lighter than proton

19.

That electron which have positive charge is called positron.

20.

Relative mass of electron is 1/1840 cannot be zero.

21.

Mass of one mol of electron is 0.55mg.

22.

Charge on one mol of electron = 96500 coulomb = 1 Farad.

̅

Note:

Milikon discovered charge on electron. 2010-125 Other Sub particle:

1. 2.

Positron Strange

3.

Charm

4.

Baryon

5.

Bozan

6.

Gluons

7.

Quark

8. 9.

Neutrino Tau Neutrinu

10.

Up quarks

11.

Down quarks

12.

Pion.

Discharge Tube Experiment:

1.

It is also called cathode rays experiment.

2.

It is also called discovery of electron.

Principle:

18



Electric current should be passed through gases at a very low pressure 0.01 mmHg because gases are bad conductor of electricity at normal pressure they become become good conductor at low pressure. Experimental Procedure:

A discharge tube is a simple glass tube having two electrodes fixed at the two ends. Cathode:

Electrode connected to the negative terminal of battery is called cathode. Anode:

That electrode which is connected to the positive terminal of th e battery is called anode. Vacuum Pump:

A vacuum pump is connected to discharge tube. The function of the vacuum pump is “to reduced the pressure of gases to 0.1 mmH g. Mechanism:

i.

When electric current of voltage of (5000 – 10,000v) 10,000v) was passed through gases at a very low pressure of 0.1 mmHg, it produces a uniform glow inside the tube. Because of ionization of gasses.

ii.

When the pressure is further reduced to 0.01 mmHg, the glow disappear and a dark space is produced in the tube.

NOTE:

Now at this stage the electrical resistance between two electrodes become very high and the discharge become very difficult, now if the potential difference (applied v) is further increased, then again rays are produced (faint yellow) moving from cathode side towards anode, and glowing the tube. These rays are called cathode rays. Properties of Cathodes Rays:

Following are the properties of cathode rays. 1.

Travelling in straight path

2.

Material nature

3.

Negatively charged charged particles

4.

Deflection in magnetic field

5.

X-rays production

6. 7.

Ionization of gasses Reducing effect

8.

Charge to mass ratio

1.

Travelling in Straight Path:

Entry Test MCQs:

i.

The colour of the glow produced in the discharge tube depends upon…..

ii.

(a) Composition of Glass (b) Nature of gass (c) Both of them (d) Non of them The voltage of the discharge tube depends upon.

iii.

(a) Length of the tube (b) Pressure inside the tube (c) Both (d) Non of them At ordinary pressure gas does not conduct electricity even at (5000v). Inside the tube, this was observed by. (a) William Croocks (b) Sir Humpry Davi (c) J.J Thamson

(d) Nasrat Ullah Katozai

Hittorf in 1869 discovered that cathode rays produces shadow of an opaque object when placed in their path, which means that cathode rays travel in a straight line and are particle not electromagnetic radiation.

2.

Deflection in Magnetic Field:

J.J Thamson in 1897 demonstrated that when these rays are passed through magnetic field they are deflected at right angle to the applied magnetic field. 3.

Cathod Rays, Casting Shadow

Production of X-Rays-Rays:

Cathoder rays can produced x-rays when have high atomic masses. 4.

Ionization of Gases:

They can cause ionization of gasses when passed through gases. 5.

stick with that metals which Cathode Rays Deflecting in Magnetic Field

Reducing Effect:

Cathode rays are reducing in nature due to which causes chemical changes. 6.

Material Nature:

Crooks in 1870 demonstrated that cathode rays are materialistic in nature because when a light pinwheel is placed in their path they rotates it. Which means; that cathode rays also posses momentum, velocit y, mass and energy. Cathode Rays, rotating a light pin wheel 7.

Negatively Charged:

19



J. Perrine in 1895 and J.J. Thamson in 1897 showed that cathode rays are when passed through an electric field they are deflected toward the positive pole. Which shows that cathode rays are n egatively charged. Cathode Rays Deflecting in an Electric and Field 8.

Charge toMass Ratio:

The e/m ratio of cathode rays is constant i.e. 1.7588 × 1011 col/kg and resembles to that of electrons. 9.

Heat Production:

They on heat a thin foil of metal (Pt). If placed in their path, it is because of the conversion of KE in to heat energy. 10.

Effect Photographic Plate: Note: The nature of the cathodes rays is independent of the

Cathode rays also effect the photographic plates.

nature of the gas used in the tube because in all gases electrons are same.

Conclusion:

These negatively charged particles were declared as electron by G.J Sto ney in 1891. Charge to Mass Ratio of Cathode Ray:

Charge to mass ratio was determined by J.J Thamson in 1897. Procedure:

A special discharge tube was used in which both electric and magnetic fields were applied simultaneously. Mechanism:

When both fields were kept (off) close the cathode rays strikes at P1 on the fluorescent screen. When only electric field is applied they strikes at “P 2” and when only the magnetic field is applied then they strikes at “P 3”. Balancing:

By applying the both field in such away that electron stricks again “P 1”.

In this way the electric field force become equal to magnetic field force.

B C

= =

Magnetic field Charge on cathode rays

V

=

Velocity of cathode rays.

Thamson Apparatus for Measuring e/m Ratio

Mathematically:

The force of attraction produced by magnetic field. Magnetic Field:

Magnetic force = Bev Due to magnetic field cathode rays moves in a circular path and thus produce an out word force called centrifugal force. Centrifugal force = Since:

 

Magnetic force and centrifugal force are equal but opposite in direction.

 Be

   =  Bev =

m

=

Mass of cathode rays

V

=

Velocity of cathode rays

r

=

Radius

Rearranging:

    

(1)

Electric Field:

Now for velocity we used electric field (Ee) and then it is adjusted in such away to balance magnetic field due to which cathode rays come to original path. Thus: Bev = Ee → (a) Rearranging

→ (b)   V = → (c) V=

20



Now:

Putting the value of (c) in equation (1)

      v = E/B      

e/m = 1.75 × 1011 col/kg Charge on an Electron:

The charge on it was determined by R.A. Milikon by an experiment called oil drop method. 19 Charge of electron = 1.622 × 10 – 19 col

OR 10 4.8 × 10 – 10 esu lelectroratic unit

Mass of an Electron:

The mass of an electron can be determined d etermined from the value of charge and charge to mass ratio of electron. e= charge on electron = 1.6022×10 -19 col

  = charge to mass ratio = 1.7588×10 Dividing charge by charge to mass ratio .× m= ⁄ = .×

Since:

So:

⁄

11

col/kg.

= m

M=9.11×10-31kg M= 9.11×10-28g Note: The mass of electron in a.m.u is 0.00055 amu. DESCOVERY OF PROTON:

1)

Proton was discovered by Gold stine in 1886

Apparatus:

For the discovery of proton Goldstine used a special discharge tube, having perforated cathode.

Mechanism:

1. 2.

When electric current is passed through gases in a discharge tube. Then not only cathode rays are produced but also some other kinds of rays are produced which are moving from anode to cathode. These rays passes from the canals of the cathode so they are called cannal rays.

Origin of canal rays:

Canal rays are produced when fast moving electric current is passed through gases, then this current, eject electron form the atom of the gases, due to which th e atoms become positively charged, which are called canal rays. Properties of canal rays:

Following are the properties 1.



Travel in straight line:

They travel in a straight line 2.

material nature:

They can also rotate a pinwheel which shows the material nature of canal rays. 3.

e/m ratio:

The e/m ratio of canal rays is always smaller, than that of electron. 4.

Nature of canal rays:

The nature of canal rays depends on the nature of the gas. 5.

Note:

21



e/m ratio depends upon the mass; as. Mass of an atom increase e/m ratio decreases decreases For example:

For hydrogen maximum e/m ratio is 9.54×107 c/kg Mass of proton:

It can be determined as;

 /  .× M= .× M=

M= 1.67×10-27 kg Discovery of neutron:

Neutron was discovered discovered by James chadwick from a nuclear reaction. Experimental procedure:

He bombarded the nucleus of a light metal like beryllium by α - particle from a polonium source due to which high penetrating

radiations were produced which were named neutron by the Chadwick. Reaction: 9 4 12 1 4Be + 2He 6C + 0 N Beryllium α-particle carbon neutron

→

Note:

These neutron produced can passed through the paraffin p araffin oil. Note:

It was believed that an atom must have same no of protons, and electrons, but the actual mass of proton & electron is always less than the actual mass of atoms. So Chadwick discovered a third sub-particle i.e. neutron. Properties of Neutron:

Following are properties of neutrons. 1. Neutrons are highly penetrating penetrating particles 2. Carry no charge. 3. Not deflected by electric electric or magnetic field. field. 4. Can knockout high speed proton from substances like paraffin, water, cellulose. 27 5. Neutron has mass 1.67×10 – 27 kg and in 1842 time heavier than an electron. 6. Neutron is also fundamental fundamental particles. Particles

Mass(kg)

9.11×10-31 1.6726 ×10-27 1.6749 ×10-27

Electron Proton Neutron

Charge unit

-1 +1 0

Charge

-1.6×10-19 +1.6×10-19 No

Relative atomic mass

0.00055 1.0073 1.0087

PLANK’S QUANTUM THEORY THEORY

This theory was presented by a German physicist max plank in 1900. Background:

When a body is heated it absorbs heat in the form of radiation. This hot bodies emit the photon in discontinues manner when it cool. Following are the main postulates: 1)

2)

Quantization:

i.

The energy emitted by a body or absorbed by a body will be in discontinues, form i.e. packet.

ii.

This discontinues form is called quanta.

Energy of radiation:

Each radiation has a definite amount of energy, which is directly proportional to the frequency of radiation. Eα

E= hv Where “h” is called planks constant 3)

Whole no:

The energy emitted or absorbed by a body is always in whole no. For example:

E= nhv

:

n= 1,2,3….

Wavelength:

22



“The distance between two consecutive crests or trough, in a wave is called wavelength” Representation:

It is represented by Lambda : “λ” and its unit is nm or pm. Wave Function:

The no of waves per un it length is called wave function. Representation:

   

̅

It is represented by . It is the reciprocal of wavelength.

̅

α

 

Unit: m-1 or cm-1

Frequency:

The no of wave passing through a point per unit time is called frequency. Representation:

It is represented by f . Unit:

Hz (Hertz) Derivation of E=hc Proof: As we know that frequency is inversely proportional wave length.

̅

V

α

(i)

Using constant of proportionality “c” called velocity of light.

̅  =

(ii)

From Flank’s Quantum theory

̅ ̅  

E= h

(iii)

Putting the values of “ ”in ”in (iii)

E=



E2 = energy of higher orbit. h = planks constant.



(iv)

As we know that

 = ̅ 

E1 = energy of lower orbit.

= frequency of electron.

(v)

Putting (v) in (iv)

̅

E = hc Bohr’s Model of atom: atom:

Hence proved

After the Ratherford atomic model the Bohr presented the structure of atom in 1913. Which was based on planks Quantum theory. Postulates:

Following are the postulates. 1.

Fixed Circular Path:

The electron revolves around the nucleus in fixed circular path, which is called shell or orbits or energy level. Note: Each orbit has a definite amount of energy. 2.

Energy in the orbit:

As long as electron revolves in a specific orbit it will neither emit nor absorb energy. 3.

Concept of Jumping:

When an electron jumps from high energy level to lower energy level it emit energy and when it jumps from lower to higher energy level it absorb energy. Change in energy can be determined as:



E1= h ΔE = E2 – E 4.

Angular momentum:

The angular momentum (mvr) of the electron in the Hydrogen atom is quantized. mvr =

 

Application of Bohr’s model:

n = 1,2,3, …………..

m = mass of electron v = velocity of electron r = radius of orbit

23



Derivation of radius, energy, frequency, wavelength and wave no. Calculation of the radius of the Bohr’s orbit:

Consider the Hydrogen atom which have one proton pro ton and one electron.

 

The charge on electron is “e”.

The charge on nucleus is ze.

Coulomb’s law:

According to coulombs law the force of attraction between two oppositely charged bodies is directly proportional to the product of the charges and inversely proportional to the square of the radius between them.

× → (i) q = e, q = ze  So (1) become as. . Fα   Fα  Now converting converting constantly constantly of proportionality proportionality to equality by by using “K” “K”  F = K  _________________________(ii)  Since “K” is proportion ality constant and K =  Fα

1

2

εο is the permittivity of free space.

12 2 – 1 – 1 Its value is 8.85×10 – 12 C J m

Since: Equation (II) become as; Fe =

 __________________(III) 

Note: This coulombic force is because of centripetal force. But: As the electrons revolves in the circle, then an outward force also act which is called centrifugal force and is given by. Fg =

– __________________________(IV) 

Negative sign show the direction away from from nucleus. Now: Comparing (III) and (IV)

  =     m =  Rearranging  V =  2

2

__________________________(V)

_________________________(VI)

Form 4th postulate of Bohr’s:

From the 4th postulate of Bohr’s model, we have;

  Rearranging for “v”  =  Squaring on both sides.  =    =    ___________________(VII)



m =

Now:

Comparing (VI) & (VII)

 =           =   

_____________________(VIII)

24



 =     Rearranging for (r)   =        r =   _____________________________(IX) Separate constant and variable    r =      constant variable

– .×–  .  × r= .×–×  ×.×–   r = 0.529Å 

For first shell: n = 1 z=1

 

r = 0.529×

r = 0.529Å

For second shell:

  

= 0.529

  

n= 2,

= 0.529×4 = 2.3Å

2=1

 0.52931

°

= 0.529 9 = 4.7

For third shell: =

Calculation of energy of the electron in an orbit:

An electron in an atom po possesses ssesses two types of energy. i.

K.E

ii.

P.E

K.E:

It is because of motion which is given as; K.E = ½ mv 2

_____________(i)

Potential energy: The energy possessed by electron due to interaction with nucleus is called P.E. Potential energy is due to position between electron and nucleus. Work done = PE = Force× distance = F× radius = F× r

–  × r – =  ________________________(ii) =

Now:

The total energy of the electron is

      –     –  

= KE + PE Putting values

_________________________(III)

= m + = m

Putting values of (m ) from equation (V). 25



            –        –   – .       –          –                         2.18 ×    =

– –

=

_______________________(IV)

–

=

= –

________________________(V)

________________________(VI)

= – putting putting the values of “r” “r” from =

.

–

z=1

.

J/atom

.

J/atom

18 10 – 18 J/atom

Thus;

18 = – 2.18×10 2.18×10 – 18

J/atom

Conversion to KJ/mol



×    .

18 = – 2.18×10 2.18×10 – 18

×

KJ/mol

 KJ/mol  Energy of the first shell = 0  = –  KJ/mol 1313KJ/mol  = – 1313KJ/mol Energy for second shell:  = –  KJ/mol 328.32 KJ/mol  = – 328.32 Energy for third shell:  = – 145.92 KJ/mol  = – 145.92 E = –

n=2 (2)2

n = (3)2, n = (9)

Note:

As the values of n in creases energy of the electron also increases.

–  –  –  –  >

>

>

985 > – 182.27 182.27 > – 63.8. 63.8. – 985 Note; Energy difference between two orbital:

As we have that;

 

  .   

–

_____________________(1)

26


 

  .   

–


_____________________(2)

Therefore;

Δ–   –    –  .    –   – –       –              =

.

_____________________(3)

–

=

18 = 2.18×10 – 18

18 ∆E = 2.18 × 10 – 18

J/atom

DE = 1313

z=1

J/atom

OR

Cosmic most energetic – 3  ays 6 × 10 X – rays rays 6 × 10 – 3 – 8 Ultra violet 8 3.8 × 10 2 Visible 3.8 × 102 – 7.6 7.6 × 102 Infra-red 7.6 × 102 – 10 106 6 Microwaves 1 × 10 – 3 × 108 Radio waves 3 × 108 least energetic

kJ/mol

Calculation of frequency, wavelength and wave no of electron.

As we have that;

ΔΕℎ    –         –   / ̅ ̅      –   ̅    1       –   ̅     –  

1 ∝ 

m –

R – Rydberge Rydberge constant and its values is 1.0974×10 7m – 1 Spectrum of Hydrogen atom:

The arrangement of waves on the basis of increasing o r decreasing order of frequency or wavelength is called spectrum. OR A band of radiations obtained from the dispersion of white light is called spectrum. Spectroscopy:

The study of spectrum is called spectroscopy. Electromagnetic spectrum:

That spectrum which contain electromagnetic electromagnetic radiations and waves is called electromagnetic spectrum. For example:

Below is electromagnetic spectrum which contain electromagnetic electromagnetic radiations? Explanation:

i.

When white light is passed through prism, then this light split into several lines.

ii.

These radiations will arranged themselves in order of increasing or decreasing frequency.

Types of spectrum:

It has two types

i.

i.

Continuous spectrum

ii.

Line spectrum Continuous spectrum:

27



That type of spectrum in which there is no clear cut boundary among the different colours of o f the spectrum is called continuous spectrum. For example:

i. ii.

Spectrum of sunlight Spectra of Electric light

iii.

Black body radiations

iv.

Bulb light

v.

Rainbow

ii.

Line spectrum:

It is also called atomic spectrum. “That spectrum in which there is a clear cut boundary among the different colours of the spectrum is called line spectrum”. Classification of Line Spectrum:

It has two types: 1.

Line emission spectrum

2.

Line absorption spectrum

1.

Line emission Spectrum:

That type of line spectrum which is obtained from a substance when it is heated in a discharge tube is called line lin e emission spectrum. Explanation:

1. 2.

When a substance is heated in a discharge tube, it will emit light radiation. When these emitted light, is passed through prism it will split into several radiations in the form of a spectrum called line emission spectrum.

3.

This spectrum contain bright line with dark background. b ackground.

For Example:

When Na is heated on a flam it gives two characteristic bright lines separated from each other by dark lines. 2.

Atomic Absorbtion Spectrum:

The spectrum obtained from that radiations from which some  have been observed called is called atomic absorption spectrum. Explanation:

When white light is passed through the cool vapours of sodium, then the sodium vapours will absorb two colours from the spectrum and the spectrum contains two dark spaces separated by bright spaces. Exactly at the same region as that in line emission spectrum. Spectral Series:

“These are different coloured lines which are obtained when an electron jumps from higher energy level to lever energy level”. Explanation:

i)

Hydrogen atom has only one electron which normally occupy the lowest energy level or ground state.

ii)

Now if energy is given to hydrogen atom then its electron will jumps from its ground state to excited state.

iii)

Now if the electron come back to the ground state it will emit radiation which are called spectral line. These no of line of different colours are called spectral series.

Classification:

Spectral series may be classified into; 1.

Lyman series

2.

Balmer series

3.

Paschen series

4.

Bracket series

5.

P-fund series

1.

Lyman Series:

These are different coloured lines which are obtained when an electron of hydrogen atom jumps from one of the higher energy level for which n= to lowest energy level for which n=1. Note:

It lies in the ultraviolet region. Mathematically:

̅      

n =  to n=1

28



̅      

n= 2, 3, 4….

2.

Balmer Series:

These are the spectral lines which are obtained when an electron jumps from highest energy level to n=2. Note:

They lies in visible region. Mathematically:

̅        3.

n=, to n=2.

Paschen Series:

These are the spectral lines which are obtained where an electron jumps from n=  to n=3. Note: It is present in near IR Region Mathematically:

̅      

n =  to n = 3

4.

Bracket Series:

These are the spectral series which are obtained when an electron jumps from n =  to n = 4. Note: Mid IR region Mathematically:

̅      

n to n = 4

5.

P-fund Series:

These are the spectral series which are produced when an electron jumps n =  to n=5. Mathematically:

It is present in for IR

̅      

n= to n=5

Spectral series of Hydrogen in various region due to different electronic configuration.

X-rays:

    

These are electromagnetic radiations of high frequency. The wavelength of x-rays ranges from 10 – 2Å to 10+2Å. X-rays (0.001nm –– 10nm). 10nm). Discovered by Roentgen

Accidently discovered discovered by studying stud ying Cathode rays. Types of X-Rays: It has two types: i. ii.

Characteristics Characteristics x-rays Continuous x-rays

Methods of preparation:

x-rays can be produced by several methods some are given below;

i.

i.

Roentgen method

ii.

Coolidge method

iii.

Betatron method Roentgen method:

a.

A very high electric current of (30,000 –– 50,000v) 50,000v) was passed through a special discharge tube at a very low pressure of 0.001 mmHg.

b.

Due to which the gas in side the tube ionizes to positive and negative charges.

c.

The heavier positive ions are attracted toward the cathodes and due to collision with cathode it emit electrons from the cathode, and these electrons are very fast moving and since it moves toward the anode and eject x-rays from anode.

Types of x-rays: It has two types 1.

Characteristics x – x – rays rays :(inner shell transition)

When a heavy atom is bombarded by high energetic beam of electron of several (Kev). Then its ejects the electron from the inner shell which is filled by the higher shell electron. Hence at that time some energy is evolved called x -rays. 29

Nasrat Ullah Katozai (Chemistry) Types:it has three types of

::  

i.



When the electrons jumps from “L” to K shell that will called K x-rays.

K :

iii.

2.

capture rule  principle = k – capture

K

ii.


..

When the electron jumps from “M” to “K” the x -rays produced that will be called K

K :



The production of x-rays when electrons jumps from “N” to “K” shell is called K x-rays.

Continues x - rays

Uses of x-rays: It is more energetic.

Following are uses of x-rays. 1.

In the field of medicine:

X-rays are power full in penetration so it is used for the detection of bone fracture. For example:

If a liquid crystal is passed through the body then it has crystalline structure when it reaches the defected region it is converted into liquid due to high temperature & since shown by b y x-rays. 2.

XRD:

It is used for the checking of crystallinity of a substance.ie by; (x-rays diffraction analysis) 3.

Inter ionic distance:

It is also used for the determination of interionic distance between ions in a compound. 4.

Gas Ionization

They can also ionize the gases and produced cation and anion. Properties of x-rays:

They are following: 1.

Traveling in straight line:

They travel in a straight line like the ordinary light. 2.

Neutral:

They are neutral having no charge because x-rays are not deflected by magnetic and electric field. 3.

Ionization:

X-rays can also causes ionization and ionizing power depends on the intensity of x-rays. 4.

Fluorescence:

They can produces fluorescence on the metal surface ie Zn detectors, rock salt, Uranium glass, compounds of Ca and Barium. 5.

Reflection:

They can be reflected by the electric field. 6.

Refraction:

They can be refracted. 7.

Note: continuesx – rays rays when a high energy

electron incident on a target and come to rest, in the first collision if lost all the energy which is appeared in the form of x – ray ray and continues xrays

Diffraction:

When x-rays passed through a small hole. It spread which confirm diffraction. 8.

Penetration:

They can also penetrate through an object. 9.

Blacking photographic plate:

They can blacken a photographic plate when placed in their path. 10. DNA helix:

DNA is composed from “Double helix” it was determined by the crick and W atson by x-rays. Moseley’s Lay: Lay:

(This law states that the square root of the frequency is directly proportional to the th e atomic number of an element) Mathematically:

√    √     

It can be also written as

30



Since b is screening constant

  √ 

“a” is a constant of the proportionality and depends on the metal.

Quantum Number:

These are four constant numbers which fully describe an electron in an atom. (i) It is also called identification no (ii) following are the four sets of quantum no. 1.

Principal Quantum mumber.

2.

Azimuth AL quantum no.

3.

Magnetic Quantum no.

4.

Spin quantum no.

1.

Principal Quantum Number.

It deals with the shells, o rbits, or energy level. Introduced by Neil Bohar. Representation:

It is represented by “n” where n = 1 , 2, 3, 4, 5 …..

But for = k L M N 1 23 4 Explanation:

It deals with 1. Distance of electron from the nucleus 2.

Energy of an electron orbit.

3.

Size of an atom (roughly)

4.

Size of electron orbit.

5.

As the values of “n” increases, the P.E of electron also increases.

6.

As the values “n” increases the kinetic energy and velocity of electron decreases.

7.

As the values of “n” increases, then the atomic radius increases i.e.

0. 0.592 

r 8.

i.e.

. n2

For energy we used the following equation.

 

28 E = – 2.18 2.18 × 10 – 28

No of electrons in the Shells:

The no of electron in the shell can be determined by the following formula; No. of es = 2n2 (i) No. of Electrons in K-Shell:

K = 1 (first shell) No. of es = 2n2. = 2(1)2 = 2 (1) K = 2 es. ii.

No. of electrons in L Shell:

forL shell n = 2 So no of electrons = 2 n 2 = 2 (2)2

 n = 2

= 2 (4) L = 8 Electrons since L-shell contain 8 electrons. Azimuthal Quatum No:

i)

It is also called secondary quantum.

ii)

It is also called subsidiary quantum number

iii)

It was presented by Arnold Somerfield.

iv)

It is also called angular momentum quantum qu antum number. 31



Representation:

It is represented by “ l” where as;

Values:

l = 0, 1, 2, 3 sp d f

Note: It deals with with i)

Shape of orbitals

ii)

No of sub energy levels levels present in the principal quantum no.

S stand for spherical P = principals D = diffused F = fundamental Note:

“l” is always one less than “n” so. l = n -1 1. No of subshell in k:

For k → n = 1 SO l = n – 1 1 l = 1 – 1 1 l = 0 → Zero means one digits [0] S Since: “k” contain only one sub she ll “S”. 2. No of sub-shell in L: for

 L  n = 2 =

So 2 mean 2 digits. l = n – 1 1 l = 2 – 1 1 l = 1 mean p-sub shell

01 

3. No of sub-Shell in M:

Since for M  n = 3, 3 means 3 digits d igits l = n – 1 1

012

 

l = 3 – 1 1 l=2 4. No. of Sub-shell in N:

Since for N: n = 4, 4 means 4 digits

01 23

 

l = n – 1 1 l = 4 – 1 1 l = 3 (f) Note:

It also shows no of electrons in the sub orbital. 1. No of electrons in S orbital:

No. of electrons = 2 (2l + 1) For s = l = 0 No of es. = 2 (2(0) + 1 No of es.= 2 (0 + 1) No of es.= 2 2. No. of es in P-Orbital:

No. of es = 2 (2l + 1)

For P , l = 1

No of es. = 2 (1) + 1) No of es.= 2 (2 + 1) 32



No of es.= 2 (3) = 6 es



3. No. of s in d-Orbital:

No. of es = 2 (2l + 1) No of es.= 2 (2 (2) + 1) No of es.= 2 (4 + 1) No of es.= 2 (5) No of es.= 10 es || || || || 3.

||

Magnetic Quantum Number:

That set of quantum nu mber which deals with orbital orientation and magnetic orientation is called magnetic quantum no. It was introduced by Stark & Zeeman. It is also called angular momentum quantum no. Representation:

It is represented by “m” and its values are; m = 0, + 1, + 2, + 3….. Mathematically:

It can be written as; m = 2l + 1.

l = azimuthal quantum number.

Note:

It also deals with the no of orbitals in sub energy level by; m = 2l + 1 i. No. of Orbitals in “S” sub Shell: MCQs:

m = 2l + 1

1.

 for “S” the “ l” value is zero

m = 2(O) + 1 m=0+1

(c) 7

(d) 9

Since:

m=1 Since S-sub shell contain only one orbital.

An electron revolving in the “P” orbital its magnetic

orientations. (a) 3 (b) 5

2.

Me = 2l + 1  P = 1 2 (1) + 1 2+1 2+1 OR (-1, 0, + 1) What will be the magnetic orientation of an electron revolving in “d” orbital

2. No. of Sub Orbitals in P:

(a) 3

(b)

(c) 5

(d)

Since for p → l = 1 m = 2l + 1 m = 2(1) + 1 m=2+1 m=3 Conclusion:

Since “P” sub shell contain 3-orbital. Px

Py

Pz

No. of Orbital in d-sub shell:

Since for d = d = 2 m = 2l + 1 m = 2 (2) + 1 Conclusion: Since at subshell contain 5 orbitals.

m=4+1

m=5

Dxy, dyz, dxz, dx2 – y y2, dz2 4.

Spin Quantum No.

That quantum no which describes the spin of electron from its own axis, is called spin quantum number. Representation:

It is represented by “s”. Values of S:

   

It has two values + , – Note:

It deals with the spinning of electron from its own axis. Clock wise spinning

Anticlockwise spinning

Note:

In the same orbital the electrons will cancel the effect of each other if they are in opposite spin. 33



Shapes of Orbital:

d-orbital contain two sets 1.

eg

2.

t2g

1. eg d-orbital

It has high energy. It contain 2 orbital. a. dz2

b. (dx2 – y y2)

2. t2g d-orbital

It has low energy. It contain 3 orbital a. dxy

b. dzx c. dyz

Electronic Configuration:

The distribution of electron in an atom is called electronic configuration. Rules:

The rules of electronic configuration are given below:

i.

i. ii.

Afbau principle Hunds rules

iii.

Pauli exclusion principle

Afbau principle:

According to this rule “The electrons will be distributed in sub energy level on the basis of increasing order of energy”. Explanation:

i.

Electron will first go to low energy level.

ii.

Filling of electrons will be on the basis of n + l rules

iii.

It also called wisvisered rule.

iv.

Afbau principle can be discussed by the following d iagram.

Note:

First we will form one ball then 2 balls upto 3 ball and then we will make 4 ball again and then 4, 3, 2and 1. Note cross the ball from above. 4P6

3d10

4s2

3p6

3S2

2P6

2S2

1S2

For example:

The electronic configuration of Hydrogen. H  Z=1 E.C =1S1 He Z=2

E.C =1S1

Na  Z=11

E.C = 1S2, 2S2, 2P6, 3S1 Cu = 29

“Cu” electronic configuration of copper.

Ec = 1S2, 2S2, 2P6, 3S2, 3P6, 4S2, 3d9 But it is not actual E.c because here the stability rules are used. ie

Completely filled > Half-filled > Partial Since: One e will go from 4s 2 to 3d9& so the new electronic configuration because as: 11

11

11

The electronic 2

>

1

1

1

>

11

configuration of “Cu” can be 6

2

6

10

1

1 written as;

2

1S , 2S, 2P , 3S , 3P , 3d , 4S

Since 3d10 is completely filled so it should be written first than 4S 2: Hunds rules:

According to this rules “The electrons will not pair up until there is an empty degenerate orbital present”. Degenerate orbitals:

34



Those orbitals which have same energy are called degenerate orbital. For example:

Consider the p-orbital having 3-degernerate orbitals ie px, py & pz. Now if three electrons electrons are present, present, then these electron will will fill up the p-orbital as; The filling will not be like as under. Since in “px” two electrons are present while “pz” is empty so this is not correct according to Hunds. Because “pz” is degenerate orbital so the electrons will first fill one orbital & then other. Pauli Exclusion Principle:

Accor ding ding to this principle “No two electrons in atom can have the same set of four Quantum N umber .”. .”. For example:

Consider the s-orbital which have two electron. i. ii.

Principle Quantum No for one electron=n=1 similarly for oth er electron=n=1 Azimuthal Quantum No for first electron=l=0 for 2 nd electron=l=0

iii.

Magnetic Quantum No for first electron = m = 0 and similarly magnetic Quantum Number = m = 0.

iv. Now Spin Quantum No:

The spin Quantum no is not same for both electrons first electron=+1/2 clockwise clockwise and for second electron= – 1/2 1/2 anticlockwise. Conclusion: Since spin quantum No. for both electron cannot be same. Angular node: That area in an orbital where there is minimum probability of existence of electron. For example: p – subshell subshell has one angular node Radial node: That area where there is zero probability of existence of electrons is called radial node.

35



CHAPTER-3: THEORIES OF COVALENT BOND AND SHAPE OF MOLECULES Introduction:

After the formation of chemical bond molecules or chemical compounds are formed. But there was a problem, that what will be the shape of molecules? Since: In order to solve the th e problem about the shape of molecules several theories was presented, presented, which are given as under; un der; Theories about shape of molecules:

Following theories are about the shape or geometry of molecules. 1. VSEPR theory 2. V.B.T 3. MOT 4. CFT (Valence shell electron pair repulsion theory) (VSEPR theory) 1.

History:

This theory was presented by Sidwick and Powell in 1940 and further modified by Gillespie and Nyholm in 1947. Objectives

The main objective of this theory was to draw the correct geometry or shape of molecules. Postulate:

It has the following postulates; 1.

Shape of Molecules:

The shape of molecules depends upon the arrangement of electrons in the valence shell of th e central atom. i.e.: CH4

Central atom

Tetrahedral Structure 2.

Best Arrangement of Electron:

The electron pairs will arrange themselves in such away to h ave maximum separation and minimum repulsion. For Example:

4 – Pairs Pairs ––  Shape will be tetrahedral. → 3 – Paris Paris –– 

Trigonal shape →

2 – Pair Pair –– 

Linear shape →

3.

Distortion in the Shape:

Lone pairs will repel bond pairs electrons and thus produce distortion

) ( in the shape of molecules, due to which bond angle will be

changed. 4.

Space Occupation:

Lone pairs will occupy more space as compared to b ond pairs. Increasing order of repulsion order: 36



L.P – L.P L.P > L.P – B.P B.P > B.P – B.P B.P 5.

Concept of double and triple bond:

Double bond and triple bond will be considered as single bond in determining the geometry or shape of molecules. For example:

Consider N2 molecule N ≡ N This triple bond should be considered as single bond Applications of VSEPR Theory:

There are three cases; 1.

Case-I

2.

Case-II

3.

Case-III

Case=I: (i) (Ax 4 type molecule)

That molecules in which 4 atoms are attached to the central atom is called Ax 4 type molecule. The Geometry of Ax4 type molecule will be; (a)

For example Note for Entry Test:

Consider CH4

Al the members of group-IV A has got tetrahedral geometry. eg. CH4, SiH4 etc. GeH4, SnH4, PbH4 etc.

If:a molecule have; Total pairs = 4 Bond pairs = 4 Lone pairs = Zero Best Electronic Geometry:

Tetrahedral Bond Angle:

109.5 Hybridization:

Structure of Methan

SP3 Conclusion:

As there are no lone pairs of o f electrons with central atom carbon, so methane has got tetrahedral geometry. ii.

̈

A 3 type molecules:

Ax3 are those molecules which have 3 atoms attached to the central atom. For example:

NH3 (ammonia) If in a molecule: T.P = 4 B.P = 3 L.P = 1 Bond angle: 107.50 Hybridization: SP3 Shape = Trigonal pyramidal

Trigonal Pyramidal Structure

Note: All the members of VA has got pyramidal geometry. geometry. i.e. NH 3, PH3. ASH3

Best Electronic Geometry:

As there is one lone pair on nitrogen so it will repell the bonds pairs due to which angle closes to 107.5. Thus NH 3 has got pyramidal geometry. iii.

̈

A 2 type molecules

Those molecules which have; T.P = 4

Note:All the member of VIA has got angular shape.

B.P = 2 L.P = 2 Shape = Angular For example

H2O 37



Bond angle: 104.50 Hybridization: Sp3 Case-II: (i) A type molecule: Those molecules which have

T.P = 3 B.P = 3 L.P = 0 Shape = Triangular planner bond angle = 1200

Triangular planner

Examples: Note: All the members of group III A has

BF3, AlCl3, BI3

got triangular planner geometry.

Conclusion:

As there is no lone pair of electrons so their geometry is triangular planner, and bond angle is 1200. Note: It is two dimensional ii.

Ax2 type molecules:

Those molecules which have. T.P = 3 B.P = 2 L.P = 1 Q = 1200, 11780 BEST ELECTRONIC GEOMETRY: Angular planner Note: Confusion should be removed by clearing the concept here;

Since For such molecule the best electronic geometry should be trigonal planner with bond angle 120 0. But as there is one lone pair of electron which repel the bond pai r and since bond angle decreases to 117.8 and the shape become “Angular planner”.

For example: i. ii.

SO2 SnCl2

Note: Alcohol have tetrahedral

structure and bond angle is 105 0.

Case-III

Ax2 Type Molecules: Those molecules which have Structure of SO2

T.P = 2 B.P = 2 L.P = 0 Q = 1800 Best Geometry:

The best electronic geometry will be linear Example:

i.

BeCl2

ii.

MgCl2

iii.

CaCl2

iv.

SiO2

v.

CO2

Disadvantages:

1.

Molecules having extensive delocalization of electrons cannot be explain by VSEPR.

2. 3.

It does not explain the mechanism of geometry. Does not give reason about the structure of a molecules.

Valence Bond Theory:

1.

This theory was presented by Helter and Fritz London and further modified by Pauling and Slater in 1927. 38


2. 3.


Helter & London explain VBT on the basis of sharing of electrons. Pauling & Slater explain VBT on the basis of directional and non-directional n on-directional characters. characters.

Postulate: V.B.T has three postulates. 1.

Half filled orbital’s: orbital’s:

Chemical bonds are formed by the overlapping o verlapping of half filled orbitals. orb itals. i.e. For example: 2.

↑

+

Bond can form

↑

Completely filled orbitals:

Completely filled orbitals will not overlaps during bond formation. i.e. For example: 3.

↑

+

No Bond will form

↑

Strength of Bond:

The nature of the strength of the bond depend upon the extent of overlapping of orbitals. i.e. i. Head to head overlap: strong bond. ii. Sidewise overlap = weak bond Explanation:

V.B.T can be explained on the basis of following examples. (1) H2 formation:

(i)

Atomic no of H2 = Z = 1 Electronic configuration = 1s1

(ii)

Second (H) atom = Z = 1 E.C

= 1s1

Bond Formation:

Since “1s1” orbital (half filled) of one hydrogen atom overlap s with 1s1 half-filled orbital of other hydrogen atom and form a single covalent bond. 3.

HF Formation: HF is composed from Hydrogen if flouring so consider the

Atomic No of (H) = Z = 1 E.C

= 1S1

Atomic No of (F) = Z = 9 E. Configuration = 1S2, 2S2, 2P5

 2P5 = 2Px 2Py, 2Pz Bondformation:

So 1s1 half-filled orbital of hydrogen atom combine co mbine with 2P21 half-filled orbital of fluorine and form sigma bond. As shown above iii.H2O Formation:

Central atom – O O2 Z=8 E.C = 1s2, 2s2, 2p4 (H) = 1 = 1s 1 (H) = 1 = 1s 1 Bond:

Since the bond is formed between the half-filled orbitals (2Py & 2Pz) of oxygen and 1s 1 half-filled orbitals of two hydrogen atoms. 2py1 – 1s 1s1 bond 2Pz – 1s 1s1 bond Plane of bond:

The plane of bond lies li es in that direction where the overlapping of orbitals occur. Conclusion:

The direction of bond is in “J” shaped manner so the “ ” should be 90 0 but the actually it is 104.50. So V.B.T fails to explain the structure of

water. Sigma Bond

PI – PI – Bond Bond ( ) 1. Definition:Theb

1. Definition: The bond formed by the lateral, or sidewise or parallel overlap of half-filled half-filled orbital will be called called -bond.

39



ond formed by the linear overlap or head to head overlap that will be called sigma bond.

2. Maximum Electron Density

2. Maximum Electron Density

The maximum electron density region is present in between the nuclei of the bonded atoms.

Maximum electron density is present above and below the bond axis or nuclei.

3. Stability

3. Unstable

Sigma bonds are more stable to external reaction

They are not stable like “ ” bond

4. Less Energetic

4. More Energetic

They are less energetic in nature. So more energy is required for

They are more energetic than  bonds so less energy is required for

breaking “” bond.

breaking “” bond.

5. Second definition

5. Second Definition

When the bond axis of overlapping orbitals lies in the same axis is called sigma bond.

Where the bond axis of overlapping orbital does not lie in the same

6. Strong Bond

axis is called “ ” bond. 6. Weak Bond

It is very strong bond when formed

It is weak bond in nature.

7. Representation

7. Representation

It is represented by “ ” sigma 8. Note

It is represented by “ ” (pi) 8. Note In double bond one bond is considered as sigma and other is “ ”

All single bonds are sigma bonds i.e. Alkane H | H – C – H |  H Single bonds ()

bond i.e. Alkene H H | | C=C | | H H

9. Note

In triple bond one bond is “ ” and two bonds are “ ” 9. Note B

The first bond formed during a chemical reaction will be -bond.

The “” bond will formed, if there is sigma bond present already otherwise no “ ” bond will formed.

10. Sigma bond formation

10. -bond formation

It is formed by the overlapping of the following orbitals S-S -bond i.e. H2 S-P -bond i.e. HF S-SP ie Alkynes S-SP2 Alkynes S-SP3 Alkynes (P-P) If linear

It is formed by the side wise overlap or parallel overlap of the following orbitals;

   ⟹    {  }

(Sidewise)

: 11. Un-susceptibility

11. Susceptibility

In “ ” bond the electrons are not susceptible or exposed to chemical

In  bonds the electrons are more susceptible or exposed to chemical attack. 12. It has a nodal plane 13. Free rotation does not occur around the bond axis.

reaction. 12.-bond has no nodal plane. 13.It can rotate freely around the bond axis. Note: (2011 Entry Test) which one is more susceptible to chemical reaction.

a. b. c.

Alkane Alkene Both

40



O2 Formation:

As oxygen molecule is composed from two oxygen o xygen atoms. So [O] = Z = 8 Electronic. Configuration = 1S2, 2S2, 2P4 Second Oxygen atom:

[O] = Z = 8 Electronic. Configuration = 1S2, 2S2, 2P4 Bond formation:



So the “2

́

” half filled orbital of one oxygen overlaps with the “2P ” of other oxygen and form “ ” bond.

́

And the 2Pz orbital overlaps with the 2P orbital orbital of one O2 will other oxygen atoms and from “” bond.

Structure of Ammonia: (NH3 formation) The central atom in ammonia is nitrogen, and the atomic no of nitrogen is; 7 Central atom = N Atomic No = Z = 7 E.C = 1S2, 2S2, 2P3 Atomic No of [H] = 1 = 1s 1 Atomic no of [H] = 1 = 1s 1 = 1 = 1s1 Bond Formation:

Three half-filled orbital’s of nitrogen (2Px, 2Py, &2Pz) overlaps with the 3 half-filled orbitals of 3 hydrogen atoms and forms 3-sigma bonds as shown above. Conclusion:

Since the plane of bond lies in that direction where overlapping of orbital occur so plane is “J” shaped since the “”should be 900 but actually the “” in NH3 is 107.5 which is & so V.B.T fail. Not explain by V.B.T. Drawbacks of V.B.T:

Valence bond theory has the following draw No space backs. 1.

Ionic Bond:

It gives no information and explanation about the ionic bond formation. 2.

Para and diamagnetic concept:

It does not explain the paramagnetic and diamagnetic substances. 3.

Note:

Moving from left to right the ionization increases but the ionization energy of nitrogen is greater than oxygen why…….?

a. b. c. d.

Nitrogen is gaseovs gaseovs Nitrogen has small small size Nitrogen has half-filled half-filled orbital None of them

Paired and Unpaired electrons:

No explanation is given by V.B.T about the paired and unpaired electrons. electrons. 4.

Spectra = 0

Does not explain spectra. Concept of Hybridization:

i. ii. iii. iv. v.

Concept of hybridization was introduced by Pauling & Slatter. It is exothermic process. It is physical process. It is hypothetical process. It is theoretical th eoretical process.

Background:

(i)

According to V.B.T only half filled orbital can form chemical bond. But in methane carbon form four bonds although carbon has two (half-filled orbital). So it can be explained by hybridization i.e. C = 1S2, 2S2, 2P2

It is because of hybridization in which carbon form 4 half-filled orbitals. General Concept:

Let TWIGSSchoolhas 4 rooms with different sizes, shape and this is general Hybridization angle if we decide to d estruct it and rebuilt it again with new same shape, size, and same angle.

Definition:

41



The process of intermixing of orbitals of different shape and energy to form a set of new degenerate orbitals with same, shape, and energy is called – hybridization. hybridization. Degenerate Orbitals:

Those orbitals which have same energy are called degenerate orbital. Type of Hybridization:

There are there types of hybridization. 1. 2. 3.

Sp3 Sp2 Sp

1.

SP3 Hybridization:

The process of intermixing of one(s) orbital orb ital and three p-orbitals to form a set of 4 new degenerate orbitals is called SP3 – hybridization. hybridization. Geometrical Shape:

SP3 hybridized orbitals will have tetrahedral shape. Bond Angle:

The bond angle will be 109.5. For Example:

1. 2.

Consider methane, in which carbon is S P3 hybridized. i.e. One “s” and 3 -p intermixes. Consider the atomic no of C: Z=6

1)

Note:

All the members members of IVA VA & VIA are sp Hybridized.

The shape is lik e “P” because p -character is maximum. S : P 1

:

3 75  p-characters is maximum.

25 : 2.

Sp2 – Sp2 – Hybridization: Hybridization:

The phenomenon of intermixing of one S-and 2-p 2 -p orbitals and form three new equivalent orbitals is called sp 2-hybridization. Shape:

The shape will be, trigonal or linear or coaxial (same axis) or coplanar. Angle:

The accepted bond angle is 120 0 or 117.80 ie (SO2) For example: 1.

Consider ethene in each corbon is sp2 cusbrized consider the atomic No of C c= 6 Ec = 1S2 2 S2 2 P2

Note

i. ii.

2.

2pz does not take part in hybridization and remain unhybridized, and lies perpendicular to the bond axis. If form bond.



Sp-hybridization

The intermixing of one “S” and one p orbital to form two new orbital is called sp hybridization. Shape:

The shape is linear. Angle:

The angle will be 1800. For example: 

In acetylene one “S” and one “P” intermix.



In butylenes each carbon is sp hybridized hybrid ized so consider atomic No of carbon. C=6



42

Nasrat Ullah Katozai (Chemistry) 


Ec = 1S2 2S2 2P2.

Note:

S

:

P

1

:

1

50%

:

50%

Note:



2py & 2 does not take part in hybridization and remain perpendicular to the bond axis. Steric No:

The no of sigma bond bo nd + lone pair-of electron is called steric number. Note: Page # 74 (ETEA) Chapter # 6MCQs No. # 43

1. 2.

It steric no = 4, hybridization will be Sp3. It steric no = 3, hybridization will be Sp2.

MCQs: Consider NH 3, H2O, H 2S, and CH4 the central atom is …….. hybridized.

3.

It steric no = 2, hybridization will be Sp.

(a) Sp3

(b) Sp2

(c) Sp3& Sp2

(d) All

For Example:

What will be the hybridization in of carbon in CH4? 1. 2. 3. 4. 5.

+ Lon pair 4 +0 = 4 Hybridization = Sp3 C2H6=  + Lon pair 4 + 0 = 4 Hybridization = Sp3 C2H2=  + Lone pair = 2 + Zero = 2 Hybridization = Sp C2H4=  + Lone pair. 3 + Zero = 3 Hybridization = Sp 2 NH3 =  + Lone pair. 3 + 1 = 4 Hybridization = Sp3 CH4=

Structure of Methane:

Note:

During hybridization the carbon changes its structures.

In methane 4 SP3 hybridized orbital react with 4 hydrogen Structure of Ethane:

In ethane two Sp3 hybridized orbital react with each other. Molecular Orbital Theory:

This theory was put forward by Hunds and Mulikan”. Objectives:

1.

It explains covalent bonding.

2. 3.

It explains spectra. It explain paramagnetism. paramagnetism.

Note:

i. ii.

Shape depends only on the -bond.  Bond only gives stability & strength.

Postulates:

It has the following postulates. 1. Bond Formation:

For the formation of a covalent bond, half-filled atomic orbital overlaps to form molecular orbital. Atomic Orbitals:

Those orbitals which have only one nucleus is called atomic orbital.

 Atomic orbital Molecular Orbitals:

Those orbitals in which two nuclei are buried. b uried.

 Molecular orbital 2. Lose of Originality:

After the formation of molecular orbital, the atomic orbitals lose their originality. 3. Molecular Orbital:

Formation of new orbitals occur which is called molecular orbitals. No o f atomic orbitals  No of molecular orbitals. Note:

Molecular orbitals are of two types: Molecular orbital

43


(a) MBO Stable Low Energy Constructive interference As the number of electron electron increases increases Stability increases


(b) ABMO unstable High E Destructive interference As the number of electron electron increases instability increases increases Lose and gain of electron occur

4. Law of Conservation of Energy:

The total energy of the Bonding Bondin g Molecular Orbital and Anti Bonding Molecular Orbital is equal to the total energy of atomic o rbitals. 5. Filling of Electrons: Electron will first go to Bonding Molecular Orbital & then to Anti Bonding Molecular Orbital. 6. Distribution of electrons: The distribution of electrons in Bonding Molecular Orbital & Anti Bonding Molecular Orbital will be according

to the general rules of electronic configuration ie. (1) Afbau principle (2) Hunds rules

(3) Pauli exlusion principle

7. Bond Order:

Bond order means the no of bonds formed i.e. (1) (2)

If bond order  1, its means single covalent bond is present If bond order = 2, its means double covalent bond.

(3)

If bond order = 3, its means triple covalent bond.

8. Formula:

Bond order = OR

 . .  −      

B.O = Total pair of es in BMO – Total Total pairs of es in ABMO 9. Valence Electrons:

Valence electrons will be under the influence of two nuclei. Energy Diagram: 1.

For Example:

Consider the formation of hydrogen molecule. Atomic no of hydrogen = 1 : Electronic Configration = 1s1 Atomic No of second hydrogen = 1 : Electronic Configration = 1s1 So we will write the atomic orbitals 1s 1 s1 of both hydrogen atoms. Bond Order:

    −  − B.O =  = 2/2 = 1 B.O =

Conclusion:

Since bond order is 1 which means single covalent is present in H2 molecules. Bond order can also be d etermined as; 1 – 0 0 = 1 2.

Example

+1

H2 is …… in nature? H 2+1 = means that only on ly one electron is present. or one electron lose. Since:

BO =

− =  = 0.5  

Conclusion: Its mean physical force is present. 3.

H2 – 1:

Its means that hydrogen has gain one o ne electron. Bond Order =

−   = 0.5  

 

One unpaired electron It is paramagnetic p aramagnetic

4.

Structure of N2 Molecules: i. ii.

N2is composed form two Nitrogen Consider atomic No of N.

Bond order = 3 – 0 0 = 3 OR

− = 3 

44



Conclusion:

i. ii. iii.

The bond order is 3- its means that triple covalent bond ispresent in N2- molecules. No unpaired electrons Diamagnetic

Order of Energy:

 px<2py = 2pz <2py = 2pz <2px 2px – 2p 2p = =  bond

́ ́ ́

2py – 2p 2p = =  bond 2pz – 2p 2p = =  bond 5.

Structure of N2+1

̅

N2+1 = means one is is lost. Unpaired e = 1

Paramagnetic in nature: B.O = Note:

−   = 2.5  

  6.

Paramagnetic One unpaired electron N2+2 +2 means that it lost 2 its. Mean that the aomic no of N 2 is 7 but 2 electron gas loose by N 2 so 2 = 5.

2

1s , 2s2, 2p1 Bond Order:

−   = 2  

Unpaired e = 2: paramagnetic Oxygen:O2 is paramagnetic in liquid state because of 2 the presence of electron elections on th e basis of MOT.

e.c = 1s2, 2s2, 2p4 Bond Order:

−   = 2  

̇

:

( 2 2 Pz)

Unpaired es: 2

 (Degenerate)

Conclusion: Since: in oxygen there are two unpaired electrons in the above so O 2 is paramagnetic in nature.

i. ii. iii. iv. v. vi. vii. viii.

Feasibility of Bond: MOT predict that either A bond is feasible or not. e.g. He 2 bond is not possible because it B.O is zero. Bond Order: MOT predict the bond order and bond order shows the following properties. Bodin g Molecular Stability of Molecules: If a molecule has greater no of electrons in the Bonding Molecular Orbital than Anti Boding

Orbital then the molecule will be more stable. Bond Dissociation Energy:AsB.O increases bond the dissociation energy will be increases. Bond Length: Greater bond order less will be the bond length. Paramagnetism: Its shows that a molecules is paramagnetic or d iamagnetic. Spectra:It also shows spectra. Dipole Moment: (Orientation)

Polarity is expressed in term of dipole dipo le moment. Polarity means “the concentratio n of positive and negative charges on the same or on each end of molecules. Definition:

The cross product of charge and distance is called d ipole moment.

MCQs:Note: entry test 2011: MCQs = 162, Paper = A

1.

Mathematically:

Dipole moment = charge  distance

 = e  d

Which one is more unstable (a) He2+1 (c) He2+2

(b) He2 – 1 (d) He2 – 2

√

The unstability depend, on the no of electros As in the Anti Boding Molecular Orbital. If the no electrons is greater in the ABMO unstability will be greater.

Representation:

The lose and gain of electron occur from the Anti bonding molecular orbital not from Bonding molecular orbital.

It is represented by “mu” “”

MCQs No. = 8, Paper A, Medical

Unit:

2.

Which compound shows the highest Boiling point (a) C2H6 (c) CH3OCH3 (d) C2H5OH

(b) C2H5

√

45



Its unit is deby (D) 1D = 3.34  10-30 C.m. MCQs:

200is equal to coulomb meter. Since 30 1D = 3.34  10 – 30 Cm 200 = x

By cross multiplication 30 X  1D = 200  3.34  10 – 30 Cm

X=? Nature:

It is vector quantity Notation:

It is denoted by an arrow directed from positive to negative end. Explanation:

1.

Suppose we have two plates one is positive and other is negative, in a container as shown above.

2. 3.

And if a bundle of molecules are dropped into it and the battery is OFF. Now if the battery is on, then all the molecules will orient itself in such such away that the positive end of the molecule come in contact contact with the negative plate and the negative n egative end of molecule come in contact con tact with the positive plate as shown. sho wn.

Conclusion:

1.

If the applied voltage decreased then its means that the molecules are polar.

2.

If the applied voltage remain constant then its means that the molecules are non polar.

3.

If the applied voltage decreases decreases more then its means that the molecules are more polar.

Applications of Dipole Moment:

There are two applications 1. Polarity determination 2.

Geometry determination 1.

Determination of Polarity:

It is used for the determination of polarity of molecules. i.e. If  = 0 then molecule will be non-polar. For Example;

H2, Cl2, Br 2, F2, CO2, CH4, BF3. If > zero Molecules will be polar For example:

HBr, HI, NH 3, HCl For Entry Test:

 × 100     % polarity =   × 100  / × 100 % ionic characters =  /

1.

% ionic character =

2. 3.

 = e × d 2.

Determination of Geometry:

Dipole moment also determines the shapes of molecules. i.e. i.

Dia atomic Molecules:

If the molecule is composed from the th e two atoms, then it will be called diatomic molecule. If  = 0 OR

> O Shape:

Then its shape will be linear 46



For Example:

HCl, HBr, HI, H2, Br 2, Cl2, CO Note:

All diatomic molecules have got linear geometry, if either they are homoatomic or heteroatomic molecules. H2 ____Homo HCl ____Hetro 3.

i.e. All the members II A form linear structure

Tri Atomic Molecules:

They have two possible structures i. ii. a.

Linear Angular Linear Shape:

i.e. members of VA and VIA form angular structures i.e. H 2O, N 2O, SO2, H2S.

If  = O Molecules should be non-polar. (i) CO2 (ii) SiO2 (iii) BeF2 (iv) BeCl 2 b.

Angular Shape:

If > Zero, molecule will be polar p olar

Tetra atomic molecules 4.

Tera Atomic Molecules:

They have also two possible structures. Trigonal planner Pyramidal

 

Triangular planner: If = 0

= 120 Hybridized: sp ion example = BF 3 all the member of IIIA. 2

Pyramidal: If

 

> 0 = 107.5 Hybridization: sp polar molecules example = NH3 all the member of VA. 3

Pentra Atomic Molecules:

Penta atomic molecules has got tetrahedral shape; For Example:

CH4, SiH4, CCl4, SiCl4, CH2Cl2, CH3Cl, CHCl3 etc. Penta Atomic Molecules:

Symmetrical

Asymetrical

If a molecules has  = 0 Non Polar

If a molecules having > o polar

For Example:

For Examples:

CH4

CH3Cl

SiOH4 GeCl4

CH2Cl2 CHCl3

SnCl4

It is also called unsymmetrical

Choose the best answer. 1)

2)

3) 4) 5) 6)

Bond is formed because an atom a) Tends to lose an electron. b) Tends to gain an electron. c) Tens to complete its octet. d) All of above. Bond formation between atoms a) Increases the energy and decreases d ecreases stability. b) Decreases Decreases the energy and increases stability. c) Increases the energy and Decreases stability. d) No change in energy. Which one of the following molecule has zero dipole moment. a) NH3 b) NF3 d)BF3 e) H2O Which one of the following compounds has the heighest ionic character? a) HF b) HCl c) HBr d) HI 2 Specie in which the central atom uses sp hybird orbital in its bonding is: a) PH3 b) NH3 c)+CH3 d) sbH3 Liquid oxygen is: 47


7) 8) 9)


a) Diamagnetic b) Paramagnetic c) Antimagnetic d) None. Which of the following atoms is reluctant to form double bond with another identical atom. a) Car bon b) Oxygen c) Silicon d) Nitrogen Carbon monoxide molecule possesses ________ covalent bonds. a) One b) two c) three d) four The bond angle in H2O is: a) 900 b) 109.50 c) 1800 d) None 1) c 4) a 7) c 10) b 2)

b

5)

c

8)

b

3)

c

6)

b

9)

d

11)

b

Short Questions:

Explain with reasons. 1.

CO2 molecule has linear but H 2O molecule in an angular structure.

Answer:

CO2 has linear structure due to zero dipole mo ment while water has angular structure due to 1.8 Debye dipole moment. a.

CO2:

i.

CO2 is a triatomic molecule. To have zero dipole moment its structure should be linear so that C=0 bond moment cancel the effect of each other. In CO2 molecule the dipoles are equal and opposite and cancel out the effect of each other.

ii.

→

O-8==C+8 b.

== O-8

H2O:

i.

Water is also a triatomic molecule but its structure is angular Reason. It is because experimental data shows that water has dipole moment equal to 1.8 D. This shows that water has an angular structure so that the two equal O-H bond moments do not cancel each other. USEPR theory: according to USEPR theory thr lone pair of oxygen repel the bond pair in H2O.

ii.

CONCLUSION:

Due to zero dipole moment CO 2 has linear structure and water has 1.8 D dipole moment, so its structure is angular. 2.

O2 molecule is paramagnetic while N2 molecule is a diamagnetic.

ANSWER:-

O2 molecule is paramagnetic due to 2 unpaired electrons in anti bonding molecular orbital’s while N 2 molecule is diamagnetic due to Nounpair electron in bonding molecular orbital. OXYGEN:

Oxygen is paramagnetic because it has two unpaired electrons in ABMO (anti- bonding molecular molecular orbital’s) the filling of electron in O 2 molecule occur according to Hund’s rule and Afbau’s Principle so oxygen get two unpaired electrons, and shows paramagnetic. NITROGEN:

According to molecular orbital theory, nitrogen do not have any unpaired electrons so it is diamagnetic in nature. CONCLUSION:

Due to the presence of unpaired electrons, oxygen shows paramagnetic behavior and nitrogen shows diamagnetism because it does not have unpaired electron. 3.

Bond angle in NH3 II 107.5 0 and H2O molecule it is 104.5 0.

Answer:

According to VSEPR theory, the overall geometry of a molecule depends upon the number of bond pair and lone pair of electrons in the valance shell of the central atom. a.

Ammonia:

In case of NH 3, there are three bond pairs and one lone pair of electron. According to VSEPR theory, the lone pair repulsion is greater, so it occupy more space. As a result the bond angle in NH 3 becomes 107.50 than normal tetrahedral angle of 109.5 0. b.

Water:

According to VSEPR theory, water molecule has two lone pair and two bond pair, so repulsion of lone pair is more greater than NH 3 and hence bond angle decreases decreases to 104.50. CONCLUSION:

Ammonia has bond angle equal to 107.50 due to presence of one lone pair and water has 104.5 0 due to having two lone pair of electrons.

48

Nasrat Ullah Katozai (Chemistry) 4.


All the four bond in CH4 are equal.

Answer:

In methane (CH4) carbon atom forms four identical bonds with four hydrogen’s using four identical sp3hybrid orbital’s. These new orbital’s are the result of sp 3 hybridization. Each sp 3hybird orbital is composed of 25% character of s and 75% character of p. the angle between any two sp3hybird orbital is 109.50. Conclusion:

Due tosp3hybridization, carbon form four identical bonds in CH4 molecule. 5.

Sigma bond is stronger than Pi bond?

Sigma Bond:

Sigma bond is formed by parallel or linear overlapping of two orbitals along the same axis. The charge density is greater between the two nuclei. Therefore it is a strong bond. PI BOND:

Pi bond (π-bond)isformed by sidewise or perpendicular overlapping of orbital. REASON:

In sigma bond the electronic density is maximum in between the two bonded nuclei. The shared pair of electron are attracted by two nuclei and as a result the decrease in energy occur, so the stability of sigma bond is more than the pi bond. In case of pi bond the orbital has two regions of electronic cloud density i-e., above and below bond axis unlike in the sigma bond where electronic density has one region around bon d axis. This overlap is not maximum, therefore Pi-bond is weaker than sigma bond. CONCLUSION:

Sigma bond is stronger than pi bond due du e to maximum overlapping of orbitals. 6.

Bond energies of polar molecules are greater than non polar molecule?

Polar Molecules:

The molecules having two poles i-e positive po sitive and negative (dipole) are called polar molecules. Non-Polar Molecules:

The molecules which don’t have dipoles are refered as Non -polar molecules. Reason:

The bond energy of polar molecules is greater than that of non-polar molecules. As the polarity increases, the bond energy also increases. In halogens the polarity decreases decreases in following order, HF is more polar than HCL and so on. HF > HCI > HBr > HI HF has higher bond energy due to higher polarity than HCl. Conclusion:

As polarity increases, bond energy also increases, so polar molecules has higher bond energies than non-polar. Resonance: The representation of a single molecule by two o r more than two possible structure is called resonance. For Example: Benzene can be represented by two possible structures as. Discovers = 0

Resonance was discovered by Hysenberge.  

It is Hypothetical process. It gives stability.

49


CHAPTER-4:


GASES

Definition of Gases:

The state of matter which has neither specific shape nor volume is called gas. Note: It is due to weak intermolecular forces. Properties of Gases:

Following are the properties of gases. 1.

Expansion:

Gases has the ability of expansion due to reduced pressure. 2.

Pressure:

i. ii. iii. 3.

Gases exert pressure on the walls of container. Pressure of the Gas is because of collision with each other and with the walls of container. Pressure of the Gas is increased by increasing Temperature. Low Density:

Gases have low density. It is 10 – 3 times less dense than liquid & solid. Example: The density of liquid O2 is 0.00142 gcm – 3 liquid 02 = 1.149gcm – 3 solid 02=1.426gcm – 3 Diffusion:

Gases can diffused into each other because of Random motion of molecules in all directions. Compressibility:

i. ii.

Gases are compressible because of greater intermolecular distances are present between them. When pressure is increased empty spaces between the molecules decreases. decreases.

KINETIC MOLECULAR THEORY OF GASES:

This theory explain the physical behavior of gases. It was proposed by “Bernoulli”. “Bernoulli”. Postulates:

Its main postulates are the following. 1.

Composition of Gases:

All the gases are co mposed from small particles called molecules. 2.

Kinetic Energy:

Gases molecules have greater kinetic energy than those of liquids and solid. 3.

Intermolecular Distance:

Gases molecules are for away from each other because of larger intermolecular distance. 4.

Random Motion:

They moves randomly because they collide with each other and with the walls of container. 5.

Actual Volume:

The actual volume of the gas molecule is very small as compare to the distance between them so therefore, actual volume of the gas is negligible. 6.

Intermolecular Distance:

The force of attraction between gaseous molecule are negligibly small. 7.

Average Kinetic Energy:

The average kinetic energy of the gas molecule is b ecause of motion. 8.

Note:

At the same temperature, molecules of every gas have same average kinetic energy. 9.

Effect of Gravity:

There is no effect of gravity on the gas molecules. 10.

Elastic Collision:

The collision of the gas molecules are perfectly elastic. Monometer:

That apparatus which is used for measuring the pressure of a gas is called monometer. 50



Barometer:

That monometer which is used for measuring the atmospheric pressure is called barometer. Torricellian Barometer:

The most common type of Barometry is called torricellian Barometer. Composition of Torricellian Barometer:

The test tube is first completely filled with Hg and the end is closed with a tumb. 1.

Then it is inverted in a tough of Hg.

2.

The coloumn of Hg stands at a height “h”

3.

The atomsophere exert pressure on the Hg due to which it goes down to 760mmHg.

4.

The substance whose pressure is to be determined is placed under the tube it will evaporate and exert pressure on the surface of Hg due to which Hg goes down and cover a distance “h” so vapour pressure can be determined as;

p = h2 – h h1. Pressure and its unit:

The pressure can be defined as a force perunit area”. P = F/A Standard Pressure:

The pressure at 273.15k is called standard pressure. It is equal to; 76 cmHg 760 mmHg 760 torr Units of Pressure:

The S.I unit of pressure is Nm – 2 or Pascal One atm:1P = 1Nm – 2

The force exerted by 76 cm long coloumn of Hg on an area of 1cm2 at 273.15k is called one atm. Ideal Gases Laws:

Note: MCQs

Boyle’s law: Boyle’s law:

This one atm can support. 760 mmHg at sea level.

This law was putforward by Robert Boyle in 1662 in order to explain the relationship

1 atm = 1.01325 Ba 1 atm = 14.7 Psi (pound per square inch)

between pressure pressure and volume at constant temperature. temperature.

1 atm = 760mmHg

Statement:

1 atm = 760 torr

The volume of a fixed mass of a gas is inversely proportional to

1 atm = 101325 pascal

the pressure if the temperature is kep t constant”.  K = boyle’s law constant.

1 atm = 760 mmHg 1 atm 101325 nm – 2

Mathematically:

V  1/p PV = k Since this law can also be defined as “The product of pressure and volume for a fixed mass of a gas remain constant at constant temperature”. For initial state: Test MCQ:

(a) for initial state Boyls law be written as: For final state:

When pressure is applied to a balloon then its volume increase

(a) P1V1 = K 1  (1)

but according to Boyle’s law when pressure increases the

volume decreases but in case of balloon we increase no of moles (mass) so volume increases?

K 1 = K 2

P2V2 = K 2 So comparing (a) and (b) P1V1 = P1= V2P2/V1

P1 =V2P2/V1Boyl’s law equation Graphical Representation:

There are three types of Graph. 1. Volume Verses Pressure:

If we plot a graph between pressure (p) on x-axis and volume v on y-axis an isotherm isoth erm is obtained at constant temperature calledhyperbola. For example 2. Volume Verses 1/p:

The volume is directly proportional to the reciprocal of (p). 3. “PV” Verses “P”

V

1/P 51



For all values of pressure the values of PV will remain constant. Units of Bayle’s Law: As we know

K = PV K = atmdm3 or atmL Note:

The values of “k” depends upon the following 3 factors.

(i) Mass (ii) Moles(iii) Temperature i.

Mass:

If mass increase, the volume will be increases due to which the value of k will be increase. ii.

Moles:

If no of moles increases in creases volume will be increases due to which the values of (k) will be increases  iii.

Temperature:

As the temperature increases the volume increases due to which the value of k will be increases. Experimental Verification:

Let us consider a cylinder which have a movable piston. The cylinder is attached to monometer which measure the pressure. 1st Experiment:

In this experiment pressure is 2 atm and volume is 1 dm3. 2nd Experiment:

In this experiment pressure is 4 atm and volume is ½ dm3. 3rdExperiment:

In this experiment pressure is 6 atm and volume is 1/3 dm3. Truth Table Exp

Patm

Vdm3

Pv = constant

1

2

1

2

2

4

½

2

3

6

1

2

/3

Charles Law:This law was put forward by Charles in 1787, in order to explain the relationship between volume and absolute, temperature. Statement:

Note:

This law states that “The volume of a fixed mass of a gas is di rectly proportional to the absolute temperature.

Boyl’s law table

Mathematically:

Numerical for for test: V1 = 5 dm3, P1 = 1atm

It can be written as; V  T

 Charles law constant

V = KT Rearranging

 = K 

V2 = 1 dm3, P1= ?

  × P =  P2 = 2

P2 = 5atm

For Initial State:

 = k 1 ––  (i) 

For final State:

 = k 2 ––  (ii) 

Since: k 1 = k 2 Comparing (i) and (ii)

 =  v1 =   T1

Unit:

As we know k = V/T



K = 

52



K = dm3 k 1 OR K = Litre. K -1 The value of “k” depends upon.

1.

Mass (quantity of gas)

2.

Pressure

3.

No of moles 1.

Mass (Quantity)

If mass is increases volume will be increases due to which the volume of k increases. 2.

Pressure:

If pressure is increases no of mole volume will be decreases due to which the values of k decreases. decreases.

↑

Mole: if n increase V due to which the values of (k) in creases. creases.

3.

Graphical Representation: 1.

High Pressure:

Line 3 is obtain at high pressure volume 2.

Intermediate Pressure:

Since 2 is obtained at intermediate pressure. 3.

Low Pressure:

Line 1 is obtained at low pressure. Experimental Verification:

Consider a cylinder having a movable piston. Suppose a fixed mass of a gas has taken in a cylinder. Its volume is V 1 and temperature is T 1. Now if the temperature temperature is increased from from normal. The new temperature temperature will be T2& volume will be V 2.

  

Truth Table

Avogadro’s Law:

This law states that equal volumes of ideal gases contain equal no of molecules at the same temperature and pressure at STP. OR Different gases of equal volume under similar conditions of temperature and pressure contain equal no of molecules or atoms at STP. Note:

A definite volume of any gas contains a definite no of molecules. Mathematically:

V  n V = kn K=

 

Unit of Avogadro’ Avogadro’s Law:

As we know k = 3

 

Note:

i. ii.

Avogadro’s law is independent of the size & shape of molecules.

Also independent of the I.M.F & intermolecular distance.

– 1

K = dm /mol or K = L mol Molar Volume: 

On mole of any gas at STP occupy a volume of 22.414dm3 which is called molar volume.



As one mole of gas has Avogadro’s no of particles so

22.414dm3 of gas at STP will have Avogadro’s no of molecules i.e. 6.02  1023 Note:

22.414dm3 of a gas at 273.15k and one atmospheric pressure has = 6.02  1023 molecules 2nd Explanation:

If we have 1dm3 of each of H2, O2, N2, and CH4 in separate vessels at STP, the no of molecules in each will be 2.68  1022 No. of molecus = NA/22.414 NA/22.414

53


=


. ×  .

No. of molecule = 2.68  1022 Now when the T & P are equally changed for these these four gases, then the new equal volumes will have the same same (2.68  1022) no of molecules. Absolute Zero:

The hypothetical temperature at which the volume of a fixed mass of a gas becomes equal to zero is called absolute zero. Absolute Zero  – 273 273 C0 Absolute Zero = 0 k Absolute Zero = – 40 40 F0 Concept:

Practically it is impossible to have zero volume o f a gas because before reaching this temperature all gases liquefy or solid ify. Note:

At this temperature all type of motion of the gases molecules ceases therefore the kinetic energy become zero. Experimental Verification of Absolute Zero:

Step – I I In this experiment Hg – is is placed in a glass tube. Sealed at one end in such away, that Hg – form form a frictionless piston. The air is intrapped below Hg – plug plug in the closed end of the tube. Step-II: Now placed the tube inhot water water bath due to which the volume of gas increases. increases. The external environment and Hg – plug plug exert pressure on the trapped air. Since:

As the temperature of boiling water increases the volume of air also increases. Note:

It has been found, experimentally that for a number of gases, for every degree rise in temperature, the volume of a gas increases by 1/273 or 0.00362 of its original volume at zero O0C at constant pressure. New volume at 100 0C = original volume at O0C + 1/273  original volume at O0C. Similarly for every degree decrease in temperature the volume decreases by 1/273 of its original volume. New volume = original volume volume + Suppose:

  × original volume 

Suppose we have a gas sample. Whose original volume is 273 ml at O0C. TC0 = 273 = TC0 = 100 = TC0 = 2 = TC0 = 1 = TC0 = 0 = TC0 = – 1 = TC0 = – 100 100 = TC0 – 273 273 = Conclusion:  

 273mL = 546 mL   273mL +  273mL = 373 mL  273mL +  273mL = 275 mL  273mL +  273mL = 2.74mL  273mL +  273 = 273 mL − 273mL +  273mL = 272 mL − 273mL +  273mL = 173 mL − 273mL +   273mL = 0mL 273 +

Decreasing Decreasing the temperature down to – 273 273 0C would make the volume of a gas Zero. But such a situation cannot occurs, because all gases liquefy or solidify at such a lo w temperature.

Ideal Gas Equation:

That equation which shows the relationship between volume, pressure, temperature and number o f moles is called ideal gas equation. Derivation:

It can be derived by combining Boyle’s law, Charle’s law, and Avogadro’s law. From Boyle’s Law:

V  1/P ––– 

(1) 54



From Charles Law: V  T

––– 

(2)

From Avogadro’s Law:

V  n –––  (3) Now combining (1) (2) & (3)

Note: Universal gas constant is independent of the

nature of the gas.

V  1/P T.n

   V =   R is universal un iversal gas constant V

Rearranging:

PV = RTn (4) Standard mole for any gas n = 1 PV = RT  (5) So equation 5 can b e written as Rearranging:

R = PV/T –– 

(6)

 –– (7)    For final state:R =  –– (8)    Since R = R So     

For Initial state: R 1 = 2

1

2

This is the generalized form of ideal gas equation. Application of General Gas Equation

It can be used to determine the

    1.

The value of “R” at STP The value of “R” in SI system

Density of gas () Mass of gas Calculation of Value of “R” at STP:

According to general gas equation: PV = nRT

 (i)  By putting these values in equ (i)  ×.  R =  × R=

(a)

Note: At STP it is known that;

Standard temperature (T) = 273k Pressure (p)= 1 atm

Unit of “R” in atm:

Amount (n)= 1 mol

R = 0.0821 dm3 atm mol – k – (b)

Volume (v)= 22.414 dm3

Unit of “R” in mmHg:

R = 62.4dm3 mm of Hg mol – 1. K – 1 (c)

Unit of “R” in Torr:

(d)

R = 62.4torr dm3 mol – 1 k – 1 Unit of “R” in cm3:

 1dm3 = 1000cm3

R = 62400torr cm3 mol – 1 k – 1

Note: In SI system it is known that

Calculation of Value of R in SI System:

Standard temperature = T = 273k

 R =    ×. R=  ×

Pressure

= p = 1 atm = 101 325 Nm – 2

Volume

(v) = 22.414 dm3 = 0.0224m3

R = 8.3143 Nm – 2 m3 mol – 1 K – 2

55


 Nm = Joul  1 calorie = 4.184 J

R = 8.3143 Nm mol – 1 K – 1 R = 8.3143J mol – 1 K – 1 (1) R = 1.98 cal mol – 1 k – 1 2.

x = 8.314J

. ×   ×. = x = 1.98 cal . .

Calculation of Value of Density of gas:

According to general equation PV = nRT 

 n =

(I)

Equation (1) become

  By rearranging  PM =  RT

PV = RT – 


 

 

 = 

(II)

m/v =  PM = RT  =

3.

 

Calculation of Values of MolarMass of a gas:

According to the general gas equation. PV = nRT Putt the volues of (n) in ––– (1) (1)

  PV = 

 n = –––  (II)

PVM = mRT Rearranging

M=

 –––  (III) 

Ideal Gases

Non Ideal Gases

1.

Those gases which obey all gas laws at all condition of temperature and pressure is called ideal gases

1.

Those gases which does not obey all gas laws

2.

Those gasses which obey ideal gases equation i-e (PV = nRT)

2.

Which does not obey id eal gases equation

3.

It does not obey Vander wall equation

3.

It obey Vander wall equation

4.

Imaginary gases

4.

Real gasses

5.

Compressibility factor is equal to (1) Z = 1

5.

Z > 1 compressibility factor greater than one.

6.

For ideal gases the graph will be straight

6.

Graph for non ideal gas is discontinues

 Unit: It has no unit: Compressibility factor = PV/nRT = Z

  

All real gasses shows marks deviation from ideal behavior The extent of deviation from ideal behavior is on compressibility factor = Z If z values is greater then gases shows more deviation:

Deviations from the gas Laws:

Gases show ideal behavior at high temperature & low pressure. Causes of Deviation:

The causes of deviation is due to two false point of Kin etic Molecular Theory (KTM). (i) No forces of attraction (2) Zero volume 56



Attractive Forces:

According to KMT there is no attractive forces between the molecules of gases. But actually attractive and repulsive forces are present between the molecules molecules of gases. Note:

At low temperature & high pressure these forces become more significant. 2.

Volume of Gases:

According to KMT the actual volume of the gas isneglibles very small as compared to the total volume”.

But actually gases molecules can have some volume which cannot be n eglected. Note:

Note: Some Special Point:

1) Polar molecules shows more deviation. 2) Non polar molecules show less deviation. Van Der Waals Equation:

This equation was put forward by a Dutch scientist J.D. van der waals” in 1873.

 

Gases show more deviation at low temperature and high pressure. Gases show more ideal behavior at high temperature & low pressure.

It is also called non-ideal gas equation. It is also called real gas equation.

Main Points:

1.

Volume correction

2.

Pressure correction

1.

Volume Correction:

According to KMT the actual volume of the gas is to zero so; V(molecule) = Zero. But When a molecule is free to move then the volume is V(free) Vfree. But when pressure is increased upto some exten t then volume become actual “V” (total). So free volume V (free) = V (total) –– V V (actual) According to K.M.T. Actual volume = 0

Vfree = Vtotal – O O –– 

(1)

According to vander wall when “P” is increased then volume of the gas decreases upto some extent (b)

Vactual = is not equal to zero But Vfree = (V-b). ––  (2) For “n” molecules. nb V(free) = V(total) – nb  (3) From ideal gas equation

PV = nRT ––  (4) Now putt (3) in (4) P (V-nb) = nRT –––  2.

(5)

Pressure Correction:

i. ii.

The attractive forces between the gas molecules in creases by increasing pressure. Suppose a molecule (A) is present in the container the resultant forces on it zero. But when the molecules goes to the wall of container. It is attracted by inword force due to which pressure decreases so. Observed pressure is less than ideal pressure. P = pi – p p/ ––––– (a) (a) Rearranging:

P = observed o bserved pressure (actual pressure) Pi = Ideal pressure P/ = decrease in pressure Pi = p + p / ––––– (b) (b)

57



Vander Vaal suggested that the part of pressure used against inter molecular attraction should decreases as the volume increases, for molecular attraction increases. increases. P/ = a/v2 ––– (c) (c) For “n” molecules

 –––  (d)  Put (d) in (b)  Pi = p +  ––  (e) Put “d” in the ( 5) nb) = nRT      (V – nb) Pi =

1)

Unit of a =

 = 

a = 2)

   a=

Unit of b:

   =    = atmdm mol   6

–

Vnb b =

 

 b = dm b =

3

mol –

Statement:

This law states that “The total pressure exerted by the mixture of gases is equal the sum of the partial pressure of the individual gas”.

OR “the total pressure of a mixture of gases is the sum of partical pressures of all the gases present in it”. Mathematically: Application:

Pt = P1 + P2 + P3 + ………… + Unit = atm Partial Pressure:

“The pressure exerted by a single gas in a mixtureis called partial pressure. Total Pressure:

1. 2. 3. 4.

Respiration Respiration at high altitude Pressure of dry gases Respiration Respiratio n under water

5.

Clinical application

The pressure exerted by the sum of all the components of a mixture is called total pressure. Explanation:

Consider 3 non reacting gases A, B & C present in individual container at the same temperature & pressure. The individuals pressures are P 1, P2& P3 1.

Consider the gas (A) in container which exert a partial pressure of 20 atm.

2.

Similarly the partial pressure exerted by gas B is 10 atm.

3.

Similarly the pressure exerted by gas (c) exert30 atm pressure.

So the sum of these partial pressure will be PT = P1 + P2 + P3 PT = 20 + 10 + 30 = 60atm

Gases 1 + 2 + 3

 CH4 = 16g Step-2:

 =1   nH =  = 8 Step-3:  P =   So = Partial pressure of CH  P (CH ) =  +  n = 1 mole 

(n) CH4 = 2

Pt = P1 + P2 + P3

(g)

Reasons:

There is no attractive as well repulsive forces. Pressure

4

4

Note:

The friction part of a gas in a mixture is equal to the mole friction. (1) If total pressure (P T) is not given then.

Note:

Equall masses of methane & H 2 are placed in closed container fraction of the pressure will be? Step-1:consider that molar mass as given mass which is greater. H2 = 16g

Now all these 3 gases A, B & C are placed in a single container & since their pressure will be 60 atm. atm.

Pgass = ng/nt  pressure gas = p =

Example:

 -------------+ +

 n1 = first gas  n2 = 2nd gas

 +  P(CH ) =   P(H ) =  =   = + P(CH4) = 2

2

P (H2) = 8/9

58



If total pressure is given than the following formula is used i.e. P = Diffusion of gases =

 + + × P Total

The spontaneous mixing of different non reacting gases gases is called diffusion. diffusion. Effusion = the controlled mixing or o r passage of gas molecules through a small orifice o rifice is called effusion. Rate of diffusion = the d istance covered by a gets per unit time is called rate of diffusion. Graham Law of diffusion:

This law was presented by Thamus Graham in 1827 . Statement:

This law states that “the rate of diffusion of two gases A & B are inversely proportional to the square root of their densiti es at the same temperature and pressure”. Mathematically:

It can be written as;

      OR

     

Where as r 1 and r 2 are the rate of diffusion of two gases, d 1 and d2 are the densities of two gases and M 1 and M2 are the molecular masses of two gases. Unit:

It has no unit because it is the ratio between two same quantities. Experimental Verification:

Take a 100cm long tube and plug a cotton swab socked in HCl at one end another socked in NH3 at the other end of the tube simultaneously. Now:

The two gases escape out from their solutions and meet at a distance of 60cm from NH 3and 40cm from HCl plug, where they combine they will form a white smoke of NH 4Cl. Reaction:

N()H3 + Hd(b) NH4Cl(Solid)

Note:

Density of Hcl Hcl = 1.66g m – 3 Density of NH3 = 0.76g – 3

Conclusion:

Since the molecular mass mass of HCl is high So HCl moves slowly slowly than NH3. Mathematical Verification:

Two gas NH3& HCl are diffuses into each other so their rate of diffusion will be;

      

OR

Putting the molecular masses;

     

Hydrogen diffuses six times faster than the gas A. The molar mass of gas A? (a) 36 (b) 6 (c) 24 (d) 72

   ×  =   

 HCl = 1 + 35.5 = 365 NH3 = 14 + 3 = 17

 1. 5 

   ..  

2  36 =

 ×1

61   2 361 × 2 72

Conclusion:

Since ammonia moves 1.5 time more faster than HCl. For H2& O2:

Which moves faster.

    .     .                4       4 

Conclusion:

Since Hydrogen moves four time faster than HCl between of low mass. Application: which are the application of Boyl law. It has the following applications. applications. 1. 2. 3. 4. 5.

Molecular weight of unknown gas. Density Rate of diffusion of unknown gas. Velocity of unknown gas. Separation of different gases from mixture.

6.

Separation of Isotopes

Conclusion:

59



Since H2 moves 4-time faster than O 2. Example No. 3:

Which of the following will moves faster. A.

H2

B.

CO2

Note: It is not possible to liquefy a gas by pressure alone if the

required temperature is not obtained.

                  22    √ 22     4. 6 9

Liquefaction of Gases

The phenomenon of conversion of gases in to liquid by low temperature and high pressure p ressure simultaneously is called liquefaction of gases. Note:

i.

If pressure is increased it bring the molecules closer due du e to which attractive forces increases and intermolecular spaces decreases.

ii.

By continuously decreasing temperature and increasing pressure finally the attractive forces overcome the free movement of molecules.

Critical Temperature

i. ii. iii.

The temperature above which a gas cannot be liquefied by pressure alone is called critical Temperature. Method for the liquefaction of gases. There are several methods but we will use the linds method.

Joule Thomson Effect:

Sudden expansion of highly compressed gas causes cooling effect called Joul Thamson effect”. Note: In compressed gas molecules are very close to each and have attractive forces. Reason:

i. ii.

In compressed gas molecules are very close to each other and have greater attractive forces increases. increases. Why Joul thamson th amson effect causes cooling effect.

When a gas is expanded suddenly molecules moves away from each other. This process required energy, which is obtained from the gas itself, that is why Joul thamson th amson effect causes cooling effect. Lends Method:

This methods is used for the liquefaction of gases on the basis b asis of Joul thamson effect. Basic Principle: Joul Thamson effect 1.

Compressor:The air is passed through compressor where air is compressed at a pressure of 200 atm.

2.

Refrigerating Liquid: Now the compressed compressed air is passed through the water water cooled pipe where heat of compression is removed. removed.

3.

Spiral Tube: Now this air is passed through spiral tube having a Jet at the end.

4.

Expansion Chamber: Now the air is passed through the expansion chamber where low pressure is present and sudden expansion

occur & because of sudden expansion the air cool down because of Joul thamson effect and converted into liquid. Repetition:

This process is repeated for several time. Fourth State of Matter:

PLASMA STATE Definition:

The fourth state of matter which is a mixture of neutrals particles, positive ions and negative electrons is called plasma state. Origin of Plasma:

This term was given by Irving Langmuire in1928 to an ionized gas at high temperature. Formation of Plasma:

i.

On heating a solid it is converted into liquid.

ii.

On further heating the liquid is converted into vapours.

iii.

Thus the phase of matter changes from solid to liquid and then liquid to vapours.

iv.

Now if vapours are heated, then on further heating some of the atoms lose electrons and produce positive ions.

Hence:

60



A mixture of neutral particles, positive ion s and negative electrons is formed,which is called plasma, state. Note:

The ionization is produced by high temperature but plasma is neutral as a whole. Occurrence of Plasma:

i. ii.

Plasma is found around the sun and stars. 99% of the universe is made up of plasma.

Note:

The sun is a 1.5 million km ball of plasma. It is heated by nuclear fusion. On Earth:

On the Earth plasma is produced depending on light. When electric current is passed passed through neon gas it produces both plasma & light. Note:

Plasma is microscopically neutral. Application:

Following are the application of plasma. 1.

Computer:

It helps in the working of computers & electronic devices. 2.

Semiconductors:

It is used in semi conductors 3.

Sterilization of Medical Products:

It is also used for the sterilization of medical products & devices. 4.

Fluorescent Bulb:

Fluorescent bulb is different from the regular bulb. It contain a long tube filled with Neon gas, when clectric current is passed through neon it produces glowing plasma. Exercise No. 4

1.

Choose the correct answer.

i.

One dm3 of hydrogen h ydrogen at STP weights approximately. (a) 0.0789g

ii.

(b) 0.0799g

(c) 0.09987g

(d) 0.0899g

Generally, a gas in a closed container will exert much higher pressure at the bottom due to gravity than at the top. (a) True (b) Untrue (c) In some cases true and in some cases untrue (d) None

iii.

In a factory producing liquid air, one of the pipes carrying dry air at 80 0C is blocked with a white solid. What is this white solid? (a) Argon (b) Ice

iv.

(c) Nitrogen

(d) Carbon dioxide

0

A gas has certain volume at 10 C. How much temperature should be raised to double its volume? (a) 566K (b) 283K (c) 293K (d) 283 0C

v.

The rate of diffusion of H2 compared with He is. (a) 0.5 times

vi.

(b) 1.4 times

(c) 2 times

(d) 4 times

Hydrogen gas possesses K.E at the same temperature as compared to oxygen. (a) More (b) Less (c) Some time less & some time more (d) Same

vii.

The molar volume of He is 44.8 dm3 at (a) 1000C and 1 atm (c) 00C and 0.5 atm

(b) 250C and 0.25 atm (d) 400C and 0.5 atm

viii.

At the same temperature and pressure which of the following gases has the greatest density? (a) CO2 (b) SO2 (c) Cl2 (d) H2O

ix.

The number of atoms in 2 moles of CO2 is;

x.

(a) 6.022 × 1023

(b) 12.044 × 1023

(c) 1.8132 × 1024

(d) 3.6132 × 1024

The value of ideal gas constant in dm3 torr K – 1 mol – 1. (a) 0.0821

(b) 1.98722

(c) 62.364

(d) 8.3143

61



Answer Keys: (i)

d

(ii)

b

(iii)

a

(iv)

a

(v)

b

(vi)

a

(vii)

c

(viii)

c

(ix)

a

(x)

c

SHORT QUESTION Answer-1:

One cm3 of H 2 and 1cm3 CH 4 at STP will have the same number of molecules although one molecule of CH4 is 8 – times times heavier than that of H2 it is because of the following reasons. 1.

Avogadro’s Law:

According to Avogadro’s law “Equal volume of all gases contain equal number of molecules at the same condition of temperature and

pressure. 2.

Note:

The mass and size donot effect the volume. Therefore, 1 cm3 of H2& 1cm3 of CH4 will have the same number of molecules at STP. 3.

Proof:

As we have that at STP. 22.4 dm3 of H2 = 6.022 × 1023 molecules. Similarly:

22.4 dm3 of CH4 = 6.022 × 1023 molecules. No. of Molecules in H2 Molecule:

As we have that; 22.4 dm3 of H2 = 6.022 × 1023 molecules. 1 dm3 of H2 =

. ×   ––  (1) . 

But 1dm3 = 1000 cm3

So converting dm3 into cm3

. ×   ––  molecules .  So . ×   1cm of H =. × 1000 = 10

100 dm3 of H2 = 3

2

– 3

1cm3 of H2 = 0.268 × 1020 molecule No. of Molecules in CH4:

As we have that at STP. 22.4 dm3 of CH4 = 6.022 × 1023 molecules

. ×   molecules .  .   ×  1000 cm = . × molecules  1dm = 1000 cm 1dm3 of CH4 = 3

3

3

1 cm3 = 0.268 × 1020 molecules

Conclusion:

Since both H2& CH4 has same no of o f molecules which has been proved mathematically. 1cm3 of H2 = 1cm3 of CH4 because both contain contain same No No of particles 0.268× 1020 = 0.268 × 1020 proved Answer-2:

Yes, two postulates of kinetic molecular are faulty which are given b elow: 1.

Actual volume

2.

Forces of attraction

Actual Volume:

According to kinetic molecular theory, the actual volume of the gas molecule is negligible i.e. equal to zero as compared to the total volume. Forces of Attraction:

There are no intermolecular forces of attraction between the gases molecule, which is faulty and incorrect. Answer-3:

The gases shows non ideal behavior because of the following two reasons. i.

Low Temperature:

At low temperature the gas molecules come closed to each other due to which, intermolecular distance between the molecules decreases, also kinetic energy decreases, 62



Due to which the intermolecular forces among the molecule increases and since gases shows non ideal behavior. ii.

High Pressure:

At high pressure the gas molecules come closed to each other oth er due to which the intermolecular distances between the molecules decreases, and gases show non ideal behavior. Sudden Expansion:

The sudden expansion of compressed gases causes coolness, according to Joul Thamson effect. Joul Thamson Effect:

According to Joul Thamson effect “Sudden expansion of highly compressed gases causes cooling effect which is called Joul Thamson effect”. Note:

In compressed gas, molecules are very closed to each other and have attractive forces” and since shows more deviated. Reason:

When a gas is expanded suddenly, molecules moves away from each other. This process require energy, which is obtained from the gas itself, due to which gas become cooled down. d own. Answer-4:

Lighter gases can diffuse more rapidly, than heavier one, b ecause of the following reasons. Graham’s Law of Diffusion:

According to Graham law of diffusion “The rate of diffusion of a gas is inversely proportional to the square root of its density at STP”. Mathematically:

Graham law can be written as;

     

     

OR

For Example:

The order of rate of diffusion of CH4, CO2, SO2& NH3 is CH4> NH3> CO2> SO2. Conclusion:

Its means that lighter gases will moves faster than heavier gases. 4.

Avogadro’ Avogadro’ss Law:

According to Avogadro’s law “Equal volume of all gases contain equal number of molecules at the same condition of temperature and

pressure. 5.

Note:

The mass and size donot effect the volume. Therefore, 1 cm3 of H2& 1cm3 of CH4 will have the same number of molecules at STP. 6.

Proof:

As we have that at STP. 22.4 dm3 of H2 = 6.022 × 1023 molecules. Similarly:

22.4 dm3 of CH4 = 6.022 × 1023 molecules. No. of Molecules in H2 Molecule:

As we have that; 22.4 dm3 of H2 = 6.022 × 1023 molecules. 1 dm3 of H2 =

. ×   ––  (1) . 

But 1dm3 = 1000 cm3 So converting dm3 into cm3

. ×   ––  molecules .  So . ×   1cm of H =. × 1000 = 10

100 dm3 of H2 = 3

2

– 3

1cm3 of H2 = 0.268 × 1020 molecule No. of Molecules in CH4:

As we have that at STP. 22.4 dm3 of CH4 = 6.022 × 1023 molecules

. ×   molecules .  .   ×  1000 cm = . × molecules  1dm = 1000 cm 1dm3 of CH4 = 3

3

3

63



1 cm3 = 0.268 × 1020 molecules Conclusion:

Since both H2& CH4 has same no of o f molecules which has been proved mathematically. 1cm3 of H2 = 1cm3 of CH4 because both contain contain same No No of particles 0.268× 1020 = 0.268 × 1020 proved Answer-2:

Yes, two postulates of kinetic molecular are faulty which are given below: 1.

Actual volume

2.

Forces of attraction

Actual Volume:

According to kinetic molecular theory, the actual volume of the gas molecule is negligible i.e. equal to zero as compared to the total volume. Forces of Attraction:

There are no intermolecular forces of attraction between the gases molecule, which is faulty and incorrect. Answer-3:

The gases shows non ideal behavior because of the following two reasons. 1.

Low Temperature

2.

High pressure iii.

Low Temperature:

At low temperature the gas molecules come closed to each other due to which, intermolecular distance between the molecules decreases, also kinetic energy decreases, Due to which the intermolecular forces among the molecule increases and since gases shows non ideal behavior. iv.

High Pressure:

At high pressure the gas molecules come closed to each other oth er due to which the intermolecular distances between the molecules decreases, and gases show non ideal behavior. Sudden Expansion:

The sudden expansion of compressed gases causes coolness, according to Joul Thamson effect. Joul Thamson Effect:

According to Joul Thamson effect “Sudden expansion of highly compressed gases causes cooling effect which is called Joul Thamson effect”. Note:

In compressed gas, molecules are very closed to each other and have attractive forces” and since shows more deviated. Reason:

When a gas is expanded suddenly, molecules moves away from each other. This process require energy, which is obtained from the gas itself, due to which gas become cooled down. Answer-4:

Lighter gases can diffuse more rapidly, than heavier one, b ecause of the following reasons. Graham’s Law of Diffusion:

According to Graham law of diffusion “The rate of diffusion of a gas is inversely proportional to the square root of its density at STP”. Mathematically:

Graham law can be written as;

     

OR

     

For Example:

The order of rate of diffusion of CH4, CO2, SO2 & NH3 is CH4> NH3> CO2> SO2. Conclusion:

Its means that lighter gases will moves faster than heavier gases.

64



CHAPTER-5:

STATES OF MATTER LIQUID

Kinetic Molecular Theory: The K.M.T about liquid is given with postulates; 1.

Kinetic energy of molecules: This theory was presented by Bernauli. The liquid molecules have high kinetic energy than solid

molecules and low K.E than gas molecules. 2.

Intermolecular forces: The intermolecular forces in the liquid molecules are stronger than gas molecules and weaker than solid

molecules. 3.

Intermolecular distance: The intermolecular distance between liquid molecules is greater than the solid and smaller than gas

molecules. 4.

Definite volume: It is because of specific intermolecular distance.

5.

n o definite shape because of no specific intermolecular int ermolecular forces. forces. No definite shape: It has no Difference between Intermolecular forces

1.

2. 3.

4. 5. 6.

Definition: The forces of attraction between

separate or individual ions or molecules is called intermolecular forces. Example: the attractive forces among the molecules of a substance. H2O ||||||| H 2O Energy: Low energy is required to break the bonds b/w molecules. H2O(L) H2O(g) H = 40.71 kJ/mol Weak forces: These are weak intermolecular attractive force. Physical forces: These are physical forces. Examples: i. Hydrogen bonding ii. Dipole dipole interaction iii. London dispersion forces iv. Ion dipole forces v. Dipole induced dipole forces

∆

Intramolecular Forces 1.

Definition: The Forces between the atoms in a

Molecules molecules is called intramolecular forces. 2. 3.

Example: the attractive forces within a molecule Energy: Greater energy is required to break the

bond b/w the atoms. 4. Strong Forces: These are strong forces. 5. Chemical forces: These are chemical forces. 6. Examples: i. ionic bond ii. covalent bond iii. coordinate covalent bond

Note: Physical and chemical properties depends upon the intermolecular forces. Intermolecular forces: They are also called weak Waal’s wall forces.

Following are the types of intermolecular forces. i. ii. iii. iv. v.

Dipole dipole forces. London dispersion forces Hydrogen bonding Ion dipole forces Dipole induced dipole forces

1. Dipole Dipole Forces: The forces of attraction between the partial positive end of one polar molecule with the partial negative end of the

other polar molecule is called dipole dipole interaction. eg Dipole: Any molecule having two oppositely charged ends is called dipole. Second definition: The (forces) of attraction between the permanent poles of polar molecules is called dipole d ipole interaction. Third Definition: The forces of attraction between polar molecules is called dipole dipole interaction. Explanation: When two polar molecules are brought closer to each other, then these molecules orient themselves in such a way that the partial

positive end of one polar molecule come in contact with the partial negative end of the other polar molecule and attract each each other with a force called dipole dipole interaction. Orientation: Examples:

HCl, HBr, HI H2S, PH3, CCl3H CH3Cl etc Factors: Dipole dipole interaction depends upon the following factors:

∝

Polarity: Polarity D.D.I

As the polarity increases the dipole dipole forces increases.

Note: increasing order of acidity. acidity. HF
Increasing order of polarity: HF>HCl>HBr>HI

65

Nasrat Ullah Katozai (Chemistry) 2. Distance between molecules: D.D.I


∝ ..

As the distance between the molecules increases dipole dipole forces decreases. Note: Gases
As the distance between the molecules in the gaseous state is greater so forces of attraction between them are smaller, in liquid these forces are relatively stronger. 3. No of Shell: It depends on the no of shells and shapes. MCQS: In gases the D.D.I become stronger at low temperature and high pressure.

5. London dispersion forces: It is also called

i. ii. iii. iv. v. vi.

MCQS: Greater the D.D.I, greater will be the values of different

Dispersion forces Momentary forces Instantaneous dipole induced dipole forces Transitory forces Short range forces Loosely Vander wall forces

thermodynamic parameters like m.p, B.P, heat of vaporization and heat of s ublimation., ublimation.,

Dispersion forces

The momentary forces of attraction that exist between instantaneous dipole and induced dipole are called London dispersion forces. OR These are weak temporary forces which are due to temporary dipoles OR

Patent  some scientific secrets which s in the form of design Axodi for the luminous of TV. Note: About (pb) lead. (MCQS) 1. Lead is a heavy metal and is more toxic. It acts as a protoplasmic poison. poison. 2. It causes weight loss and osteoporosis. 3. Lead upto 0.3 ppm in blood shows its harmful effects and upto 0.8 ppm causes Anemia. Note: Leafy vegetables accumulate more lead than fruity vegetables, so it is necessary to grow leafy vegetable away from road side.

These are the only inter molecular forces present among symmetrical, non polar substances such as CH 4, SO4, O2, H2, Cl2, F2, I2, Br 2, CO2 and mono atomic noble gasses. ( He, Ne, Ar, Kr, Xe & Rn) PRODUCTION: It can be produced due to “The dispersion of electronic cloud from its normal path.

Atoms consist of two parts 1. Central positive part 2. Negative charge electron around it i.

When the two atoms of the same substances are brought closer.

ii.

Then the electron cloud of one (1) atom repels the electron cloud of the other atom as shown.

iii.

As electron cloud is pushed backward due to which which negative charge density increase on one side of the molecule and the (+) charge density increases on the other side of the th e molecule As a result temporary dipole is developed which is called instantaneous dipoles?

iv.

Now if this temporary dipole is brought near to another atom i.e. (neutral (neutral atom), then that neutral atom also become induced dipole. v.

Induced dipole: The poles produced on a neutral atom due to instantaneous dipole is called induced dipole.

vi.

So this neutral atom which is represented by (III) become polar which will be called (induced dipole).

vii.

These forces are called temporary or instantaneous forces because they produce for a small interval or instant of time.

viii.

These ae the only forces present in all substances But more significant in non polar substances.

Factors: It depend upon the following factors.

̅



1. Size of -cloud: When the -cloud -cloud increases the chances of polarizability increases due to which L.D.F increases.

̅

That is why the L.D.F in creases down the group because No of increases increases down the group. also depends upon L.D.F. If L.D.F increases increases then the physical state of a substance also also changes. Note: The physical existence of a substance also ie 2. Number of Atoms in Molecules: As the number nu mber of atoms in a molecule in creases L.D.F also increases. increases.

For example: 3. Size of the atom: If the size of the atom increases then the L.D.F will be increases.

d epends shape of molecules. 4. Shape of molecules: L.D.F also depends Linear molecules Angular molecules Regular shape

Irregular Molecules

Straight

branched

MCQS: The existence of iodine in solid form is due to strong stron g dispersion forces. Strong Dispersion Force is due to greater No of electrons.

So, The boiling point of linear molecule will be high. Reason: greater chances of overlapping due to which polarization increases & Hence L.D.F also increases due to which boiling point of linear hydrocarbon is greater than branch chain hydrocarbon. 3. Hydrogen bonding: The force of attraction between positively charged hydrogen of one molecule with more Electronegative/ Element like

Flourine, Oxygen and Nitrogen of other oth er molecules is called Hydrogen bonding. 66



Concept:(i) If we go from top to bottom in a group VA, VIA & VIIA then the MP and BP of the binary compounds of H 2 with each element increases gradually.

Due to London dispersion forces. (ii) But as we know that the BP & MP increases from top to bottom in the above table so B.p & Mp of the first three molecules ie NH 3, H 2o & HF is exceptionally very high why? (ii) This can be explained by H-bonding means that there is some special force of attraction between these molecules which is called Hydrogen bonding. Types: There are two types

(i) Inter Inter hydrogen bonding (ii) Inter a hydrogen bonding.

Characteristic properties:

MCQS: The strength of Hydrogen bond is generally twenty

times less than that of a covalent bond. (kips). B.p of H2O = 100 C0It is due Hydrogen bonding. B.p of NH3 = – 33 33 C0 B.p of HF = 19.5 C 0 The boiling point of water is high than HF because one water molecule form 4 hydrogen bond while one HF form one hydrogen bond.



It is a strong type of dipole-dipole interaction/ electrostatic force of attraction.  It is denoted by dotted line.  H-bonding exist between substances in which at least one H-atom is directly attached to Flourine, oxygen and nitrogen.  Energy is from (25-33.1) kJ/mol  It is 20 time weaker than simple covalent bond.  In H-bonding the H-atom is squeezed in between the two fluorine.  Fluorine has small size due to which it cannot form bond?  Chlorine also form hydrogen bond in rare cases. Demerization: The process in which two molecules of carboxylic acid live combinely is called demerization. For Example: The two carboxylic molecules live combine as Note: Protein has special structure due to Hydrogen bonding. Examples of Hydrogen Bonding:

i. ii. iii. iv. v. vi. vii. viii. ix. x.

HF, NH3, H2O, H2O2 Alcohol Amine (primary) Protein Glycol Amino acid Glycerol Carboxylic acid

Note: All compounds having “OH” group are responsible for H -

bonding formation. formation.

Hair’s fiber

Silk, sucrose, paint, dyes, glue, food materials, carbohydrates, carbohydrates, Glucose, fructose, Honey.

Applications of Hydrogen Bonding Strength of Acid:

1.

1. 2. 3. 4. 5. 6. 7. 8.

Acidity: Those substances which have hydrogen bonding will be less acidic because it will give “H” with more difficulty. Solubility: Those substances which form Hydrogen bond with each other are soluble in each other. For example: Alcohol are soluble in H2O because of H-bonding. Cleaning action: The cleansing action of water is because of hydrogen bonding. Biological compounds: It is also present in many biological compounds like, amino acid and DNA. DNA: The DNA strands are held together by H-bonding. Paint and dyes: the sticking ability of paint and dyes is because of hydrogen bonding. Food material: The food materials also contain Hydrogen bonding ie Honey, Glucose, Sucrose etc. Structure of ice: The hexagonal structure of ice is because b ecause of hydrogen bonding due to which its volume increases.

Comparison between Dipole-Dipole Interaction and London dispersion forces: Any substance may or may not have dipole – dipole

interaction but will always have London dispersion d ispersion forces. Flouro methane No of electrons C = 6 electrons F = 9 electrons H = 3 electrons No of = = 18 Polar molecules D.D.I is present B.P = 194k But London dispersion is also present.

̅ ̅

Ethane No of electrons C = 6 x 2 = 12 electrons H = 3 x 2 = 6 electrons

̅ ̅

No of = = 18 Nature Non polar B.P = 184k L.D.F is present only

67



Conclusion: London dispersion forces are present in all molecules / compound whether polar or nor polar but more prominent in non-polar

compounds. Second example

Tetrachloromethane Non polar L.D.F Present No D.D.I No of electrons C=6 s Cl = 17 x 4 = 68 68+6 No of electron = 74

̅

̅

B.P = 61.2 Co

Trichloromethane Polar D.D.I L.D.F No of electrons C=6 Cl = 17 x 3 H=1 6+51+1 No of = = 58 B.P = 76.8 Co

̅ ̅

As the molecular mass increases then the velocity of the molecule decreases and hence they can easily attract each other by strong intermolecular forces & Bp increases. Evaporation: The spontaneous conversion of liquid into vapours is called evaporation. Explanation:

i. ii. iii. iv. v. vi. vii. viii. ix. x. xi. xii. xiii. xiv.

Spontaneous process Occurs at all temperature Silent process Surface phenomenon Occur from all liquids Endothermic process Occur in open container Rate of evaporation is different for different liquids. Higher KE molecules escape, leave behind the molecules with low kinetic energy and hence causes cooling effect. Causes cooling effect Note: Each liquid has its own vapour pressure Require no activation energy. temperature. No need of catalyst catalyst Continuous process. Physical process

at a given

Examples:

i. ii. iii.

Cooling of water in earthen port is because of evaporation. Perspiration causes cooling effect because of evaporation. Cooling effect on a shiny day morning is because of evaporation.

Factors affecting evaporation:Following are the factors which affect evaporation.

i. ii. iii. iv.

Pressure Temperature Surface area I.M.F

i. ii. iii. iv.

Intermolecular forces:As the IMF increases the evaporation decreases. Pressure: As the pressure increases the evaporation increases. Temperature: As the temperature increases evaporation also increases increases because the kinetic energy of the molecules increases. increases. Surface Area: As the surface area increases, rate of evaporation will be increases, because greater number of molecules will

evaporates. Vapour pressure: The pressure exerted by the vapours of a liquid at equilibrium is called vapour pressure. Unit: Its unit is mmHg. For example:

i. The vapour pressure of water is 18 mmHg at 20c0 ii. The vapour pressure of ether is 537 mmHg at 25co.

Explanation:

i. ii. iii. iv.

Consider a liquid in a closed container, which is heated. The molecules are continuously converted into vapours by evaporation. After some tome the vapour are converted into liquid by condensation. In the beginning the rate of evaporation is greater than the rate of cond ensation. But after some time the rate of evaporation become equal to the rate of condensation.

Equilibrium: That point at which the rate of evaporation becomes equal to the rate of condensation is called equilibrium. Note: Equilibrium always occur in a closed container. Factors of vapour pressure: Following are the factors.

MCQs:

All the organic matter in all forms in the soil material is called Humus. Polyhydroxy material Urea.

68


i. ii. iii. iv.


Intermolecular forces of attraction. Temperature Nature of liquid Intermolecular distance

1. Nature of Liquid:

It depends upon the nature of liquids i.e. weaker the intermolecular forces forces of attraction greater will be vapour pressure. 2.Temperature:As temperature increases vapour pressure also increases. 3.Intermolecular Forces: As intermolecular forces increases vapour pressure decreases. 4. Intermolecular Distance: As intermolecular distance increases in creases vapour pressure decreases. Note:

Vapour pressure is independent of the amount of liquid. If the amount of liquid is greater or less the vapour pressure will b e same. Because if area is double force will also become b ecome double so vapour pressure remain constant.

   P =  P=

P = 10 atm

   P =  P=

P = 10 atm Measurement of Vapour Pressure:

There are two methods used for measuring vapour pressure. 1.

Barometric method

2.

Monometric method

1.

Barometric Method

(1) In this method one meter long glass tube is filled with Hg. (2) It is in verted in Hg – dish. dish. (3) When the Hg – tube tube is inverted Hg-dish then the Hg – level level in the tube tub e decreases and falls until it is stopped by b y atmospheric pressure. (4) Then its height is measured which is 760 mmHg which is called atmospheric pressure. Procedure:

Now a small amount of ether is taken and placed at the lower end of the tube, with the help of a dropper whose vapour pressure is to be measured. The ether will rise above the surface of Hg because of its low density. Above the surface of Hg ether will evaporate & the vapours of ether will exert pressure on Hg and push the Hg downward. This fall of Hg in the Hg- tube was measured. It gives the vapoure pressure of liquid. For Either it 537 mmHg. 2.

Monometric Method:

1. It is the accurate method for the measurement of vapour pressure of a liquid. 2. The apparatus consists of a flask with T-shaped glass tube. 3. One end of the tube is connected to u-shaped Monometre and other end is connected to a vaccume pump as shown. Procedure:

1. 2.

The liquid in the flask is prozen with the help of freezing mixture. The space above the liquid is evacuvated using vaccum vaccum pump. pu mp. The Frozen liquid is melted again to release the entrapped air. This process is repeated for a number of times so that the entire air is removed. 3. Now close the vaccum vaccum pump end, and open the monometric end. 4. Set the thermostat at a particular temperature, on the evaporation of the liquid, the vapour exert pressure and vapour pressure is measured as: V.P = P + l Boiling Point:

Boil

OC0

The point at which the vapour pressure of a liquid become equall to atmospheric pressure is called called boiling point. Note: Each liquid has got its characteristics Boiling point at constant pressure. Explanation:

1.

When a liquid is heated its vapour pressure increases. At certain temperature the vapour pressure become equal to atmospheric pressure.

2.

At this stage the liquid start just Boiling. At the boiling point the K.E of the molecules of liquid becomes maximum. 69



3.

At this stage any further heating will not increase the temperature. This heat will be used to break the intermolecular forces and convert the liquid into vapours.

4.

A point will reach at which the vapour pressure of liquid become equal to atmospheric pressure, pressure, that point is called boiling point.

5.

At boiling point the temperature remain constant.

Example:

The boiling point of (carbondisulphid) CS2 is 46.30 c0 at one atm

     

The B.P of CCl4 is 76.50 C0 at one atm The B.P of ethanol (C2H5OH) is 78.26 C0at 1 atm. The B.P of Benzene is (C 6H6) is 80.15 C0 at 1atm. The B.P of H2O is 100 C0 at 1 atm. The B.P of acetic acid (CH 3COOH) is 118.50 C0at 1atm.

Molar Heat of Vaporization ( Hr):

The amount of heat required to vaporized one mole of a liquid at its boiling point is called molar heat of vaporization. The pressure, p ressure, during the change is Kept one atmosphere. The molar heat of vaporization of H 2O is+ 40.7 Kj/mol Factors Affecting Boiling Point:

Following are the factors. (i) External Pressure:

Boiling point depends upon the external pressure. If external pressure is high boiling point will be high. Note: If external pressure is low B.P will be low. e.g.

i.

Water Boils at 100 C0 at 760 mmHg

ii.

Water Boils at 25 C0 at 27.0 mmHg

iii.

Water Boils at 0 C0 at 4.5 mmHg

Note:

1.

At sea level the external pressure is high therefore the water boils at high temperature i.e. 107 C 0 E.g. if the chicken cooked at 100 C 0 then the chicken will cooked before the Boiling point. Less time is taken. It will consume less fuels.

2.

At the Hilly area i.e Swate, Murre the external pressure is low so the boiling point will be consumed. is low. And the substance will not cook soon and will consume more time and more fuel will be used. cooking point

50 C0 (B.T)

External Pressure

90 C0 BP

:

Note:

Once a liquid start boiling its temperature remain remain constant. Although heat is continuously supplied to it now given heat perform two functions. 1.

To increase the KE of molecules.

2.

Breaking of intermolecular bonds.

2. Nature of liquid:

Boiling point of a liquid with stronger intermolecular forces will be higher than liquid with weaker intermolecular forces at the same pressure. For example: The BP of H2O is 100 0C & diethyl ether is 35.5C0 at same atmospheric pressure because of nature. Applications of Boiling Point: Pressure Cooking:

When external pressure is increased the boiling point of a liquid increases so that food is cooking quickly and less fuel is consumed. Vacuum Distillation: (An example of reduced pressure)

The distillation carried out under reduced pressure is called vacuum distillation. This process is carried out for those liqu ids. i.

Which have high boiling point.

ii.

Which are thermally unstable.

Glycerin = The B.P of glycerine is 290 0C but at 290 0C the glycerine decomposes therefore it is distilled at low temperature i.e. 210 0C by

decreasing external pressure to 50 mmHg. Surface Tension:

Stretchiness in the surface of liquid is called surface tension.

70

Nasrat Ullah Katozai (Chemistry) Representation:

It is represented by Gamma “ ” Unit:

Master Coaching Academy Note:

It is due to surface tension that small object floats on the surface of water.

Nm – 1 OR Jm – 2 1.

It is the force per unit length not area

2.

It is the force acting at right angle on per unit length of the surface.

Explanation: Cohesive Forces: Forces between water molecules or same molecules of a substances is called cohesive forces. Adhesive Force: Forces of attraction between the water molecule and the walls of container is called adhesive forces. Concept:

1.

Liquid molecules have the ability of cohesion.

2.

Cohesive forces of attraction between water molecule is due to H-bonding.

3. 4.

The molecules of the surface of water are attracted by the molecules from of below & sides but not from above. The molecules at the surface, therefore, feel a net (inword) attraction this attraction create surface tension in the form of network.

5.

“An unbalance force on the molecule of surface produce an inword force called surface tension.

Note:

The molecules of the surface are attracted by those molecules which are present in the middle, due to which the potential energy of surface molecules increase (like rubber). Work must be done to pull it to the surface. Since: “The amount of energy required to expand the surface of liquid by a unit area is called potential energy”. Note:

1. 2.

Detergent: It reduces the surface tension by breaking the H-bond. Detergent: Those substances which reduces the cohesive forces between water molecules is called detergent.

The reduction of cohesive forces increases the ability of water to wet a solid surface by adhesion. Thus its c leaning action is increeases. Measurement of Surface Tension: Following methods are used for measuring surface tension.

1.

Torsion method

2.

Capillary method

3.

Drop method of stalagmometre

Calculation:

Note: MCQs:

   = ƞ  × 

Copper sulphate is also called blue vitrioel.

Factors: Following are the factors:

1.

IMF (Inter Molecular Forces)

2. 3.

Temperature Density

1.

intermolecular forces (IMF):

CuSO4. 5H2O The forces of attraction between CuSO 4 and 5H2O molecule will be. (a)Ionic (c)Intra

(b)Covalent (d)Coordinate

(e)All of them

As the intermolecular forces of Attraction between the molecules increases surface tension also increases. increases. 2.

Temperature:

As temperature increases surface tension will be decreases. decreases. 3.

Density:

As density of molecule increases surface tension will be increases. increases. Note: (MCQ)

Unit of Noise or sound is hertz. Energetic Phase Change:

The conversion of one physical state in to another physical state is called energetic phase change. Following are the types: Molar Heat of Vaporization ( Hr):

It is the amount of heat absorbed when one mole of a liquid is changed into vapours at its boiling point. The pressure, durin g the change is kept one atmosphere. 1 mole of liquid H2O(L)

vapours –––––  1 mole of vapours –––––  H2O(v)

71



For example:

Heat of vaporization for water is 40.7 kJ/mol

H = 40.7 kJ/mol Note:

It depends upon the strength of intermolecular forces of attraction. Molar Heat of Fusion (

H0):

It is the amount of heat absorbed by one mole of a solid when it melts into liquid form at its melting point. The pressure, during the change is kept one atmosphere. For example:

When one mole of solid H2O is converted into liquid H2O, 61 kJ/mol of heat is evolved. H2O (ice) –––––  H2O (Liquid)

H = 61 Kj/mole Molar Heat of Sublimation (

Hs):

It is the amount of heat absorbed when one mole of a solid sublimes to give one mole of vapors at particular temperature and one atmospheric pressure. For example: Iodine (solid) _________ Iodine (gas)

i. ii. iii.

Iodine Naphthalene parachlorobenzene

Note: Liquid Crystal:

1.

The intermediate state between solid and liquid is called liquid crystal.

2.

Arrangement of atom is like crystalline solid and motion is like liquid.

3.

Those semisolids which have the characteristics of both solid as well as liquid.

4.

They have optical properties like crystalline solid and surface tension & viscosity like liquids.

Discovery:

It was discovered by W.Heitz in 1850 “that stearin at 52 0C converted into cloudy liquid (opeaque liquid) and after 62.6 0C convert into true liquid”. An Austrian botanist F. Reeintzer also reported an organic liquid crystal “Cholesteryl benzoate”. He studied that this compou nd becomes

milky at 1500C and becomes clear at 179 0C. Types of Liquid Crystal:

It has 3 types.

i. ii. iii.

Smectek liquid crystals. Nematic liquid crystals crystals Cholestric liquid crystals i.Smectek liquid crystal: That liquid crystal in which the atoms are arranged in regular pattern is called smectek liquid crystal. ii. Nematic liquid crystals: That liquid crystal in which the th e atoms are arranged in irregular pattern is called nematic liquid crystal. iii. Cholestric liquid crystal: That liquid crystals in which the atoms are arranged diagonally is called chole stric liquid crystal. Example: Concentrated solution of soap. Properties: It has the following properties.

1.

Long cylindrical molecules

2.

More sensitive to pressure & temperature

3.

Weak positional and orientational order

4.

Anisotropic in nature

5.

Some degree order like solids

6.

They flow like liquids.

7.

They have surface tension and viscosity

8.

Optical properties like crystalline solids.

Application: It has the following applications

1. 2.

Electrical research work Medical research work

3.

Skin thermograph i.e.

4.

Finding blockage in the vein & artery. 72


5.

In display of electrical devices.

6.

Watches and calculators.

7.

Computer screen

8.

As a screen in oscilographs and TV

9.

In liquid solid chromatograph as a solvent

10.

As temperature sensors

11.

Radiation and pressure sensor

Viscosity:

The resistance to the flow of fluid is called viscosity. OR


ƞl = Viscosity of liquid te = flow time liquid dw = density of water

Ɩ = density of liquid ƞ = viscosities water

The resistance offer by the liquid to its flow is called viscosity. Unit: OR

i. ii. iii.

Its S.I unit is kgm-1 s-1 Poise 1 poise = 10-1 kgm-1 s-1

Mathematically:

       

Note: Honey and glycerine has high viscosities as compared to ether & water.

Explanation:

1.

Liquid has the ability to flow because molecules of liquid can slide over each other.

2.

Velocity of liquids near the side of tube is low.

3.

Velocity of liquid at the centre is high.

4.

The resistance to the flow of fluid is due to the internal resistance among different layers of liquid.

5.

The side molecular layer decreases the velocity o f the central layers.

Factors Affecting Viscosity: Following factors effect viscosity.

1.

Molecular size

2.

Molecular shape

3.

Intermolecular forces

4.

Temperature 1. Molecular Size: As molecular size increases viscosity increases. 2. Molecular Shape:

I.

Irregular molecules offer more resistance.

II.

Ring shape molecule offer less resistance.

III.

Regular molecules have less resistance.

3. Intermolecular Force: As intermolecular forces increases viscosity also increases.

1.

Water is more viscus than alcohol.

2.

Glycerine and H2SO4 have more viscosity because of greater intermolecular forces.

4. Temperature:

As temperature increases viscosity decreases. Measurement of Viscosity:

1.

It is not possible to measure the absolute viscosity.

2.

We can measure only relative viscosity.

Relative Viscosity:

It can be define as “The ratio of the viscosity of a liquid to the viscosity of water taken as standard is called relative viscosity”. Note:

 

The viscosity of water is taken as 1 centipoise at 25 0C. The device which is used to measure the relative viscosity of liquid is called Ostwald’s viscometer.

Coefficient of Viscosity ( ):

The force required to maintain a difference of velocity of one meter per second between two parallel layers of liquids one meter apart from each other is called coefficient of viscosity.

73



Short Questions: i.

Give at least two of the effects on our lives if water has weak hydrogen bonding among its molecules.

Ans.

If water has weak hydrogen bonding among its molecules, then it has the following effects.

(a)

Evaporation: Higher rate of evaporation would result in excessive accumulation of water vapours in atmosphere. It would cause

humidity, global warming and rain fall that would badly affect the life on earth. (b)

Vapours form: Water would evaporate rapidly and would not be available in liquid state for drinking, irrigation etc. It would badly

effect animals, plants and the environment. (c)

Solubility: Water would not dissolve certain substances like ammonia, amines, alcohol, phenol, carboxylic acid and sulphates etc.

(d)

Universal Solvent: Water will not be b e considered as universal solvent.

ii.

HF is a liquid at ordinary temperature while HCl is a gas.

Ans.

(i) Hydrogen – bonding: bonding: HF is a liquid at ordinary temperature due to strong H-Bonding among its molecules – (ii) D.D.I.: HCl is a gas at room temperature due to weaker dipole – dipole dipole attraction among its molecules.

iii.

H2O has high boiling point than HF, although fluorine is more electronegative than oxygen?

Ans.

(i) The The strength strength of H-bonding increases increases with decrease in size and increase increase in electronegativity electronegativity of the atom covalently bonded to hydrogen. (ii) Although fluorine is more electronegative than oxygen but still water has high boiling pint than HF due to more extensive H bonding in water than HF – water water has two H – atoms atoms per molecule and forms four H – bonds while HF has one H – atom atom per molecule and form one H – bond. bond.

iv.

Water and Ethanol can easily mix in all proportions.

Ans.

(i) Polar Nature: Both water and ethanol are polar liquids. liqu ids. (ii) Hydrogen Bonding: They can mix easily with each other in all proportions due to H – bonding bonding among. (iii) Like Dissolved like principle: Water & ethanol can easily mix into each other because of like dissolve principle.

v.

Neon and Argon both are mono-atomic noble gases of the same group – group – Neon Neon has – has – 248 248 0C boiling points while Argon has – 189 0C why?

Ans.

(i) Both neon and Argon are mono-atomic noble gases which have only London forces – strength strength of London forces in noble gases increases down the group in the periodic table due to increase in size of electronic could. (ii) Greater the size of electronic cloud, More easily it polarizes to form dipole and thus stronger are the London forces. (iii) As compared to Neon (B.P – 248 248 0C), Argon has higher boiling point of – 189 189 0C due to its larger size and stronger London forces.

vi.

Different liquids have different rate of evaporation even at the same temperature?

Ans.

(i) At same temperature, different liquids have different rates of evaporation due to difference difference in strength strength of intermolecular forces. (ii) Stronger the intermolecular forces, lesser number of molecules change into vapours, hence lower is the rate of evaporation and vice versa. (iii) For example ethers have higher rate of evaporation than water at same temperature du e to extensive H – bonding bonding in water while ethers have weaker dipole – dipole dipole and London forces.

vii.

Earthenware vessels keep water cool even in hot summer days? Vacuum distillation can be used to avoid decomposition of a sensitive liquid?

Ans.

(i) Earthenware Earthenware vessels keep water cool in hot summer summer days days due to combined effect of capillary capillary action and and evaporation. evaporation. Earthen Earthen (clay) 1 pots have very small pores po res which act as capillary tubes. (ii) The water molecules of higher K.E move out through these porces to the outer surface of the pot due to capillary action where evaporate. (iii) Average K.F of the remaining molecules decreases, hence temperature decreases and cooling is produced. (iv) During vacuum distillation, B.P is decreased decreased by decreasing pressure on liquid surface. (v) A liquid boils at lower temperature than its normal B.P. Certain compounds decompose at their normal boiling points. Vacuum distillation is used to distill sensitive liquids without decomposition. (vi) For example glycerin decomposes at its normal B.P (290 0C) when pressure is reduced form 760mmHg to 50mmHg, it boils and distill without decomposition at 210 0C.

viii.

A liquid boils at different temperature at sea-level and at mountains.

Ans.

(i) Boiling point of a liquid directly depends upon the eternal pressure. (ii) Greater the external pressure on liquid surface, more heat is required to make equal the vapour pressure of liquid to the external pressure, hence high high is the boiling point and vice versa. versa. (iii) A liquid boils at different temperature at sea – level level and at mountains mo untains due to different external pressures. (iv) For example boiling point of water at sea – level (1 atm pressure) is 100 0C but it becomes 98 0C to mountains when the pressure is 0.98 atmospheres. atmospheres. 74

Nasrat Ullah Katozai (Chemistry) ix.

Evaporation of a liquid causes cooling.

Ans.

(i) Evaporation of a liquid causes cooling.


(ii) During evaporation molecules of higher K.E (hotter) of a liquid change into vapour state and the average K.E of the remaining molecules decreases – As average K.E is directly proportional to absolute temperature, so, temperature of remaining liquid molecules decreases decreases and cooling is produced. x.

Temperature of the liquid remains constant during boiling although heat is being supplied continuously.

Ans.

(i) At boiling point temperature of a liquid remains remains constant although heat is being supplied. (ii) The heat supplied is not used to increase the temperature of a liquid. (iii) It is partly used to break the intermolecular forces and separate the molecules while partly taken away by the outgoing molecules.

xi.

Why water droplet is spherical?

Ans.

(i) Water droplet is spherical in air due to the surface tension. (ii) Sphere has the least surface to volume ratio. (iii) It experiences minimum force per unit area in air and becomes more

stable. Boiling Point and External Pressure

As we know that boiling point depends on external pressure, so when external pressure is change, its boiling point will also be changed. A liquid can be b e made to boil at any temperature by changing the external pressure. When the external pressure is high the liquid require greater amount of heat to equalize it to external pressure. In this way bo iling point is raised. Similarly, at a lower external pressure a liquid absorb less amount of heat and it boils at a lower temperature. Surface Tension: Molecules at the surface of a liquid experience attractive forces

downward, toward the inside of the liquid an d sideways, along the surface of the liquid. On the other hand, molecules in the center of the liquid experience uniformly distributed attractive forces. This imbalance of force at the surface of a liquid results in a property called surface tension. Extra information: Surface tension also explains the beading of raindrops on the shiny surface of a car. iii.

H2O has high boiling point than HF, although flouring is more electronegative than oxygen?

Ans. H2O has high boiling point than th an HF is because of hydrogen bonding. Water can form 4 hydrogen bond but HF can form only 1 hydrogen bo nd. ix.

Evaporation of a liquid cause cooling.

Ans.

(i) Evaporation causes cooling.

Instantaneous Instantaneous dipole induced dipole forces.

When two atoms of same substance are brought closer to each other, the electron of one atom influence the moving electron of another atom. Electron repel each other and tend to stay as far apart as possible. When electron of one atom come close to the electron of another atom, they are pushed away from each other. In this way, a temporary dipole is created in the atom. As a result the electron density of an atom is no more symmetrical. It has more negative charge on one side and positive charge on another side. At that particular instant the atom becomes a dipole. This is called instantaneous dipole. Then instantaneous dipole then disturb the electronic cloud of other nearby atom, so, a dipole is induced in the second atom. This is called induced dipole.



The momentary force of attraction between instantaneous dipole and the induced dipole is called dipole induced dipole interact or London forces.



It is a very short times attraction because the electrons keep moving.



These forces are present in all types of molecules.

Factors Affecting London Dispersion Forces

i.

ii.

Size of electronic cloud of atom or molecule. When size is large, dispersion is greater and so London forces are more prominent. Number of atoms in a non-polar molecule, molecule, greater the number of molecule, greater is the polarizability and so greater is the London dispersion force.

Note: Molecules with large chain length experiences stronger

London dispersion forces. The reason is that longer molecules have more places along its length where they can be attracted to each other.

(ii) The reason is that when high energy molecules leave the liquid and low energy molecules are left behind, the temperature of the liquid falls and heat moves from the surrounding to the liquid and then the temperature of the surrounding also falls. (iii) Thus in this way it causes cooling.

75


CHAPTER-6:


STATES OF MATTER III

Solid: The state of matter which has specific shape and volume is called solid. Kinetic Molecular Interpretation: The kinetic interpretation of solid is given below;

i.

I.M.F: The intermolecular forces of attraction are very strong therefore the molecules of solid are very closely compact, cannot move freely and can vibrate only. ii. I.M.D: They have very small intermolecular distance due to which the intermolecular forces are strong and so they are in solid state. iii. Specific Shape: They have specific shape because of strong intermolecular forces of attraction. iv. Specific Volume: They have specific volume because of smaller intermolecular distances between the particles as a result they can vibrate only and are incompressible. Types of Solid: There are two types of solids. 1. 2.

Crystalline solid Amorphous solid

1. Crystalline Solid: That solid in which the atoms, ions or molecules are arranged in three dimensional pattern is called crystalline solid.

Properties: They have; i. ii. iii. iv.

Definite Geometry True solid Crystalline in nature Examples: NaCl, CaCO 3, CaO, CuSO4.5H2O, Graphite etc.

2. Amorphous Solid: Those solids in which the atoms, ions or molecules are arranged in irregular pattern are called Amorphous solid. Features:

i.

No specific geometry geometry

ii.

No sharp melting point

iii.

No specific heat heat of fusion

iv. v.

Also called super cooled liquid Look like liquids.

vi.

Psudosolid

Examples: glass, plastic, rubber, coal tar and gemstone. Properties of crystalline solid: Following are the properties of crystalline solids; 1.

Symmetry:

 

a)

The repetition of edges, faces and angles when a solid is rotated at 36 00 at its own axis is called symmetry. There are several types of symmetry; 1. Centre of symmetry 2.

Plane of symmetry

3.

Axis of symmetry

Plane of symmetry: The phenomenon in which a crystal can be divided by imaginary plane into two equal halves in such a way that

one half is the exact mirror image of other that will be called plane symmetry. b)

Axis of symmetry: It is an imaginary line drawn through the crystals, such that rotating the crystal through 360 0 the crystal represent

the same appearance more than once. c)

Centre of symmetry: The point at the center of a crystal which is equidistant from all faces is called centre of symmetry.

Note: A crystal may have a number of planes of symmentry or axis of symmetry but it can have only one centre of symmetry.

2.

Geometrical shape: Crystalline solid have definite geometrical shape because of regular arrangement o f atoms, ions or o r molecules in

the 3D space and strong IMF. Note: Upon grinding to powder still crystalline solid retain specific geometrical shape.

3.

Melting point: All crystalline solids have specific melting point, which is called sharp melting point.

4.

Cleavage plane: The breaking of a crystalline solids at specific planes is called cleavage plane.

It is anisotropic in nature. 5.

Habit of crystal: The shape in which a crystal grows is called habit of crystals OR The formation of a crystal of a pure compound is

same when they are prepared at different places. For example: NaCl has a habit of cubic growth. Its means everywhere Nacl will grows in cubic shape.

6.

Crystal Growth: The phenomenon in which crystals are prepared by slow cooling of a substance in liquid state or cooling a hot

saturated solution is called crystal growth. 76



Explanation: The shape of a crystal depends upon the methods and condition of preparation. For example: Nacl grows in three habits.

i. ii. iii.

Three dimensional ie (cubic) Flate shape (two dimensional) two dimensional One dimension (needle)

Note: 10% urea present in a solution compel the NaCl to grow in one dimension (needle like)

7.

Anisotropy: The phenomenon in which the physical properties of a substance are different when measured from different direction

for a crystal is called Anisotropy.

OR

Some of the crystals show variation in properties depending upon the direction. Such properties are anisotropic properties and the process is called anisotropy. OR The phenomenon in which the intensities of a property is different from dif ferent direction is called anisotropy. For example: Electrical conductivity of graphite is greater in one direction than in the other direction.

o f electron is parallel to the layers so the electrical conductivity is of graphite is greater in the parallel direction than Mechanism: The mobility of perpendicular direction. direction. Similarly cleavage are also anisotropic behaviour. Note: All crystal are anisotropic except (cubic crystal). Other example:

i. ii. iii. iv.

Refractive index Thermal conductivity Electrical conduction Coefficient of thermal expansion

Isomorphism:

i.

The phenomenon in which two different d ifferent compounds exist in the same crystalline form is called isomo rphism.

For example:

NaNo3 and KNO3 CaCO3 are different compounds but they have same s ame crystalline form ie they are Rhombohedral Rho mbohedral structure.

ii. NaF, MgO, NaCl have cubic crystals. iii. K 2SO4, K 2SeO4 form orthorhombic crystals. iv. Ag2SO4, NaSO4 form hexagonal crystals. 1. Same ratio: 2. Isomorphism have usually same ratio but it is not necessary ie A. (1) Na:F (2) MgO (3) NaCl (cubic) 1:1 1:1 1:1 B. (1) NaNO3 (2) CaCO3 (3) KNO3 (Rhombohadral) 1:1:3 1:1:3 1:1:3 C. K 2SO4 K 2SeO4 (rhombohedral) 2:1:4 2:1:4 Note: Some time ratio ratio is not same but structure is same. same. i.e. D. SnSO4, BaSO4.4H2O (Tetrahedral) Different ratios: Some time two crystals have some ratio but they are not isomorphic i.e. Zno & NaCl has same ratio but they have different

crystals i.e tetrahedral and cubic. 9. Polymorphism: The phenomenon in which a compound exist in more than one crystalline form is called polymorphism. For example: i. polymorphic substances Polymorph

i. ii. iii.

KNO3 AgNO3 CaCO3

a. Rhombohedral b. orthorhombic a. Rhombohedral b. orthorhombic Trigonal orthorhombic

Note:

1. 2. 3. 4. 5.

Those substances which shows polymorphism are called polymorphic substaces ie, KNo 3, AgNo3& so on. Polymorphs: The different crystalline form of the polymorphic sub stace are called called polymorphs. Same chemical properties: All the polymorphs have same chemical properties because of same molecular formulas. Inter-conversion: They are inter-convertible into each other under different environmental conditions. Physical properties: They have different physical p hysical properties.

10. Allotropy: The existence of an element in more than one crystalline form is called allotropy. Explanation: The various crystalline forms are called allotropes.

For example: 1.

Sulphur exists in two allotrophic forms i.e. a. Rombohedral b. Monoclinic 2. Similarly: oxygen exist in two allotrophic forms i.e. 77



a. O2 b. O3 3. Carbon exists in three allotropic forms i.e. a. Diamond (cubic form) b. Graphite (hexagonal) c. Bucky ball (pentagonal)

MCQS: Order of strength of Hydrogen bonding. H2o>HF>NH3

100B.P 34C o – 33C 33Co

Note: When we discuss the polymorphism of element it will be allotrophy. 12. Transition Temperature: The temperature at which the two crystalline forms of a substance co-exist in equilibrium with each other is

called transition temperature. Explanation: The temperature at which one crystalline form of a substance changes into another forms is called transition temperature. For example:

95.⇌5

Sulphur S8

sulphur S8

Rohombic Monoclinic Note: Above and below 95.5c0 one form of the crystalline sulphure will be present only ie below 95.5c 0 Rhombic sulphur will be present only and above the 95.5c0 the monoclinic will be present only. Other example: Grey tin (cubic)

13.⇌2  128⇌

KNO3 (orthohombic)

white tin (tetragonal) KNO3 (rhombohedral)

Entry test: The term “Isomerism” was first used by Berzilius.

The overall arrangement of particles in a crystal is called crystal lattice. ion s or molecules of crystal at different sites in three dimensional spaces is called crystal lattice. Crystal lattice: The arrangement of atoms, ions o f crystal lattice which possess all the properties of the crystal lattice is called unit cell. Unit cell: The smallest unit of Explanation:

i. ii. iii.

Unit cell is the building block of the crystal. It is the structural unit of crystal. It has complete information about the crystals.

Composition:

i. ii. iii.

Unit cell consists of three angles r, B & r. Unit cell has three axis a, b &c. These six parameters are called unit cell parameters or crystaqllographic elements.

The study of the structure & properties of crystals with the help of x -rays is called crystallography. Nacl crystal: crystal: Sodium chloride is a typical example example of crystal lattice (cubic (cubic system). Cubic system: A cubic system is one in which; a. b.

All the edges are equal ie (a=b=c) All the interfacial angles are equal to 90 ie (

90

)

th ree types Classification of crystal lattice (cubic): The cubic crystal lattice are of three i. Simple cube ii. Face centered iii. Body centered 1. Simple cubic lattice: Those crystals which have eight atoms at the corner, and six forces, will be considered as simple cubic lattice system. Explanation:

     2.

The no of atom per unit simple cube is 1 = i.e. 8 × 1/8 = 1

Maximum empty space Points are present at corners only Rarely found in nature Each point is shared by the eight unit cell For example polonium. Body centered cube: That simple cubic crystal which have 8 atoms at the system corner and one atom at the center.

Explanation:

i. ii. iii.

Maximum space is occupied by particles. Particles are more paced Each point in the center of the cell is part of only one unit cell.

For example: CsCl2 Face centered cube: They have one point on each faces of the cube, in

The no. of items per unit body center cube are 2 i.e. 1 + 8 × 1/8 =2

The no. of items per unit face center cube are 4 i.e. 8 × 1/8 + 6/2 =4

78



addition to the points at the corners. Explanation:

i. ii. iii. iv.

Maximum space is occupied by the particles. Each point at the corner is shared by eight unit cell. The point at the center is shared by two unit cells. For example NaCl

Lattice energy: Lattice energy can be defined as “The energy required to break one mole of an ionic crystal into its constituent ion in the g as phase is called called lattice energy”. energy”. Unit: Its unit is KJ

–

Representation: Lattice energy is represented by For example:

1Nacl

∆. + – ∆787/ .

+

Sign of energy: The sign the energy will be positive in this case.

OR Second definition: The amount of energy released when one mole of an ionic crystal is formed from its gaseous ion of opposite charges is

called lattice energy. For example:

∆.

Note:

+ – +

will be negative in this case ie

∆–   .



Note: As crystal energy increases, stability also increases i.e. NaI=690KJ/mol. So it is stable than Nacl. Types of crystalline solids: Crystalline solids are of four types;

i. Ionic crystals ii. Covalent crystals iii. Metallic crystals iv. Molecular crystals 1. Ionic crystals: Those crystals which are formed by the interaction of oppositely charged ions are be ionic compound. For example: NaCl, Mgo, NaBr KBr, CsF etc. Properties:

i. Specific geometrical shape because of regular packing. ii. Solid and hard at room temperature. iii. Never exist in liquid or gaseous gaseous form at NTP. iv. Soluble in polar solvent. v. Do not conduct electricity. vi. High M.P vii. High heat of vaporization. 2. Covalent crystals (network covalent solid): Those crystalline solid in which the atom, ions or molecules are held together by covalent bond called covalent crystals. For example:

    

Diamond Bucky ball Graphite Carborandum Silicon carbide

Properties:

Electron Pool Theory:

In metal electron are very loosely bound, and hence move freely from one place to another inside the metallic crystal lattice, like molecules in a gas and are called conduction electron or free electrons, these electrons are loosed by the atom and hence atom become positive charge & electrostatic electrostatic forces of attraction are produced.

i. Very hard ii. Cannot be cleaved easily iii. High melting point iv. High heat of fusion v. Poor conduction of heat and electricity vi. Insoluble in polar solvent vii. Hardest in the word viii. Valuable like diamond ix. Used for ornamental purposes i.e. i .e. Palaces, Necklace. 3. Metallic crystals: Those crystalline solids in which the atoms are held together by a special type of bond called metallic bond & the called metallic crystal. Concept: The metallic crystals can be explained by three theories:

i.

Electron gas theory (pool theory) 79


ii. iii. iv. v.


Valence bond theory Band theory or MOT No cleavage plane May be liquid

Properties:

i. Hard because of compact structure ii. High melting point iii. Few of them are soft and can be cut with knife ie Na iv. Malleable and ductile v. Metallic luster. (because of oscillation of ) vi. Good conductor of electricity & heat. vii. Examples: Silver, Cu, Al, Na, Fe etc. 4. Molecular crystals: Those crystals in which the atoms, ions, or molecules are held together by weak Vander wall forces are called molecular crystal.

̅

Explanation:

i. ii. iii. iv. v. vi. vii.

Molecules are packed tightly and regularly. Force of attraction is non-directional On melting only weak Vander wall force shown May be polar or non-polar. Polar molecular crystals: i.e. ice, sugar, sugar, high B.P & M.P Non-polar molecular crystal crystal i.e. solidified noble gases, gases, CO2, S, I & P. Sublime easily

Properties:

i. ii. iii.

They have low M.P Soluble in non-polar solvent Soft crystal because of weak Vander waal forces.

Structure of Ice: Solidified water (ice) is example of po lar molecular crystal.

The molecules of water have tetrahedral structure. Two lone Paris of electrons on oxygen atom occupy two corners of the tetrahedron. When temperature of water is decreased and ice is formed then the molecules become more regular and this regularity extends throughout the whole structure. Empty spaces are created in the structure. That is why when water freezes it occupies 9% more space and its d ensity decreases. As a result ice floats on water. The structure of ice is just like diamond because each atom of carbon in diamond is at the center of tetrahedron just like the oxygen of water molecules in ice. Exercise

1.

Choose the correct or most probable answer for each of the following.

i.

Which type of motions are present in solids. (a) Translational

ii.

(b) Rotational

(b) Strong intermolecular force (d) Low rate of diffusion

Solid particles do not diffuse into each other, it is due to ________ (a) Little empty space

(b) Lack of translator motions

(c) No intermolecular forces iv.

(d) None of these

Amorphous means (a) Without any specific shape

v.

(b) Two (c) Three (d) Six

A crystal conduct heat and electricity with different magnitude in different directions. This property is called. (a) Allotropy

vii.

(b) Without any force

(c) Without transition temperature (d) Without motions A cubical crystal has ______ centrels of symmetry. (a) One

vi.

(d) All of the above

The property of solids responsible for their rigid and definite shapes is their __ (a) High density (c) Anisotropy

iii.

(c) Vibrational

(b) Anisotropy

(c) Isotropy

(d) Polymorphism

The geometry of NaCl is (a) Simple cubic

(b) body center cubic

(c) Face centered cubic

(d) Hexagonal

ii)

C

(i)

B

(iii)

C

v)

A

vi)

B

viii)

C

iv)

A

80


Briefly Answer the following

viii.

Define the terms with reference to NaCL (a) Space lattice


(b) Unit cell

Space lattice is a three dimensional geometric arrangement of atoms or molecules or ions composing a crystal. Space lattice is also khown as crystal lattice. Unit cell

The smallest part of the crystal lattice which has all the characteristics features of the entire crystal is called unit cell. It is the smallest block or geometrical figure form which the entire crystal is can be build up. b.

Center of symmetry

A point at the center of the crystal which is equidistant from two opposite faces of a crystal. iii.

Define and differentiate between amorphous and crystalline solids definition

Crystalline Solids: The solids in which the structural unit is arranged in a regular manner which repeats itself in three dimension. Amorphous Solids

The solids in which the constituent particle (atoms, ions or molecules) are packed together randomly and lack ordered structure Difference between amorphous and crystalline solids Crystalline Solids

Amorphous Solids

(i)

Internal arrangement arrangement of particles is well defined

(i)

Internal arrangement of particles is not well defined.

(ii)

Posses sharp melting point

(ii)

Do not posses sharp melting point

(iii)

These have characteristic heat of fusion

(iii)

Do not have characteristic heat of fusion

(iv)

They give regular cuts when cut with knife

(iv)

They give irregular cuts

(v)

True solids

(v)

Super cooled liquids

(vi)

Generally incompressible

(vi)

May be compressed

(vii)

Break down along definite cleavage plane

(vii)

They have no symmetry elements

(viii)

These can be converted into amorphous solids

(viii)

These cannot be converted to crystalline solids

(iv)

Why crystalline solids have sharp melting points while a morphous solids don’t.

i.

Ans: Crystalline solids have sharp melting point due to specific orderly arrangement of its particles throughout the crystals. Therefore, require a specific amount of energy to break the bonds in these crystalline structure. That is why a crystalline compound melts on a constant temperature called sharp melting point.

ii.

Amorphous solid:

On the other hand, the particles in amorphous solids are at random, with no specific geometry. There is no uniformity in the arrangement of particles throughout the substance. Therefore, amorphous solids, melt over wide range o f temperature. (v)

Why metallic crystals are good conductors of electricity but ionic crystals are not.

i. ii. iii.

In metals, mobile free electron exists in the form of Electron Sea. These electrons are readily available for th e conduction of o f electricity. When electric current is applied to one end of an iron rod, it will conduct electricity in microseconds.

Note:

iv. v.

On the other hand, iron not readily available. Ionic compounds conduct electricity either in the molten form or in the form of aqueous solution. Even then there conduction is slow. Because the oppositely charged ions move to the electrodes slowly. Therefore conduction is slow as compared to metals.

(iv)

Cleavage is an anisotropic behavior discuss

Ans.

See the text

(vii)

Isomorphic substances have usually the same atomic ratio.

Ans. The phenomena in which which two two different different substances substances exist in the same crystalline form is known known as as isomorphism. isomorphism. The The isomorphic substances usually have same atomic ratio. e.g

NaNO3, KNO3 1:1:3

1:1:3

K 2SO4, K 2CrO4 2:1:4 (viii)

2:1:4

What is axis of symmetry. Describe with simple diagram.

Ans. If a crystal can be divided by an imaginary. imaginary. Plane into two equal halves halves such such that one half is the exact mirror image image of the other, as shown in the figure below. 81



Plane of Symmetry (x)

Describe why covalent crystals are hard while molecular crystals are soft?

Atoms are arranged in a face centered crystal structure. Its hardness is due to tightly packed carbon atoms in the face centered cubic system. Carbon atoms are arranged as rigid network called giant covalent structure in which atoms cannot move, having a high melting point. Corpulent bonds are relatively weak bonds. As compare to ionic bonds. Covalent bonds are formed by sharing of electrons. Covalent bonding, generally occurs in bonding non-metals. For example iodine crystal, water methane, naphthalene, scion carbide sic pcl 5 are all covalent compounds having lower melting and boiling point. That’s why covalent compounds are soft. 3.(a)

What is a solid state of matter. How it differ as from the gaseous and liquid state.

Ans. The state state of matter which which has definite shape shape and volume is called solid. It differ from from liquid and and gases gases due to its properties such as intermolecular distance, volume, shape etc. which are different from that of liquid s and gases. (b)

Explain with reason, that why amorphous solids are also termed as super cooled liquids.

Ans. Super cooling cooling is also known known as under cooling. cooling. IN this process a liquid is cooled cooled below its freezing freezing point without becoming a crystalline structure. Water normally freezes at O 0C. However it can be super cooled at about – 48.3 48.30C. It has been has been found that changing a liquid into solid, liquid transition is required. Without proper cooling liquids are ju st like glass in structure. 4.

How kinetic molecular theory (KMT) explains the behavior of solids substances. Discuss in detail.

Ans. The particles (atoms, molecules molecules or ions) are closely packed packed in solids. These These particles particles are are close enough enough and are tightly packed 12 forming a rigid structure. In fact, they are separated by a distances of only a few pico meter (1 pm = 1 × 10 – 12 m). The three dimensional regular arrangement of the particles in a solid, is called a lattice. Each point in the lattice is occupied by a particle of the substance. The nuclei do not move freely in space and stick to their fixed positions due to the attractive forces of the electronic clouds. They, however, vibrate about their fixed position. 5.

Define and explain the following properties of crystalline solids

Ans.

(i) Geometric shape (iv) Anisotropy See the text

6.

Explain latticeenergy with example

Ans.

See the text

7.

How solids are classified on the basis of forces present between them. Discuss their properties briefly.

Ans.

See the text

8.

Compare the following

(ii) Cleavage plane (iii) Habit of crystal (v) Crystal growth (vi) Transition temperature

Polymorphism

Allotropy



The existence of a compound in more than one crystalline form is called polymoophism





Different crystalline forms of a compound are called  polymorphs e.g. AgNO3 exist in rhombohedral and  orthorhombic form

Different crystalline forms of an element are called allotropes Sulphur exist as rhombic and monotonic cry stalling form



Silica (SiO2) exist in the forms quarts, tridymitie and  oristobalite

Carbon cubic and hexagonal form

The existence of an element in more than one crystalline form is called allotropy.

Covalent Crystal

Molecular Crystal



Atoms are connected by covalent bond network



Polar or nonpolar molecules may be atoms or noble gases form molecular crystals

    

Covalent bond is present

   

Vander wall’s forces are present

These are very hard These have high melting point Poor thermal and electrical conductance E-g diamond, quarts silica (SiO2), SiC

These are soft There have moderate to high melting point. Poor thermal and electrical conductance. E.g. Dryice, Iodine sugar, Sulphur etc.

Melting temperature



iv.



The temperature at which a substance melts is called melting temperature Density of water

Density of water is greater then density of ice

Transition temperature



The temperature at which two different crystalline forms of a substance co-exist in equilibrium with each other is called transition temperature

Density of ice



Density of ice is less than the density of water

82




CHEMICAL EQUILIBRIUM

If the reaction is reversible then equilibrium will be present. Reversible Reaction:

Those reactions which proceed in both direction backward as well as forward direction that will be called reversible reaction. Second Definition:

The conversion of reactants into products and produ cts into reactants simultaneously is called reversible reaction. Third Definition:

Those reactions in which the products may also change into reactants simultaneously. Fourth Definition:

Those reactions which donot reach to completion are called reversible reaction. reaction. Representation:

It is represented by double arrow;

⇋

A+B“ ”C+D Dynamic Equilibrium:

When the rate of reactants concentration become equal to the rate of products concentration in a reversible reaction is called dynamic equilibrium. Note: At equilibrium the reaction is continued not stop. Concept:

When a reversible reaction starts. Then; 1.

Initially reactants concentration is maximum concentration means quantity (mol/liter)

2.

The concentration of reactants decreases

3.

The concentration of products increases increases

Note:

When the reactants convert into products at th e mean time the product will also change into reactants and hence reverse reaction will also start. Initially the forward reaction is greater than reverse reaction and after some time, the rate of forward reaction become equall to backward reaction. R f f R b Equilibrium Constant:

The concentration of reactants and products at equilibrium is called equilibrium constant. Representation:

It is represented by capital Kc Types of Equilibrium:

There are two types of equilibrium.

MCQs:

(i) Homogeneous (ii) Heterogeneous

Equilibrium is more stable but products may also convert into reactants depending upon (a) Favourable condition (b) Stability of products

i. Homogeneous Equilibrium:

In homogeneous equilibrium the components are present in only one phase.

At equilibrium the concentration of products and reactants are constant.

For Example:

⇋

N2g + O2(g) 2NO(g) (gaseous equilibrium) CH3COOH(l) + 2C2H5OH(l) CH3COOC2H5(l) + H2O(l) (liquid) ii. Heterogeneous Equilibrium:

That equilibrium in which in which the reactants and products are present in different phase. For Example:

⇋

CaCO3(s) CaO + CO 2(g) Graphical Representation of Equilibrium:

Following reaction; A  B Initially A = 100 mol/L 83



B = 0 mol/L After some time 60  40 (R) (P) (i)

This graph shows that (R) is decreased from 100 to 80m/L

(ii) (i)

Curve “2” shows that product concentration has increased from zero to 50 mol/L

This graph shows that the reaction is continues and is in equilibrium

(ii) (R)  (P) This graph shows that more products has formed at equilibrium and less (R) are present. A  B Initially: 100  0 Finally: 30  70 Law of Mass Action:

This law was presented by Gulberge and peter wage” Statement:

The rate of reaction is directly proportional to the product of reactant concentration raise the coefficient to their power”. Mathematical Explanation:

Let us consider the following general reversible reaction. aA + bB

⇋

lL + mM

Rate of forward reaction  [A]a [B] b Rf  [A]a [B] b Using constt “kf”

Rf = kf (A)a (B) b ––– (1) Now the rate of backward backward reaction reaction  (L)l (M)m Rb  (L)l (M)m Using constt Rb  kb [L]l [M]m –––  (2) Since: At equilibrium the rate of forward reaction b ecome equal to the rate of backward reaction. R f  R b So comparing (1) & (2) Kf [A]a [B] b kb [L]l [M] m ––––  (III)

   ––––  (IV)    Since: the kf/kb is called “Kc” equilibrium constant. So (iv) can be written as;    ––––  (V)

Kc: The concentration of reactions and products at equilibrium is called equilibrium constant.

Note:

i.

Kc is called equilibrium constant

ii.

The value of Kc is constant at specific temperature

iii.

Its value depend upon temperature.

iv.

Independent of initial concentration

v.

No unit

vi.

Doesnot depend on moles

vii.

Depends upon stability of products.

Note:

If product formed is stable then no backward reaction will occurs.

Gaseous Equilibrium: Equilibrium constant in term of partial pressure:

For gaseous equilibrium we will use “p” with the concentration. For example:

H2(g) + I2(g) 2HI(g) Kp =

   

84



Equilibrium Constant for MOLE Fraction

If the concentration are expressed in term of mole fraction then the “kc” can be expressed as;

Kx =

   

X = mole fraction Relation between Kp and Kc:

To find the relation between Kp and Kc consider the following reaction for ideal gases.

⇌

aA(g) + bB(g)

lL(g) + mM(g)

According to law of mass action

 ––––  (1)   Now for concentration concentration use “C” in equation (1)    ––––  (2) Kc =         Since there is gaseous equilibrium so using “P” for partial pressure equation (1) can be written as    ––––  (3) Kc =         But Pv = nRT   P =  ⇒  = c Kc =

p = CRT

Putting the values of “P” in equation (3)

 ––––      Kp=   Kp =

(4) Rearranging

Taking [RT] common.

 .      Base is same:   Kp =  .     Kp =  .   . RT+m−a+   Kp =

OR

Kp=

∆ ∆ ∆

Substituting the values of “Kc” from equation (2)

l + m = np

Kp = Kc

a + b = nR

RT∆

Conclusion: i.

∆ ⇌

If

= +ive then its means

np nR Kp  Kc Example:PCl5 PCl3+ Cl2

1mol 1mol 1mol

∆ ∆ ∆ =



= (1 + 1) – (1) (1) = 2 -1

∆ ii.

= +1 If

∆ ⇒ = -ve

Its means that; np nR 85



Kp Kc

⇌

Example: N2+ 3H2 2 NH3

1mol 3mol 2mol

∆ ∆ ∆

∆ ∆ ∆

=  = 2 – (1 (1 + 3) = 2 – (4) (4) = -2

∆ ⇒ ≅ ≅ ⇌ ∆ ∆ ∆ iii.

= zero

If

It meansnp nR

Kp Kc

Example:H2+ I2

∆ ∆ ∆

⇌ ∆ ∆ ∆

2HI

=  = 2 – (1 (1 + 1) = 2 – 2 2 = 0

N2+ O2 2NO

∆ ∆

=



= 2 – 2 2 = 0

Unit of “KC” Equilibrium Constant:

Consider the following reaction

⇌

H2(g) + I2(g) 2HI(g) -

(mol L – )2[HI] = mol L Kc = mol mol L – 1 mol L – 1 = Kc 

mol L  mol L  mol L  mol L 

Note: when the no of moles of [R] = [P] then kc has no unit

Case-II: when the no of moles of reactants are not equal to the no of moles of product then Kc has a unit of mol L ie.

⇌

PCl5 PCl3 + Cl2

Kc  Kc 

[ pcl3 ][cl 3 ] [ Pcl 5 ] [mol L ] mol L  mol L 

Kc = mol L – (i) Conventionally “Kc” has no unit but if no moles of Reactants are equal to no of moles of product s. Then Kc has a unit of mol L – 1

Note:

(ii) According to Federal test Nust test Kc has a unit of mol L – 1. Equilibrium Constant for Partial pressure:

The equilibrium constant for partial pressure can be written as; consider the following reaction. aA + bB

⇌

cC + dD

equilibrium constant can be written by adding “p”.

Kc =

[C ]c [ D]d [ A]a [ B]b

Adding “P” for partial pressure:

Kp =

[C p ]c [ P o ]d [ P A ]a [ P B ]b

Equilibrium Constant for Mole Fraction:

The equilibrium constant for mole fraction can be written as; aA + bB

⇌

cC + dD 86


Kc =


[Cx ]c [ D ]d [ A]a [ B ]b

Using “X” for mole fraction

K x =

K x =

[Cx]c [ Dx]d [ Ax]a [ Bx ]b [ xC ]c [ xD ]d [ x A ]a [ xB ]b

Relation between Kp and K x : x

Consider the following rxn; aA + bB Kp =

⇌

cC + dD

[C p ]c [ p D]d [ P A ]a [ p B ]b

But: Mole friction in term of A, B, C & D. According to the Dalton law of partial pressure; PA = XAP PB = XBP PC = XCP PD = XDP Squation (i) can be written as;

[ X C P]c [ xD ]d [ X A P]a [ xB P ]b Kp =

Kp =

Kp =

Kp =

( X C )c ( P )c .( X D )d ( P )d

( X A )a ( P )a .( X B )b ( P )b ( X C )c ( X D )d .( P )c ( P )d

( X A )a ( X B )b .( P ) a (P )b ( X C )c ( P ) c P ( c  d ) ( X A ) a ( P) a P ( a b ) ( X C )c ( X D ) d a

b

( X A ) ( X B )

. P ( c  d )( ab)

Substituting the value of “Kx” from to ––– (3) (a+b) – (a+b) Kp = Kx P(c+d) –

K p = K x pn ––  (4) Putt the values: (P) Pv = nRT

 n = 1   ∆ Kp = Kx (   P=

Relation between Kp and Kn

When the concentration is expressed in term of moles then Kp can be expressed in term of “Kn”.

According to Dalton law: law: P(A) = XAP –––– (a) (a) But: XA =

 ––––– (b)  (b)

Where nA = no of mole of a single gass. n = No of moles of total gass 87



So equation (b) should be putt in (a) PA =

  ––––– (c) (c) 

Consider the following Rxn; aA

+ bB

Kc =

⇌

cC

+ dD

(.C )c (.D)d ( A) a ( B )b

But for the no of moles add (n) and replace “Kc” by “Kn”.

Kn =

(nC )c (nD)d (nA)a (nB )b

–––––  (1)

Now: for partial pressure pressure the equilibrium can be written written as;

( PC )c ( P D )d Kp =

( P A ) a ( P B )b ––––  (2) n

But P =

N  (c)

Putting (c) in (2) C

Kp =

d

n  n   cP   DP   N   N  a b n  n   AP   BP   N   N 

(nC )c ( P / N )c (nD)d ( P / N ) d

a a b b ( ) ( / ) ( ) ( / ) N P N nB P N A Kp =

(nC )c (nD )d (P / N )c ( P / N ) d

a b a b ( ) ( ) ( / ) ( / ) n n P N P N A B Kp =

c

(nC ) ( nD ) Kp =

d

(n A ) a ( nB )b

c  d

 N   P N  P

(nC )c (nD ) d P a b N ( n ) ( n ) A B Kp





a b

( c  d ) ( a b)

Substitute the values of “Kn” in equation (3)

Kp = kn (P/N)n Note: If the no of moles of reactants are equal to the no of moles of pr oducts then i.e.

nP nR then; Kp Kc = Kx = Kn Note:

“Kc” is independent of the molar concentration of the solid. For Example:

Which one will be the Kc for the given reaction;

⇌

2Fe(s) + 4H2O(aq) Fe3(s)O4 + 4H2(g) 88



[ Fe3O4 ][ H 2 ]4

Kc =

[ Fe]3 [ H 2O ]4

Since:

We take the molar concentration zero for the solid so Fe3O4 = Zero Fe = Zero So equation (1) can be written as;

[ H 2 ]4 [ H 2O ]4

Kc = Note:

The molar concentration of solid remain constant and varies for gases. Molar Concentration of solid:

  = 100.  

If we have a container of 10 dm3 volume and no of mole are 1000 then its molar concentration is given by: c = = For Half Molar:

Since for half moles the volume also become half so the molar concentration will be;

 

C =  n = 500 V = 5L C = 500/5 = C = 100 ––––  (2) For One third:

n = 250 V = 2.5 C = 250/2.5

C = 100 –––  (3)

Conclusion:

Since from (1) (2) & (3) it is clear that the molar concentration of solid remain constant. Molar Concentration of Gases:

As the volume of the gases are not constant and expands so their concentration varies i.e. consider a container having 1000 moles of gas and have 10 litre volume so it concentration will be. i.

Concentration for 1000 Molar:

 = C = 1000  (1)  Concentration for Half:  C =  = C = 500  (2) Concentration One third:  C =  = C = 25  (3) C=

ii.

iii.

Conclusion:

Since the molar concentration varies for gases. Application of Kc:

Following are the applications of the Kc. 1. 2.

Extent of chemical reaction Prediction of direction

3.

Equilibrium concentration of [R] and [P].

Extent of Chemical Reaction:

As we know that Kc = a.

 

If Kc is very Larger:

1.

Kc larger means that the values of Kc 1

2.

Kc will be greater than one.

3.

Its means that more products has formed at equilibrium. 89



4.

Forward reaction is more favourable.

5.

Reactants are unstable easily converted into products.

6.

Products are stable.

7.

Activation energy of forward reaction is low.

For Example:

2Cl

⇌ ⇌

Cl2 (Kc = 1  1038 at 25 C0)

2O3 3O2 (Kc = 1  1055 at 25 C0) Conclusion:

High values of Kc means that more products has been formed. b.

If Kc is very Small:

i. ii. iii. iv. v. For Example:

N2 + O2 2HF c.

⇌

⇌

Kc very small means that Kc <<< 1 Kc smaller means Less product has been formed. Forward reaction is not favourable. Reactants are stable, therefore does not easily change into products.

30 2 NO Kc = 1  10 – 30

F2 + H2

13 Kc = 1  10 – 13 at 2000 C0

If Kc is neither very small nor very large:

Its means that moderate or appreciable amount of products and reactants are present at equilibrium. i.e. Kc = (10 – 3 – 10 10+3) R f f R B For Example:

1.

⇌

N2O4 2NO2 Dinitrogen tetraoxide Kc =

⇌

  Kc = 0.36 at 25 C  

0

N2 + 3H2 2NH2 Kc = 10 at at 300 C0

2. Since:

It is neither very smaller nor very larger but moderate. 3.

Prediction of Direction:

Kc gives us information about the direction of a chemical reaction, that in which direction it moves to reach equilibrium. Note:

The values of “Kc” which has been given in the literature book is experimental values i..e Kc or Kp. But:

The values of “Kc” determined by the students in the lab from the concentration or fr om experiment is Qc & Qp. Case-I:

If

Kc >Qc OR

Kp> Qp its means; i.

No equilibrium has been attained yet.

ii.

And the reaction is in forward direction.

Case-II:

If

Kc
Then its means; i.

No equilibrium has been attained yet.

ii.

The reaction is in backward direction.

Case-III:

If

Kc >Qc and Kp> Qp

Then its means; i.

Reaction is in equilibrium

ii.

Rf = Rb 90



For Example:

Given data: H2 = 0.01M I2 = 0.01 M HI = 1.02 M Kc = 1.6  10 – 2 Solution:

2HI

⇌  

H2 + I2

 .. Qc = . Qc =

Qc = 9.61  10 – 5 Since:

Qc< Kc Result:

The reaction is in forward direction. Since reaction is in forward direction. 4.

Calculation of Equilibriums concentration of Reactants and Products:

When Kc values of reaction, as initial concentration of reactants and products are given, then their equilibrium concentration can be calculated from “kc” expression. The unknown equilibrium concentration of each reactant and product is denoted by “x” which is then calcul ated. For Example:

Consider the following reaction for which Kc = 64 at 400 C0 Reaction:

Initial conce: Change in con:

H2g +

I2g

2.0

4.0

– x

(2.0 – x) x)

⇌

2HI 0

x

2x

(4.0 – x) x)

2x

Equilibrium conce:

     64 = −−  64 = −−+  64 =  −+  64 =  −+

Kc =

Rearranging

64 (x2 – 6x 6x + 8) = 4x 2 64 (x2 – 6x 6x + 8) – 4x 4x2 = 0 64 x2 – 384x 384x + 512 – 4x 4x2 = 0 64 – 4x 4x2 – 384x 384x + 512 = 0 64(a)x2 – 384 384(b)x + 512(c) = 0 Now using quadrating equation

  −  −  −  −  −     

x = – 6 +

x = – ( – 384) – 384) x = 384+ x = 384+

91


x=


±. 1. 9  

Le-Chatelier Principle:

If a system in equilibrium is disturbed by any change then it will shift in that direction, in order to minimize the effect of that change” is calle d Lechatelier principle. Changes means

(i) concentrations (ii) Pressure (iii) Temperature

OR: If a system at equilibrium is subjected to a stress, by a change in concentration, temperature or pressure the system shift t o counteract the stress and re-establish a new equilibrium”. Applications:

Followings are the applications. a.

Effect of change in Concentration:

Suppose the following system is in equilibrium A + B i.

⇌

C+D

If we disturbed or change the concentration of any one of the reactants or products of the system, then the system will be no longer in equilibrium. Now according to Le-chatelier Principle this system will go either in backward or in forward direction, to gain equilibrium state. If we increases the concentration of reactants the reaction will go in forward direction. Similarly if we increase concentration, of product then reaction will go in backward direction.

ii. iii. iv. Note:

At new equilibrium the “Kc” value will remain constant. For Example:

⇌

N2 + 3H2 2NH3 (i) (ii) (iii)

If we increase the [N 2] the reaction will go in forward direction If we decrease the concentration of NH 3 then reaction will go in forward direction. If we increase the product concentration then reaction will go in backward direction. b.

Effect of Temperature:

Equilibrium as well as Kc depends on temperature. Therefore by changing the temperature equilibrium and Kc will also changes. There are two types of reaction. 1.

Endothermic

2.

Exothermic

Exothermic Reaction:

i. ii. iii. iv.

Exothermic reaction are favourable at low temperature It means that more product will be formed at low temperature They are represented as: H = – ive ive In exothermic reaction the Kc values will be very high at low temperature If in a reaction the p roduct are stable it will be exothermic exo thermic

⇌ ⇌

(i) N2 + H2 2NH3H = – ve ve e.g: 2NO(g) N2(g) + O2(g) Kc decreases as temperature increases 2. Endothermic Reaction:

1. 2.

For endothermic reaction H = + ive Endothermic reactions are favourable at high temperature

3.

Means more products are formed at high temperature.

4.

If we increase the temperature more product will be formed and the reaction will go in forward direction.

For example: KNO3 K+ NO3

If products are not stable, reaction will be endothermic.

⇌

N2 + O2 2NO

H = +ive

6. Effect of Pressure

Pressure has got no effect on the equilibrium of solid and liquids. 

Pressure will change only gaseous equilibrium system

92



Gaseous equilibrium systems are of two types.



Case I in which the number of moles of reactants and products are equal.

np nR Vp vR Such systems are pressure independents e.g.

⇌ ⇌

H2 + I2 2HI N2 + O2 2NO 2 mole reactants  2 moles of products Case II the Equilibrium system in which the number of moles of reactants and products are not equal i.e.

≠ ≠

nR np Vp vR Such reactions are pressure dependent If we increase the pressure the volume of the system will be decrease & system will shift in the direction of decrease volume. 

But if we decrease the pressure the volume will.



Shift in the direction of increases in volume.



If pressure is increased then those things will be farmed which h ave low volume.

⇌

N2 + 3H2 2NH3 Effect of Catalyst:

Any substance which alters the rate of a chemical reaction in called catalyst. 

Catalyst has got no effect on the th e equilibrium.



Only decrease the time required or necessary for reaching equilibrium.



It decreases the energy of activation for chemical reaction. Catalyst decrease the energy of activation for both forward and backward reaction due to which rate of forward reaction and backward reaction reaction increases in the same same proportion.





Therefore, equilibrium is established soon



Catalyst doesnot change equilibrium concentration therefore the value of “Kc” remain constant.

Solubility Product:

Sparingly soluble

(dissolve)

salt .

Consider a saturated solution of sparingly soluble salt like AgCl.

⇌

AgCl

Ag+ + Cl –

It has more tendency to recombine to form salt. Every lionization process is reversible. According to law of mass action, Kc = [Ag +] [Cl – ] AgCl  solid = contant. Kc =

   

= Kc[Ag]=[Ag +] [Cl-]

Ksp = [Ag+][Cl-] Since:

The product of ionic concentration of a sparingly soluble salt is called ksp. OR The product of ionic concentration in a saturated solution where ions are in equilibrium with solid. Note:

i. ii.

Value of ksp depend upon temperature. Independent of the amount of solution.

10 Ksp for AgCl = 1.8  10 – 10 at 25 co 13 Ksp for AgCl = 5.0  10 – 13 at 25 co

Ksp for PbCl2 (s) –––––––– 1.6  10 – 5 at 25 c0 –––––––– 1.6

⇌

PbCl2(s) Pb+ + Cl Ksp = [Pb+2] [Cl – 1] Application of solubility products :

93


1.

Product solubility

2.

Determine that weather a solution is a.

Saturated

b.

Un-saturated

c.

Super-saturated

3.

More salt can be dissolved or not no t

4.

Ppt formation formation occurred or not


Case#1

If ksp is greater than ionic product i.e. ksp > ionic product Then the solution will b e un-saturated More salt can be dissolved No ppt formed  Case 2: If ksp < Ionic product i.e. 50 60 

Then solution will be: i) ii)

super saturated ii) ppt formation occur.

No more salt can be dissolved. dissolved. Case 3: then the solution will be:

If ksp = Ionic product i. ii. iii.

Saturated No ppt can be formed. formed. No more salt can can be dissolved at specific temperature

Common Ion Effect

The same ion which is produced produ ced in a solution from two different d ifferent substances Definition:

The phenomenon in which the solubility of an already soluble salt is decreased by adding a strong electrolyte due to common Note:

The ionization of weak electrolyte decreases by adding another strong electrolyte having common ion. Consider a following reaction H2S

⇌

2H+ + S –

If we add small quantity of HCl. It will ionizes and will give H + ions. Due to which equilibrium is shifted to backward direction and due to which unionized molecules of H2S will formed. Since the ionization of H2S is suppressed. Consider the 2 nd example:

⇌ ⇌

NH4+ + OH –

NH4OH

Na+ +OH –

NaOH

Application:

Following are the application of common ion effect. 1)

Affect of Solubility:

Consider a saturated solution of sparingly soluble salt like calcium oxalate;

⇌ ⇌

CaC2O4 C+a+ + C2O4 – CaCl2 Ca++ + Cl2 Now if we add calcium calcium chloride then it will decrease decrease the solubility of CaC CaC2O4 Conclusion:

Since the solubility of sparingly soluble salt is decreased by adding strong electrolyte. 2)

Purification of Salt:

By passing HCl through NaCl solution, the impurities like MgCl2 & CaCl2 will be settled down b ecause of common ions. Procedure:

1.

We will make a saturated. Solution of NaCl

Now we will pass HCl gas through it.

⇌ ⇌

NaCl HCl

Na+ + Cl –

H+ + Cl – 94



Now because of common ion the NaCl will be settle settle down and will be collected collected at the bottom: Detection of Group IIA & IIB : it is uses for the detection of group IIA and IIB. Detection of Group III Basic Radical: it is uses for the detection of group IIIA. Exercise: 1)

⇌

What is the unit of kp in 2SO 2(g) + O2(g) 2SO3(g).

a) 2)

3

1/atm

b) 1/atm c) atm

d) No units.

In which of the following values of Kc, the reaction goes to completion in the forward direction:

a) 3)

102

b) 1030

c) 10-30

d) 1.

The appreciable amount of product and rea ctants is present in a reaction if its equilibrium constant value is:

a) Negative and large b) Negative and small small c) zero d) Neither small nor large 4)

⇌

Consider N2(g) + 3H2(g) 2NH3(g)H = - 92.46 kJ | mol the optimum temperature for the production of ammonia is:

a) b) 5)

50000C b) 4500C c) 00C d) constant temperature

In production of SO 3 from SO2 and oxygen. The yield of SO3 is increased by:

a) Increasing temperature c) Adding SO2 6)

b) adding catalyst d) removing O2

Le chatelier’s principle applies to a:

7)

a) Mechanical System

b) Physical System

c) Chemical System

d) Both b and c

The unit of Kc for following system is :

8)

a) mol2/ dm6

b) dm3 / mol

c)mol / dm6

d) mol / dm3

forward reaction goes virtually to completion when Kc is:

a) Positive and small c) Negative and large 9)

b) Positive and large d) Unity

Endothermic reaction are faroured in forward direction by:

a) Cooling c) Heating

b) Freezing d) Adding a catalyst

10) molecules of chlorine decomposes to a very small extent into atomic

⇌

chlorine i-e;

Cl2 2Cl This is because Kc for reaction is:

a) Very large b) very very small c)zero d) 1 11) how much reaction is complete w hen Kc=1 for the system A+B

a) 10%

b) 25%

c) 50%

d) 100%

⇌

C+D

12) HCl when added to H 2solution

a) Suppress ionization of H 2S

b) Enhances the ionization

c) Solution become colored

d) Does not affect

13) the following reaction goes to completion because: AgCl(s) + 2NH4OH(aq) [Ag(NH3)2]Cl (aq) + 2H2O(l)

⇌

a) H2O is evaporated c) The product is removed

b) The complex is formed d) Silver amine is hydrated

14) Kp is more than Kc when the difference of moles of the procucts and reactants is:

a) zero

b) Positive

c) Negative

d) one

Write answer for given short questions. i.

The change in temperature changes the equilibrium position of the reaction. N2(g) + O2(g) 2NO2(g) but change in pressure does not.

⇌

The reaction is endothermic. Increase in Temperature will favair the forward reaction. Good yield of No is obtained at temperature of 30000C. the change in pressure has no effect on equilibrium position. Change in Temperature:

i.

The above reaction is endothermic reaction. If we increase the temperature amount of product is increases. The optimum temperature for this reaction is 3000 0C. upto this temperature, rate of reaction will in crease.

Change in Pressure:

95


i. ii.


The volume on the both side of the equation are same so pressure will not effect equilibrium position. so this reaction is independent of pressure.

Give the concentration units for the following reversible reactions? a) PCl5(g) PCl3(g) + Cl2(g)

⇌   Kc = . . Kc = .− Kc = .   When gases are expressed in unit of pressure, then  =   = atm Kp =   N + 3H ⇌ 2NH    Kc =      Kc = ..  . Kc =   .    Kc =  = atm  H + CO ⇌ CO + H O    Kc =  . . Kc = . . 

b)

2

2

3

– 2

c)

2

2

2

= No units 4NH3 + 502

+ 6H O ⇌ 4NO        Kc =  . . Kc = ..

d)

2

Kc = mol. mol. dm-3

When gases are expressed in unit of pressure, then

         Kc =    = atm. Kp =

Q: 03 the value of Kc falls with rise in temperature for the synthesis of SO 3

⇌ ∆ ∆  . .   ∆

2SO2 + O2

2SO3

-1 -1

Because the value of

is negative for the forward direction in exothermic reaction, therefore a rise in temperature will not favour the formation of SO3.on the other hand, the dissociation of SO3 will occur in reverse direction. EXPLANATION:-

According to Lechatlier principle, “ if a system s ystem in equilibrium is distrub by change in temperature, pressure or concentration, the system tends to adjust itself as to minimize the effect of change” Now if we increase the temperature the backward reaction w ill be favored so, as to minimize the effect of temperature. This reaction will be favored at low temperature. 4) There is a dynamic not static equilibrium present between liquid and vapour at a c onstant temperature.

“The pressure exerted by the vapours of a liquid at equilibrium is called vapour pressure.

(i) (ii) (iii)

⇌

H2O(L) H2O(g) By sealing the container, after sometime the vapour molecules become saturated and then the reverse precess starts at a constant temperature. That is some molecules of the vapour start reconverting into the liquid state. It is called condensation. The process of evaporation and condensation will continue till a stage is reached at which rate of evoporation become equal to the rate of condensation. It is called dynamic equilibrium.

5) The equilibrium constant of all reactions may be equal to one when they are 50% complete? Kc =

  

Let the reaction be complete 50 percent. It means 50% will be the concentration of the reactants and 50% that of the products. The above equation will become.

      % %  % %  Kc = % %% % Kc =

Kc = 1 6) The change in concentration of reactants does not change the value of equilibrium constant per manently.

96



The equilibrium constant of concentration gives the ratio of concentrations of products over reactants for a reaction that is at equilibrium. REASON:-

By changing the concentration of reactants, the reaction will shift in that direction to cancel the effect of that change. It is according to Lechetlier’s principle. Therefore the ratio of o f two values i-e i -e concentration of product divided by b y concentration of reactants will remain be same (constant) 7) Discuss the equilibrium of sparingly soluble salt.

For a sparingly soluble salt such as AgCl dissolution occurs saturated solution is formed. The solution contains Ag + , Cl- and undissolved AgCl. It means that equilibrium establishes between dissolved Ag +,Cl- and undissolved AgCl. AgCl Ag+ + ClAppling law of mass action, the equilibrium constant is

⇌

Kc =

 + +− 

For a saturated solution the concentration of undissolved solute [AgCl] is constant, so combining its concentration with Kc. Kc [AgCl] = [Ag +] + [Cl-] Kc [AgCl] = Ksp Ksp = [Ag+] + [Cl-] 8) Common ion effect operates best in purification of certain substances. Common Ion Effect:-

The shift of equilibrium, caused by addition of an electrolyte having an ion in common with the dissolved salt is called common ion effect. Purification Of Nacl:-

Purification of NaCl is carried out by p assing HCl gas through a saturated solution of NaCl. NaCl Na+ + ClHCl H+ + ClCl-is a common ion due to which equilibrium shifts to the backward direction where by NaCl precipitates.

⇌⇌

9) The ionization of calcium oxalate is suppressed by adding CaCl 2 to it.

Definition: the shift of equilibrium, caused by addi tion of an electrolyte having an ion in common with the dissolved salt is called common ion effect. REASON:-

Adding CaCl2 to the saturated solution of common oxalate, the reaction proceeds in the reverse direction.

⇌

Ca(COO) 2 Ca2+ + 2COO2+

CaCl2 Ca + 2ClCONCLUSION:-

Due to the presence of common ion, io n, the precipitation of calcium oxalate occurs. 10) The solubility of sparingly soluble substance is calculated from the solubility data.

For sparingly soluble salt of AgCl in aqueous solution can be represented represented as AgCl Ag+ + ClAppling the law of mass action, the equilibrium constant is:

⇌

Kc =

     

For a saturated solution the concentration of undissolved solute AgCl is constant, so combining its concentration with Kc. Kc [AgCl] = [Ag +] + [Cl-] Kc [AgCl] = Ksp Therefore Ksp = [Ag+] + [Cl -] Ksp is called the solubility product. In general, the solubility product is the product of molar concentrations of ions in the saturated solution each raised to exponent equal to the coefficient of the balanced equation. MCQ Key

1) 2) 3)

b b d

4) 5) 6)

b c d

7) 8) 9)

d b c

10) 11) 12)

b c a

13) 14)

b b

97



CHAPTER-8:

ACIDS, BASIS AND SALTS

(a) Acid

Those substances which give Hydrogen ions in aqueous solution is called acid. For example:

1. 2. 3.

HCl H2SO4 HNO3

Reaction:

− 

HCl

(b) Bases

  +

+

Those substances which give hydroxyl ions “OH” in aqueous solution is called bases. For Example:

1. NaOH 2. KOH 3. Ca(OH) 2 For example: NaOH (c)

  Na + OH +

–

Amphoteric Substances:

Those substances which behave both acidic as well as basic substances are called amphoteric substances. For example:

H2O NH3 H2O + HCl  [[H3+O]] + Cl – base acid Similarly H2O + NH3 [NH+4] + OH Concepts about acids, bases:

There are several concepts about the acid and basis; 1. Arhenius Concept: This is the first concept which was presented by Robert Boyle. This is also called “Theory of electrolytic dissociation”. a. b.

Acid: Those substances which produces Hydrogen [H +]ions or hydronium ion [H3+o] in aqueous solution is called acids. ion in aqueous solution is called base. Base: Those substances which give

  Na +     KOH  K +  

e.g NaOH



Advantages: 1. It show strength of acid & bases

2. Ionization constant of acid & bases. Limitations: Following are the limitations: aqueous medium: medium: It is only limited to aqueous medium. ie (H2O) i. Only aqueous

ii. Non aqueous medium: It does not explain the acidity or basisity non aqueous medium. iii. It cannot explain the acidity of gases. i.e. Liquid HF, NH3, SO2 e Failure: It does not explain the acidity of those substances which have no Hydrogen. For example: CO2 is acidic or basic it cannot be explain by Arheneus because it has no Hydrogen or hydroxyl ions “OH”.

Since: Arheneus concept fail. (2) Lowery and Bronsted Concept: According to this concept acid and basis can be define as; a. Acid: Acids are proton donor. For Example:

HCl + NH3 NH+4 + Cl acid base b. Basis: Bases are proton acceptor

HCl + NH3 NH+4 + Cl Advantages: (1) It explain the acidity and basisty in non aqueous solvent.

(2) It also explain the acidity of gases; g ases; ie NH3 + HCl  NH4Cl 98



Disadvantages: (1) It cannot explain the acidity of oxides ie Mgo

(2) Does not explain the acidity of AlCl3, FeCl3 means those substances which have no proton ie AlCl3. 3. Lewis Concept:Concept acid base can be explain by Lewis concept as: a. Acids:Those substances which have no lone pair of electron is called Lewis Acid. ie CH 4, . b. Basis: Those substances which have lone pair of electron is called Lewis Bases.

 ̈

̈

Example: H3, H2O, ether, etc.

CH3 - - CH3

Advantages:It explain the acidity of those substances which have no protons i.e AlCl3 FeCl3. Disadvantages:It does not explain the acidity of the metal oxide. i.e MgO, CO 2. 5) Luxflood concept: According to this concept acid base can be defined as; Acid: The substances which accept oxide is called acid. For example: CO2 +

 

 −  C

CO2 is acidic because it take oxide. Base: Those substance which gives oxides is called base.

For example: Mgo  Mg



6. Usanuvich Concepts: According to this concept. Acid: any substance which;

i. ii. iii.

Produce cation. Can combine with anion. Form salt when combine with base.

For example

Na  Na+ + ie Na+ + 1 + + NaCl Base: Any Substance which gives anion Combine with cations Form salt when combine with acids.

̅

i. ii. iii.

For example:

̅

Cl + 1 Cl – + Na+  NaCl. 7. Shab Concept:

Note: A reactant and product which

It is the recent concept of acids bases. b ases.

is differ by a proton (H+) is called conjugate acid base pair.

Conjugate acid base pairs: Consider the following reaction between acid base pair.

 ̅

⇌

H3O+ + Cation Anion (Conjugate acid) (Conjugate Base) Conjugate Base: The anion derived from Bronsted acid is called Conjugate base. i.e. is is the conjugate base of the HA. Bron sted base is called conjugate acid Conjugate Acid: The cation derived from Bronsted + i.e. H3O derived from the H2O is the conjugate acid of H2O. Acid base pair: /HA. /HA. Acid base pair: H2O/H3O+ HA (B.A)

+

H 2O (B.B) (B.B)

 ̅

 ̅

Self ionization of water:

It is also called auto ionization. Pure water are bad conductor of electricity because of the below reasons: H2O + H2O H3+O + H As in the above reaction H3+O is more acidic than H2O (acid) and H is more basic than H2O. Therefore The reverse reaction reaction is more favrable as compared to the forword reaction and due to which the quantity of product decreases ie (H3+O & H) and since the ionization of H2O is less. So according to law of mass action,

⇌







   +  Kc [H O] = 3  Kw = +  ∴ [H O] = Kw Since: Kc =

2

2

As the molar concentration of water remain constant so it has been experimentally determined that . Kw = 10-14 at 250C Note:

Self ionization of water depends upon the temperature and is independent of the amount of H2O. 99



As pure water is neutral substance therefore; [H3+O] [ H] are equal [H3+O] [ H] Thus [H3+O] = 1 x 10 -7 [ H] = 1 x 10 -7 Note: As the temperature increases the ionization will be increases. Prove that; 1. pkw = 14 Pkw = -logkw 2. As we know 4. pkw = -log (10-14) kw = 14 5. pkw= -14 (-log(10)) 3. taking negative log 6. pkw= +14 log (10) 7. pkw = 14 log 10 log10 = 1 pkw = 14 (1) pkw = 14 Sorenson seale: Sorenson expressed the (H +& H) ie acidity and basisity in the form of pH & P OH. PH: It is the log of reciprocal of active H + concentration. OR It is the negative log of the active hydrogen ion concentration. PH = log 

  ̃ 





 H

Note: In PH “P”

PH = -log [H+] PH = -log [H+] PH 

stands for “Potenz”

 H

Potenz mean = potential to be lose H+.

Conclusion:

Small PH = strong acid Large PH = weak acid

i. ii. Note:

  

MCQS: The reaction of Increase in P H by one unit mean decrease in [H+] by 10 time. strong with strong base is H + Increase in P by two unit un it mean decrease in [H ] by 100 time. irreversible weak acid and Increase in P H by three unit mean mathematically decrease in [H+] by 1000 time.

POH = It is the negative log of Prove that P H + POH = 14

 

POH = - log

 

weak base reversible.

reaction

is

As we know that the [H+] [ H] =10-14 Taking negative log on both sides -log [(H+) (OH-)] log a × b = log a + b + -log(H ) + - log (OH ) = – 14 14 ( – log) (10) – log)



∴

Note: For entry test only;

pH + pOH = +14 (log (10) pH + pOH = 14 (1) log (10) = 1 pH + pOH = 14 acids and Ionization Constant for Acid (Ka): Numerically the strength of acids bases can be determined determined by Ka and Kb. For example: Consider the ionization of weak acid in H2O. HA+ H2O H3+O + Acid base According to law of mass action.

 ̅

⇌

) ̅   (  = Kc [H O] =   ⌈⌉  +  Conclusion: Ka =   Kc =

2

 Kc=

In H2O the molar concentration of [H+] is 10-7 so PH will be; PH = -log [H+] PH = -log (10-7) PH = -7(-log (10)) PH = -7(log (10)) log(10) = 1 H P = 7(1) PH = 7

∴

  [H O] = constant   2

i. If Ka increases acidity will be increases. If Ka decreases acidity will be decreases. ii. strength of base can be measured measured by kb, Ionization constant for base: Numerically the strength

Consider the following reaction; B + H 2O BH+ + H Accord to law of mass action

⟶



100


Kc =

  

= Kc [H2O] =

Conclusion:


  = Kb =   

i. As the value of kb increase Basicity will be increases. As the value of Kb decrease basisity decreases. PKa and PKb: As the values of pka and pkb are very small therefore it is convenient to express it in log. a. Pka:It is the log of reciprocal of Ka. b. Mathematically: pka = log Or. pka = -logka

 

Note: MCQS:

Application:

If ka = 10-5 Then: Pka = -log (ka) Putting values Pka = -log(10-5) Pka = (-5) – log log (10) Pka = +5 log (10) Pka = 5(1) pka = 5

i. ii.

Small pka  strong acid Large pka  weak acid c. Pkb: It is the log of reciprocal of Kb. Mathematically: pkb = log (1/ka . Pkb = -log (kb)



Application:

i. ii.

Small pkb: strong base Large pkb: weak base

Relation between ka and kb:

OR Prove that; Ka.kb = kw OR prove: Ka Proof: Let us consider the ionization of Acetic acid in its aqueous solution; CH3COOH+ H2O CH3CO + H+

 

 ⇌        Ka =   

(A)



Now consider the reaction reaction of conjugate base in water. CH3CO + H2O [CH3CO H] + [H+]

  ⇌



 (B) Now multiplying A with B      Ka.kb =     ×        Ka =

Since ka.kb = [H+][ ] © but [OH][H] So equation © can be written as; ka.kb = kw (D) Equation (D) can also be written as; Ka =



 1

Ka 



kw (ionization constant for water)

hence proved

Leveling effect: The phenomenon in which the cation of an acid become in equilibrium with the cation of a solvent th at will be called leveling

effect. Explanation: All the strong acids like HClO4, HI, HBr and HCl have nearly same pka values because of same leveling effect.

They appear to have nearly same strength, because of their strength is due to hydronium ion with each other and with the contain of water and this phenomenon is called leveling effect. Note:1. All acids are completely dissociated in aqueous solution and are suppressed by

+

3 

mean their cation are in equilibrium

ions.

2. It is not possible to find the strength of each acid b ecause they are completely completely ionized. 3. Strength of acid can be determined by dissolving them in acetic acid (anhydrous).

Buffer Solution: The solution which have the ability to resist any change in its P H by adding strong acid or base upto some extent is

called buffer solution. Buffer action: The ability of buffer to maintain its P H constant is called buffer action. Composition: Buffer solution can be formed from 1.

Weak acid and it salt 

i.

Buffer solution always formed from weak acid and its salt with strong base CH3COOH/CH 3COOK

101



CH3COOH/CH 3COONa Weak base and its salt with strong acid: NH4OH/NH4Cl NH4OH/NH4 NO3 NH4OH/(NH 4)2SO4 Ca(OH) 2/CaCl2 3. Salt of weak acid & weak base: ii. 2.

Note: Entry Test:

HCl/NH4is not buffer solution because strong acid will never be used for Buffer solution formation.

CH3COONH4/(NH4)2CO3 Mechanism of Buffer solution:

 

If we take buffer solution of CH3COOH/CH3COONa in a beaker. If we add small quantity of HCl (Strong Acid) then because of the [H +] of HCl the concentration of H + in the buffer solution will be increase and P H will decrease. But as the acetate ions are present in the solution so it will accept the Hydrogen and the P H will be maintained. Now if we add small small quantity of NaOH (strong (strong base) then the pOH will be increase increase because of H ions. NaOH N + H Since as there are (H +) ions in the Buffer solution and will combined with the [ H] and form water and since P H will remain constant.

   

⇌

+







PHOf Buffer solution: MCQS: Which one is a correct sentence?

It is also called Handerson equation. OR prove that PH = Pka + log[ Proof:

a) b) c) d)

] 

Acid should be added to water. Base should be added to water. Water should be added to acid. All are correct.

Let us consider the ionization of weak acids in water;

̅ ⇌   ̅   

HA + H2O

+ + H3+O

   Equation (i) can be also written as;  ̅  KC [H O] =   ̅  Ka =  Rearranging (2)     [H O)  Taking log on both sides;   Log [H O] = log     ̅   Log [H3+O] = Log Ka + Log   ̅ Kc =

(1)

2

3

+

3

(2)

(3)

+

(4)

Log [A – ]/[HA] Note: Inverting [HA]/[A] due to which sign will become negative Log [H3+O] = Log Ka – Log

Now: Multiplying ( – – ) sign on both sides.

 ̅   ̅ log [H O] = – (log (log Ka – log log  – log  ̅ pH = Pka + log  Equation (5) can be written as;  ̅ pH = pka + log   pH = pka + log    log [H3+O] = – (log (log Ka – log log – log 3

+

(5)

Conclusion: This mean that pH of the suffer solution is controlled by the pha value and conce of acid and base.

If log(1) = 0

̃

Then acid base

̃

PH = pka

Application of Buffer: It has application allot but some of the most important

Note: Order of acidity HclO4>HI>HBr>HCl>H2SO4>HN03

Small pka

large pka

102



are: i. Industrial process: The use of buffer is very important in some industrial process and is managedby P H.i.e manufacture of leather, ii.

Photographs. ii. In Bacteriological research: The culture of bacteria is maintained for researches at specific P H. iii. Biochemical reaction:It is important in biological system because biochemical biochemical reaction in both plant and animals are sensitive to PH. iv. Protein study: The study of protein should be occurred in buffer media. Exercise: Choose the correct one?

1.

Unit of kw is? a) mole dm-3

b) mole-2dm-6

c) mole2 dm-6 d) mole2 dm-3

2. conjugate acid base pair differ by: a) A proton

b) A proton pair

c) An electron

d) An electron pair

3. 1M solution of Ca(OH) Ca(OH)2 is mixed with 1M solution of HCl. The solution formed is a) Acidic

b) Basic

c) Neutral

d) Amphoteric

-

4. Cl is the conjugate base of: 5. 6.

a) AlCl3 b) NaCl c) HCl H PH of an aqueous solution is 9. Its PO is

d) KClO3

a) 11

d) 5

b) 9

c) 7

salt of weak base and strong acid has a pH approximately a) 8

7.

b) 6

c) 7

d) 9

very large ka value means that the solution is a a) strong acid

b) weak acid

8.

A solution with pka = + 9 is

9.

a) strong acid b) weak acid Example of a buffer solution is a) HCl / NaCl

10.

c) weak base

d) strong base

c) Neutral

d) strong acid

b) NH4OH / NH4Ol c) NaOH / H2CO3 d) NaOH / NaCl

which onw of the following is strongest acid a) HF

11.

b) HCl

c) HBr

d) HI

b) Base

c) Amphoteric

d) None

An anion is a | an a) Acid

12.

which one of the following statements is false false for acids acids that these these

13.

a) liberate H+ b) Accept electrons c) Have high pH – 3 10 moles of HNO3 is dissolved/ L its PH is.

d) Turn blue litmus to red.

(a) -3

(d) 1

(b) 5

(c) 3

Answers:

1. c

2.) a

3.) b

4.) c

5.) d

6.) b

7.) a

8.) b

9.) b

10.) d

11.) b

12.) c

13.) c (Short Answers):

1)

What are conjugate acid and bases? Give the conjugate bases of the following acids; HC10 4, HCN, H2CO3, NH4+

CONJUGATE ACID: A specie formed from a base after gaining a proton called conjugate acid of that base. EXAMPLES: - HC10 4, HCN, H2CO3, NH+4 CONJUGATE BASE:A specie formed from an acid after losing proton is called conjugate base of that acid.

  ̈

EXAMPLES: - ClO4, CN-, HC

3,

H3

 ̈  ̈

Classify an acids and bases basis giving reasons; BF 3, H3, H4, Ag+, CAO, KCN, H 2S, SO4-2, Na+, Cl-? BF3:- (ACID): Boron is an electron pair deficient atom in BF3, so it can accept a pair of electrons pair accepter species is called an acid.

 ̈

H3(BASE): due to present of lone pair on N in NH 3. According to Lewis concept Lone pair electron doner specie is caused base.

+

N H4 [ACID]: It is proton doner. Ag+[ACID]: Electron deficient specie, or electron accepter specie. It is an acid.

CaO[BASE]: it is metallic oxide. Metallic oxides are when dissolved in water they give bases, so basic in nature.

103



KCN[BASE]: when KCN is dissolved in water, reacts with water and produce a strong base (KOH), and weak acid )(HCN)

KCN + H----- OH

→

KOH + HCN

specie which donated proton called called acid. H2S(ACID):H2S is proton donor. According to Bronsted - lowery those specie SO4-2is proton accepter According to Bronsted lowery concept those specie which accept proton called base

SO4-2(BASE):

N+a(ACID):Because ofelectron accepter specie. According to lewis.

accepter According to bronsted lowery concept those species species which accept accept proton called base. base. Cl-(BASE): because of proton accepter Q3)

Classify the following as lewis acid or lewis base (CO2,SO2,BCl3,H2O,I, NH3,OH);

LEWIS ACID: Acid is a specie which can accept pair of electron. EXAMPLE: CO2, SO2, BCl3, are lewis acid LEWIS BASE: Those species which can donate pair of electron called lewis base.

 ̅  ̈ 

EXAMPLE: H2O, , 4)

H3, O

What is the pH of 0.0001M



[O ] = 1x 10-4

→   logO log 4 =

= 1x 10-4

Taking – tive tive log at both sides of equation -

log

PoH=-(-4

)

PoH = 4x1

PoH = 4 We also know that P H =PoH =14 pH = 14 – Po PoH pH = 14-4 pH = 10 What is [H +] ions concentration of solution which has a pH = 4.87?

5)

SOLUTION:-we know that

pH = -log [H+] -log [H+] = pH

log [H+] = -pH multiplying on both sides by -1 -1(-log [H+]) = -pH log [H+] = -pH Taking antilog at both sides [H+] = Antilog – pH pH [H+] = Antilog (-4.87) [H+] = 0.00001349 [H+] = 1.349x10-5 [H+] = 1.35x10-5 We also know that

O− O H+  − O .×××− × × × [H+]

= 10-14

=

=

= 0.740

10-14+5

= 0.740

10-9

= 7.40

7)

10-10

Explain how a buffer solution, resists changes in pH when a small amount of an acid or base is added?

BUFFER SOLUTION: a solution which resists changes in pH when small amount strong acid and strong base is added to called buffer

solution offered by a buffer solution to change in pH p H on the addition of acid or base called buffer action. BUFFER ACTION: the resistance offered ACIDIC BUFFER:consider an acidic buffer solution consisting of acetic acid and sodium acetate. When strong base such as NaOH is added,

the added

O

ions are removed by reaction with acetic acid molecules.

CH3COOH

O → (aq)



CH3CO

(aq) + H2O

104



When a strong acid is added, H+ ions of the acid reacts with acetate ion are of buffer

→  O

H+(aq) + CH3CO Hence the added

CH3COOH

(aq) +

O

(aq)

and H ion are removed and pH of buffer solution remains constant BASIC BUFFER: solution containing a weak base and its highly ionizable salt such as ammonium hydroxide and ammonium chloride can be explained on adding a strong acid such as CHl the added hydrogen hydr ogen ions are removed by rection with NH4OH

→

H+(aq) + NH4OH(aq) NH4+ + H2O(l)

 O O →

Where as, the

ions of the base added are removed by reaction with ammonium ions of the buffer solution

NH4+(aq) +

NH4OH(aq)

(aq)

CONCLUSION: By adding the small amount of acid and base in solution, then solution become buffered and no change in pH occur. 8) Explain solution of Na2CO3is alkaline, and solution of ferrous sulphate (Fesa) is acidic? Na2CO3 (BASIC): A solution of Na2CO3is alkaline because when sodium carbonate is added into water, it ionizes into its ions, Then its cations and anions reacts with water to form a strong base NaoH and a weak acid “ Na2CO3”. As the base is strong so it will provide large amount of ions in solution so the solution will be alkaline.

O   – → →

2 Na+(aq)+ CO3-2(aq)

Na2CO3(s)

2Na+ + 2H OH -2

CO3 + H2O

2NaOH + H 2

H2CO3 + O-2

FeSO4 (ACIDIC): FeSO4is acidic because when Feso 4is dissolved in water, give weak Fe(OH) 2 and a strong acid [H2SO4] As the



acid is strong so the concentration of H +ion in solution will be greater than the concentration of O ion so the solution as a whole will be acidic

→

4−

FeSO4 Fe+ + SO

– → →

Fe + 2H OH -2

Fe (OH)2

SO4 + H2O H2SO4+ O-2 CONCLUSION: in case of Na 2CO3; (H2SO3) is strong base which is produce when CO 3-2react with water so because of this solution Na2CO3is basic 

In case of FesO4;H2SO3is strong acid is product when SO 4-2 is react H 2O to form acidic solution. Because of H 2SO3solution of FeSO4 is acidic.

9)why solution of NaCl is neutral? NEUTRAL SALT: The salt which is formed by strong acid and strong b ase called neutral salt NaCl: NaCl is neutral salt which is formed by a stong acid [ HCl] and strong base NaOH

 ̅

ions. These cations and anions donot react with water becomes EXPLANATION: when NaCl is dissolved in water it provide Na + and C ions. hydrated so there is no formation of any acid or base. This is the reason that NaCl is neutral. CONCLUSION:

HCl + NaOH ↓

→

NaCl + H2O

→

Strong acid + strong base Neutral salt + water 10) GIVE SHORT REASONS: i) Buffer solution resists resists changes in PH? H BUFFER SOLUTION :- The solution which has the capacity to maintain its P constant resistance offered by solution to change in pH on the addition of acid acid and base is called called buffer action. BUFFER ACTION:- The resistance EXAMPLE:- ACIDIC BUFFER;CH3COOH + O CH3CO + H2O CH3COOH is weak acid it produce CH 3CO but when it CH3CO react with H + it produce acetate. H + + CH3COO CH3COOH BASIC BUFFER: NH4OH + H+ NH4+ + H2O Where as, the O ions of the base added are removed by reaction with ammonium ions of buffer solution. NH4+ + O NH4OH CONCLUSION:-when small amount of an acid is added to a buffer solution the acid are neutralized by the anions present in buffer solution. OR when a small amount of base is added to buffer solution, the O ions of the base are neutralized by the cations of buffer solution and hence no change occurs in the pH of a buffer solution.

 →

→  →





→



ii) NH3is base according to lewis concept, comment? INTRODUTION OF LEWIS:- lewis concept was given by lewis in 1923. STAEMENT:- acid is a species which accept the pair of electrons and a base is a species which donate th e pair of electron. REASON:-According to Lewis concept, base is that substance which donate a lone pair of electrons to electron deficient specie. As

 ̈

ammonia ( H3) posses lone pair and donate the pair of electron so a|c to lewis concept, it is base. iii) water act as either weak acid or weak base?Give reason?

105





NEUTRAL MOLECLULE:- Those molecule molecule which have same number of H+ and O mean having same number of positive and

negative ion called neutral molecule. REASON:- water molecule contain equal number of H + ion so as a whole it is neutral in natural in nature. When water ionizes, it provide H+ ions and O ions. The ionization io nization constant kw of water at 25 C is 1 10-14. This value is very small. H2O H+ + O [H+] [O ] = Kw [H+] [O ] = 1 10-14 at 25 C + -7 3 [H ] = 1 10 mol || dm [O ] =1 10-7 mol || dm3 It is now clear that are dm 3 solution, contain little amount of H+ ion [1 10-7] so it is a weak acid similarty l - dm 3of water contain little amount of O ions (1 10-7) so it act as weak base. CONCULSION:- small amount of H+ and O shows that water is weak acid and weak base

   ⇌  × ×  ×  ×

° ×

°

×



iv) if the dissociation constant (Ka) of substance increases, the value of the dissociation constant of it conjugate base (Kb) decreases. Why? Ka: ka is ionization constant of an acid suppose simple rxn:-

⇌  ̅ H+  ̅

H2O + HA

H3O +

HH+  ̅  ̅ Kc[H O] = H  ̅ H+   ̅ Ka = H  ̅ Kc =

2

It is the ratio of product of concentration of dissociated ion to dissociated acid molecule in aqueous solution base called Kb suppose a simple simple reaction: Kb: dissociation constant of base + + H2O HA + O

 ̅

 ⇌   H H+

  +HH H+ Kc[H O] =     H H + Kb =   Kc =

2

Kb is the ratio of product of concertration of dissociated ion to un-dissociated base molecule in aqueous solution and Kb REASON:-There is inverse relation between Ka and Kb. Ka When Ka increases Kb decreases decreases it means that solution will be either acidic or basic.

 

 

v) The sum of PKa and PKb is alw ays equal to 14? PKa:- Negative logarithm of Ka called PKa. Greater the value of P Ka weaker would be th e acid. PKa:- Negative Negative logarithm of Kb Kb called PKb. Greater Greater the value of PKb weaker weaker would be the base. base. REASON:-we know that the PH of solution ranges from 0-14, similarly the POH of solution ion ranges from 0-14.

For every type of solution PH + POH = 14 Acidic strength can be represented by PKa and basic can be represented by PKb. Ka=

H+  ̅ ----------- (i) H̅ ̅HH + + H+O ⇌ HA + O Kb = (ii)  H+ ------------- ̅ H H+ Ka × Kb = H  ×  ̅ 2



×× × ×

Ka Kb = [H] + [O ] Ka Kb = Kw Kw = 1 10-14 Taking – log log at both sides of equation -log Ka Kb = -log 10-14 -log Ka Ka + ( – log Kb) = -( 14 log 10) – log PKa + PKa = 14 vi) A strong acid has always a weak conjugate base and vice versa? STRONG ACID:-strong acid is that which ionizes upto maximum extent in equeous solution. CONJUGATE BASE:-when strong acid dissolved in water in provide a conjugate HCl ↓

  H

×

+ 3O +

̅

C ↓

Strong acid conjugate base The base of strong acid is always weak REASON:-conjugate base a strong acid is stable in solution. Its reactivity is low and bence de not accept proton from the conjugate acid and thus the revere reaction do not takes place or very slow. We know that Ka. Kb = Kw Ka =

KwKb

106



Strong acid will have weak conjugate base and vice versa. vii) Justify your answer with equation that th at CH3COONa gives a basic solution while NH 4Cl an acidic solution with water. SODIUM ACETATE:-As we know that sodium acetate (CH 3COONa) is made up of a weak acid (CH 3COOH) and a strong base NaOH. When it is dissolve in water it produces it constituent acid and base again CH3COOH + NaOH CH3COONa + H2O

↓

⇌ ⇌

↓↓

Weak acid + strong base salt + water CH3COONa + H 2O CH3COOH + NaOH REASON:Sodium acetatewill produce the basic solution Ionization of NaOH = 84% Ionization of CH3COOH = 1.3% Ammonium Chloride:ammonium chloride is made up of strong acid and weak base

⇌ ↓↓

HCl + NH4OH NH4Cl + H2O

↓

Weak acid + strong base salt + water When NH4Cl dissolve in water it will produce p roduce strong acid and weak base. Reason:-Solution of ammonium chloride as a whole will be acidic

HCl ionize up to 92% NH4Cl ionizes up to 1.4%

̅

viii) Why do you call AlCl 3 and BF 3a lewis acids and C and and NH 3as lewis base? Lewis Acid:Acid is a substance which can accept a lone pair of electrons Example:AlCl3 and BF3are electrons deficient compounds and both Al and B has 6 electrons in their lost shells. As they can accept a lone pair

of electrons so, these are acid A|c to lewis concept. Lewis Base:A specie which can donate a lone pair of electron called lewis base.

̅  ̈

Example: C and and Numerical:

3 both

are lewis bases bases because they can can donate a lone pair of electron. electron.

1. A buffer solution contain 1 mole dm3 each of acetic acid sodium acetic acid is 4.74. Solution:

[CH3COOH]= 1 mol. dm-3 [CH3COONa] = 1 mol.dm-3 PKa = 4.74 Using Henderson’s equation

   PH = 4.74 + log  PH = PKa + log

PH= 4.74 + log 1 PH = 4.74 + 0 PH = 4.74 2. Calculate the PH of a buffer solution containg 0.04M, NH4Cl and 0.02M NH4OH Kb for NH 4OH is 1.8 10-5

×

Solution:

[NH4Cl] = 0.04M [NH4OH] = 0.02M Kb = 1.8 10-5 POH = ? First we should find PKb Kb = 1.8 10-5 Taking – tive tive log at both sides -log Kb + - log 1.8 10-5 PKb = -0.25-(-5 log 10) PKb = -0.25 + 5 PKb = 4.75 Using Henderson equation

NHC  . PoH = 4.74 + log .

× ×

PoH = PKb + log

×

PoH = 4.74 + log

∴

log 10 = 1

Poh = 4.74 + log0.02 Poh = 4.74 + 0.301 Poh = 5.05 PH = 14 – PoH PoH = 14 – 5.05 5.05 = 0.95

 

107



CHAPTER-9:

CHEMICAL KINETICS

Definition:

The branch of chemistry which deals with the; i.

Rate of reaction.

ii.

Mechanism of reaction and

iii.

Factors affecting the rate of reaction is called chemical kinetics.

Rate of reaction:

i.

Change in concentration or amount of reactants or product per unit time is called rate of reaction. OR Reactant Product

Decrease Decrease in the concentration of reactant per unit time is called rate of reaction. Reactant

Product

OR Increase in the concentration of products per unit time is called rate of reaction. Mathematically: Rate of reaction can be written as;

Note for Entr y Test: Test:

   Rate of reaction =    K= 

Kc: Capital “Kc” represent equilibrium

constant. ate constant. kc: small “kc” represent r ate

Graphical Representation:

This is the graphical representation of the rate of reaction. a.

This graph shows that; 1. Rate of reaction decreases as the concentration o f reactants decreases. 2. Rate of reaction is not uniform. 3. Products concentration in creases. 4. Rate of reaction decreases as the reaction proceeds

Unit:

The unit of rate be determine as. K= K=

    

K= mol

 ̅

FOR GASEOUS REACTION:

For the gaseous reaction the unit of o f rate is atms -1. Rate Law:

 

It is also called rate expression. It is also called rate equation.

Statement:

“The experimental relationship between the rate of reaction and concentration of reactants is called rate law”.

OR That equation which shows the relationship between the rate of reaction and concentration of reactants is called rate law. For Example:

Consider the following general reaction. aA + bB = cC + dD According to rate expression Rate of reaction  [A]a [B] b ______________________(1)

OR



K= rate constant

 AaB ______________________(2)   kAaB ______________________(3) 

Conclusion:

Since rate expression shows that rate of reaction is directly proportional to product of the concentration of reactant raise the coefficient to their power. Specific Rate Constant:

When the molar concentration of reactants species are unity then it will b e called specific rate constant. 108



For Example:

Consider the following general reaction aA + bB  products According to law of mass action  

If

 AaB → 1  kAaB → 2  A=1

B= 1 Then equation (1) become as:

 k1a1  k1a+    k

 (1)a+b = 1

Conclusion:

Since “k” is called specific rate constant Unit of Specific rate constant

As we have

     k=  k= mol –  – k=

Difference between between molecularity and order: 1.Molecularlity:

1.Order:

The total number of atoms, ions or molecules present in a balance chemical chemical equation is called molecularity. molecularity.

The total number of reactant species whose concentration changes during a chemical reaction is called order.

2. Coefficient:

2.Sum of Exponent:

It is sum of the coefficient of all reactants present in a balance chemical equation.

It is a sum of the exponents of reactant species in the rate equation. i.e. aA + bB ––  products

i.e. aA + bB



products

a + b  molecularity

  

3.Theoritical:

Experimental:

It is theoretical

It is experimental.

4. Not fraction.

4. Fraction:

It may not be in friction.

It may be in friction. i.e. Frictional Order

5. Mechanism:

5. Mechanism:

It does not deals with mechanisms of reaction.

It deals with the mechanism of reaction.

6. Never be Zero:

6. May be Zero

It may never become equal to zero.

It may or may not be zero i.e. Zero order reaction.

7.Greater Than 3:

7. Never be greater than 3:

It may be greater than three.

It never be greater than th ree

Example:

i.e.N2+3H22NH3

Moleculairty

1+3 = 4  greater than three Order of Reaction:

The sum of the exponent of the reactant species present in the rate expression is called order of reaction. For Example:

A + B  products 109

Nasrat Ullah Katozai (Chemistry) dx dt dx dt

'

x  A  B


'

Order r  k(A)1 (B)1 11 2 Orde

Classification:

Order has been classified into the following types; 1. 2. 3. 4. 5. 6.

zero order. first order. 2nd order. 3rd order. Fractional order Pseudo order.

1.

Zero Order Reaction:

That reaction in which sum of the exponent is Zero in the rate equation is called Zero order reaction. OR “That reaction in which rate is independent of the reactant concentration is called Zero order Reaction. Mathematically:

Consider the following general react. A

→

Product

dx 0  A dt

dx dt

dx dt

0

 K  A

 Anything raise to the power zero is equal to one.

 K 1

dx  K dt

Note: Zero order reaction does not depends on slow step.

Unit of Zero order:

Aswe have know

 = k  − k = 

k = mol L – S-1 For example:

1. 2.

Those reactions which are catalysed by enzyme already obey zero order reactions. 2 HI

Au

H 2  I2

3. photochemical reaction reaction also obey zero order reaction. reaction. i.e; 4.

H2 c

2

Tungston

HCl

5.

Photosynthesis reaction between mercury chlorid & oxalate is zero order. 2Hgcl 2 + C2O4 + 2CO2 + Hgcl 2

6.

Reverse Haber ’s ’s process is zero order 2NH3

2.

N2 + 3H2

First Order Reaction:

That reaction in which sum of the exponent in the rate expression is one on e is called first order reaction. Or “That reaction in which the rate of reaction is directly proportional to the concentration of reactants raise to the power one”. For example:

Consider the following general reaction A B dx 1 [ A] dt dx  k[A] dt

Note: First (order reaction) is unimolecular unimolecular

Note: that reaction in which order and

molecularity is same is called simple reaction. That reaction in which order and molecularity is not same is called complex reaction.

110



Since order is one K Sec Sec1

Unit of first order of reaction : As we know Note:

dx

Rearrangin ging g  k [A] Rearran dt dx   k dt[A] molL  k  SmolL  1 k  S 1 S

The unit of first order rate constant is Sec -1 2009-15Med. Note for Entry Test:

In case of 1 st order reaction. If the concentration of reactants is double the rate of reaction will also become double. i.e.

Example of first order reaction:

1.

Br2 2 Br Br 226

2 22 22

88

86

Ra Rn  2He 4

2.

 = K [A]   =  = K (2)  ∝ 2K  =

∴

A = 2

Note:

All radioactive disintegration are first order reactions. 3. 4. 5.

Hydrolysis of sucrose to give glucose and fructose is first order reaction: ie. C12H22O11  H 2O  2C6H12O 6 Hydrolysis of tertiary butyl bromide to tertiary alcohol is first order. SO Cl2  SO2  Cl2

6.

N 2O5

2NO

7.

H 2O2



CCl4

1

2

H2 O 

 O2  (1) 2

1

02 2 Second Order Reaction:

3.

That reaction in which sum of the exponent are two in the rate expression is called second order reaction. Mathematically:

Consider the following general reaction. A  B  product dx dt dx

1

1



k[A] [B]



k[A]

dt

2

Note for Entry Test:

In case of second order reaction if the th e concentration of reactants are double, then the rate of reaction increases four times. Unit: as we know

dx dt dx dt

 k 



k[A]2



k[A]2 molL

molL

1 1

S

1

molL1

k  mol1S1L Example of Second Order:

1.

Thermal decomposition of nitrogen dioxide is second order. 2 NO2

2.

N

2

2 O2  2O

Thermal decomposition of hydrogen iodide.

H

2HI

2

 I2

Note: For Entry Test

There are 3 patterns of MCQs on third order reaction. Proof: First pattern:

In case of third order reaction if the concentration of reactants (A&B) are double, then rate of reaction increases eight time i.e.

111


3.


Decomposition of Ozone. 2O3

4.

3O

2

Decomposition of Aldehyde.

CH

CH3  CHO

5.

4

 CO

Oxidation of nitric oxide with ozone. NO  O3

6.

NO

 O2

2

Formation of Urea in solution phase. p hase.

(NH ) CO

NH 4  CNO 4.

2 2

Third Order Reaction:

That reaction in which sum of the exponent is three in the rate expression is called third order reaction. Mathematically:

Consider the following general reaction. 2A  B  product dx dt dx



k[A]2[B]



k[A]3

dt

Unit of third order = As we know

dx dt k 

k 

k 

k 



k[A]3 dx

dt  [ A] A]3 molL molL1Sec Sec 1 (mol (molL L1 )3 molL1 S1 ( molL molL1 )(molL )(molL1)(mol )(molL L1) S1 mol2L2

k  mol 2L2S1

Example of 3 rd Order:

1.

Gas phase reduction of nitric oxide 2 NO  H 2

2.

NO

2

 H2 O

Reaction between ferric chloride and potassium iodide in solution. 2FeI 2Fe

6FeCl3  6KI 6KI NOTE :

dx dt

2

6KCl  I 2 (2 (2015)125 Entry Test  6KC

k[FeCl3 ][KI]2  k[FeCl

It is slow step and order can be determined from slow step. 3.

Formation of Nitrosyl Chloride is also third th ird order reaction.

5.

Factional Order

That reaction in which sum of the exponent is in fraction in the rate expression is called fractional order, reaction. Mathematically:

Consider the following r eaction between trichloromethane and chlorine free radical. CHCl3 

1 2

Cl2

CCl

4

 HCl

112


dx dt dx dt


 k[CHCl3]1[Cl].5  k[CHCl3]1[Cl].5

Order=1+0.5=1.5 Note: order = 1.5

i.

There is no linear relation between rate of reaction and concentration.

ii.

If concentration of [Cl2].5 is double, then th en the rate of reaction increases one time.

6.

Pseudo order Reaction:

That reaction in which the excess is not written in the rate expression is called Pseudo order reaction. Note: In these reaction, the excess is not brought under consideration. For Example:Consider the following reaction.

CH3COOH  H2O dx dt

(Excess)

CH 3COO  H3O

 k[CH3COOH]

Conclusion:

Since this reaction is seem to be second order but actually it is first order reaction so such types of reactions are called pseudo order reaction. Experimental Determination of Order:

The order of reaction can be determined by t wo methods. 1) 2) 1.

Physical Method Chemical Method Physical Method:

In this method no sample is drawn from the reaction mixture. Following are the method. (i) Gravimetric. (vi) Dialometry 1.

(ii) Spectrophotometric (vii) Polarimetry

(iii) Conductometric (viii) Refractometry

(iv) PH metric (v) Pressure measurement. (ix) Conductivity.

Chemical Method

In this method sample is drawn from the reaction mixture. Steps involved in the determination of order. Step-I:



First of all we will determine the rate o f reaction.



As we know that rate is th thee change in concentration per unit time.

We will determine the relationship between concentration and rate of reaction and th is relation is a clue to the order of reaction. Step-II:



In this step we will determine the order of reaction by changing the concentration of reactant.

Step-III: For Example:



If there is only one reactant, and if its concentration is double and rate also become double then its means that reaction is first order.



If the concentration is double and rate increases in creases 4 times its means the reaction is second order.



If the concentration is double, & rate increase 8times its means that reaction is 3rd o rder.

Factors Affecting the Rate of Reaction:

Following factors affects the rate of reaction. 1. The Nature of reactant. 4. Temperature

2. The Concentration of reactants. 5. Catalyst

3. The Particle size of a solid reacting with gases

1. Nature of Reactant:

i.

The rate of reaction depend on the nature of reactants i.e. Mn react with oxygen faster than Cu in the p resence of flame. Because rate of Oxidation is different for different for d ifferent metals.

ii.

Acid base reaction, formation of Salt andexchange of ions are fast reaction. 113



iii. Reaction in which large molecules are formed or boken are usually slow reaction. 2. Effect of Concentration:

The rate of reaction depends upon the concentration of reactants. i.

With increase in concentration chances of collision increases among the molecules and since rate of reaction increases.

ii.

For a chemical reaction, it is necessary for molecules to collide with each other.

iii.

A piece of wood burns more rapidly in excess of oxygen i.e. 100%. Oxygen than in limited oxygen

3. Particle size:

There is an inverse relation, as the particle size decreases, the rate of reaction increases. Reason:

Because the surface area of the particle increases and since active site increase. For example:

Lumps of soft coal do not burn easily but when it is divided into smaller pieces, then it burns explosively. 4.


As the temperature increases the rate of reaction also increases. Reason:

It is because of the following reasons.



Kinetic energy of the molecule increases.



Chances of collision increases.



Average kinetic energy of the molecule increases.



Molecules possess activation energy also increases.



For each degree rise in temperature the rate of reaction become double.



The quantitative relationship between temperature and rate of reaction is given by Arrhenius equation.

Arrhenius Equation: Ea

k  Ae

RT

This exponential equation shows. a.

As temperature increases activation energy d ecreases. ecreases.

b.

As activation energy increases, the value of “k” will be decreases and thuss low will be the rate of reaction.

Logk

 logA

 Ea

2.303RT

If logk is plotted verses 1 " " T

Note For entry test:

Then it will give a straight line.

Slope 

Slop of straight line is negative.

Ea

log k = intercept

2.303R Ea  slopx2. slopx2.30 303x 3x R

Ea = It can be determined from the slope of the line.

Ea  slopx slopx 2.303x8.3 2.303x8.314 14

Slope  4.

 Ea

2.303R

Effect of Catalyst:

(i) (ii)

The rate of reaction depends on catalyst. Catalyst decreases the Ea for a reaction.

Ea  tan  x 2.303x 2.303x 8.314______ 8.314__________ _______ ______ ____(N) _(N) concentration ion   changein concentrat Conclusion: Activation energy can be calculated from equation.

For Example:

The thermal decomposition of potassium chlorate (KClO3) is very slow. KClO3 ––––→ 2KCl + 3O2 But if we used (MnO2) catalyst then the rate of reaction increases. increases. Activation Energy:

Reactants does not change into product directly.

114


(Re (Reactant) ctant)


(Produ uct) (Prod

because because theyfirst forman activated activatedcompl complex ex Reacta Reactan n ts  Activ ctivaatedCo tedCom mplex plex  Produc Productt

    

Morereactive species Weak bonded species Short lived species Unstable species Transition state

Definition:

The MINIMUM amount of energy required for a molecule to initial a chemical reaction is called activation energy. i. ii.

Reactant does not change in to product directly Because first they form an activated complex.

Explanation:

i. ii. iii.

If a reaction does not occur, its means that its activation energy is higher. There is not a single molecules which posses its own energy of activation. For spontaneous reaction activation energy of forward reaction is minimum

Mathematically:

Ea  slop slopee x 2.303 2.303 x R

Log K

THEORIES OF RATE OF REACTION:

i. 1. 2. 1.

According to kinetic studies, there are two theories about th e rate of reaction, Collision Theory Transition State Theory.

1/T

Collision Theory:

For a chemical reaction it is necessary for molecules to colli de with each other. Postulates:

It has 3-postulates i.

Collision:

For a chemical reaction the molecules must collides with each other. ii.

Energy of Activation:

Only those molecules will collides which possess activation energy. iii.

Effective Collision:

The molecules must collide with each other in proper orientation because only those collisions will be affective which occur in proper orientation. Explanation:

1.

According to first postulate, if we increase the number of molecules then the rate of collision will be increases due to which rate of reaction will be increases. Because according to kinetic molecular theory 10 32 molecules can collide per liter per second at STP.

2.

But according to third postulate those collision will be effective which occur in proper orientation.

For example:

(1) When CO combines with NO2 to forms CO2 and NO, only (2) When the carbon atom of “CO” come in contact with the oxygen atom of the NO2. Then reaction will occur because of effective collision. Transition State Theory:

This theory was presented by Henry Eying and Michal Polany in 1935. Background:

1. 2.

This theory is based on statistical mechanics mechanics but reach to the kinetics of reaction by an alternative approach. This theory gives its full attention to an activated complex. (A short lived li ved unstable species).

Postulates:

It has the following postulates; 1.

Activated Complex:

The reactants molecules before forming products must form an activated complex [R] → activated complex → [P] 2.

Properties of Activated Complex:

The activated complex are assume to have thermodynamics like ordinary molecules except one lose vibration. 115



Equilibrium:

The activated complex is in equilibrium with reactants molecules i.e.. R 4.

Tran Transi siti tions onsta tate te prod produc uctt

Rate of Reaction:

Rate of reaction is directly proportional to the concentration of activated complex i.e k Activ Activate atedComp dComple lex x  Explanation:

Consider the bimolecular reaction as; AB

K 1

k 3

*

[A B] Product

K 2

The rate of reaction depends upon the following factors; a. b. c.

Concentration of transition state. Rate at which activated complex break into products. According to law of mass action the equilibrium is given as; *

Keq



[AB]

But according to transition state they

[A][B]

Rateof Rateof reac reactio tion n  [AB] [AB]* Conclusion:

Since the rate of reaction is directly proportional to the concentration of activated complex. Catalysis:

The speeding of a chemical reaction by specific catalyst is called catalysis. Definition of Catalyst:

That substance which facilitate a chemical reaction is called catalyst. Explanation:

Catalyst has the following properties.

        

Does not take part in chemical reaction. Shows no chemical change in itself. Show some physical change. Unchanged after chemical reaction. Increases the rate of reaction. Do not consume in chemical reaction. Recovered chemically unchanged after reaction. Decrease Decrease the activation energy. Increase the speed of reaction.

Types of Catalysis:

It has two types; i. ii.

Homogenous catalysis Heterogeneous catalysis (1) Homogeneous Catalysis:

That type of catalysis in which the reactants, products and catalyst are present in the same phase is called homogeneous catalysis. For example:

Liquid ethyl acetate when combine with water to form acetic acids and ethanol, all the species are in a liquid state. So this is homogeneous catalysis. H3 O

CH3COO2 H5  H2O  CH3COOH C2 H5OH ( L)

(L)

( L)

( L)

(2) Hetrogeneus Catalysis:

That type of catalysis in which the reactants, products and catalyst are not present in the same phase is called hetrogeneuscatalysis. For example:

If the products and reactant are gases and the catalysts is in solid state in a reaction then it is the example of hetrogeneus catalysis.

Reaction:

116



Ni

C2H4  H2  C2 H6 ( L)

(g)

N 2( g )  3H 2( g )

(g) Fe2O3 (s (s )

 2NH 3(g )

Enzymes Catalyst:

It is protein with h igh molecular weight and act as catalyst for Biochemical reactions occurring in all living organisms. OR These are biological catalyst which is present in the body o f living organism. EXPLANATION:

Following are the p roperties of enzyme catalyst. 1)

Present in yeast.

2)

Organic in nature.

3)

Act as biochemical catalyst.

4) 5)

Protenic in nature. Found in living things.

6)

Specific in nature.

7)

Larger molecular size.

8)

Molecular mass 10 5-107amu

9)

More reactive in nature.

10)

More effective at body temperature.

11)

At low temperature innocent.

12)

At high temperature destroy (rapture)

13)

Optimum temperature 37C0 / 98.6F0 for most animal cells.

14)

Enzyme = Apo-enzyme +co-enzyme Active part Proteinic part

Example:

Ptyalin is an enzyme found in saliva Accelerate the conversion of starch into sugar. Exercise

1.

Activated complex is unstable.

2.

A reaction is first order with respect t o “A” and second order with respect to “B”, the rate of react = K [A] [B] 2 will be

3.

For a reaction A  product, doubling doubling the concentration concentration of “A” “A” quadruples quadruples the rate. The The react is …..Second order.

4. Q-2

Due to increase in temperature the rate of reaction increase s it is due to ….increase in collisions. Give brief and appropriate answers to the following questions.

i.

Determine the overall orders from the following rate equations. (a) Rate = k [NO 2] [O2] (b) Rate = k [N]2

(a)

Order:

It is the sum of exponent of the reactant species in the rate expression Rate equation: Rate = k [NO2]2 [O2] This equation is 3rd order with respect to [NO 2]2& [O] because order = 2 + 1 Order = 3 (b)

Rate Equation:

Rate = k [NO2]2 (i) (ii)

i.

This rate equation is second order with respect to [NO2]2 So order is 2 A molecular collision is sufficiently energetic to cause collision because of the following reasons.

Breaking of Old Bond:

117



When the molecules colloids with each then their KE is converted in to vibration energy and since they vibrate strongly & breaking of old bonds occur. ii.

Formation of New Bond:

When the old bond breaks then new bonds are formed. Note:

According to collision theory only those collision are effective which occur in proper orientation. Q-3

Study text

Q-4

As temperature increases rate of reaction also increases because of the following reasons:

i.

Rate of Collision:

As Temperature increases rate of collision increases. increases. ii.

Kinetic Energy:

iii.

Average KE of the molecules increases. Activation Energy: Molecules posses activation energy also increases.

iv.

Arhineus Equation: The quantitative relationship between temperature and rate of reaction is given by Arhineus equation: Ea/Rt K = Ae – Ea/Rt

∝ ∝

K - 1/T K T Conclusion:

Since as T ses “k” will be increases. increases. v.

Consider two gases A & B

In a container at room temperature. What effect would the following changes have on the rate of the reaction between these gases. (a) The pressure is double

(b) No of molecules of gas (A) is doubled

(c) The Temperature is d ecreased. ecreased. Reasons:

Following are the reasons. (1) Pressure:

As the pressure increases, the rate of reaction between to gases A & B increases at Room temperature. Reasons:

As the pressure increases, the molecules come closer to each other due to which changes of collision increases & rate of reaction increases. (2) No. of Molecules of Gas-A

As the no of molecules increases increases rate of reaction also increase because chances of collision increases. increases. But there are two cases (i) Case-1, (ii) Case-2 Case-1:

If the reaction is first order and the concentration of [A] is double then the rate of reaction become double. Mathematically:

A + B ––––  product According to rate expression

 ∝ [A] [B]  (i) 

But we take about A only so we will consider [A] only

 ∝[2]   = 2K 

 ∝ [A] 

 A = 2

Since rate become double.

Case-2:

If the reaction is second order and if the concentration [A] is doubled the rate of reaction increases 4-time. Mathematically:

A + B ––––  products According to rate law

 ∝ [A] 

2

 B = Constant 118



 ∝ [2]   ∝ 4K 

2

Conclusion:

Since the rate of reaction increases 4 time. Q1: The rate constant for a reaction below is CO + NO 2 CO2 + NO. At 4000C is 0.50 Lit/mol S, the reaction is first order with respect to both CO & NO2. a)

What is the over all order of the reaction

b)

What is the rate of the reaction at 400C0 when the concentration of CO is 0.025 mol/L & that of NO2 is 0.04 mol/L?

Overall order of reaction: for blew reaction:

 = K [CO] [NO ] ––  (i)  2

Equation (1) is the rate equation which shows the overall order of the reaction Order = sum of exponents

= k [CO] [NO ]  1

2

1

Order = 1 + 1 Order = 2 Conclusion:

Since the over all order is 2. b.

Given Data

Rate constant = (k) = 0.50 L.mol – 1 S. [NO2] = 0.040 molL – [CO] = 0.025 molL – Required Data?

 

Rate = =? Solution:

As we have that rate = k [CO] [NO 2] Rate = 0.50 × 0.025 × 0.040 Conclusion:

Rate = 5 × 10 – 4 mol L – Se – All collision between reactant molecules donot lead to reaction because of the following reason. Reasons:

Following are the reason i. No. Proper Proper Orientation: Orientation:

According to collision theory those collision will lead to chemical reaction which occur in proper orientation. So those collison which are not in proper orientation will not lead to reaction. ii. Activation Energy:

According to collision theory only those molecules will colloid which posses it own activation energy. So those molecule which have no activation energy, their collision doesnot leads to chemical reaction.

119




SOLUTION AND COLLIDES

Solution:

The homogenous mixture of substances is called solution or when two or more than two non-reacting substances forming a homogeneous mixture is called homogeneous mixture or solution. Composition of Solution:

Solution is composed from two things; 1.

Solute

2.

Solvent

1.

Solute:

The component of solution in less quantity is called solute. i.e. sugar is solute in sugar solution. 2.

Solvent:

The component of solution which is in greater quantity is called solvent. i.e. In sugar solution water is solvent. SOLUTION

1.True solution

Colloidal small particle size is

Particle size is smaller than 1 m

(1-100m)

Suspension particle greater size than 100m

Colloidal Solution:

That type of solution which is intermediate between solution and suspension is called colloidal solution. History:

The name colloid was coined by Graham. Colloid is a Greed word which mean glue like. Kolla=glue Eides=like Composition of Colloides:

It is composed from two media;

 

Dispersion medium Dispersed Phase

Dispersion Medium:

The continues homogeneous medium in the colloidal solution is called colloidal solution. it is also called outer phase. Dispersal Phase:

The particles of a discontinues medium termed as d ispersed phase. It is called inner phase. For example:

The milky dispersion of Sulphur (dispersed) in water. COLLOIDAL (ppt)

   

Emulsoid ppt Hydrophilic (Water like) Levophillic Gell

   

Suspeniod ppt Hydrophobic (water hate) Lyphobic Solution

Nature of Solution in Liquid Phase:

When the Solute come in contact with solvent then ion dipole interaction is produced between the solute and solvent. The ion dipole interaction depends upon. 1. 2.

Nature of Solute Nature of Solvent

Like Dissolve Like principle

According to this principle polar substances are soluble in polar solvent and non -polar substances are soluble in non-polar solvent.

For example:

120



CaCl2, NaCl, are polar so they are more soluble in polar solvent like H 2O mean ion dipole force will be stronger due to which solute will split easily and completely. Solutions of Liquid in Liquid:

There are three types of solution

  

Completely miscible liquid Partially miscible liquid Completely immiscible liquids

1.

Completely Miscible Liquid:

Those solutions in which the liquids are completely soluble in each other in all proportions are called completely miscible liquid s. For example:

Alcohol and water are completely soluble in each other because both are polar. Note:

 

After mixing two substances the volume decreases generally. In such cases heat may be evolved or o r absorbed.

Separation:

Such mixture can be separated by fractional distillation. 2.

Partially Miscible Liquid:

Those liquids which are not completely miscible with each oth er but upto some extent are called partially miscible liquids. For example:

i) Ether dissolve in water upto 1.2% and water dissolve in ether upto 6.5%. ii) On shaking equally volume of o f ether and water two layer are formed. Conjugated Solution:

That solution in which two layers are formed and both are the saturated solution of each other are called conjugated solution s. For example:

  

Phenol water system. Tri-ethylamine water system. Nicotine water water system.

Phenol Water System:

If equal volumes of water and phenol are mixed together then they show partial miscibility. There are two layers.

 

Upper Layer Lower Layer

i) Upper Layer:

In the upper layer 5% p henol solution is present at room temperature. ii) Lower Layer:

In the lower layer 30% water solution is present in phenol. Note:

i) These two solutions are conjugated solution of each other. ii) The lower has greater density because of phenol iii) In the lower layer water acts as a solute and phenol as solvent. iv) In the above layer phenol act as solute & water act as a solvent. Upper Consulate Temperature:

It is also called critical solution temperature. “That temperature at which two conjugated solutions merge into each other is called upper consulate temperature”. It is 65.9C 0 for water

phenol system. Note:

When the temperature of phenol water system is increased then the composition of both layers changes, water starts travelling from upper layer to lower layer and phenol travel from lower to upper layer. At 65.9C0 both layers merge into each other and form a homogeneous mixture which contains 34% phenol and 66% water. iii) Completely Immiscible Liquid:

Those liquids which are completely insoluble in each other are called completely immiscible liquids. For example:

121


 


Water and benzene Water and carbon-disulphide

Solubility:

“The number of gram of gram of solute dissolved in 100 gram of solvent at specific temperature” is called solubility. Mathematically:

So lub ility 

massof solute mass mass of solvent solvent

x100

Explanation:

The solubility occur in three steps; 1. 2. 3.

First the solute come in contact with solvent. The ion dipole is developed d eveloped between solute and solvent. In the last the solute p articles are separated separated from each other.

Factors:

Following factors affects the solubility. 1. 2. 3.

Nature of solute and solvent. solvent. Pressure. Temperature.

1.

Nature of Solute and Solvent:

It will be discusse d in the class but however it depends upon “like dissolve like principle”. For example:

NaCl in H 2 O CaCl2 in H 2 O NaOH in H 2 O etc 2.

Pressure:

1. 2.

The pressure has very less effect on the solubility o f liquid and solids. The solubility of gases is affected by pressure.

3.

The solubility of gases is directly proportional p roportional to temperature at constant temperature.

For example:

CO2 is filled up to 3-5 atm, in Pepsi and when it is open pressure decrease and and gas escape out. 3.

Temperature:

There is diverse effect of temperature on the solubility o f solids. Case-I:

When temperature increases the solubility increases. For example:

KNO3 AI2 (SO4)3 Endothermic reaction. Case-II:

Sometime solubility decreases with increase in temperature. For example:

Ce2 ( SO4 )3 Li2CO3 Exothermic Exothermicreactio reaction n Case-III:

Sometimes temperature has no effect on the solubility. If temperature is increased or decreased decreased the solubility remains consta nt. For example:

NaCl KBr Special Cases:

The solubility of sodium Sulphat Sulph at increases upto 305.4k and on raising the temperature its solubility decreases. Note:

122



Sodium sulphate is decahydrated (Na2SO4, 10H2O) below 305.4k but above it become Anhydrous (Na2SO4). Maximum Solubility range is 305.4k 313k — 313k Concentration Unit: i.

Percentage by Weight (W/W%)

It is the number of grams of solute dissolved in sufficient amount of solvent to make 100g of solution e.g. 10% solution of glucose by weight means that 10g of glucose glucose are dissolved in sufficient sufficient water so that solution weights weights 100g.

 %     ×100    

ii.

Percentage Weight by Volume (W/V%)

It is the weight of solute dissolved per 100 parts by volume of solution. 10g of NaOH dissolved per 100cm3 of solution is 10% w/v solution of NaOH.

 %     ×100  .  

iii.

Percentage Volume by Weight (V/W%)

It is the number of cm3 of a solute dissolved per 100g of solution. If we dissolve 10cm3 of alcohol in water and the total weight of solution is 100g then it is 10% v/w solution of alcohol in water.

 .    %  .   ×100

iv.

Percentage by Weight (V/V%)

It is the volume of the solute present per 100cm 3 of the solution e.g. 10cm3 of alcohol is dissolved in water so that the volume of solution is 100cm3. It is 10% V/V solution of alcohol in water.

 %     ×100    

Molarity:

The number of moles of solute per dm3 of solution. Representation:

It is represented by “M”. Mathematically:

n

M 

volume of solution

M 

n v

1

but n

mass mass of solute solute

(2)

molar molar mass mass of solute

Put (II) in (I) M 

wt. of solut solutee

x

1

Mol.mass of solute solute Volume Volume of solution solution

For example:

One molar solution of sugar C 6H12O6 contains one mole or 342g of sugar per dm3 of solution. Put 342g of sugar in one dm3 volumetric flask. Then adding water for further dilution.. Now add distal water water for further dilution until reach reach the mark.

1. 2. 3. 4. Note:



It depends upon temperature.

Molality:

The number of moles of solute present in 1Kg of solvent. Or The number of moles of solute present in 1000g of solvent. Representation:

“m” m

No.of moles 1 kg of solvent solvent

123


m

wt. of solute solute Mol. mass of solute

x


1 Kg of solvent

Procedure:

   

Take 1000grams of water in flask. Put 1 mole (58.5)grams of NaCl into it. Shake it well This is called 1 molal solution.

Example:

If 180g of glucose is dissolved in 1000gm of solvent. The so lution prepared will be 1.0 molal solution. Mole Fraction:

The number of moles of a particular component divided by the total number of moles of all the components in the solution. Mathematically:

Mole friction of solvent Mole friction of solute x 1 x2

n1



n1  n 2

 x1 

 x2 

n2



n1  n 2

n1 n1  n2

If n1 is the number of moles of solvent. And n2 is the number of moles of solute. x1 and x2 are the mole friction of solvent and solute.

n2 n1  n2

1

Note:

The sum of the mole friction is equal to zero i.e., i) Parts Per Million (ppm):

The number of parts of solute per million parts of the solution. Note:

  

It is used for very very low concentration of solution. The impurities in water can be expressed in ppm. It may be in weight or volume.

Mathematically:

ppm 

wt. of solute solute wt. of soluti solution on

x106

ii) Parts Per Billion (ppb):

The number of parts of solute per billion parts of the solution. ppb 

wt. of solute solute wt. of solution solution

x109

Parts Per Trillion (PPT):

The number of parts of solute in trillion parts of solution. It is very lowest concentration. ppt 

wt. of solute solute wt. of soluti solution on

x1012

Roults Law:

The quantitative relationship between between the vapour pressures of solution and the composition of solution is given by Roult Law. Note:

This law was given by F.M Roult in 1847. Statement:

According to Roult Law “ The partial vapour pressure of any volatile component of a solution is directly proportional to the mole friction of

that component in a solution P A  X A .

Or

Partial vapour pressure of each volatile component of a solution is equal to the partial vapour p ressure of the pure component multiplied by b y the mole fraction. P A  PAO X A Explanation:

Let us consider a solution containing two volatile components “A” and “B”. Now according to Roult Law Law for component A.

124



P A  X A Or P A



     

PAo X A

Similarly for component B P B



X B

Or P B



XA = mole friction of component A XB = mole friction of component B P0A = V.P of component A (pure) P0B = V.P of component B PA= V.P of component A in solution. PB= V.P of component B in solution.

PBo X B

Now: According to Dalton Law of partial pressure vapour pressure of a solution is given as; PT

 PA  PB  C

Putting the values of (a) and (b) in (c) PT  PAo x A  PBo x B  D

But as we know that sum of the mole friction is = 1 X A  X B

1

( X A  1  X B )  E

Putting the values of (e) in equation (d) PT  PAo (1  X B )  PBo X B PT  PAo  PAo X B  PBo X B Re arranging PT  PBo X B  PAo xb  PAo  F Tak Taking ing " X B " Comm Commo on PT  X B ( PBo  PAo )  PAo  FA

If X B



zero

X A

1

and

Then (f A) become as PT  (PBo  PAo ) X O  P Ao PT  P Ao

In the same:

X B  1

and

X A  zero Then

PT



P Bo

Graphically:

The composition of component and vapour pressure is given as this graph shows that; 1.

O “PA” increases as “XA” incre ases and become P A at X A  1 and X B  0

2.

O “PB” increases at “XB” increases and become P B at X B  1 and X A  0

3.

The upper solid line shows vapour pressure.

Note: O

O

1.

Vapour pressure of solution will not be less than P A and not be greater than P B .

2.

Vapour pressure will be in between P A and P B .

3.

Ideal solution is that which obey Roults Law.

O

O

125



Roult’s Law for a solution of Non volatile non electrolyte solute dissolved in a volatile solvent.

1.

The vapour pressure of a pure solvent decrease d ecreasedd when a nonvolatile, non volatile, none electrolyte solute is dissolved in it.

2.

The vapour pressure is because of the evaporation of molecules from the external surface (of pure solvent).

3.

If nonvolatile solutes are added to the solvent it will decrease evaporation because it will block the surface molecules of solvent from escaping.

Supposition:

PO

 V .P

of pure solvent

P  V .P of solution

Pure  p0> p→ solution Lowering in Vapour Pressure:

This difference between the vapour pressure of pure solvent and solute is called lowering in vapour pressure ( P) and since P  PO  P Relative Lowering in Vapour Pressure

P  PO  P :

The ratio between P and P O is called relative lowering in vapour pressure

P P O

 X 2

Now according to Roults Law “The relative lowering in vapour pressure is equal to the mole fraction of solute.  P O

P

 X 2  1  X 2  mole friction of solute

rea rearran ranging ing

O P  X 2 P

 P  PO X 2  2 But as we know that

 P  PO P  3 Putting 3 in 2 O 0 P  P  P X 2

rearranging O

P P O

P

 X 2  4

Now subtracting subtracting both sides sides from one one “1”

1



( P O

 P )

O

1

P

 1  X 2 

5

Taking Taking LCM of LHS P O

O  ( P  P )

P O P P O

 1  X 2 

 1  X 2

6

But as we know that x1  x2  1 x1   1  x2  7 Putting 7 in 6 P O

P

 x1 or P  PO X 1

Conclusion:

This equation shows that vapour pressure of solution is equal to the product of vapour pressure of pure solvent and mole friction of solvent. Note:

P P O

 X 2

P  PO X 2

126



Conclusion:

This equation shows that lowering in vapour pressure of solvent depend on both the vapour p ressure of pure solvent and mole friction of solute in solution. * Coligative Properties of Solution:

Those physical properties of a solution which depends upon the number of solute particles and is independent of their nature (i.e. shape, size and chemical nature) that will be called colligative properties. For example:



Lowering of vapour pressure.



Elevation of the B.P.



Depression of freezing point.



Osmotic pressure.

Conclusion:

Since both solutions “B” and “C” contain same number of solute particles (i.e 1 mole=6.02 ×1023molecules). So both shows same increase in

its B.P therefore, we can say that colligative properties is independent of the nature of the solute. a.

Lowering of Vapour Pressure:

The vapour pressure of pure solvent is always higher than the vapour pressure of solution containing non-volatile, non-electrolytic solute dissolved in a volatile solvent. This means that the vapour pressure of a solvent decreases by adding a non -volatile, non-electrolyte solute. PO = Vapour pressure of pure solvent. P = Vapour pressure of solution.

P=PO – P P = Lowering in vapour pressure.

Whereas:

Note:

n2 = number of moles of so lute

Since the term relative is used to avoid the role of temperature.

n1 = number of moles of solvent.

P

P is a colligative property whose values depends u pon the number of solute particles and is independent of their nature. If P is divided by P O then it is called relative lowering in vapour pressure. i.e. PO  P P

O



P

 Re lative lative Vapour pressure pressure

P O

Roult’s Law:

According to the Roult Law “The relative vapour pressure of the solvent to solute is equal to the mole friction” of the solut e. Mathematically:

P P O

 X 2  1

Whereas X 2 is the mole friction of the solute. X 2 

n2 n1  n2

 2

Now, putting the values of 2 in 1. P O

P



n2 n1  n2

1

Now Roult’s Law can be also written as; as; P O

P

 X 2

Subtracting from “1” both sides.

127


1

P


 1  X 2

O

P

O P  P 1 ( )  1  X 2 O P LCM taking O O P  P  P )  1  X 2 O P

P

 1  X 2  4

O

P

But we have that X1  X 2

1

X 1  1  X 2



5

Putting 5 in 4 P

 X 1 

P O

6

Or P  P O X 1 P  X 1 Conclusion:

This equation shows that the vapour pressure of solution is directly proportional to the mole friction of solvent. Second Form of Roult’s Law:

This law also states that “lowering in the vapour pressure ( P) of a solution containing a non volatile, non electrolyte solute is directly proportional proportional to the mole mole friction of solute”. Mathematically:

P  X 2 P  Constt X 2 O

P  P X 2 Molecular Weight Determination from Lowering in Vapour Pressure:

According to Roult’s Law; P

 X 2  a

P O

X 2 

n2 n1  n2

Comparing (a) and (b) P



O

P

n2 n1  n2

c

∵

So equation (c) can be written as; P



P O

n2 n1

If the solution is very very dilute the n 2<<< n1

 n2 = 0

 d

But here n2  n1 

mass ass of so solute lute Molecula Molecularr mass mass



W 2 M 2

 (i)

mass of solvent Molecula Molecularr mass mass of solvent solvent



W1 M 1

 (ii)

Putting the values of (i) and (ii) in (d).

128



w2 P



O

P

M 2  E w1 M 1

Rearranging P



O

P

w2 M 2

x

M 1 w1

 F

Rearranging again. M 2



PO .w2 xM 1 P.w1

M2 is the molecular weight of non volatile solute. b. Elevation in Boiling Point:

The increase in the boiling point of the solvent due to the addition of non volatile, non electrolyte solute is called elevation of boiling p oint. Explanation:

As the vapour pressure of the pure solvent is always higher than the vapour pressure of solution containing a non volatile, non electrolyte solute, therefore, a solution boil at higher h igher temperature than pure solvent. Supposition: Note:

Suppose;

The elevation in the boiling point depends upon;

T1 = Boiling point of pu re solvent. T2 = Boiling point of solution. Since T2> T1Tb = T2 — T1

   

Nature of solvent. solvent. Concentration Concentration of solute. Independent Independent of t he nature of solute. All

Tb:

T6 is the difference in boiling point. Since, this difference in the boiling point of the solution and pure solvent that will be called elevation in boiling p oint. For example:

1.

When one mole (NA) of solute which is non volatile, non electrolyte solute is added to 1Kg of solvent (H2O).

2.

Then the temperature of the water increases 0.52 0C, which is called a.

Molal point elevation or

b.

Molal point constant or

c.

Ebullioscopic constant “kb”

3.

If the concentration of the solute increases, elevation of bo iling point will be increases, also

4.

Thus, elevation in boiling point  molality of solution.

Tb  m Tb = constt m Tb =kbm Where “kb” is called ebullioscopic constant.

If m = 1 i.e. solution is 1 molal. Then Tb = 1 x kb

Tb = kb Note:

Thus the mole boiling constant is the elevation (increase) in the boiling point which is produced due to dissolving 1 mole of non volatile, non electrolyte solute in one kg of volatile solvent. Graphically:

Graphically elevation in B.P is given as; 1.

Curve AB is the vapour pressure of pure solvent.

2.

Curve CD is the vapour pressure of solution.

3.

The AB lies is above the th e solution which shows that vapour pressure of solvent is greater than solution at any temperature.

M.Wt Determination of Boiling Point Elevation:

As we have; 129



Tb  kbm  i

but m

solute mass M . solvent in kg

Moles of solutes  n2 

 a massof solute solute

Molecular Molecular mass of solute

w2 M 2

Similarly masso massoff solv solven entt in gm

n1 

1000



w1 1000

b

Putting (a ) and (b ) in ii w2 m

M 2 w1 1000

Tb



kb.

W2 x1000 M 2 xW 1

Re arranging Tb .M 2



kb .W2 x1000 W 1

Tb .M 2 .W 1  M 2



kb .W2 x1000 Tb .W 1

kbw2  1000 Tb.w1

Since molecular weight can be determined from it. c. Depression in Freezing Point of Solution:

The freezing point of a system is the temperature at which the solid and liquid phases of the substance coexist, having the same vapour pressure. Explanation:

1.

A solution which contain a non volatile, non electrolyte solute dissolved in a volatile solvent. It will freezes at lower temperature then pure solvent.

2.

The difference in the freezing point of pure solvent and solution is called depression in freezing point.

Representation:

It is represented by Tf. Let T1 = Freezing point of pure solvent. T2 = Freezing point of solution. As OC

1C

T1  T2  i

T1  T2  Tf  ii

Note:

Tf depends upon the; a. b.

Number of solution solution particles. Independent Independent of t he nature.

Tf is called colligative property. For example:

When one mole (Avogadroes number of particles) of any non volatile, non electrolyte solute is added to one Kg of water, then the decrease in the freezing point is 1.860C which is called mole freezing point constant or cryoscope constant (Kf). Note:

If the concentration of solute is increased increased then depression in Freezing Point will be in creased. Since depression in the freezing freezing point is directly proportional to the molality of the solution. Since; 130



Tf  m Tf = constt m Tf = Kfm Kf is called mole freezing point constant or molar depression in freezing point or cryoscopic constant. If

m=1

There

Tf = Kf

Molal Freezing Point Constant:

It can be defined as “The decrease in freezing point which is produced by dissolving one mole of a non -volatile, non-electrolyte solute in one

kg of volatile solvent. Graphically Representation:

Graphically depression in freezing point is shown by; 1.

Curve AB is the vapour p ressure of pure solvent.

2.

At point “B” solvent freezes.

3.

BC shows the frozen solid which have very very lo w vapour pressure.

4.

T1 is the freezing point of pure solvent.

5. 6.

Curve DEC is for the solution which lie below the pure solvent curve. Curve DE meet with BC at point E at which freezing temperature is T 2.

7.

Vapour pressure of solution is P

8.

Vapour pressure of solvent is P O

9.

T2 (depression in freezing point). Tf = T1 – T

M.wt Determination from Depression in Freezing Point:

As we have that; Tf  kfm  1 Where; m

mass mass of solute solute mass of of solvent solvent n2

m n2 

n1

mass of solute solute M .mass of solute

n2 

n1 

 2

W 2 M 2

M 1 1000

 (i)

 (ii )

Putting (i ) and (ii ) in 2

m

W 2 M 2 M 1 1000

Re arranging m

n1 

W2 x1000 M 1M 2 M 1 1000

 (ii )

 (ii )

131



Putting (i ) and (ii ) in 2

m

W 2 M 2 M 1 1000

Re arranging m

W 2 1000 M 1M 2

 (ii )

Putting (ii ) in 1 Tf



M 2



W  1000 kf 2 M 1M 2 kf  W 2  1000 Tf  W 1

This equation can be used to calculate the M.Wt of solute. Osmosis:

The flow of solvent from the region of higher concentration to the region of lower concentration through semipermeable membraneis called osmosis. Explanation:

1.

Osmosis means “to push”.

2.

Any membrane which is permeable to solvent and not to solute is called semi-permeable membrane.

3.

The flow of solvent from pure pu re solvent towards the solution through semi-permeable is called osmosis.

Types of Osmosis:

It has two types; 1. 2.

Endosmosis. Exosmosis:

Endosmosis:

The flow of solvent molecules to the solution is called endosmosis. Exosmosis:

The flow of solvent molecules from the th e solution to pure solvent is called exosmosis. It is also called reverse osmosis. Experiment:

a.

A concentrated solution of sugar has taken in the thistle funnel.

b.

Its mouth is tightly bound b ound by egg membrane which acting as semi-permeable membrane.

c.

Now it is inverted in a beaker beaker of water (pure (pure water).

d. e.

The level of sugar solution is marked on the thistle funnel i.e. first level. After sometime the level of sugar solution in creases in the thistle funnel (second level).

f.

This is due to the moment of water from high concentration to the lower concentration and is called osmosis.

Note:

If pressure is applied on the sugar solution in the tube, the water molecule will move from sugar solution to water. Since; This movement is called Reverse Osmosis. Osmotic Pressure:

The external pressure exerted exerted on a solution in o rder to stop the flow of solvent into the solution is called osmotic pressure. pressure. OR The hydrostatic pressure built up on the solution which just stops the flow of pure solvent in to the solution is called osmotic pressure. Representation:

It is represented by . Mathematically:

 Concentration.  C  = CRT Explanation:

132



1.

Consider a container containing two compartments, A and B.

2.

A-containing solution and B containing solvent.

3.

A semi-permeable membrane is present in between the two compartments.

4.

Now due to Osmosis the solvent solvent molecules enter from B to A and due to which the piston go upward. upward.

5.

Now pressure is required to stop the movement of H2O molecules from pure solvent to solution which is called Osmotic Pressure.

Impact of Osmosis on Daily Life:

Osmosis has many important impacts on our daily life, some are given below; 1.

Industries:

It is used in many industries process. 2.

Purification of Sea Water:

It is used for the purification of sea water and make it drinkable. 3.

Support to Plant:

By osmotic pressure plant cell become turgid and these turgid cell gives support to the weak parts of the plants. 4.

Transportation in Plant:

The accent of sap by Osmotic pressure from roots to the leaves is also the impact of Osmosis. 5.

Isotonic solution:

Isotonic solution are those which have same osmotic pressure, so isotonic solutions have been prepared to prevent plasmolysis (shrinkage of cytoplasm) and heamolysis (deficiency of red blood cells). Solvation:

The process of addition or combination of solvent molecular with solute molecules or particle is called salvation and if water is u sed it will be called hydration. OR The phenomenon in which water molecules surrounds the solute particles is called hydration. The compounds to which water combine are called hydrated compounds. For example: H 2O

Mx  M  X g

g

g

Dissolution is followed by ionization according to Arhenious. There is no time for in teraction with each other due to the presence of water. Note:

Ion dipole interaction is present between water and solute particles. 



g

g

NaCl  Na  Cl S

H 



 qq 



g

g

g

Na  Cl  Na  Cl g





NaCl  Na  Cl

Ionization Ionization  Endothermic Endothermic

H  45 H  3.4

Exothermic Exothermic

Heat of Hydration Depends Upon:

1.

Charge density.

2.

Size of cation and anion.

3.

Number of water water molecule.

4.

Small atom has greater charge density.

Salvation:

If solvent other than water is used u sed is called solution. Heat of Solution and Its Applications:

The amount of heat or evolved absorbed when a definite amount of solute is dissolved in excess of solvent is called heat of solution. Representation:

Heat of solution is represents by H:

H may be positive or negative. Explanation:

1.

The formation of solution is always accompanied by change in temperature.

2.

If NaOH is dissolved in water the temperature will rises.

3.

If KI is dissolved in water the temperature will be decreases. 133



4.

Either heat is evolved or absorbed.

5.

The evolution or absorption of heat depends the intermolecular forces i.e. if weak bonds are broken and strong bonds are formed then it will emit energy.

For example:

i.

When KI is dissolved in water then it absorbed 20.33kjmol.

ii.

While If NaOH if dissolved is in it will emit -44.51kj/mol of heat and H= -ve Exothermic reaction.

Ammonium Nitrate (NH4NO3)

1.

It is an electrovalent compound.

2.

On dissociation in water it ionizes into Ammonium (NH+4) ions and (NO-3) ions.

3.

H = 25.69 kj/mol

reaction = Endothermic

-4

NH4NO3

-3

NH +NO +NO

Classification of Colloids:

Colloides have been classified on the basis; a. b.

Physical state. Behaviour towards liquids.

1.

Classification on the Basis of Components:

It will be discussed in the class during lecture in TWIGS. Ser

1

Dispersed

Dispersion

Phase

medium

Gas

Special Name

Example

Gas Liquid

Foamic Soap Leathr

Cannot excise because it form homogeneous mixture

Solid

Solid Form

Air dispersed in solid such as metal and glass.

Gas

Fog, Aerosol

Cloud, fog, moist

4

Liquid

Emulsion

Homogenised milk H2O in oil or benzene

5

Solid

Gel

Halwa, Jellies

Gas

Smoke

Coal smoke

7

Liquid

Sol or Collidal solution

As2S3 or Lassi

8

Solid

2 3

6

Liquid

Solid

Coloured glases

Properties of Colloides:

The characteristic properties of colloidal solution are; 1.

Optical Properties

When light is passed though the colloidal solution, then the colloidal particles scattered the light in all direction, this property is called optical property of colliodes. This is due to the large size of colloidal colloidal particles. Note:

The phenomenon of scattering of light by solute particle is called Tyndale effect. 2.

Brownian Movement:

The continues rapid zig zag movement of colloidal particle in the dispersion medium is called Brownian movement. Tyndale effect:

The particles of the colloide are always in zig zag movement, this is the due to the collision of molecules of the dispersion medium with dispersed phase. 3.

Filterability:

The colloidal ppt are filtered by ultra filter paper and this p rocess is called ultra-filtration. Note:

Colloidal ppt cannot be filtered by ordinary o rdinary filter paper because the pores of ordinary paper are too much large. 4.

Diffusibility:

 

Low rate of diffusion.

5.

Colour:

Because of greater molecular size.

134



The colour of the colloids depends upon the wavelength of the light scattered by the dispersed particle and it depends upon the size of the particles and shape i.e. organ yellow colour = 6 x 10-5mm. 6.

Osmotic Pressure:

Osmotic pressure is given as; 



RTC M

Osmotic pressure is directly proportional to the temperature and concentration but inversely proportional to molar mass. Since; Colloidal solution have low osmotic pressure. Note:

Osmotic pressure is measured by Osmometer. 7.


Temperature has diverse effect on the colloidal solution. For example:

If we heat egg then th e small colloidal particles convert into Lumps of colloidal big particles. 8.

Stability:

The stability of the colloides depends upon; a.

Charge on solution particles.

b.

Solvation.

c.

Brownian motion.

i.

Charged on Colloidal Particles:

The charged colloidal particles attract the solvent molecules and form a layer around itself and this layer protect and prevent them from dispersion. 2.

Solvation:

The solvation occur only of the solute have some affinity with solvent. If solute stabilize itself by forming layer around itself then who can it be dispersed. 3.

Brownian Movement:

It reduces the gravity force on the th e colloidal particles and make it stabilize. Exercise: Choose the correct answer.

i.

In atmospheric gaseous solution, Nitrogen is a: (a) Solvent

ii. iii.

In ice cream, sugar is: (a) Solute (b) Mixture

vii.

(b) Constitutive (c) Colligative (d) None (b) Volume

(c) Temperature (d) Number of moles

Freezing point of solution as compared to the solvent is (a) Higher

vi.

(d) None

Enthalpy is a measure of the heat of solution at constant (a) Pressure

v.

(c) Solvent

(d) None

Elevation of boiling point is a property. (a) Additive

iv.

(b) Solute(c) Mixture

(b) Lower

(c) Variable

(d) Remain the same

Phenol-water system is the example of: (a) Completely miscible liquids

(b) Completely immiscible liquids

(c) Partially miscible liquids

(d) None

Fog is the example (a) Solution

(b) Colloid

(c) Suspension (d) None

(i) Solvent

(ii) Solute

(iii) Colligative

(v) Lower

(vi) Partially miscible liquids

Correct Choices:

(iv) Pressure (vii) colliods

135



Short Question 2.

How a given mixture can be differentiated into a true solution or coarse suspension?

Ans.

Solution:

Definition: the homogenous mixture of substance is called solution or when two or more than two non-reacting substances forming a homogenous mixture is called homogeneous mixture or solution. Composition of Solution: solution is composed from two things:

1.

Solute

2.

Solvent

1. Solute: The component of solution in less quantity is called solute i.e. sugar is solute in sugar solution. 2. Solvent: The component of solution which is in greater quantity is called solvent. Example: In sugar solution water is solvent. True Solution: In true solution particle size is smaller than 1um. Example: A true solution obtained when a substance like sugar dissolves in water. Coarse suspension: Suspension have greater size than 100 um. Example: When ground day is mixed with water it is an example of suspension. 3.

Classify colloids on basis of their behavior towards medium and the physical states of mutter.

Ans.

Colloidal Solution: That type of solution which is intermediate between solution and suspension is called colloidal solution.

Composition of Colloides: It is composed of two media.

Dispersion medium Dispersed phase. Dispersion medium: The continues homogeneous medium in the colloidal solution is called colloidal solution. Outer phase: It is also called colloidal solot outer phase. Dispersed phase: The particles of a discontinues medium termed as dispersed phase. Inner phase: It is also called inner phase. Example: The milky dispersion d ispersion of Sulphur (dispersed) in water. Classification of Colloid Solution:

a.

Components (Physical State of Matter)

b.

Behavior towards liquids.

a.

Classification on the basis of components

One basis of components there are eight types of colloidal suspension. S.#

1.

Dispersed Phase

Gas

Dispersion Medium

Special Name

Gas

Typical example

Cannot exist. Both components diffuse into each other to form a homogenous phase Foam

Soap lather, beer, foam, whipped cream.

Liquid

Solid foam

Air dispersed in solids, such as metals, glass etc.

Solid

Fog, aerosol

Clouds, fog, moist, aerosol sprays

4.

Liquid

Emulsion

Homogenized milk, H2O in oil or benzene.

5.

Solid

Gel

Halwa, Jellies, Gelatin Ointments, silicagel, Fe(OH)3 gel.

Gas

Smoke

Coal smoke (carbon dispersed) NH4Cl fumes, I2 vapours in air suspended dust.

7.

Liquid

Sol or colloidal solution

AS2S3, Gold or Fe(OH) 3 Sol solution of high polymers milk, Lassi

8.

Solid

2. 3.

6.

Liquid

Solid

Coloured glasses, minerals, gems alloys and mixed crystals.

b. Behaviour towards liquids: It depend on the force of attraction or repulsion between the dispersed particles and dispersion medium, a

colloidal suspension may be lyophilic and lyophobic. Lyophilic: If a force of attraction exist between the particles of dispersed phase and dispersion medium the solution called lyophilic. Example: If the dispersion medium is water then sol is term as hydrophilic. E.g. milk. Lyophobic: the existence of repulsive force between dispersed phase and the dispe rsion medium terms the solution is called lyophobic. Example: If water is used as dispersion medium solution is term as hydrophobic. E.g. Lassi. 4.

Calculate Molarity (M) of the following solutions.

136



(i) 2.0g of H2SO4/ 2dm3 of H2O (0.01) Ans (ii) 0.4g of NaoH/ 100cm3 of H2O (0.4) Ans (iii) 0.5g of Na2CO2/ 250cm3 of H2O (0.02) Ans Ans.

Given: Mass of H2SO4 = 2g

Required: Volume of sol = 2dm 3

Volume of sol = 2dm3 Required: Molarity of solution (M)=? Solution: Molecular mass of H2SO4 = 98g. mol – 1 we have that,

               ×          

Molarity = Molarity =

= b.

   ×  = 0.01 mol.dm

3

Given

Mass of NaOH = 0.4g

Volume of solution = 100 cm – 3= 0.1 dm3 Required Molarity of solution (M) = ?

4 0g.mol – 1 we have that Solution: Molecular mass of NaoH = 40g.mol Molarity = Molarity =

= c.

               ×          

. ×  = 0.1 dm  .

3

Given:Mass of Na2CO3 = 0.5g

Volume of solution = 250 cm – 3= 0.25 dm3 Required: Molarity of solution (M) = ? Solution: Molecular mass of Na 2CO = 106g.mol – 1 we have that

               ×  Molarity =          .  =  × . = 0.18 mol.dm Molarity =

3

5.

Calculate mole fraction of each component in the following solution.

(i) 2.0 moles of water (H 2O) + 1.5m moles of Nacl. (10.625, 0.375) (ii) 500g of H2O + 600g of glucose (C6H12O6). (0.835, 0.107) (iii) 800g of H2O 120g of acetone (CH3)2 Co. (0.9445, 0.0555) Ans.

Given: water = 2.5mol

Solution: Total moles of H2O + sodium chloride = 2.5 + 1.5 = 4.0ml Mole fraction of water (x1) =

of ethyl alcohol (x2) = b.

    . 0.375   

    . = 0.625 Mole fraction   

Given:Water H2O = 500g = 500/18 mol Glucose (C6H12O6) = 6500g = 600/180mol = 3.38mol

Solution: Total moles of water + glucose = 27.78 + 3.33 = 31.11 Mol

    . = 0.89   .    . Mole fraction of glucose (x ) =    . = 0.107

Mole fraction of water (x 1) = 2

c.

Water = 800g = 800 / 18mol = 44.4mol

Acetone(C3H6O) = 120g = 120/58mol = 2.2 1mol Solution: Total moles of H2O+acetone = 44.44 + 2.21 = 46.65 Mol

     . = 0.95   .    . Mole fraction of glucose (x ) =    . = 0.05

Mole fraction of water (x 1) = 2

6.

A 3.0 dm3 cylinder contains a mixture of oxygen, nitrogen and chorine gases. If their concentrations are 4:64 × 10 – 5g, 3.5 × 10 – 4g and 2.3 × 10 – 5g respectively. Calculate the concentration of each in (i) Parts per million (ppm) (ii) Parts per billion (ppb) (iii) Parts per trillion (ppt)

137


Ans.


Mass of O2 = 4.64 × 10 – 5g

Given:

Mass of N2 = 3.5 × 10 – 4g Mass of Cl 2 = 2.3 × 10 – 5g Solution: Volume of mixture = 3dm3 = 3000cm3 we have that

          × 10 . × O =  × 10 = 0.0155ppm . × N =  × 10 = 0.1167ppm . × Cl =  × 10 = 0.0077ppm      ppb = =      × 10 . × O =  × 10 = 0.0155ppb . × N =  × 10 = 116.7ppb . × Cl =  × 10 = 7.667ppb      ppt = =      × 10 . × O =  × 10 = 1.547 × 10 ppt . × N =  × 10 = 1.167 × 10 ppt . × Cl =  × 10 = 7.667 × 10 ppt

(i)

ppm = = 2 2

2

(ii)

2 2

2

(iii)

4

2

4

2

3

2

7.

Give Give the statement of Roult’s law. Explain the low ering of vapour pressure of a solution base o n this law.

Roults law: The quantitative relationship between the vapour pressure of sol and the composing of components of solution is given by Roult

law. Note: This law was given by F.M Roult in 1847. Statement: According to Roult law “the partial vapour pressure of any volatile component of a solution is directly proportional to the m ole

friction of that component in solution PA×XA0 or partial vapour pressure of each volatile component of a solution is equal to the partial vapour pressure of the pure component component multiplied by the mole fraction fraction PA = PA0× XA Explanation: Lets us consider a solution containing two volatile components “A” and “B”

     

XA = mole friction of component A XB = mole friction of component B PA = Vop of component A (pure) PA = Vop of component B (pure) PA = V.P of component A in solution PA = V.P of component B in solution

According to Roult law for component A.

PA× XA OR

P A



PAoC  x A

o

PB× XA P B  PB C  x B Dalton law of partial pressure of solution given as:

PT  PA  P B C putting value of a & b in c

PT



PAo x A  PBo x B  D

Sum of mole fraction is = 1 XA + XB = 1 XA = 1 – XB XB – E Putting values of (e) in eq (d) PT P A (1  xB ) PB xB o

o

138

Nasrat Ullah Katozai (Chemistry) o

o


o

PT P A PB xB PB xB (Rearranging) o

PT P A xB

 P Ao – F

Taking XB common PT

fA  xB ( P Bo  PAo )  P Ao – fA

If XB = Zero and X A = 1 Then fA become as

PT  ( PBo  PAo ) X 0  P Ao PT  PBo  PAo xB  PBo xB PT



P Ao

If XB = 1 and X A = 0 So

PT



P Bo

Lower in vapour pressure = P = P0 – P The ratio between P and P0 is called relative lowering in vapour pressure p ressure

∆ = X 

2

Now according according Roult law law “The relati relati ve lowering in vapour pressure is equal to the mole fraction of solute.

∆ = X –– (1) (1)  X = mole friction of o f solute rearranging  = X P  2

2

P

2

0

P = P0X2  (2) But as we know that P = P0P  (3) Putting (3) in (2) P0 – P=Px P=Px2 rearranging

P 0 P

 X 2

P 0

(4) Now substracting both side from one 1. – (4)

1 ( P 0  P ) P 0

1

 1  X 2  (5)

Taking L.C.M of LHS

P 0  ( P 0  P )

 1  X 2

P 0 P 0  P 0  P P 0 P P 0

 1  X 2

 1  X 2  (6) As we know that

X1 + X2 = 1 X1 + = 1 – X2  (7) putting (7) in (6)

  = X or P = P X 0

1

1

Conclusion: This eq show that vapour pressure of solution is equal to the product of vapour pressure of pure solvent and mole friction of

solvent. Note:

∆ = X P = P X  2

0

2

Conclusion: This equation shows that lowering in vapour pressure of solvent depend on both the vapour pressure of pure solvent and mole

friction of solute in solution. 8.

0.0874 grams of ethyl alcohol (M.Wt=46) dissolved in 20.0g of H2O, produced a depression of 0.177K in the freezing point of solvent. Calculate the cryoscopic constant of w ater.

Ans.

Given: Mass of water (W1) = 20.0g Mass of ethyl alcohol (W 2) = 0.0874g

Freezing point depression (Tf) = 0.177k Required: Cryoscopic constant (kf) = ? 139



Solution: We have that

Molecular mass (M2) of ethyl alcohol (C 2H5OH) = 46g .mol – 1 Now:

  ∆ ×  × 1000 .  Kf = ∆ × .× 46 = 1.863 K.g – 1. 1. Mol M2 =

– 1

9.

Explain the phenomenon of osmosis and the pressure exerted in this process.

Osmosis: “The flow of solvent from the region of highe r concentration to the region of lower concentration through semi permeable membrane” CID osmosis. Examination:

1.

Omosis means “to push”

2.

Any membrane which is permeable to solvent and not to solute is called semi-permeable membrane.

3.

The flow of solvent from pure solvent towards solution through semi-permeable is osmosis. Types of Osmosis: It has 2 types

 

Endosomosis Exosmosis

Enodosmosis: The flow of solvent molecules of the solution is called endosmosis.

o f solvent molecules from solution to pure solvent is exosmosis. It is also called reverse osmosis. Exosmosis: the flow of Experiment: As concentration sol of sugar has taken in the thistle funnel.

    

Its mouth is tightly bound b ound by egg membrane. Which acting as semi-permeable membrane. Now it is inverted in beaker beaker of water The level of sugar solution is marked on the thistle funnel i.e. i. e. first level. After some time the level of sugar solution solu tion increases in the thistle funnel (second level) This is due the moment of water from high concentration to the lower concentration and is called osmosis.

Osmotic Pressure: The external pressure applied to a solution in order to stop the flow of solvent into the solution separated by semi-

permeable called called osmotic pressure. OR The hydrostatic pressure built up on the solution which just stops the flow of pure solvent into in to the solution is called osmotic pressure. Representation: It is represented by . Mathematically: concentration

 C  = CRT Explanation:

    

Consider a container containing two compartments, A and B A-Containing solution and B containing solvent. A semi-permeable membrane is present in between the two components Now do to osmois the solvent molecule molecule enter from B to A and due to which the pistongo upward. Now pressure is required to stop the movement of H2O molecules from pure solvent to solution which will be called osmotic pressure.

10.

In an experiment 1.0 0C depression of freezing point was calculated of a solution of 25.6g of of benzene containing 6.4g of an organic substance. Calculate M.Wt of substance (kf for C 6 H6 = 5.12) .1298

Ans.

Given: Mass of solvent (W1) = 25.6g Mass of solute (W 2) = 6.4g Cryoscopic constant (kf) = 5.12 0C.g-1. Mol – 1

Freezing point depression (Tf) = 1 0C Required: Molecular mass (M 2) of solute = ? Sol:

We have that

  M = ∆ ×  × 1000  .   .  M =  × .× 1000 = 1280g . Mol 2 2

– 1

– 1

140



11.

Calculate M.Wt of iodine (I2) for a solution containing 1.19g of I 2 in 35.0 grams of ether. The raise in boiling point was observed by 0.26 0C (Kb for ether = 2.22)

Ans.

Given: Mass of ether (W1) = 35.0g Mass of iodine (W2) = 1.19g Ebullioscopic constant (Kb) = 2.22 0C. g – 1. Mol – 1

Boiling point elevation (Tb) = 0.2 0C Required: Molecular mass (M 2) of solute = ? Solution: We have that:

M2

=

M2

= =

 ×  × 1000 ∆ . ×. × 1000 .  255g – 1. Mol – 1

141



CHAPTER-11:

THERMOCHEMISTRY

Definition:

The branch of chemistry which deals with the heat evolved or absorbed during a chemical reaction is called Thermochemistry. OR Each and every physicochemical change is accompanied by some new changes in which evaluation or absorption of heat occur that will be called Thermochemistry. Example of Thermochemistry:

1. 2. 3. 4.

All those reaction in which heat evolution or o r absorption occur are the examples of Thermochemistry. On the basis of heat evolved or o r absorbed, there are two types of reactions. Exothermic reaction. Endothermic reaction.


Those reactions in which the evolution of heat h eat occur is called exothermic reaction. For Example: When CaCO3 is dissolved in water, it evolves heat which is an example of exothermic reaction and we can say that it is

Thermochemistry. i.e.

CaCO3 CaO+CO2+H

Endothermic Reaction:

Those reactions in which the absorbtion of heat h eat occur is called endothermic reaction. For Example: When KNO3 is dissolved in water it absorb heat from the surrounding which is an example of endothermic reaction. Other Examples of Endothermic Reactions:

1.

Decomposition of KI.

2.

Decomposition of KNO3

Note:

Chemical change is nothing but the breaking of old bonds and formation of new bonds. Energy Change in Chemical Reaction:

It can be explain as;

MCQs:

1.

Heat evolved or absorbed during a chemical depends upon;

Thermochemical Equation:

That equation which shows; a. b. c. d.

Nature of reactants reactants and products. Physical state of reactants and products. Enthalphy Changes of reaction is called thermochemical equation.

a.

Number of bonds bonds formed.

b. c.

Number of bonds bonds broken. Strength of bonds.

Energy Changes:

During a chemical reaction heat must be evolved or absorbed. i.e.

A + B  C+ D

H= +

If H= + ve:

It means that; 1.

Strong bonds are broken and weak bonds are formed. A – A A + B – B B  A – B B + A – B B H= + Strong bond weak bond

2.

Reaction is endothermic.

3.

The energy of reactant is less than the energy of the products.

4.

Product is less stable.

5.

Activation energy of forward reaction is greater.

6.

Reactants are stable.

If H= - ve:

It means that; 1.

Weak bonds are broken and strong bonds bo nds are formed.

2.

Reaction is exothermic.

3.

Energy of reactant is greater than the energy of p roduct.

4.

Products are more stable.

5.

Reactants are unstable. 142



Reaction:

A  A  B  B  A  B  A  B Weakbo Weakbond nd

stro strongbo ngbond ndss form formed ed

Difference between Spontaneous and non- Spontaneous process Spontaneous

NON- Spontaneous

i. Definition:

i. Definition:

That process which do not require an external assistant or energy for operation is called spontaneous reaction

That process which required a specific assistant for operation is called non spontaneous reaction.

ii. Natural Tendency

ii. No Natural Tendency

It has natural tendency to occur.

It has no natural tendency to occur.

iii. Unidirectional

iii. Bidirectional

It is unidirectional in nature.

It is bidirectional in n ature.

iv. Non Equilibrium

iv. Equilibrium

It goes from non equilibrium state to equilibrium state.

It goes from equilibrium state to non equilibrium state.

v. Irreversible.

v. Reversible.

It is irreversible process.

It is reversible process.

vi. Real Process.

vi. Un-Real Process

It is real process in nature.

It is unreal process.

vii. Unstable.

vii. Stable.

It goes from unstable state to stable state.

It goes from stable state to unstable state.

viii. Gips free Energy.

viii. Gips free Energy.

“G” should be negative, “The us eful part of the energy is called Gips free energy”.

It should be positive for non-spontaneous reaction. G = +ve

ix. Enthalpy.

ix. Enthalpy.

The enthalpy of a spontaneous reaction should be negative. H = (-) negative.

The H of non-spontaneous reaction is positive. H = (+) positive.

x. Entrophy.

x. Entrophy.

It should be positive. S = + ve

It should be taken negative. S = - ve

xi. Energy of product.

xi. Energy of product.

The energy of the product is low.

The energy of the product in non-spontaneous reaction is always high.

xii. Examples.

xii. Examples.

It has the following examples.

It has the following examples.

xiii. Evaporation.

xii. Bond Breaking.

It is an example of spontaneous process but evaporation is endothermic process. Note:Spontaneous process are mostly exothermic except evaporation

It is an example of non-spontaneous process. It is mostly endothermic.

xiv. Melting of ice

xiv. Freezing of water .

Melting of ice is spontaneous process.

Freezing of water is non-spontaneous process.

xv. Cooling of Hot Tea.

xv. Heating of cool tea.

xvi. Note.

xvi. Note.

Escape of steam from pressure cooker when the safety value of the cooker is opened is spontaneous process.

Decomposition of water by electrolysis is non spontaneous process.

xvii. Flow of water from high potential to low potential is spontaneous.

xvii. Lifting of water from lower to higher potential is non spontaneous.

xviii. Burning of natural gas is spontaneous.

xviii. Compressing of gas by applying pressure spontaneous process

is non

MCQs:

When “Zn” pieces are dipped into the solution of CuSO 4, the blue colour of (CuSO4) solution disappear which is an example of spontaneous

process. System:

A substance undergoing a physical or chemical change and is under scientific study is called system. Classification:

143



System has been classified into three types; 1.

Open system.

2.

Close system.

3.

Isolated system.

1. Open System:

That system from which the free exchange of both matter and energy occur with surrounding is called open system. For example:

Boiling of water is open container is an example of open system. 2. Close System:

That system from which only the exchange of energy occur with the surrounding is called closed system. For example:

Tea port. 3. Isolated System:

That system from which neither the exchange of matter nor energy occur is called isolated system. For example:

Thermoflask. Surrounding:

Each and every thing around the system is called surrounding. Or The environment in which the system is present is called surrounding. Boundary:

That line which separate system from surrounding is called Boundary. Bulk Modulus:

Bulk modulus means that “not only a single molecule of a system should be mentioned but all should be mentioned”. Note:

1.

It is not necessary to mention the values of all state function.

2.

A state property has definite values in a given state independent of the path followed.

3.

It is not necessary to know all state function.

4.

If we know the values of some state function, the values of other can be found by the help of the following formula; P1V1 = P2 V2

State:

The specific condition of a system is called state. Explanation:

1.

2.

State of a system is described by; a.

Temperature.

b.

Volume.

c.

Pressure.

d. e.

Internal Energy. Entropy.

MCQs:

Heat and work are not state function but thermodynamic function and path function.

The state of a system is determined by b y certain macroscopic properties which are called state properties.

State Function:

Those properties which describe the state of a system is called state variable or state function.

 

It is independent of the path followed. It depends upon. i. Initial State. ii.

Final State.

Initial State:

The condition of a system before a change occur is called initial state. For initial state the state properties can be written as; i.e.

Temperature Pressure

= =

T1 P1

144


Volume

=


V1

FINAL STATE: Numerical Type

Internal Energy:

∵

E = Change in internal energy.

The total energy contained in a system is called internal energy.

Pattern of MCQs Test:

OR Sum of all possible kinds of energy present in a system is called internal energy. It is the sum of potential energy and kinetic energy of a system.

OR

Explanation:

The internal energy is because of two types t ypes of energy. 1.

Kinetic Energy.

2.

Potential Energy.

The initial energy of a system was 10kj/mol after heating its internal energy become 15kj/mol. So find change in its energy? So, E1 = 10 kj/mol E2 = 15 kj/mol So

E = E2 – E E1 E = 15 – 10 10 E = 5kj/mol

Kinetic Energy:

1.

The energy produced due to the motion of electron is called Kinetic Energy.

2.

Kinetic Energy is produced because of the following motion. i.

Translational Motion.

ii. iii.

Vibrational Motion. Rotational Motion.

Potential Energy:

 

It is produced due to the position of particles and the position of particles is because of bonding forces. The bonding forces are of two types; typ es; i. Intermolecular Forces. ii.

Intramolecular Forces.

Note:

It is impossible to determine the internal energy of a system but however change in internal energy can be determined. Mathematically:

E1 E = E2 – E Note:



Internal energy is independent of the path i.e. that change has brought by whatever source i.e heating by coal, petrol or by some other means.



It only depends upon the initial and final state.

Results of Internal Energy:

1.

If internal energy increases the temperature and kinetic energy will be increases.

2.

If the internal energy increases energetic phase occur.

3.

If the internal energy increases a chemical reaction must occur.

4.

If the internal energy increases the old b ond breaks.

First Law of Thermodynamics:

This law is also called law of conservation of energy because in this law the total energy remain constant. Statement:

The total energy of the system and surrounding remain constant. OR The energy lost by one system is equal to the energy gained by the surrounding. Note:

Energy inters or go out from a system in the form of heat or energy. Explanation:

Let us consider a system; i.

In the initial state its energy is E1.

ii.

Now if heat heat “q” is added added to it then according according to first law of thermodynamics thermodynamics it total energy energy become. become.

E2=E1+q iii.



(i)

Due to heat the piston is pushed upward against the atmospheric pressure and since work is done by the system. So the energy of the system in the final state will be; E2=E1+q-W



(ii)

∵

W = Work done by the system. 145



Rearranging E2=E1+q-W W E = q – W

Where;

 (a)

q = Heat evolved.

OR

W = Work done.

q = E + W  (b) Equation (a) an d(b) are the two form of first law of thermodynamics.

E = Change in energy.

Note:

For small change first law of thermodynamics can be written as; dE = dq + dW Special Forms of First Law of Thermodynamics:

Under specific condition the first law of thermodynamics can be written as; 1.

Work done at Fixed Volume:

If heat is given to a system and the piston is fixed then the work done is zero because the piston do not cover di stance in the direction of force. And the heat given to a system is converted into internal energy only and since the internal energy of the system increases. Since; According to first law of Thermodynamics.

E = q – W

∵

Sign Convention for Work and Heat:

W = 0

- ve q = Heat go out from system. + ve q = Heat enter to system. + ve W = Work done on system.

E = q – 0 E = q

- ve W = Work done by the system.

Since; Change in energy is equal to the heat given. 2.

If No Heat Enter or go Out:

If neither heat enter nor go out from a system then the work is done on the system. So, According to First Law of Thermodynamics

E = q – W E = + W

q = 0

Since this is called adiabatic work. 3.

∵

Heat Given to System:

If heat is given to a system then at the same time work is done on system.

E = q +W 4.

Heat Go Out from System:

If the heat is given out from a system then the work will be negative. q – W E = – q Expansion Work (Pressure Volume Work):

Consider a cylinder having movable piston containing gases. The cross sectional area is “A”.

The pressure on the piston is given by; P 

F



A

I

OR F



PA  II

Now if the piston moves upward upward by a distance “l ” then the work done on the system will be; W  F

III

Now Putting the values of (II) in equation (III). W  PAl

l .A  V

W  PV  IV

Now from first law of Thermodynamic 146



 E  q  W  V Put thevalue of W from IV to V

 E  q  PV  E  q  PV

PV  W

CONCLUSION: This is called expansion work.

Since from this equation we can calculate

E = ? q=? W=? Application of First Law of Thermodynamics:

There are two applications; i.

Isochoric Process.

ii.

Isobaric Process.

i.

Isochoric Process:

The heat evolved or absorbed during a chemical process at constant volume is called isochoric process. Explanation:

Let us consider a system enclosed in a cylinder having immovable piston or fixed. Now If heat q is given to a system then according to first law of thermodynamics.  E  q  PV q  E  PV  1 But the piston is immovable immovable so changein volumei volumeiss zero.

V  0 So (1) become q  E  P (0) q  E  0 Since qv  E Conclusion:

In isochoric process the internal energy will be very high because the heat given is converted into internal energy. ii.

Isobaric Process:

  

The process occur at constant pressure called isobaric Process. The work done under atmospheric pressure that will be called isobaric process. The process occur in the lab in an open vessel is also called isobaric process.

Explanation:

Let us consider a cylinder having frictionless, weightless and movable piston. Initial state:

Initially the energy of the system is E1 and the volume of the system is V1. Since; Initial volume = V 1 Initial Energy = E 1 Final State:

Now if heat is given to a system the piston will will go upward and the energy energy of the system will become become “E 2”and the volume become “V 2” due to

which work is done; Final Energy = E 2 Final Volume = V 2 Form First Law of Thermodynamics:

Now from first law of thermodynamics we have that;

147



 E  q  W  1 from pressuer volume work W  PV Soequa Soequaito iton n (1)beco becomeas meas ;

 E  q  PV  2 Re arranging equation ( 2) q p  E  PV  3 But ;

V  V2  V 1  E  E2  E 1 Putting the values of " V "and " E " in equation (3) So q p  ( E2  E1 )  P(V2  V1 ) q p  E2  E1  PV P V2  PV1 )  4 Re arraning arraning eqution (4) q p  ( E2  PV PV2 )  ( E1  PV1 )  5 From Enthalyphy Definition:

From the definition of enthalphy we have that; Enthalphy = Internal Energy + Work

H = E + P V Since; Now putting the values in equation equation (5) So, equation (5) become as; H 2 q p  H 2  H 1  6



E2  PV 2

H1  E1  PV 1 H 2  H1   H

So (6) become as; q p  H  7

proved

Now comparing comparing (3) and (7)

 H  E  PV  8 Conclusion:

1. 2.

H can be find out experimentally. H is gr eater eater by a factor “P V” from E.

MCQs:

For liquid and solid V  0 so  H   E Since

MCQs:

For solid and liquid equation (8)

The enthalphy of the element present in their elemental state is arbitrary taken as zero.

can be written as ;

i.e Standard enthalphy state of C = zero

 H   E Standard Enthalphy Change:

The enthalphy change of a substance present in its standard state is called standard enthalphy change. OR The change in enthalphy measured at room temperature and one atmospheric pressure when the reactants and products are present in their standard state is called Standard Enthalphy Change. Standard State:

The most stable physical state of a substance sub stance in which it exist at room temperature and pressure. 148


 


Temperature = T = 298k or 250C Pressure = P = One atm.

For example:

Most stable physical state of water is “liquid state”. Representation: 0 Standard enthalphy change change is represented represented by  H 298

Heat Capacity:

The amount of heat absorbed by a system to raise its temperature by one degree is called heat capacity. Unit:

Its S.I unit is JK -1/mol Note:

The molar heat capacity of water is 4.2 kJ/mol The capacity or ability of a system to absorb heat and store energy is called heat capacity. Note:

As the system absorb heat then the (KE) of the particles of the system increases and since the temperature of the system will be increases. Expression of Heat Capacity:

Heat capacity has two types

 

Specific heat capacity. Molar heat capacity.

Specific Heat Capacity:

It is the amount of heat absorbed by one gram of o f a substance to raise its temperature by on e degree that will be called specific h eat capacity. i.e.

T 10 c

1 gram  1 gm  Heat

Unit:

J/ gm.k Molar Heat Capacity:

It can be defi ned as “The amount of heat required to raised the temperature of one mole of a substance by one degree is called molar heat capacity. Unit:

Its unit is KJ/mol Expression of Molar Heat Capacity:

It can be expressed as; i.

Molar Heat capacity at constant volume.

ii.

Molar Heat capacity at constant pressure.

i.

Molar Heat Capacity at Constant Volume:

It is the amount of heat required to raise the temperature of one mole of a substance by one degree at constant volume. It is represented by C v. ii.

Molar Heat capacity at constant pressure.

It is the amount of heat required to raise the temperature of one mole of a substance by one degree at constant pressure is called molar heat capacity at constant pressure. It is represented by C p. Hess Law of Constant of Heat SUMMA TION: Back Ground:

This law was presented by G.H Hess in 1840. Statement:

This law states that “the amount of heat evolved or absorbed during a chemical reaction will be the same, either if the react ion occur in one step or in several steps”.

Or The heat of reaction depends only on the initial and final state of the system and is independent of the intermediate steps between reactants and products. Or The enthalphy changes of a system depends upon its initial and final state and is independent of the path followed. Mathematically:

Hess law can be written as; 149



H = H1+H2 + H3…………..Hn General Verification:

Since the heat evolved from A to D will be the same as that of heat evolved from A to B and B to C and C to D. H A D (Direct)

H1

H3 B

H2

So

C

(Indirect)

According to Hess Law H = H1+ H 2 + H 3

( AD )

Experimental Verification:

Let us consider the conversion of carbon into CO2. This may occur in two ways;

 

Indirect Method

i.

Direct Method:

 

This is one step method.

Direct Method

In this step excess oxygen reacts with carbon to form CO2.

Reaction:

C  O2  CO2

( s )

( g ) excess

 H  393kJ / mol ii.

Indirect Method:

It is two step methods.

 

Second Step

a.

First Step:

First Step

In this step carbon react with limiting oxygen to form carbon monoxide. Reaction:

C  1 O2 2

 CO CO

lim ited

 H1  110kJ

b.

/ mol

Second Step

In this step carbon monoxide is again reacts with limited oxygen to form CO2. Reaction:

CO  1 O2  CO2 2 lim ited

 H 2  282.8 kJ / mol Conclusion:

Since according to Hess law heat evolved must be the same, so;  H  H1  H 2 393.8  110.8   282.8 393. 393.8 8  393. 393.8 8 Conclusion: Since the heat evolved or absorbed is the same in both methods. Second Example:

Sodium bi-carbonate can be prepared by two methods. i.

Direct Method:

In this method CO2 is reacted with sufficient sodium hydroxide to from Na2CO3 the heat evolved is in this step is – 89KJ/mol. 89KJ/mol. Reaction:

150



2 NaOH  CO2  Na2CO3  H 2O

 H  89 kJ / mol ii.

Indirect Method:

NaOH  CO2  NaHCO3 Limite

 H  48.4 kJ / mol Step – Step – II: II:

In this step CO2 is again reacted with NaOH (limited) and heat evolved is -41kJ/mol. Reaction:

NaHCO NaHCO3  NaOH NaOH  Na 2CO3  H 2O (lim ited )

Conclusion:

According to Hess Law the heat evolved or absorbed is the same in both steps.  H  H1  H 2  H  48  (41 41) 89.0   89 Born Haber Cycle:

It is the experimental verification of Hess Law for the calculation of Latie energy of Binary Ionic Compound, indirectly. Statement:

“That Cycle which is used for the experimental determination of Latice Energy of an Ionic Crystal is called Born Haber Cycle”. Principle:

It is based on the principle “that net energy change during cyclic process will be zero”. Note:

Born Haber Cycle is the practical application of Hess Law. Explanation:



Latice energy is the amount of energy “when one mole of an ionic compound is formed from its component in gases from”.



NaCl is an ionic compound. compound.



It can be formed by two methods.



Direct method.



Indirect method

Direct Method:

In this method NaCl is formed in a single step and the heat evolved is 410.03 kJ/mol. Na 1 Cl2  NaCl 2 ( s ) lim ited

410.03 .03 kJ / mol  H  410 Indirect Method

In this method NaCl can be b e prepared in several steps. Step – Step – I (Sublimation):

In this step solid sodium is converted into gaseous form. Na  Na ( s )

( g )

 H  108.84 kJ / mol ( s )

Step – Step – II II (Heat of I.P):

In this step the gaseous sodium is converted into gaseous ions. The energy required is ionization energy i.e. Na  Na   1e ( g )  H  ( s )

443. 443.7 7 I .P

Step – Step – III III (Heat of Disassociation):

151




The conversion chlorine molecule to atomic chlorine.



Disassociation Disassociation occur. (1) Endothermic Endo thermic  H  242


Step – Step – IV IV (Heat of Electron Affinity):

In this step the chlorine again electron from the sodium. 



Cl  ie  Cl Step – Step – V (Latice Energy):



In this step Na+ react with Cl - to form NaCl.



The energy liberated called lattice energy.

So;  Hf  H  H  H  H ( s )

( I .P )

(D)

( E . A)

Note:

Hess Law can be applied to those reactions which are very very fast or slow (very very slow). Application:

Hess Law can be used for the determination d etermination of the following reactions. 1.

Heat of formation.

2.

Heat of combustion.

3.

Head of Neutralization.

4.

Heat of Reaction.

5.

Heat of Solution.

6.

Lattice energy of compound.

Heat of Formation:

The amount of heat evolved or absorbed when one mole of a compound is formed from its element in its standard form is called standard heat of formation. Hess Law can be used to determined the heat o f formation for those reactions which cannot be determined directly. Reason:

Because; i.

Such reactions does not occur easily.

ii.

Formation of bi-products occur.

iii.

Occur very very fast.

iv.

Occur very very slow.

In such cases the reaction is carried out in different steps, and for each steps the heat is determined and then added with each other by Hess Law; Application of Hess Law:

The following are the application. 1.

Heat of Formation:

The amount of heat evolved or o r absorbed during a chemical reaction when one mole fo products is formed. For Example:

C  H 2  CH 4 s

g

Hf  74kJ / mol

g

C  O2  CO2

Hf  393kJ / mol

Note:

2 H 2  O2

 2 H 2O

Hf  ?

This is not the heat of formation because 2 mole of product has been formed, where as in the light of definition one mole should be formed. 152



Standard Heat of Rxn:

The amount of heat evolved or absorbed in 2 chemical reaction in which the reactants and products are present in standard form is called heat of rxn. Mathematically:

 H rxn 





H products products



H reac tants

Example:

CaO  CO2  CaCO3

H rxn ?

 H rxn  (H CaCO3 )  (H 0CO2 ) Heat of Neutralization:

When an acid react with base it is called neutralization”. ‘The quantity of heat evolved when one mole of hydrogen ion from an acid react with one mole of OH - negative ion from a base to form one

mole of H2O is called heat of neutralization. Unit:

Kilo Joul/mol

=

kJ

Reaction:

H- + OH-



H=( – – )

H2O

Note:

The heat of neutralization of strong acid with strong base is always constant. Because in all strong acid and base same reaction occur. i.e.

HCl + NaOH



NaCl + H2O

Standard Heat of Combustion:

The amount of heat evolved or absorbed when one mole of a substance is burnt in the presence of sufficient oxygen is called standard heat of combustion. Unit:

kJ/mol Example:

C + O2



CO2

H = -393.8

Note:

It is both heat of combustion as well as heat of reaction but not heat of formation. Heat of Solution:

When one mole of a compound is dissolved in sufficient amount of water, so that no heat is evolved or absorbed on further dilution is called heat of solution. Unit:

Kilo Joul =

kJ

Example:

NH4 Cl =



H = + 160kJ/mol

Calorimetry: Measurement of Enthalphy Changes:

The enthalphy of system can be checked by two t wo methods.

 

Direct Method Indirect Method

Direct Calorimetric:

It is used for those reaction which occurs smoothly till completion and cleanly occur. No bi-product are formed, formed, no side reaction e.g. combustion and neutralization. Heat evolved or enthalphy

H = m x s x T Bomb calorimetry is used for the heat measurement in a rxn. Rxn chamber is made from copper. Initial temperature = T 1 Final temperature = T2 T1 T = T2 – T 153




The reaction carrying on in the bomb b omb chamber will be exothermic. If heat is evolved i.e. the c hange in temperature is raised. Endothermic Reaction:

If the heat is not evolved i.e. the temperature decreases then reaction will be endothermic. The water should be stroked because if some heat is absorbed by the stirrer or glass will be released. Calorimetry:

The experimental procedure used to measure the heat evolved or absorbed during a chemical reaction is called calorimetry. Bomb Calorimeter: Note: 1.

Washing soap:

They are made from sodium and fats. So it is also called sodium soap. 2.

Bath Soap:

They are made from potassium and fats. So it is also called bath soap. Exercise:

1. i.

Choose the correct answer. H per mole is expressed in units of: (a) KJ (b) 0F (c) 0C (d) K Which one of the following is not a state function. Heat (a) Enthalpy (b) (c) Temperature (d) Pressure for solids and liquids

ii. iii.

(a) H = E (b)H >E (c)H <E (d) E = O 1 k-cal is equal to: (a) 41.8 × 103J (b) 418 × 103J (c) 4.18 × 103J (d) 0.418 × 103J Standard enthalphy (H0) for 1 mole of a substance which exists in its natural state of 1 atm pressure is measured at: 298L (a) OK (b) (c) 273K (d) O 0C H can be measured indirectly by applying: (a) Avogadro’s law (b) Gas laws (c) Hess’s law (d) Faraday’s law Enthalpy mans: (a) Disorder (b) Transition state (c) Rate constant (d) Heat content No work is done at constant: (a) Pressure (b) Volume (c) Temperature (d) Mass

iv. v.

vi. vii. viii. 2.

Appling Hess’s law calculate

Given: i) H2(g) + I2(g) ––  2HI(g)

ii) H2(g) + I2(g) ––  2HI(g)

H0 for the sublimation of one mole of iodine from the following equation:

H0 = 51.8 KJ/mol 10.5 KJ/mol H0 = – 10.5

Solution:

I2 ––  I2(g) Hsub = ? Reversing the given equation (ii) we get equation 2HI(g) H2(g) + I2(g) H0 = + 10.5 KJ/mol Now add this equation (iii) to equation equation (i) H 2

( g )

i)

 I 2

2 HI ( g ) I 2

( g )

 ( g )

2 HI ( g )

 H 2( g )  I 2 ( g )

 I2

( g )

H0 H

 51.8 KJ

 10.5 KJ

/ mol

/ mol

H sub su b   62.3

3.

Calculate the heat of formation of an aqueous solution of NH 4Cl from the following data.

i)

NH3 + aq ––  NH3 Hcl(j) + aq ––  Hcl NH3 + Hcl ––  NH4cl

35.1 KJ/mol – 1 H0 = – 35.1 72.41 KJ/mol – 1 H0 = – 72.41 51.48 KJ/mol – 1 H0 = – 51.48

Solution:

154


H 3



( g )

Hcl ( g )

i)



aq



aq



2 HI 3


( aq )

Hcl ( aq )

H 0

 35.16 KJmol

1

H 0

 72.42 KJmol

1

NH 3  Hcl( aq )  NH 4cl ( aq )

H0

NH 3( aq )

H 0f  159.05KJmol 1

 Hcl

 ( aq )

NH 4cl( aq )

1

 72.41KJmol

Liquid ethanol when burnt n oxygen at 250C H0 = -1402.14KJ mol – 1. The heats of formation of H 2O are – are – 33.50 33.50 and – and – 285.81 285.81 l – – 1 0 KJ mo respective at the same temperature. Calculate the heat of formation of ethanol at 25 C.

4.

Required:

H0f of formation of ethanol, C2H5OH=? Solution:

2C + 3H2 + ½ O2 C2H5OH Hf 0 = ? Data Given:

i)

C2H5OH + 3O2 2CO2 + 3H2O

ii)

C+O2 CO2 –––––

iii)

H2+ ½ O2 H2O by reversing equation (i), we we get equation (iv)

iv)

2CO2 + 3H2O  C2H5OH H = + 1402.14 Multiplying equation (ii) by 2, we get equation (v);

v)

2C + 2O2 2CO2

vi)

3H2 + 3/2 O2 3H2O By adding (iv), (v) and (vi)

iv)

2 CO 2  3 H 2O  C2H5OH  3O2

v)

2C  2O2  2 CO 2

vi)

2 H 2  3 O2  CH5O 2

2C + 2H2 + ½ O2 3H2O

1402.14Kg H = – 1402.14Kg 33.5 Kg H = – 33.5 857.42Kg H = – 857.42Kg

393.5) = – 787.5Kg 787.5Kg H = 2 ( – – 393.5)

H  1402.14 H   787.5

H  875.42

H = +1402.14-1644.3 = 242.79

9.

Explain the following short questions with reasons. Total energy of the system and its surrounding remains constant.

Ans.

According to first law of thermodynamics. Total energy of the system and surrounding remains constant.

Let us consider a gaseous system with energy E1 at its initial state. Now q amount of heat is supplied to the system from outside the system. The total energy will become ‘ E + q’. If the piston moves up, the work W will be done by the system. Energy is consumed is doing this work. The energy at final state of the system will be E 2. E2 = (E2 + q) – w E2 – E E1 = q – w E2 = q – w w – (i) (i) q= E + W – (ii) (ii) i. The heat supplied to the system, increased the internal energy of the system and did work of expansion. ii) Ans.

It is essential to mention the physical state of the reactants and the product in thermochemical thermochemical equation. Melting of a solid (heat of fusion)

During melting of ice the temperature remain constant at O0C until all the ice has melted. During changes of state from ice to water, heat is obsorbed to over come the force of attraction. The heat content of water is higher than the ice, all thought both has the same temperature of O0C. In thermochemistry the heat content of a substance is very important. For example

H2O(g) ––  H2O Hf = 334KJ/mol b.

Boiling of liquid (heat of vaporization)

The heat obsorbed to convert one mole of liquid into vapours at its b oiling point is called heat of vaporization. Explanation

During the process of change of physical state from liquid in to vapour requires energy that cann’t observed by rise of tempe rature. It means

that heat content of vapours is higher than for liquid. E.g. H2O1 –––  H2O(g) (Ionic reaction are very fast)

Hf = 40.65 KJ/mole

155



Ans. The ionic reaction reaction are are fast fast because because the appositely appositely charged charged ions are attract each other. When When an ionic compound is dissolved in water. It dissociated into ions which are free to move. Due to effective collision the reactions among the appositely charged ions is very fast. The work done has positive and negative n egative values. Ans. According to the sing convention convention (i) work done by the system on the surrounding surrounding will will e-w (b) work done on the system system by the sourrounding will be tw. (v) the exothermic reaction are spontaneous. Ans. A process that occurs occurs by its own own without any help are called spontaneous changes. These change have a natural natural tendency tendency to occur with out supply of external energy. These change are unidirectional flow of water form higher level to lower level, colling of tea. Buring of natural gas and rusting of iron. Burning of carbon: C + O –––  lor + heat burning of natural gas CH, + 2O2 CO2 + 2HO + Heat. Standard heat of combustion is always Ans. According to sign convention the heat evolved during a reaction will be denoted by H = KJ/Mole. When combustion occurs at standard conditions of O0C and one atmosphere pressure, is called standard of combustion. Enthalpy change is state function

A quantity is called a state function when its value depends only on that state and not on the path or route used to go from one state to another. Since enthalpy is state function. This statement is known as Hess’s law. For exemption the heat of formation of corbon dioxide will be 393.5 KJ/mole weather the reaction occur in one step or in servel step. Heat is not a state function.

Mechanical work and heat are quantitative process because their values depends on the specific transition or path between two equilibrium state. The heat content of a system can be b e calculated by q = MC T. Where M is the mass, C is the specific heat capacity and T is the change in temperature. Ans. Change in enthalpy H occurs at constant pressure, in case of solid or liquid occurs. It means no work is done. The heat absorbed increased only the internal energy of the system. E of endothermic reactions are always written with a po sitive sing. According to the sign convention the heat absorbed during reaction will be denoted by H = + KJ/mol. For example: Freezing of water pumping of water from lower to higher level compressing a gas by Appling pressure, boiling of water and decomposition of water by electrolysis. Ans.

H2O (i)  H2O(g)Hf = 4065J/M HZQ(s) H2O(1) Hf = 334KJ/Mol. 1.

State and explain Hess’s Law of constant heat summation. Show that it is a direct consequence of the first law of thermodynamics.

Ans.

Hess’s Law: the heat evolved or absorbed will be the same if the reaction compl etes ibn one step or several steps.

It means it is another form of the first law of th ermodynamics which is based on the law of conservation of energy. Determine H0 for the following reaction.

b)

2C + 2H2 C2 H4(g) i)

C(s) + O2(g) CO2(g)

ii)

H2 + ½ O2(g) H2O(g)

iii)

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(g)

H0 = ? Ho1 = -393.29 KJ mol – 1 285.7 KJ mol – 1 H02 = – 285.7 1430.2 mol – 1 H03 = – 1430.2

Solution:

2C(s) + 2O2(g) 2CO2(g) 2H2 + O2 2H2O Reversing equation (iii) iii)

2 CO2 + 2H2O  C2H4(g) + 3O2 Now adding equation (i), (ii), & (iii) (iii)

756.58 KJ mol – 1 H01= – 756.58 571.4 KJ mol – 1 H02 = – 571.4

H03 = + 1430.2

6.

2C(s) + 2H2(g) C2H4(g) H03 = + 72.14 KJ mol – 1 a) What do you mean by state of a system? What are state function?

Ans.

See text Q. No. 7

b)

A chemical reaction takes place in a container of crossectional area 100cm 2 filled with a weightless and frictionless piston. The piston is moved up through 10 cm against an external pressure of 1 atmosphere as a result of the reaction. Calculate the work done by the system. Area A = 100cm2

Sol:

Distance covered by the piston l = 10cm pressure p = 1 atm Change in volume V = A ×l = 100 × 10 = 1000 cm3 Work done W = PV = 1 × 1000 atm. cm3

156


Enthalpy of neutralization of stronger acid and strong bases has always the same value.

Ans.

Heat or enthalpy of neutralization:


This is amount of heat evolved when one mole of H+ ions from an acid reacts with one mole of OH – ions from an alkali. Thus the heat of neutralization of NaOH by HCl is – 57.36 57.36 KJ. This amount of heat is evolved when 40g (1 mole) of NaOH reacts with 36.5g (1 mole) of HCl. NaOH + HCl HCl  NaCl + H 2O H+ + OH –  H2O H = – 57.35 57.35 KJ/mol 11.

H0 and E have the same value value for the the reaction taking taking place in solution.

Ans. Change in enthalpy H occurs at constant pressure. In case of solution change in pressure is not important because no change in volume of solution occurs. It means no work (PV) is done. The heat h eat absorbed increases only the internal energy of the system, E.

157



CHAPTER-12:

ELECTROCHEMISTRY

Definition:

The branch of chemistry which deals with the relationship between between chemical changes and electricity is called electrochemistry. OR The branch of chemistry which deals with the inter conversion of chemical energy into electrical energy and electrical energy into chemical energy is called electrochemistry Conduction:-

Conduction depends upon ions; It is a process of splitting up of an ionic c ompound due to the presence of charge d particles when fused or disclosed in H 2O Ionization:

The degree of ionization of a substance sub stance depend upon the following factors; 1. Nature of electrolyte 2.

Temperature

3.

Degree of dilution (i) Nature of Electroplate: Electroplate:

If Ionic more ionizes e.g.

If covalent less Ionizes

NaCl

Acetic acid CH 3COOH

(ii) Temperature  as the temperature increases atoms gain energy and ionization increases.

∝

Ionization Temperature (iii) Degree of Dilution

Dilute more Ionizes

Concentrated less Ionizes

In dilute solution there th ere is free, path for effective collision and thus the extent of dissolution increases. Electrolyte

Those substances which conduct electricity in molten form or in solution form is called electrolyte. Types:

1.

Strong electrolytes

2.

Weak electrolytes

3.

Non electrolytes

i. Strong electrolyte:

That electrolyte which dissociate completely is called stron g electrolyte. 1.

NaOH

2.

NaCl

3.

H2SO4

ii. Weak Electrolytes:

Those substance which partially dissociated into its ions is called weak electrolyte. electrolyte. 1.

NH4OH

2.

H2CO3 etc.

iii. Non-electrolyte:

Those substances which does not dissociate into ions is called non-electrolytes. Example: - sucrose, urea 1.

Decomposition of an electrolyte into its components by the passage of electricity is called electrolysis.

2.

Electricity does not forms ions it only push the ions towards their respective poles (electrode).

Oxidation Reduction:

Oxidation reduction can be explain on the basis of two concepts; 1.

Classical concept

2.

Electron transfer concept

158



1. Classical Concept:

According to classical concept oxidation and reduction can be explain as;

All EN Substances are oxidizing agent 

Oxidation: (In term of Oxygen)

The addition of oxygen is called oxidation.

According to classical concept: An oxidizing agent is a substance which give oxygen or any EN element like chlorine to a compound.

For Example:

SO2 is oxidized to SO3 by gaining oxygen.



2SO2 + 3O2 –––→ 2SO3 (Oxidation)  

Reduction:



Note:electron is called oxidizing Any substance which gain agent. All the members of VIIA are oxidizing agents. All the cations are oxidizing agent.

The removal of oxygen is called reduction. For Example:

Question:

CO2 –––→ C + O2 (Reduction)

Q: Which one is oxidizing agent?

(In term of Hydrogen)

a) Zn+2

b) Na+1

c) Mg+2

d) All

Oxidation:

The removal of H2 is called oxidation. 

For example

Partial filled orbital are oxidizing agent.

e.g. 2NH3 –––→ N2 + 3H2 Reduction:

Addition of H2 is called reduction For example:

N2 + 3H2

⇌

2NH3

2) Electron Transfer Concept:

According to this concept the oxidation and reduction can be explain on the basis of electron transfer; Oxidation: (LEO): (L: Loss of E: Electron O: Oxidation)

The phenomenon in which element loses electrons is called oxidation.

̅

e.g. Na ––––→ Na+ + 1 (oxidation) (oxidation) The element is said to be oxidized G = Gain, E = Electron, R = Reduction

Reduction: (GER)

The phenomenon in which an element gain electron is called reduction. For Example:

̅

Cl2 + 2 ––––→ ––––→ 2Cl (Reduction) Oxidizing Agent:

Those substances which oxidizes other and itself reduces is called oxidizing agent. The oxidation number of oxidizing oxid izing agent decreases in a reaction. e.g.

̅

 ̅

oxidizing agent Cl2 + 2 –––→ –––→ 2 l (zero)

(Negative) decreased

Reducing Agent:

Those substances which reduces other substances and itself oxidizes that will be called reducing agent. Oxidation no of reducing agent increases during reaction. For Example:

°

̅

Na ––––→ Na+ + 1 (oxidation) (oxidation) Oxidation No

Oxidation no

Zero ––––→ +1 (positive) The apparent positive or negative change on an atom in a molecule is called oxidation state. Note:

It is different from valency It shows loss of electron It show gain of electron. 159



Rules:

i. ii. iii. iv. v. vi. vii. viii.

The oxidation number of a free element is zero, e.g. oxidation number of H2, O2 and Mn is zero. The oxidation number of hydrogen in its compounds is +1 but in metal hydrides it is – 1, 1, e.g. NaH and MgH2. The oxidation number of oxygen in the compounds is – 2 but in peroxides it is – 1 and in OF 2 is +2. The oxidation number of elements of Groups I, II and III in the compounds is +1, +2 and +3 respectively. The oxidation number of the halogens of Group VII in the binary compounds is – 1. 1. The algebraic sum of the oxidation numbers of all the atoms in a compound is zero. The algebraic sum of the oxidation numbers of all the atoms in an ion is equal to the charge on the ion. When an element is oxidized its oxidation number increases and when an atom is reduced, its o xidation number is decreased.

(a) calculate the oxidation no of Cr. in K 2 Cr 2 O7

Putting these values in the formula;

Oxidation no of K = + 1

2 (+1) + 2x + 7 (-2) = 0

Oxidation no of O = – 2

+ 2 + (2x) + (-14) = 0

Oxidation no of Cr = x K 2 + Cr 2 + O7

2x = +14 – 2 2  2x = 12  = X = 12/2 = 6

   

Balancing red ox-equation by oxidation number M ethod:

Following steps are involved. Oxidation and reduction o ccurs simultaneously. Principle: “law of conservation of charge”

If one substance lose electron other should gain electrons, the lose and gain will be b e the same and occur simultaneously. Steps:

They are the following: i.

Skeleton:

First we will write the skeleton (general reaction) of a chemical reaction: ii.

Identification of reducing and oxidizing agent:

Now we will identify the oxidizing and reducing agent agent by change in oxidation number. iii.

Oxidation Number:

Now we will write write the oxidation no on each element. iv.

Indication of reducing and oxidizing agent byan now indicating the change in oxidation number by means of an arrow .

v.

Multiply the formula of the oxidizing and the reducing agents by a number such that the number of electrons lost during oxidation becomes equal to the gain gain of electrons during reduction. reduction.

vi.

Balance the rest of the equation by simple inspection.

For Example:

Let us balance the following equation by oxidation number method, by using the above mentioned steps. i. Zn + HNO3 ––––→ Zn Zn (NO3)2 + NO + H2O ii.

Oxidation number of Zn is 0, Oxidation number of Zn in Zn(NO3)2 is +2. Oxidation number of Zn increases from 0 to +2 so it is oxidized and is a reducing agent. Oxidation number of N decreases from +5 in HNO 3 to +2 in NO so it is reduced and acts as an oxidizing agent.

iii.

Write the oxidation numbers over the symbols of the elements oxidized and reduced. Zno + HNO+5O3 ––––→ Zn+2(NO3)2 +NO+2 + H2O

iv.

Indicate the change in oxidation numbers of means of arrows. Reduced Zn + HNO3 ––––→ Zn(NO3)2 +NO + H2O Oxidized Hence, the substances reduced and oxidezed can be written as follows, a. b.

v.

Zno ––––→ Zn+2 + 2e – HNO+5O3 + 2e – ––––→ NO+2

(Oxidized) (Reduced)

Multiply equation (a) by 3 and (b) by 2 so that the number of electrons gained is equal to the number of electrons lost. 3Zn + 2HNO3 ––––→ 3Zn (NO3)2 +2NO + H2O Add six more NO3 – in the form of HNO 3 to the reactant’s side, to balance 6NO 3 – of Zn(NO3)2 in the products. 3Zn + 2HNO3 + 6HNO3 ––––→ 3Zn(NO3)2 +2NO + 4H2O

vi.

Now balancing H and O atom by inspection. 160



Balanced equation 3Zn + 8HNO3 + 6HNO3 ––––→ 3Zn(NO3)2 +2NO + 4H2O Balancing the Redox Equations by the Half Reaction Method

This method of balancing is also called the ion electron method. No oxidation numbers are assigned in this method. It applies to redox reactions taking place in aqueous medium. The following steps are used for balancing redox equation by this method. Split the equation into two half cell reactions, one for oxidation and the other for equation. The reactions should show only the oxidizing and reducing agent.

1. 2.

Balance number of atoms on both sides of the two half reactions independently. In neutral medium H2O and H+ can be added on either side.

3.

In acidic medium the H+ ion may be used for greater number of oxygen and H 2O can be added to th e other side.

4.

In alkaline medium H2O may be used, for removing oxygen on one side of the equation while OH – ions can be added to the other side.

5. 6.

Balance the charge by adding electron (e – ) to the side deficient in negative charge. Multiply each half reaction by a number chosen so that the total number of electrons lost by the reducing agent equals the number of the electrons gained by the oxidizing agent.

7.

Add the two half reactions, resulting from the multiplications. Cancel anything appearing to both si9des in the net equation.

8.

Check the final equation by counting the number of atoms in the net charge on either side.

Example:

Balance the following reaction taking place in acidic medium by ion electron method. I – + NO2 ––––→ I2 + NO 1. Split the reaction into tow half cell reactions, one for oxidation and the other for reduction. i.

I – ––––→ I2

ii.

NO2 ––––→ NO (Decrease (Decrease in oxidation number shows reduction)

(Increase in oxidation number shows oxidiation)

Balance the number of O – atoms atoms on each half reaction by adding H + (oxygen excess side) and H 2O to other side, also multiply I by

2.

“2” in equation.

3.

i.

2I – ––––→ I2

ii.

NO2 + 2H+ ––––→ NO + H2O

Balance the charge on both sides of each half reaction by adding electrons to either side. i.

2I – ––––→ I2 + 2e –

ii. NO2 + 2H+ + 2e – ––––→ NO + H2O The total number of electrons lost and gained in the two half reactions are the same. so adding the two half reactions after cancelig the electrons gained with the lost, we get a balanced net reaction. i. 2I – ––––→ I2 + 2e –

4.

ii.

Balanced equation

NO2  2H  2e  NO  H2 O

2H   NO2  2I  NO  I2  H2 O

Reaction of oxidizing agents:-

(K 2 Cr 2 O7 and KMNO4are the strong oxidizing agents) 1)

Potassium dichromate is a strong oxidizing agent. In presence of dilute sulphuric acid it also acts as an oxidizing agent and oxidizes a number of compounds. In all these reaction potassium dichromate is reduced from +6 oxidation state to +3 oxidation state. (i) It oxidizes iodide to iodine K 2 Cr 2 O7 + 6KI + 7 H 2SO4 Cr 2 (SO4)3 + 4K 2SO4 + 3I2 + 7H2O (ii) It oxidizes ferrous salt to ferric salts.

→

2)

→

K 2 Cr 2 O7 + 6 FeSo 4 + 7H2SO4 Cr 2 (SO4)3 + 3Fe2(SO4)3 + K 2So4 + 7H2O KMNO4 is a strong oxidizing agent. In presence of dilate sulphuric acid it also acts as an oxidizing agent and oxidizes a number of a compounds. (i) KMNO4 oxidizes KI in presence of (dil) H 2SO4 oxidizes to I2

→

2KMnO4 +10KI + 8H2SO4 2MnSO4 + 5I2 + 8H2 (ii)

KMnO4 oxidizes oxalic acid in the presence p resence of sulphuric acid to CO 2. 2KMnO4+ 10 (COO)2H2 + 3H2SO4 2MnSO4 + 2K 2SO4 + 5I2 + 8H2 (H2S) and (SO2) are reducing agents Reaction of reducing agent 1.

→

H2S is reducing agent in acidic medium (i) It reduces halogen to halo acids H2S + Cl2 2HCl + s H2S + Br 2 2HBr + s (ii) It reduces ferric salt to ferrous salt

→→

161


2.


→

H2S + 2Fecl3 2Fecl2 +2HCl + s So2 is reducing agent in the acidic medium (i) SO2 react with KIO 3 and reduces I2 (ii) 2KIO3 + 5SO2 + 4H2O K 2SO4 + 2MnSO4 + 2H2SO4

→

Electrode:

“An electro -conducting substance when placed in an electrolyte develops a certain electrical potential with re spect to the bulk of the solution”

OR It is an electrical conductor used to make contact with a non-metallic from of a circuit e.g (a, semiconductor, an electrolyte, a vacuum, or air) Origin of Word Electrode

The word was coined by “William whe n well” at the request of scientist “Michael Faraday” form the Greek word electron amber, and hodos a

way. Types:It has two types: (i) Cathod

Fardays law related

 that electrod at which reduction occur.

Reduction occur at cathode. It has high h igh reduction potential. (ii) Anode

If molar mass of the substance is M and substance gain or loss n number of electron per ion at any electrode then equivalent weight of the substance can be find by using formula. e =

 

 that electrode at which oxidation occur is called anode. It has

oxidation potential. Faradys Laws of Electrolysis:

In 1813 Michael Faraday studied the quantities relationship between the; The quantity of electricity passed & quantity of a substance deposited on different d ifferent electrodes electrodes * *

Faraday’s gives two laws

*

Second law

1.

First Law:

First law

According to law “The amount of any substance (W) deposited at an electrode is directly proportional to the quantity of electricity Q passed”. Mathematically:

W  Q

If Al charges to Al +3 in a reaction equivalent weight of Al will

 Q = It

Eqvilant weight  be Eqvilant

W  It W = ZIt

Molar ma mass Number Number of Ionlose



27 3

=9

Where: W = weight of substance Q = Quantity of electricity passed. Second Faraday Law:

This Law states that “ The amount of electricity passed through different electrolytes then amount of different substances deposited are in the ratio of their chemical equivalents.” Mathematically:

W  chemical equivalent  (i)

Chemical equivalent (e)

W  I t from first law  (ii) Comparing (i) & (ii) W α I te  (iii) Using constant of proportionality “1/F”

W = I te/F Electrode Potential (E)

Zinc rod

o wn Ions is called electrode “The potential developed when a metal is in contact with the solution of its own potential. OR Zinc sulphate solution Explanation:

The potential difference measured between two electronic conductors is called electrode potential. In the external circuit connected to an electrochemical cell the electrons will flow as they do in electronic conductors from the most negative point to the most positive point.

162


This

means

that

̅

s

in

the


external

circuits

connected

to

an

electrochemical

cell

will

always

flows

from

most negative to most positive. Since the potentials of electode can be either positive or negative. Note:

Electrons in the external circuit can also flow from l east positive to most p ositive electrode. Note:

Volt meters are commonly used u sed to measured the potential difference across electrical circuits of electrochemical cells. Standard Electrode Potential:

When a metal is dipped in a solution of its own ions (1 Molar) at 298k, then the potential acquired under these standard condition is called standard electrode potential. Standard Hydrogen Electrode SHE:

Standard hydrogen electrode is a reference electrode with which other electrodes are compared. Composition

(1) (2) (3) (4)

standard hydrogen electrode is used as a standard It is consist of a piece of platinum (pt) foil, foil, which is coated electrolytically electrolytically with finely divided platinum platinum black, black, to give it a large surface area and suspended in one molar solution of HCl. Pure hydrogen gas at one atom pressure is continuously bubbled into IM HCl solution. The potential developed on the platinum electrod is assume to be zero.if any potential is developed at the surface surface of pt then then its means that it will from from that of other measuring electrode electrode

SHE can act as cathode and anode

The electrodes above SHE in electrochemically series have negative reduction potential and positive oxidation potential. They have greater tendency to deliver the electrons to SHE. Zn  Zn+2 + Ze – (oxidation at anode) 2H+ + 2e –  H2 (Reduction at cathode) The electrodes below SHE in electrochemical electrochemical series have negative oxidation potential they accept the electrons from SHE. Cu+2 + Ze –  Cu0 (Reduction at cathode) H2 2H+ + 2e – (oxidation at anode) Measurement of electrode potentials:

It is the actual cell potential difference measured in reversible cells under standard conditions. Chemist have universally agreed to select the electrode reaction of th e simplest element Hydrogen, as the reaction and electrode against which all will be compared. Which will be called standard hydrogen electrode The physically measured potential difference across reversible cell made o f any electrode and a standard hydrogen electrode, That hydrogen electrode which is being operated under standard conditions of pressure of that electrode, E. If electrode other than hydrogen is also being operated under standard condition of pressure and concentration. Then the reversible potential difference across the electrode is the standard electrode potential or standard potential, E0 of the electrode other than Hydrogen. “In an electrochemical cell, an electric potential is created between two dissimilar metal”. This potential is a measure of t he energy per unit charge which is available from redox reaction to drive the reaction”. SHE, has been elected as reference electrical. The electrocle potential of SHE is zero. When another electrode is connected with SHE, then the emf of the cell is the electrode potential of that electrode. The hydrogen electrode is a also referred as “SHE” and some time (NHE) – normal

hydrogen electrode. Some Important Features of SHE:

(i) In SHE the pressure of the H-gas is 1 atm and concentration of hydrochloric acid is 1.00M. (ii) The potential between two electrodes i.e. hydrogen and test electrode is measured by using a digital voltmeter. This draws a negligible current, and so no electrolysis occurs during the potential measurement and there is no change between conce. Of H +& H2. When the H-electrode is operating under standard conditions the value of the potential measured is known as S.E.P of the test electrode. An important feature of the structure of cell u sed with the standard electrode. Electrochemical Series:

Electrochemical Electrochemical series tells us about the distinction between the oxidizing agent and reducing agent. The arrangement of elements on the basis of increasing order of oxidation potential or reduction potential is called electrochemical series. Those elements which lie above the (H 2) have low reduction potential and high oxidation potential. 163



Electrochemical Series

Those elements who have greater tendency than hydrogen to lo se e – to their solution are taken as electropositive and vice versa. Element

Electrode

Standard reduction potential (E0)

Li

Li++e  Li

E0 = – 3.05 3.05

K

K+

E0 = – 2.2 2.2

̅ ⇌

k

Low reduction potential (less tendency to gain electron High oxidation potential (greater tendency to lose e)

H2

2H+ + 2e  H2

E0 = 0.000 (Reference electrode)

̅ ̅

High reduction potential. ter tendency to gain High Less tendency to lose (cathode (cathode reduction occur at cathode) F2

F+1e  F

E0 = + 2.87

Applications:

Following are; 1. Spontaneity of chemical reaction:

If E0 =- +ve ten the reaction will be spontaneous. 3.

Non-Spontaneity:

0

If E = – ve ve then the reaction will not be non-spontaneous. Explanation:  

 

Different elements have different abilities to lose (oxid) or gain electrons (reduce) Metals at the top of th e series are good at giving away . therefore they are good reducing agent i.e. they lose electrons with easily. Metal Ions at the bottom of o f the series are good at picking up es. i.e. to gain es easily. The oxidizing ability of the metal ions ses as you go down the series.

̅

Note:

i)

The more negative the E0 values the more rapidly / readily the metal loses es. And the stronger reducing agent is the metal will be. Or and the metal be stronger reducing agent.

ii)

The more (+) E0 values, the less readily metal loses es (do not lose e) and is strong oxidizing agent.

Cell reaction and cell voltage

The difference which develop between two electrodes due to which electrons flows from one electrode to another electrode is called E.P. OR The potential developed when a metal is in contact with a the solution of its own ions. Is called electrode potential. Representation:

It is represented by E cell Unit:Its unit is

Volt (v) Mathematically:

Ecell = Eoxid + EReduction Cell potential depends upon; (1) Nature of electrode (2) Concentration of ions (3) Temperature (4) pressure (5) Morality Standard cell potential If Morality is

M = 1M T = 25 C0 and p = 1atm then the cell potential is called standard cell potential. Represented:

E0cell = E0oxid + E0Reduction E0cell = E0Zn/Zn++ + E0Cu++/Cu Note:

(i) If cell potential is positive = reaction will be spontaneous. (ii) If E 0 is negative = then reaction will be non spontaneous. Cell Representation: Sign Convention

Zn + Cu++ Zn++ + Cu Correct form of E 0cell

E0cell = E0oxid + E0Reduction 164


E0cu = E0 Oxid


Reduction

// + E0 Cu++/Cu(1M)

++ Zn//Zn (1M)

Correct form of E 0 for the following reaction is

⇌

cu + pb++ pb + cu ++ oxid red cu/cu++

pb++/pb

1M

1M

Type of electrochemical Cells:

“It is a cell in which the electrodes are dipped into electrolyte in which non -spontaneous redox reaction take place by the passage of an electrical current”.

OR “An arrangement in which electrical current is produced by a chemical reaction. a)

Electrolytic Cell

“The movement of ionic charges through the liquids brought by the application of electricity” Example: Charging of batteries b)

Voltaic or Galvanic Cell

A voltaic or galvanic cell consist of two half cell that electrically commented. Example: Dry cell and lead storage battery. Electrolysis of Aqueous NaCl:

1.

Caustic soda, NaOH is very important industrial chemical, manufactured on a large scale by the electrolysis of aqueous solution of NaCl. The electrolysis electrolysis is carried out in a cell called Nelson’s cell. It is an oblong steel tank containing containing a concentrated concentrated aq ueous solution of NaCl. The graphite anode is suspended in the solution. Cathode is made of a sheet of perforated steel. When connected to the battery, the half reactions taking place at the electrodes.

2.

At the anode, chloride (Cl – ) is oxidized to chlorines and at the cathode, water is reduced to hydroxide and hydrogen gas. The net process is the electrolysis of an aqueous solution of NaCl into industrially useful p roducts sodium hydroxide (NaOH) and chlorine gas.

2NaCl  2 Na+ + 2Cl At anode 2Cl –  Cl2 + 2e –

(oxidation)

at cathode 2H 2O + 2e –  H2 + 2OH – Overall reaction

2Cl – + 2H2O  H2 + Cl2 + 2OH –

⇌

2Na + 2OH – 2NaOH Advantages of the Electrolytic Cell:

1.

Charging of the lead storage batteries is done electrolytically by passing the current through the discharged battery so that the reactions are reversed and the battery is recharged.

2.

Various types of electrolytic cells are employed on industrial scale for the manufacture/extraction manufacture/extraction of industrially important metals, and gases e.g. manufacture of NaOH, Na, Mg, Al, Cl 2 etc

3.

Electrolytic cells can be used for the purification of some important metals like Cu etc.

4.

Cooper, Silver, chrome, nickel and tin plating is done by various types of electrolytic cells, for the purpose of protection, beauty and repair of the metal.

Electro plating: electrostatic Coating

In the a manufacturing process that employs charged particles to more efficiently paint conscience paint in the form of either powered particle or atomized liquid, is initially projected toward an inductive work piece using normal spraying methods, and is then accelerated, toward the work piece by a powerful electrostatic charge. Voltaic Cell Voltaic Cell: In the voltaic cell a spontaneous redox reaction is carried out and electrical current is

produced. Construction: It consists of two half cells externally connected with a metallic wire acting as a

conductor. Taking Place of Reaction: At each half cell on e half of the total cell reaction take place. Reduction Reaction: At one electrode electron enter resulting in reduction reaction. Oxidation Reaction: While at other end they leave the solution and oxidation reaction take place. Example: Daniel cell is an example of voltaic cell.

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Daniel Cell: This cell has a Zn electrode dipped into 1M ZnSO4 solution and a Cu electrode immersed in a 1M solution of Cu2+ ions. Construction: These two half cells are externally connected through a metallic wire while internally they are connected by a salt bridge.

The salt bridge contains an aqueous KCl solution in a gel. At Anode: Zn tends to lose electrons more readily than Cu giving its electrons to the electrode electron flow from Zn electrode to Cu electrode

through external circuit making it negatively charged. At Cathode: the Cu+2 surrounding the cathode pick the electrons and get deposited as neutral metal. Reduction: Reduction take place at copper. Oxidation: Oxidation take place at Zinc.

̅

E00x = + 0.76v At anode: Zn  Zn2+ + 2 E At cathode: Cu2+ + 2

̅

 Cu E0Red = + 0.34v Net reaction: Zn + Cu 2+ Zn2+ + Cu E 0 Cell = 1.10v Cell Representation: Zn(S)/ Zn2+(aq) (1M) // Cu 2+(aq) (1M) / Cu(S) Batteries:

There are four major types of batteries 1.

Primary batteries:

These batteries are not reversible and once discharged are discarded. Example: Dry cell 2.

Secondary Batteries:

They are reversible and can be recharged. Example: Lead storage battery 3.

Solar Batteries:

They are photoelectrical cells and generate energy. 4.

Fuel Batteries:

They are super batteries and have high charge density. (a) Primary batteries DRY CELL:

An electrochemical electrochemical cell in which electrolytes are present as solid or in the form of paste. Laclanche Cell: This cell is also called laclanche cell. It was designed by George Laclanchi in 1887. Construction:

(1) It consist of an outer Zinc causing which act as anode. (2) It is lined inside with a moist paper which prevent Zinc from fro m coming in contact with other reactants, but allow diffusion of ions. Carbon Rod: A carbon rod is placed in the centre which act as cathode. Ammonium Chloride: A paste of ammonium chloride (NH2Cl) is used as an electrolyte MnO 2, ZnCl2 and powdered carbon is also used. As

Zinc case serve as anode so Zinc atoms oxidizes and dissolves in the electrolyte as positive ions. At Anode:

̅

Zn(S) Zn+2 + 2 s At Cathode:

1) 2) 3)

̅

2MnO2 + 2NH+4 + 2  Mn2O3 + 2NH3 + 2H2O The electron lost at anode passed through an external circuit and do some useful work. Its voltage is 1.5V. The net reaction can be written as.

̅

At anode: Zn  Zn+2 + 2

̅

At Cathod: 2MnO2 + 2+ NH4 + 2

 Mn2O3 + 2NH3 + H2O Net Reaction: Zn + 2MnO2 + 2NH+4 Zn+2 + Mn2O3 + 2NH3 + H2O 1)

It is irreversible because Zn and NH +4 are consumed during the process of working and cannot be reversed back to their initial states, by the application of the external external electrical electrical potential.

SECONDARY BATTERIES Lead Storage Battery:

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Discovery: It was demonstrated by French academy of science by Gasten plante in 1860. Definition:when the cells are connected in series is called battery. Anode: it is made from Pb Cathode:It is made of PbO 2. Mechanism

The electrodes are kept immersed in aqueous solution of H 2SO4. When anode and cathode are connected the current flows.

̅

At Anode: Pb + H2SO4 PbSO4 + 2H+ + 2

̅

At Cathod: PbO2 + H2SO4 + 2H+ + 2

 PbSO4 + 2H2O

Net Reaction: Pb + PbO2 2H2SO4 PbSO4 + 2H2O

When cell works insoluble PbSO 4 goes on depositing on both the electrodes, H2SO4 is consumed and water is formed: Reversible Battery:

i. ii.

It is reversible battery. If the current from an external source is allowed to flow in the opposite direction during researching all chemical changes are reversed. The PbSO4 and H2O reforms Pb, PbO2 and H2SO4 and the battery is again ready for generating useful electricity.

iii.

FUEL CELLS: Definition: An electrochemical device used for continuously converting chemicals into direct current (D.C) is called fuel cell. Construction: It consist of two electronic electrodes separated by an ionic electrolyte.

 

The fuel can be gas, liquid and solid or solution.

The electrodes may be solid or porous and may contain a catalyst. Bacon Cell: It is the best known fuel cell. Reactants: It involve hydrogen and oxygen gases as the reactants (fuels).

    

The product of the cell reaction is water. Platinum is used as a catalyst. KOH is used as electrolyte. Electrodes are hollow tubes made of porous, compressed carbon, impregnated with platinum. outer ter circuit, while the electrons are accepted accepted at the cathode, where reduction Hydrogen is oxidized at anode giving electrons to the ou occur and current flow.

Reaction:

2H2 4H+ + 4es (oxidation) O2 + 4H+ + 4es  2 H2O (reduction) (i) (ii) (iii) (iv) (v)

2H2 + O2 2H2O Fuel cells are very efficient and convert about 75 % of the fuel into electricity. The cell is operated at high temperature. Thus water formed evaporates. These vapour may be condensed to produce drinking d rinking H2O. However this cell is costly as pure H 2, pure O2& platinum are expensive.

Corrosion and its Prevention

Definition: the slow and continuous eating away of the metal surface by the action of environment is called corrosion. Explanation: i. ii. iii. iv. v.

It is continues process It is chemical process It irreversible process It is spontaneous process It is exothermic process

Rusting: Corrosion of iron is called rusting.

Conditions for Rust Formation:

  

Thin film of water on the surface of iron. Air or oxygen in the environment. Weakly acidic atmosphere i.e. 167



SO2, CO2, Gases. Mechanism:

i. ii. iii. iv. v.

Pure iron is silvery white but when exposed to moist air, its surface is corroded and converted to a reddish brown mass known as rust. Chemically rust is hydrated iron (III) oxid e. The impurities are responsible for the formation of small electrolytic cells with anode of pure iron and cathode of impure iron. iron . +2 Iron is oxidized at anode producing Fe ions [Ferrous ions] and electrons. Electron moves along the surface of metal to cathode where it reacts with water and oxygen to form hydroxyl ions. ion s. Fe+3

 +2

 ̅

Fe (OH)3

2Fe  2Fe + 4 (+) Anode

    ̅ 



H



2H2O + O2 + 4  4O

̅ ̅

( – – ) Cathode

Steel Rod

Reaction at Anode: 2Fe  2Fe+2 + 4 s Reaction at Cathode: 2H2O + O2 + 4

 4OH

Ferrous ions are further oxidized by atmospheric oxygen to form, Ferric ions Fe+2 Fe+3 +

̅

Fe+3 + 3OH –  Fe (OH) 3 Thus rust is soft and porous in nature and therefore cannot prevent go further deeper by the atmospheric action. The result is that, once corrosion starts it continuous until the whole iron piece is rusted.

Note:

Prevention of Corrosion:

Corrosion cannot be completely eliminated or prevented, however a number of methods have been devised to protect metals from corrosion to a certain extent. Most of the methods are aimed at preventing rust formation. Some methods are as follow:

1.

Coating the metal surface with paint, oil, coal tar and grease.

2.

Coating the metal with sacrificial metal such as Zinc called Galvanizing or tin called tinning.

3.

Electroplating with nickel or chromium.

4. 5.

Dipping the iron into a phosphate bath (orthophosphoric acid with Zinc and manganese phosphates) Alloying the metals, steel formation.

6.

Cathodic protection.

7.

Blowing steam over red hot iron so that an adherent film or Fe3O4 gets deposited on the iron surface.

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1st Year Chemistry (Nasrat Katozai Nawaz Usb)

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