WAJA CHP.4 Thermochemistry

WAJA F5 Chemistry 2010

Chapter 4: Thermochemistry

CHAPTER 4 : THERMOCHEMISTRY A

ENERGY CHANGES IN CHEMICAL REACTION

Learning Outcomes : You should be able to: • State what exothermic reaction is, • State what endothermic reaction is, • Identify exothermic reactions, • Identify endothermic reactions, • Give examples of exothermic reactions, endothermic reactions, • Give examples of endothermic • Construct energy level diagrams for exothermic reactions, endothermic reactions, • Construct energy level diagrams for endothermic • Interpret energy level diagram, • Interrelate energy change with formation and breaking of bonds, • Describe the application of knowledge of exothermic and endothermic reactions in everyday life.

Activity 1: (a)

What is an exothermic reaction? An exothermic reaction is a chemical reaction that gives out heat to the surroundings. surroundings. Heat energy given out from the reaction is (1)…………………………to (1)…………………………to the surroundings

(b)

The temperature of the surroundings (2)………………………. What is an endothermic reaction? An endothermic reaction is a chemical reaction that absorbs heat from the surroundings . The reactants (3) …………………. heat energy from f rom the surroundings. The temperature of the surroundings (4) ……………………….

Activity 2 (a)

Identify the following reactions as exothermic or endothermic reaction by writing a ‘ √’ at the appropriate box as shown in Question (a). (a) (b) (c) (d) (e) (f) (g) (h) (i) (j)

Reaction Combustion of ethanol Burning of magnesium Neutralisation between acid and alkali alkali Adding water to concentrated sulphuric acid acid Photosynthesis Reaction between acid and magnesium Reaction between between acid and and calcium carbonate carbonate Dissolving ammonium salt in water Thermal decomposition decomposition ooff copper(ll) copper(ll) carbonate Thermal decomposition decomposition of zinc nitrate 1

Exothermic √

Endothermic


(b)


The amount of heat energy released or absorbed during a chemical reaction is called the heat of reaction.

(c)

It is given a symbol (1)………………. and the unit is (2) ………………….. The heat of reaction , Δ H = H products - H reactants (a) Exothermic reaction : The reactants lose heat energy to form the products . Thus the products

formed have less energy than the reactants, Therefore, Δ H is (3)………………… (b) Endothermic reaction : The reactants absorb heat energy to form the products . Thus the

products formed have (4) ……………energy than the reactant. Therefore, Δ H is (5)……………. (d)

List two other examples of exothermic and endothermic reaction Exothermic reaction

Endothermic reaction

1) 2) Activity 3 : (a)

Energy level diagram (i) Energy level diagram for exothermic reaction : The products have less energy than the reactants, Energy reactants ΔH = negative products Construct the energy level diagram based on the given chemical equation, Mg(s) + H2SO4 (aq)  MgSO4 (aq) + H2 (g) (reactants)

( products )

(ii) Energy level diagram for endothermic reaction : The products have more energy than the reactants, 2

ΔH = -467 kJ



Energy products

ΔH = positive

reactants (iii)

Construct energy level diagram based on the given chemical equation CaCO3(s)



CaO(s) + CO2 (g)

ΔH = + 178 kJ

Information that can be obtained from the energy level diagram ,

Energy Zn + CuSO4

ΔH = - 190 kJ Cu + ZnSO4 Figure 1 Figure 1 shows the energy level diagram for the reaction between zinc and copper(ll) sulphate, •

The reaction between (1) ……………and (2)…………………. is an (3) …………………….. reaction.

•

During the reaction, the temperature of the mixture (4) ………………………

•

The total energy of one mole of (5)……………. and one mole of (6) …………… is (7) …………. than the total energy of one mole of copper and one mole of zinc sulphate by (8)……………………. kJ

•

When one mole of (9)………………….. reacts with one mole of (10)…………to form (11)…………….. .. mole of copper and (12)……………….. mole of zinc sulphate, (13) …………………….. kJ of heat is (14) ………………………… Energy

3



2HI (g) ΔH = +53 kJ H2 (g) + I2(g) Figure 2 Figure 2 shows the energy level diagram for the reaction between hydrogen gas and iodine •

The reaction between (15)……………and (16)…………………. is an (17) ……………… reaction.

•

During the reaction, the temperature of the mixture (18)……………………

•

The total energy of one mole of (19)……………. and one mole of (20)……………. is (21) ………than the total energy of two moles of hydrogen iodide by (22) ……kJ

•

When one mole of (23)………………….reacts with one mole of (24) ……………… to form (25) …………….mole of hydrogen iodide (26)………..kJ of heat is (27) …………………

(b)

(i)

A chemical reaction involves (1) ....……… (2) ............... of the reactants and (3)…………….. (4) ........................of the product

(ii) Bonds breaking always(1) …………. energy and bonds formation always (2) ………………… energy

Bond breaking

Bond formation

(iii) If the energy absorbed during bonds breaking is less than the energy released during bonds formation ,energy is (1)………………….. to the surrounding. The reaction is an (2) ……………….. reaction

(iv) If the energy absorbed during bonds breaking is more than the energy released during bonds formation, energy is(1)…………………………. from the surrounding. (c)

Hence it is an (2) ……………………reaction Explain the application of exothermic and endothermic reaction in our daily lives. (i) Instant cold packs : Instant cold packs are used to treat (1)…………………………………………………, have separate compartments of (2) …………………and (3)……………………in a plastic bag. When the barrier between the two is broken by squeezing the outer bag, the (4)………….. …………………… dissolves in the(5) ……………….endothermically to provide instant coldness. (ii)

A reusable heat pack:

4



…………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… B HEAT OF PRECIPITATION Learning Outcomes You should be able to: • State what heat of reaction is, • State what heat of precipitation is, • Determine the heat of precipitation for a reaction, • Construct an energy level diagram for a precipitation reaction, • Solve numerical problems related to heat of precipitation

HEAT OF REACTION

HEAT OF REACTION The change in the amount of heat in a chemical reaction Symbol : ΔH The

HEAT OF PRECIPITATION

HEAT OF DISPLACEMENT

HEAT OF NEUTRALIZATION

HEAT OF COMBUSTION

Activity 4: Match the chemical equation with the type of reaction below :

Chemical equation HNO3 + KOH  KNO3 + H2O C2H5OH + 3O2  2CO2 + 3H2O Mg + CuSO 4  MgSO4 + Cu Pb(NO3)2 + 2KI  PbI2 + 2KNO3

Type of reaction Precipitation Displacement Neutralization Combustion

The heat change in a reaction can be calculated using the formula , H = mcӨ m = mass of solution, g c = specific heat capacity of the solution, J g -1 oC-1 Ө = temperature change , 0C Assumption : i) The solution is dilute, it has the same density as water , 1 g cm -3

( 1 cm3 = 1 g ) ii) The solution has the same specific heat capacity as water, 4.2 J g -1 oC-1 Example :

Calculate the amount of heat change when the temperature of 200 cm3 of water is raised from 28 0C to 400C, Q = mcӨ = 200 g x 4.2 J g-1 oC-1 x 1250C = 10080 J = 10.08 kJ

m = 200 g c = 4.2 J g-1 oC-1 Ө = 40 - 28 = 12 0C



Guidelines for the calculation of the heat of reaction

Steps to follow: Step 1 - Calculate the heat change using the formula, H = mcӨ ( p Joule ) Step 2 - Write chemical equation or ionic equation for the reaction that occurs Step 3 - Calculate the number of moles of reactant that reacts using either the following

formulae :

Number of moles

=

mass = q mol molar mass

Number of moles = MV ( q mol) (Solution)

or

M : Molarity of the solution V : volume of the solution in dm 3

Step 4 - Link the number of moles of reactants ( step 3 ) with the heat change ( step 1 )

q mol of reactants react  heat lost/gain is p J ∴ 1 mol of reactants react  heat lost/gain is p J q -1 Heat of reaction , ΔH = +/- p kJ mol 1000 x q = +/- r kJ mol-1 Note : “ + “ is used for endothermic reaction and “-“ is used for exothermic reaction Step 5 - Draw the energy level diagram

Activity 5 : (a) What is meant by heat of precipitation ?

………………………………………………………………………………………………………….. ………………………………………………………………………………………………………….. (b) To determine the heat of precipitation of silver chloride, AgCl

25 cm3 0.5 mol dm-3 sodium chloride solution 25 cm3 0.5 mol dm-3 silver nitrate solution

Procedure : 1. Measure 25 cm 3 0.5 mol dm-3 silver nitrate solution and pour it into the polystyrene cup, 2. Put the thermometer into the silver nitrate solution. Record the initial temperature, 3. Measure 25 cm 3 0.5 mol dm-3 sodium chloride solution and record the initial temperature, 4. Pour the sodium chloride solution quickly into the silver nitrate solution in the polystyrene cup. 5. Stir the solution mixture with the thermometer and record the highest temperature achieved. 6



Result : Initial temperature of silver nitrate solution = 28.5 0C Initial temperature of sodium chloride solution = 29.5 0C Highest temperature of the mixture = 32.0 0C

Average initial temperature = 29.0 0C

Calculation Step 1 :

Temperature change, Ө = 32.0 0C - 29.0 0C = 3.0 0C

Calculate the heat change using the formula H = mcӨ

Heat change , H = mcӨ

m = ( 25 + 25 ) g = 50 g c = 4.2 J g-1 oC-1 Ө = 3.0 0C H =(1) ……….........J

Step 2:

Chemical equation : AgNO 3 (aq) + NaCl(aq)  AgCl (s) + NaNO 3 (aq) Precipitate Write balanced chemical equation or ionic equation for Ionic equation : (2)………………………………………………………. the reaction that Deduce the mole ratio from the ionic equation : occurs (3)…………mol silver ion, Ag + react with (4) ……… mol of chloride ion, Cl- to produce (5)……………. mol of silver chloride , AgCl. Step 3 :

Number of moles of silver ion = the number of moles of silver nitrate

Calculate the number of moles of = (6) …………………….. mol reactant that Number of moles of chloride ion = the number of moles of sodium reacts, chloride Use the formula : n = MV ( V in dm3 )

Step 4

= (7)…………………….. mol Number of moles of silver chloride formed = (8) ………………….. mol When (9) ….. mol of silver chloride formed, heat released is (10) .……..

Calculate the heat of precipitation of silver chloride, ΔH

When 1 mol of silver chloride formed, heat released is (11)......................

Step 5

Draw the energy level diagram for the reaction that occurs in this experiment (13)

Energy level diagram

(c)

Heat of precipitation of silver chloride, ΔH = (12) - ……………kJ mol – 1

Calculate the heat change when 200 cm 3 of 0.5 mol dm -3 calcium chloride, CaCl 2 solution is added to 200 cm 3 of 0.5 mol dm-3 sodium carbonate, Na 2CO3 solution if the heat of 7



precipitation of calcium carbonate, CaCO 3 is +12.6 kJ mol-1 [ Specific heat capacity of solution : 4.2 J g -1 0C-1 . Density of solution : 1 g cm -3 ]

(d )

The thermochemical equation for the precipitation of silver chloride is as follows : Ag+ (aq) + Cl- (aq)



ΔH = –65.5 kJ mol-1

AgCl

Calculate the temperature change when 100 cm 3 of 0.5 mol dm -3 silver nitrate, AgNO 3 , solution is added to 100 cm3 of 0.5 mol dm-3 potassium chloride, KCl solution

C HEAT OF DISPLACEMENT Learning Outcomes, You should be able to: • State what heat of displacement is, • Determine the heat of displacement, • Construct an energy level diagram for a displacement reaction,

• Activity 6 (a) What is meant by the heat of displacement ?

………………………………………………………………………………………………………….. ………………………………………………………………………………………………………….. (b) To determine the heat of displacement of copper from a copper (ll) sulphate solution by

zinc.

zinc powder 25 cm3 0.2 mol dm-3 copper(ll) sulphate solution

Procedure : 1. Measure 25 cm 3 0.2 mol dm -3 copper(ll) sulphate solution and pour into a polystyrene cup. 2. Put the thermometer into the copper(ll) sulphate solution. Record the initial temperature, 3. Add half a spatula of zinc powder (in excess) quickly into copper(ll) sulphate solution. 5. Stir the mixture with the thermometer and record the highest temperature achieved. Result : Initial temperature of copper(II) sulphate solution

= 30.0 0C 8



Highest temperature of the mixture

= 40.0 0C


Calculate the heat change using the formula H = mcӨ Step 2:

Write balanced chemical equation or ionic equation for the reaction that occurs

Changes of temperature, Ө = 40.0 0C - 30.0 0C = 10.0 0C Heat change , H = mcӨ

m = 25 g c = 4.2 J g-1 oC-1 Ө = 10.0 0C H = (1)…………………………….. J Chemical equation : Zn (s) + CuSO 4(aq) Deduce the mole ratio from the equation :



Cu (s) + ZnSO 4 (aq)

copper displaced

(2)……… mol copper metal, Cu is displaced from (3) …………….. mol of copper(ll) sulphate solution, CuSO 4 by zinc

Step 3 :

Calculate the number of moles of reactant that reacts

Number of moles copper(ll) sulphate = (4)…………………….. mol

Step 4

When (6) ….. mol of copper is displaced, the heat released is (7) .…….. ∴ When 1 mol of copper is displaced, the heat released is (8).................

Calculate the heat of displacement of copper, ΔH Step 5


(c)

Number of moles of copper = (5)…………………….. mol

Heat of reaction , ΔH = (9) ……………… kJ mol – 1 Draw the energy level diagram for the reaction that occurs in this experiment (10)

Figure shows an experiment carried out to determine the heat of displacement of silver from silver nitrate by copper metal.

Based on figure above, calculate the heat of displacement for the reaction.

9



In an experiment, excess magnesium powder is added to 50 cm 3 of 0.25 mol dm -3 iron(ll) sulphate solution at 29.0 0C. The thermochemical equation is shown below, Mg(s) + Fe2+ (aq) (d )



Mg2+ (aq) + Fe (s)

ΔH = -80.6 kJ mol - 1

What is the highest temperature reached in this experiment ?

D HEAT OF NEUTRALIZATION Learning Outcomes : You should be able to:  State what heat of neutralization is,  Determine the heat of neutralization,  Construct energy level diagrams for various types of neutralization reactions, 

Compare the heat of neutralization for the reactions between a strong acid and a strong



alkali with the heat of neutralization for a reaction between a weak acid and a weak alkali,



Explain the difference of the heat of neutralization for a strong acid and a strong alkali with heat of neutralization for a reaction involving a weak acid and/or a weak alkali.

Activity 7 (a) What is meant by the heat of neutralization

………………………………………………………………………………………………………….. …………………………………………………………………………………………………………… (b) To determine the heat of neutralisation between a strong acid ( hydrochloric acid ) and a strong alkali ( sodium hydroxide ) 50 cm3 2.0 mol dm-3 sodium hydroxide solution

50 cm3 2.0 mol dm-3 hydrochloric acid solution

Procedure : 1. Measure 50 cm 3 2.0 mol dm-3 sodium hydroxide solution and pour it into the polystyrene cup, 2. Put the thermometer into the sodium hydroxide solution. Record the initial temperature, 3. Measure 50 cm 3 2.0 mol dm-3 hydrochloric acid solution and record the initial temperature, 4. Pour the hydrochloric acid solution quickly into the sodium hydroxide solution in the polystyrene cup. 5. Stir the mixture with the thermometer and record the highest temperature achieved. Result : Initial temperature of sodium hydroxide solution = 29.0 0C Average initial temperature = 29.0 0C Initial temperature of hydrochloric acid solution = 29.0 0C Highest temperature of the mixture = 42.0 0C

10




Calculate the heat change using the formula H = mcӨ

Changes of temperature, Ө = 42.0 0C - 29.0 0C = (1) ………………. 0C Heat change , H = mcӨ

Step 2:

m = ( 50 + 50 ) g = 100 g c = 4.2 J g-1 oC-1 Ө = (2) ……………….0C H = (3)…………………………….. J Chemical equation (4) ………………………………………………………….

Step 3 :

to produce (8)……………. mol of water , H 2O Number of moles of H + = the number of moles of hydrochloric acid

Write balanced chemical equation Ionic equation (5) ………………………………………………………. or ionic equation for the reaction that Deduce the mole ratio from the ionic equation : occurs (6)……mol hydrogen ion, H + react with (7)……mol of hydroxide ion, OH -

Calculate the number of moles of reactant that reacts, Use , n = MV ( V in dm3 )

Step 4

Calculate the heat of neutralisation of hydrochloric acid and sodium hydroxide, ΔH Step 5 : Energy

level diagram

(c)

= (9)…………………….. mol Number of moles of OH - = the number of moles of sodium hydroxide = (10)…………………….. mol Number of moles of water formed = (11) ………………….. mol When (12) ….. mol of water formed, heat released is (13) .…….. ∴

When 1 mol of water formed, heat released is (14) .................

Heat of neutralisation , ΔH = (15) ……………… kJ mol - 1 Draw the energy level diagram for the reaction that occurs in this experiment (16)

The thermochemical equation for the reaction between ethanoic acid and sodium hydroxide is given below, CH3COOH (aq) + NaOH (aq)



NaCH3COO (aq) + H2O (l)

ΔH = -55 kJ mol - 1

Calculate the heat given out when 200 cm 3 of ethanoic acid 0.5 mol dm -3 is added to 200 cm3 of sodium hydroxide 0.5 mol dm -3

11



The energy level diagram of a neutralization reaction is shown in figure below (d)

Energy H2SO4 + 2NaOH

ΔH = - 114 kJ Na2SO4 + 2H2O 3 -3 When 100 cm of 1.0 mol dm sulphuric acid is added to 100 cm 3 of 1.0 mol dm-3 sodium hydroxide solution. What is the change in temperature?

E : HEAT OF COMBUSTION Learning Outcomes You should be able to : • state what heat of combustion is, • determine heat of combustion for a reaction, • construct an energy level diagram for a combustion reaction, • compare the heats of combustion of various alcohols, • state what fuel value is, • describe the difference between heats of combustion of various alcohols, • describe the applications of fuel value, • compare and contrast fuel values for various fuels, • solve numerical problems related to heat of combustion.

Activity 8 : (a)

What is meant by the heat of combustion ……………………………………………………………………………………………………….

(b) To determine the heat of combustion of ethanol

Experiment to determine the heat of combustion. Procedure : 1. Measure 200 cm 3 of water and pour it into a copper can. Record the initial temperature of the water and place the copper can on a tripod stand. 2. Fill a lamp with ethanol and weigh it. Record the mass of the lamp together with its content. 3. Light up the wick of the lamp immediately. Stir the water continuously until the temperature 12



of the water increases by about 30 0C. 5. Put off the flame and record the highest temperature reached by the water 6. Weigh again the lamp and record the mass of the lamp. Result :

Mass of lamp and ethanol before burning Mass of lamp and ethanol after burning Initial temperature of water Highest temperature of water

= 190.55 g = 189.80 g = 29.0 0C = 59.00C


Calculate the heat change/heat absorbed by the water using the formula, H = mcӨ

Changes of temperature, Ө = 59.00C - 29.00C = 30.0 0C Mass of water, m = 200 g Heat change/ heat absorbed by water , H = mcӨ c = 4.2 J g-1 oC-1 Ө = 30.0 0C H = (1)…………………………….. J

Step 2:

Write a balanced chemical equation for the combustion of ethanol Step 3 :

Calculate the number of moles of ethanol that is used in the experiment, [ Relative atomic mass : H, 1 : C,12; O,16 ] Step 4

Calculate the heat of combustion of ethanol, ΔH

C2H5OH + (2)……………= (3)……………+ (4) ……………………….

Mass of ethanol burnt/used = (5) ………………. g Number of moles of ethanol burnt = (6) __________ molar mass of ethanol = …………………….. mol

When(7) ….. mol of ethanol is burnt in oxygen heat released is(8) .…… ∴

When 1 mol of ethanol is burnt in oxygen, heat released is (9) ..........

Heat of combustion of ethanol , ΔH = (10) ……………… kJ mol Step 5


-1

Draw an energy level diagram for the combustion of ethanol in this experiment (11)

(c) The following results are obtained by a student in an experiment to determine the heat of

combustion of alcohols.

Complete the table and calculate the heat of combustion of methanol, propanol and butanol based on the data given in the table below . [ Relative atomic mass : H,1 ; C,12 ; O,16 . Specific heat capacity of water , 4.2 J g -1 0C-1 ] 13



Alcohol

Methanol

Propanol

Butanol

Volume of water/cm3

200

200

200

Initial temperature of water / 0 C Final temperature of water / 0 C Mass of alcohol burnt/g Changes in temperature / 0 C Molecular formula

28.0

28.0

28.0

60.0

58.0

59.0

1.184

0.750

0.719

(1)

(2)

(3)

(4)

(5)

(6)

Number of carbon atoms Molar mass

(7)

(8)

(9)

(10)

(11)

(12)

Number of moles of alcohol burnt

(13)

(14)

(15)

Heat change/ absorbed by the water / J Heat of combustion of alcohol / kJ mol -1

(16)

(17)

(18)

(19)

(20)

(21)

(i) State the relationship between the number of carbon atoms in an alcohol and the heat of combustion, ………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………. (ii) Predict the heat of combustion for pentanol ……………………………………………………………………………………………………………….

(d )

(e)

When 1 mole of butanol, C 4H9OH is burnt in excess of oxygen, 2600 kJ of heat is produced. Calculate the mass of butanol needed to burn completely in oxygen in order to raise the temperature of 500 cm 3 of water by 300C ( Relative atomic mass : H , 1 ; C , 12 ; O , 16, Specific heat capacity of water , 4.2 J g -1 0C-1)

The fuel value of a fuel is the amount of heat energy given out when one gram of the fuel is completely burnt in excess of oxygen. Fuel value ( kJ g -1 ) = heat of combustion of alcohol ( kJ mol -1 ) 14molar mass



The heat of combustion of propanol, C 3H7OH is -2016 kJ mol -1 What is its fuel value ? Solution : (i) Calculate the molar mass of propanol, C 3H7OH.

(ii) Calculate the fuel value of propanol, C 3H7OH.

ANSWER Activity 1

(a) (1) transferred , (2) increases , (b) (3) absorbs , (4) decreases Activity 2

15



(a) Identify the following reactions as exothermic or endothermic reaction (a) (b) (c) (d) (e) (f) (g) (h) (i) (j)

Reaction Combustion of ethanol Burning of magnesium Neutralisation between acid and alkali Adding water to concentrated sulphuric acid Photosynthesis Reaction between acid and magnesium Reaction between acid and calcium carbonate Dissolving ammonium salt in water Thermal decomposition of copper(ll) carbonate Thermal decomposition of zinc nitrate

Exothermic √ √ √ √

Endothermic

√ √ √ √ √ √

(b) (1) ΔH , (2) kJ, (3) negative sign , (4) higher , (5) positive sign (c) Two other examples of exothermic and endothermic reaction Exothermic reaction 1) Physical changes : gas  liquid

Endothermic reaction Physical changes : solid  liquid ,

liquid  solid 2) Reaction between reactive metals

liquid  gas Dissolving crystalline salts in water

(sodium and potassium) with water //

example : CuSO4.5H2O

Rusting of iron Activity 3

(a) (i) Mg + H2SO4 Energy



MgSO4 + H2

ΔH = -467 kJ

Mg(s)+H2SO4(aq) ΔH = -467 kJ MgSO4(aq) + H2(g)

(ii) CaCO3



CaO + CO2

ΔH = + 178 kJ

Energy CuO + CO2

ΔH = + 178 kJ CuCO3

(iii) (1) zinc , (2) copper(II) sulphate , (3) exothermic , (4) increases , (5) zinc , (6) copper(II) sulphate , (7) more , (8) 190 kJ , (9) zinc , (10) copper(II) sulphate, (11) one , (12) one , (13) 190 , (14) produced // released 16



(15) hydrogen gas, (16) iodine gas, (17) endothermic , (18) decreases, (19) hydrogen gas (20) iodine gas, (21) less , (22) 53 , (23) hydrogen gas , (24) iodine gas , (25) two , (26) 53 (27) absorbed. (b) (i) (1) bond(2) breaking , (3) bond , (4) formation (ii) (1) requires / absorbs (2) releases (iii) (1) releasing , (2) exothermic (iv) (1) absorbing , (2) endothermic (c) (i) (1) sports injuries , (2) water , (3) solid ammonium nitrate , (4) ammonium nitrate , (5) water (ii) Uses a sodium acetate crystallization . By bending the metal disc in the bag, the sodium acetate crystallizes and gives off heat. Activity 4

Match the chemical equation with the type of reaction below : Chemical equation HNO3 + KOH  KNO3 + H2O C2H5OH + 3O2  2CO2 + 3H2O Mg + CuSO 4  MgSO4 + Cu Pb(NO3)2 + 2KI  PbI2 + 2KNO3

Type of reaction Precipitation reaction Displacement reaction Neutralization Combustion

Activity 5

(a). The heat of precipitation is the heat change when one mole of a precipitate is formed from their ions in aqueous solution. (b) (1) 630 J , (2) Ag+ + Cl-  AgCl (3) one , (4) one , (5) one , (6) 0.0125 , (7) 0.0125 , (8) 0.0125 , (9) 0.0125 , (10) 630 J , (11) 630 = 50400 J = 50.4 kJ 0.0125 -1 (12) - 50.4 kJ mol (13) Energy level diagram Energy Ag+ + Cl-

(c) 1.26 kJ // 1260 J

ΔH = - 50.4 kJ mol -1 AgCl

(d) 3.9 0C Activity 6

(a) The heat of displacement is the heat change when one mole of a metal is displaced from its salt solution by a more electropositive metal. (b) (1) 1050 J , (2) one , (3) one , (4) 0.2 x 25 = 0.005, (5) 0.005 mol , (6) 0.005 1000 Energy (7) 1050 J , (8) 1050 = 210000J = 210 kJ , (9) - 210 kJ mol -1 0.005 Zn + CuSO4 (10) Energy level diagram ΔH = - 210 kJ mol -1 17

Cu + ZnSO4



(c) ΔH = - 16. 8 kJ mol -1 (d) 33.8 0C Activity 7

(a) The heat of neutralization is the heat change when one mole of water is formed from the reaction between an acid and an alkali. (b) (1) 13.0 , (2) 13.0 , (3) 5460 J , (4) HCl + NaOH  NaCl + H2O , (5) H+ + OH-  H2O (6) 1 , (7) 1 , (8) 1 , (9) 0.1 mol , (10) 0.1 mol , (11) 0.1 mol , (12) 0.1 mol , (13) 5460 J, (14) 5460 = 54600J = 54.6 kJ , (15) - 54.6 kJ mol -1 0.1 (16) Energy level diagram Energy HCl + NaOH

ΔH = - 54.6 kJ mol -1 NaCl + H2O

(c) 5.5 kJ (d) 6.8 0 C Activity 8

(a) The heat of combustion is the heat change when one mole of a substance is completely burnt in oxygen under standard conditions. (b) (1) 25200 J, (2) 3O2 , (3) 2CO2 , (4) 3H2O , (5) 0.75 , (6) 0.75 = 0.0163 mol , (7) 0.0163 46 (8) 25200 J, (9) 25200 = 1546012.2 J = 1546 kJ , 0.0163 (10) Energy level diagram Energy C2H5OH +3O2

ΔH = - 1546 kJ mol -1 2CO2 + 3H2O 18



(c) (1) 32.0 , (2) 30.0 , (3) 31.0 , (4) CH3OH , (5) C3H7OH , (6) C4H9OH , (7) 1 , (8) 3 , (9) 4 (10) 32 , (11) 60 , (12) 74 , (13) 0.037 , (14) 0.0125 , (15) 0.0097 , (16) 26880 J (17) 25200 J , (18) 26040 J , (19) 726.5 kJ , (20) 2016 kJ , (21) 2684.5 kJ (i) The more carbon atoms in the molecular formula of an alcohol, the higher is the heat of combustion // The heat of combustion is proportional to the number of carbon atom per molecule in alcohol , (ii) 3300 kJ mol -1 to 3400 kJ mol -1 (d) 1.793 g (e) (i) 60g mol-1 (ii) 33.6 kJ g-1

19

WAJA CHP.4 Thermochemistry

Recommend Documents