ThermoDynamics Lecture Notes

CHEMISTRY LECTURE NOTES

COURSE - VIKAAS (A) (LECTURE No. 1 TO 13)

TOPIC : THERMODYNAMICS

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LECTURE#1 THERMODYNAMICS

Thermochemistry (application of I law in chemical reactions)

First law of Thermodynamics

Second law of Thermodynamics

Thermodynamics : The branch of science which deals with different forms of energy & their interconversion. In chemistry using Thermodynamics. (a) (b) (c)

(1) (2)

Q.1

Ans.

 

We can predict that if two substances are mixed then the reaction between them will takes place or not.(feasibility of the reaction) If Reaction does take place then what are the energy changes involved during the reaction. If in a chemical reaction, equilibrium is going to get attained then what will be the equilibrium concentrations of different reactants & products, can be calculated with thermodynamics. Limitations of Thermodynamics : Laws of thermodynamics are applicable to matter in bulk or on system as a whole, these can not be applied on individual particles(temperature, pressure, enthalpy etc have meanings only for system as a whole). Using thermodynamics we cannot calculate the time taken for completion of a reaction or for attainment of chemical equilibrium. A+B

C+ D

K1, eq = 103

E+F

G+H

K2, eq = 10–3

Which of these reaction will attain equilibrium earlier ? We cannot predict, because value of equilibrium constant has no relation with time taken to attain equilibrium. More the equilibrium constant K, more will be the concentration of products at equilibrium state. Smaller the equilibrium constant K, lesser will be the concentration of products at equilibrium state.

Terms to be used in Thermodynamics : (1)

System, surroundings and boundary. (a) System : A specified part of the universe selected for the study of effect of certain properties such as P,T etc.

Ex. Air in a room, water in a bottle, any living body. (b) Surrounding : Rest of the universe. System + Surroundings = universe. (c) Boundary : Anything which separates system & surroundings is called boundary. Boundary can be real or imaginary. Boundary can be flexible or rigid(for example - air in a flexible balloon(flexible boundary) while air in a room (fixed boundary). Boundary can be adiabatic(non-conducting) or diathermic(conducting). (d) Types of system : (1) Open system : System which can exchange energy & matter both with the surroundings. Ex. Living systems( any living organism) are open systems, air in an open room. (2) Closed system : System which can exchange only energy but cannot exchange matter with the surrounding is called closed system. Ex. any matter in a closed container.

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(3)

Isolated system : System which cannot exchange energy and matter both with the surroundings. Ex. (1) Water in thermos flask.(Though not a perfectly isolated system but can be taken as, for small interval of time as the energy exchanges are negligible). (2) Whole of universe is a perfect isolated system.

Q.

Human body is an example of : (A*) open system (B) closed system (C) isolated system (D) none of thse Human body can exchange energy & matter with surroundings. This represents open system.

Sol. (2)

State of a system : It means the condition in which the system is present and it can be specified/defined by measuring/ specifying some observable/measurable properties of the system. Ex. For an ideal gaseous system state of the system can be defined by specifying volume, temperature and pressure. For other systems we may have to specify more properties of the system depending on the complexity of the system.

(3)

State functions & path functions : State function : Any property of the system which is dependent only on the state of the system and not on the way that state has been achieved. Ex. ( In Mechanics, Displacement of any object will a state function but distance travelled by the object will be a path function.). for any thermodynamic system, Temperature, Pressure, Volume, Total internal energy (E or U), Enthapy(H), Gibbs free energy (G), Entropy (S) are all state functions.

Ex.



(4)

In the above example the final temperature, pressure, and the volume will be same in both the above ways but the work involved and the heat exchanged during the processes will be different. Path function : Quantities which are dependent on the path/way the system has achieved a particular state. e.g. Heat, work, Heat capacities(Molar heat capacities, specific heat capacities etc.). State functions have a fixed value (measurable or unmeasurable) in a particular state of the system while the path functions have values or are defined only when there is some process or when system changes its state from one thermodynamic equilibrium state to another, these can not have any particular value in any particular state of the system. Extensic & intensive functions : Extensive functions : Functions or properties of the system which are dependent on mass or on size of the system. If a system in a particular state is divided into two equal or unequal parts, the properties which have value equal to the original value of that property for the whole of the system is called an Intensive property. While the properties which have values different from the values for whole of the system are called Extensive Properties. For example consider air in a room at temp of 300K, 1 atm pressure. Now if the room is divided by some boundary( imaginary or real) into two parts( equal or unequal) then in these two parts the temperature, pressure, density of the gas, concentration of gaseous molecules etc. will have the same value as that of for whole of the system. While the volume of two parts, mass of gas in two parts, total energy of the gaseous molecules in the two parts, entropy the two parts etc. will be different from the values of these properties as for the whole of the system.

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Ex. Volume, Mass, Total heat capacity, Total internal energy (E), Enthalpy(H), Gibbs Free Energy(G), Entropy(S).



Extensive functions are additive in nature( The addition of the volumes of the two parts equals the volume of the whole of the room.) Intensive functions : Functions or properties which are not mass dependent or size dependent are called intensive function. Ex. Temperature, pressure, molar heat capacity, specific heat capacity, density,concentration, vapour pressure. Intensive properties are not additive in nature.

(5)

Thermodynamic equilibrium : When there is no change in any observable or measurable property of a system with time then the system is said to be in thermodynamic equilibrium.

Thermodynamic equilibrium consist of three types of equilibrium. (a) Mechanical equilibrium (b) Thermal equilibrium (c) Chemical equilirbrium (a) Mechanical equilibrium : There should not be any pressure gradient (with time or with space(for any ideal gaseous system, for a liquid system there can be pressure gradient with space as pressure at the bottom of the container in which a liquid is filled will be greater than the pressure at the surface of the liquid.)) in the system.

(b) Thermal equilibrium : There should not be any temperature gradient.

Temperature may have different values at different places/locations in a system but it should remain constant with time (c) Chemical equilibrium : There should not be any concentration gradient of any of the species in the system. (6)

Thermodynamic process : Any method/process by which system can change its state from one state of thermodynamic equilibrium to another state of thermodynamic equilibrium. There can be infinite type of thermodynamic processes, out of these the following are important ones: 1. Isothermal process : T = constant dT = 0 T = 0 2. Isochoric process : V = constant dV = 0 V = 0 3. Isobaric process : P = constant dP = 0 P = 0 4. Adiabatic process : q=0 or heat exchange with the surrounding = 0(zero)

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LECTURE#2 Classification of thermodynamics processes on the basis of the way these processes are carried out: 1. 2.

Reversible process Irreversible process/spontaneous process/Natural process. Reversible process : If a process is carried out in such a manner so that the system is always in thermodynamic equilibrium at every stage of the process. OR If the process is carried out such that the difference in driving force and opposing force is infinitesimally small so that process takes place at infinitesimally slow rate. Fdriving = Fopposing – dF and dF  0 An ideal reversible process will take infinite time to get completed. To get an idea of a reversible process we can consider the following system. An ideal gas is enclosed in a container and a massless piston is put on the gas on which a pile of sand is placed having particles of negligible mass. To carry out a reversible expansion we will slowing decrease the mass of the sand lets say by removing the particles one by one, so the expansion of the gas will take place at infinitesimally small rate and we can always assume the system to in thermodynamic equilibrium. So the expansion will be of reversible type.

FBD of piston

for piston to be in equilibrium : Pgas = Patm + Mg/A Irreversible process : If a process is carried out in such a manner so that the system is in thermodynamic equilibrium only at initial & final state of the process but not at the intermediate stages.(or system should be in thermodynamic equilibrium state at some finite number of intermediate stages only - for example - n step irreversible expansion of a gas) OR If during the process there is a finite difference in driving force and opposing force so that process takes place with a finite rate. Irreversible processes will get completed in finite time. In case of any irreversible process at intermediate stages of the process the different state function such as Pressure, temperature etc. are not defined. Consider the above system. If the stopper placed over the piston is removed then the piston will move with almost infinite acceleration and will keep moving to a position where the pressure of the gas becomes equal to the external pressure. Since the process will get completed in finite time and there was a finite difference between the driving force and the opposing force so process is irreversible.During the process the pressure of the gas can not be defined as it will be having different values at different locations.

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(7)

Heat & work : These are two ways by which a system can interact or can exchange energy with its surroundings.  Heat & work both are forms of energy. Heat : When the energy transfer both system & surroundings takes place because of temperature difference. Work : Any energy transfer which is not heat or which is not because of temperature difference is called work.Work can be of many types such as Mechanical work, Electrical work, Magnetic work, Gravitational work etc. The same energy transfer can be called work or can also be called heat depending on choice of the system. To understand this consider a system shown below in which water is taken in a closed container at 25ºC, the surroundings is also at temperature of 25ºC and there is a heater coil in the dipped in the water which is connected to a battery through a switch S.Heater coil is also at 25ºC initially.

Now there are two ways in which system can be chosen –System : All contents of the container (water + Heater coil). When switch is turned on there will be increment in the temperature of the system. Since the temperature of the surroundings was equal to temperature of the system so heat can not flow but still there is increment in the energy of the system and hence there is temperature increment. This must be because of electrical work done by the battery on the system not because of the heat transfer as initially temperatures were equal. –System : Water only is our system. Heater coil will be part of the surroundings. In this case when switch is turned on the temperature of the heater coil will increase first so there will be a temperature difference between system & surroundings. Hence this energy transfer will be called heat. 1.

IUPAC Sign convention about Heat and Work Any energy given to system is taken positive so heat given to system = positive heat given out from system / taken out from system = Negative Work done on the system = Positive Work done by the system = Negative The sign convention is different from physics, but the meaning always comes out to be same only in equation we have to use a different sign convention for work. Do not get panicked - take it lightly it is not a big difference. You can easily absorb it. So if in any problem, w = – 103 It means system has done work of 10 J on surroundings.

1st Law of Thermodynamics Law of energy conservation : -Energy of total universe is always conserved. oR Total energy of an isolated system is always conserved. OR To develop a mathematical expression for 1st Law of thermodynamics first we have to define a from called total internal energy of system (E or U symbol) E - summation of all type of energies. E = Translational KE + Rotational KE + Vibrational KE + electrostatic PE + nuclear PE + bonding energy + ............................ (energy term of undiscovered forces) Hence absolute value of E can never be calculated only change in value of E can be calculated for a particular process.

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Mathematical form of 1st Law of thermodynamics. If a system is intially in a particular state in which its total internal energy is E1. Now q amount of heat is given to it and w amount of work is done on it so that in new state its total internal energy becomes E2. Then according to 1st Law of thermodynamics. E2 = E1 + q + w so E = (E2 – E1) = q + w Q. Sol.

A system does 100 J work on surroundings by absorbing 150 J of heat. Calculate the change in internal energy. w = – 100 J, q = 150 J So, E = q + w = 150 – 100 = 50 J Calculation of different quantities in equation of 1st Law of thermodynamics. E = Calculation for an ideal gas. For an ideal gas E is a function of temperature only - Also known as Thermodynamic definition of an ideal gas  E   E   =0,   =0 so   V  T  P  T Further, with change in temperature only the kinetic energy get change, not any of the potential energy, So E = (Translational KE) +  (Rotational KE) +  (Vibrational KE) + 0 + 0 + 0 .....................

Now according to law of equipartition of energy, each degree of freedom contributes

1 RT (per mole) 2

Translational degree of freedom = 3 for all type of gases. Rotational degree of freedom =0 – monoatomic gases. =2 – diatomic or linear polyatomic gases. =3 – non-linear polyatomic gases. Law of equipartition of energy : *

Energy equal to

1 KT is associated with each degree of freedom per ideal gas molecule 2

U/molecule = f x

1 KT 2

 U/mole =

f x KT × NA 2

 U/mole =

f RT 2

 For n moles = U=

f nRT 2

f nRT only for ideal gas. 2

Vibrational energy changes for Law Temperature increments can be neglected. so E =  (Tranlational KE) +  (Rotational KE) 3   fR   nRT  =    RT  +   2   2 

=

f nR (T) 2

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f nR (T) 2 where of - initial degrees of freedom for that gas. f=3 for monoatomic =5 for diatomic or linear polyatmic =6 for non - linear polyatmic E =

Calculation of heat (q) heat is a path function and is generally calculated indirectly using Ist law of thermodynamics first of all calculate E and W & then q. Heat can be caluclated if heat capacity of any process is given to us. Heat capacity is a path function and different type of heat capacities are defined [Remember heat capacity of a substance is not fixed it is dependent on type of process which is being performed on that substance] Total heat capacity – Heat required to raise the temperature of system by 1ºC under the given process. q dq  = J/ºC T dT It will be extensive path function So dq = cdT

CT =

&q=

 cdT

Molar heat capacity – heat required to raise temperature of 1 mole of a subtance by 1ºC q dq  = J mole–1 K–1 nT ndT So. dq = ncdT

C=

q=

 ncdT  ncT

for an ideal gas CP and CV are very much used. Specific heat capacity (s) : heat required to raise temperature of unit mass (generally 1 gm) of a substance by 1ºC. q dq  = Jgm–1 K–1 m T mdT so dq = msdT

S=

q=

 dq  ms dT msT

We can define specific heat capacity at constant pressure and at constant volume.(sP & sV ) Total heat capacity, molar heat capacity & specific heat capacity of a process on a substance are related as CT = nC = ms & C = Ms Where m – mass of substance M – molar mass of substance n – no. of moles of the substance For For For For Q. Sol.

Note:

isothermal process isobaric process isochoric process adiabatic process

C=  C = Cp C = Cv C=0

During an expansion of ideal gas the work done by gas is 100 J and the heat capacity of process is found to be + 2 J/ºC final E of gas if the final temperature of gas is 25ºC higher than its initial temperature. T = 25 w = –100 , q = CT = 2 × 25 = 50 E = q + w = 50 + (–100) = – 50 J Hence we can see that heat capacity (C) is a path function, but remember Cp and CV are not path functions these are constants. Heat capacity can have volue from –  to +  as depending on process.

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LECTURE#3 WORK DONE (w) : Energy that is transmitted from one system to another in such a way that difference of temperature is not directly involved. This definition is consistent with our understanding of work as dw = Fdx. The force F can arise from electrical, magnetic, gravitational & other sources . It is a path function . Note : You will frequently encounter (litre. atmosphere) term which has units of energy . It is useful to remember the conversion : 1 litre. atm = 101.3 Joules = 24.206 cal Units : Heat & work both are forms of energy . Hence, their units are units of energy. i.e.S system: Joules (J). Much data is available in the old units of calories (cal) as well. Hence it is also used sometimes. For irreversible processes, state parameters such as P,T etc cannot be defined. Hence work cannot be estimated using Pgas. But by the work energy theorem W gas = – W ext + Kpisston When the piston comes to rest again Kpiston = 0

 W gas = – W ext = –

P

ext dv

as the external pressure is always defined hence for all processes work can be calculated using W ext = – Pext .dv



Work : dW = Fext dx P-V Work : Assuming that under an external force Fx, the piston moves down by a distance ‘dx’.

Fext dx

Fext ( A dx )  dW = Pext (dV) A As the work done in the above case by the external agent is +ve and as the expression conveys otherwise, hence a -ve sign is introduced. dW = – Pext. dV  dW 

Q. Sol.

A gas abosrbs 200 J of heat and expands against the external pressure of 1.5 atim from a volume of 0.5 Lt to 1.0 Lt. Calculate the change in internal energy. w = – PV = – 1.5 (1–0.5) = 0.75 Lt atm. = – 0.75 × 101.3 J = – 75.975 J Now E = 200 – 75.975 = 124.025 J.

Calculation of work for different type of process on an ideal gas. (a) Isothermal Reversible expansion of an ideal gas. Let initial pressure of gas be Pi and initial mass of sand placed to keep equilibrium be m0, so  m0g   pi = patm +   A  A = Area of cross section of pistion

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Now af some intermediate stage of expansion let the volume has become V and mass of remaining sand be m, then as process is reversible, so there is equailibrium (always). so

mg = pgas = p A

pext = patm +

nRT V where ‘n’ is mo. of moles of gas enclosed and T = temperature of gas = temp. of constant temp. both Now if volume of system is increased by dv amount by displacing a small mass of sand from above the pison so Pext – Pgas = dp  0 then work done in this slight expansion dw = – Pext. dv = = – Pgas . dv (with IUPAC sing conversion) To obtain the total work done. We have to do the integration so As process is isothermal so P =

Vf

W=



dw   Pext dv = –



Vi

Vf



Vf

Pgas dv = –

Vi



Vf

P dv so

W =–

Vi



Vi

nRT . dv V

W = – nRT ln (Vf/Vi) Pi Pressure

dw

P

P –dp)

This process can also be repressnted in a PV digram as follows Patm = Pf Vi

V(v+dv)

Vf

Here we have assumed pf = Patm need not become for a given problem Total work done is area under the pv diagram only in case of reversible process the pressure of gas at intemediate stages is difined and hence the process can be repressnted in pv diagram (b) Irreversible isothermal expansion of a gas There are many ways in which a gas can be expanded inersible and siothermally. (i) Single step irrersible isothermal expansion here we are assuming expansion against atmospheric pressure which need not be the case in a given problem. A mass equal to m0 is placed on piston initally to maintain equilibrium jnitially Pi – Pgas Patm + mog/A

for expanstion to take place, m0 mass is snddenly removed so gas expands against constant external pressure of Patm

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In this case the pressure of gas will not be defined as the sndden expansion has taken place so all the molecules of sampel will not get the infamation of expansion simultaneasly, there will be a time gap and hence there will be a state of turbulence. But just below the piston the pressure of gas will be equal to external pressure, Patm (using work energy thearm, or total force on a massless piston must be zero) so from some intemediate state of volume v the work done is slight expansion from V  (V + dv) dw = – Pext . dv

(IUPAC sign conrention) Vf

so

W=



dw   Pext .dv



Vi

W = – Pext (Vf – Vi)

PV digram Repressentation

Only initial and final states can be located (as at intemediate stages pressure of gas is not defined) work done by gas is less in this case in comparision to work done during reversible expansion (ii) Two step imersible isothermal expansion of mass m0 is divide into hw mass (may be equal or unequal)

M0  (m1  m 2 ) Now if m1 only is removed than the expansion of gas will take place against constant extermal pressure Pext = (Patm + M2g/A) and this expansion will take place only upto expansion will take place only upto volume V1 such that Pext. V1 = Pi Vi (isothermal) Now if second mass m2 is also removed then expansion Vi  Vf will take place against constant pressure Pext = Patm So

Patm M1

Vf

M0

V work done is expansion Vi  V1

Vi

W 1 = – (Patm +M2g/A) (Vi – V1) & work done is expansion V1  Vf W 2 = – Patm (Vf – Vi) Toral work = W 1 + W 2

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Pi

Patm + M 2g/A)

(

PV Diagram representation

Patm

W1

Vi

W2

V1 Volume

Vf

Note that m1 & m2 are symbolic only there need not be actual m1 & m2 just have to do expansions in hw steps. Again note that work done is this ineresible expansion is smaller than work done by gas during the rerersible expansion of gas. And so on for three step expansion we divide the mass m0 into three masses m1, m2 and m3 and remove these step by step and so on.

(iii) for n step expansion n  

Pressure

Pi

Vi

Vf

Volume

In the limit n   . This irrevesible process will become reversible one Conclusion : Whenever system/gas does work on. surroundings it will be maximum in case of reversible process. (c) Reversible isothermal compression of an ideal gas This can be achieved by placing particules of equal one by one at a very slow take in the assembly which keeps the temperature of gas constant in this case the expression of work done will be exacthy similar to as obtained in case of reversible expansion of gas W = – nRT ln(Vf/Vi) This will automatically come out to be +ve as Vf < Vi

(Patm + M0g/A) = Pf Patm

Vi Vi

Patm Vf

Vi

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(d) Irreresible isothermal compression of an ideal gas (i) single step compression : To compress gas irrersible a mass m0 is suddenly placed on massless piston dw = – Pext. dv = – (Patm + m0g/A) dv so. to calculate total work done on the gas W=

Vf

 dne  V

Pext. dv

i

W = – Pext = – (Vf – Vi)

Pressure

Pext

Vf Volume

Vi

Pressure

(ii) Two step compression : Place mass m0 in hw fragments (m1 + m2) the graphical representation will make the colmlation of work done clear

W2

W1

Patm = Pext Vf Volume

Vi

If m 1 is placed first then the first compression has taken taken place aganist external pressure of (Patm + m2g/A) so W 1 = – (Patm + M1g/A) (V1 – Vi) Simlarly, W 2 = – (Patm + M0g/A) (Vf – Vi) and n step compression in the limit n   will be like reversible compression Conclusion : Whenever work is done on the gas then it will be minimum in case of reversible process. Thats why different machines/engines are designd to work reversibly so maximum output can be obtained but minimum input is given to it output – work done by engine/machine/system input – work done by system us a sumoundings



Wherever expanstion/compression takes place against constant external pressure then the process is reversible.

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Q.3 Sol.

Calculate the work performed when 2 moles of hydrogen expand isothermally and reversibly at 25ºC form 15 to 50 litres. We have,

V2 50 W = – 2.303 n RT log V = – 2.303 × 2 × 2 × 298 × log = – 1436 calories. 15 1 Q.4 Sol

If a gas at a pressure of 10 atm at 300 k expands against a constant external pressure expands against a constant external pressure of 2 atm from a vol. of 10 litres to 20 litres find work done ? [Isothermal process] Process is irreversible 20

w  – 2dv  –[20 – 10] = – 20 J.



10

1 litre atm = 101.3 J (e) Reversible and irreversible isochoric processes. Since

dV = 0

So

dW = – Pext . dV = 0.

(f) Reversible isobaric process : Isobaric,

Pgas = constant

Since pressure of gas is defined during whole of the process and hence the process must be reversible So due = – Pext dV = – Pgas dV =–

Vf

V

i

dV

W = P (Vf – Vi)

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LECTURE#4 (g) Adiabatic reversible expansion –



dQ = 0 (no heat changed b/w) system and sorrods du = dQ + dW nCV dT = – PdV



 nc V dT   

nRT . dv V

T2

V2





T1



C v .dT T =–

V1

T2 V2 Cv ln T = – P ln V 1 1 T2 ln T = ln 1



R

 V . dT

 T2   T1

 V2     V1 

  V1        V2 

R / c v

 1

T2 V2 –1 = T1 V1 –1 or T V–1 = constant PV = constant This is only valid when The quantity Pv  or Tv –1 is contant only for a quasi static or reversible process.



For inreversible adeabatic process these equations are not applicable

Operation of adiabatic process (a)

Reversible Adiabatic

PAGE _ 15

Operation wise adiabatic process and isothermal process are similar hence all the criteria that is used for judging an isothermal irreversible processes are applicable to adiabatic process also volume in case of isothermal volume is more than that of adiabatic at contant pressure and no of moles, V  T w=–

 Pext. dv

w=– 

but Pext = Pint =

K VY

K

 V Y . dv – K V2  1  V1  1 / 1   





as K = P2V2 = P1V1 

=



P2 V2 .V2

1 

 P1V1 .V11   1

P2 V2  P1V1 = work done  1

(b) Irreversible Adiabatic (h) Adiabatic irreversible expansion – W=

P

ext.dv

= – Pext (V2 – V1) or 

 du   dw w = u w = nCv (T2 – T1) =

nR(T2  T1 ) P2 V2  P1V1 ) =  1  1

If the two states are linked by an adiabatic reversible and irreversible path then wrev. = urev. But as u is a state function  urev. = uirrev.  wirrev. = wrev. as work is a path function. If we assume that wirrev. = wrev.  It implies that urev.  uirrev.. which again is a contradiction as U is a state function. Two states A and B can never lie both on a reversible as will as irreversible adiabatic path. In other words if two states A and B are connected by a reversible path then they can neven be connected by an irreversible path. There lies only one unique adiabatic path linkage between two states A and B.

PAGE _ 16

Adiabatic Expansion using single stage & Infinite stage P2 V2 T2 reversible single stage wgas P1 V1 T1

P2 V2 T2 (reversible process wg as )10v P1,V1, T1

(W gas)rev > (W gas)irrev ,urev. > uirrev.  (T2)rev. < (T2) irrev  as the internal energy of the gas decrease to a larger extent in the case of reversible expansion there by temperatures 

Pirr > Preve at V2

If the same pressure gradient (P1 – P2) is creaded in a single step then the piston will come to rest at piont 2’ on its our. If piston in stopped at the volume V2 then the single stage expansion will be denoted by the point 2”. (As the decrease in tampreatrue will be lesser



Comparision of single stage Vs two stage expansion(adiabatic). If the expansion is carried out in two stages then work done in two stage by the gas > workdone in one stage by the gas.  Tf in stage D < Tf in single stage because decrease in internal energy in two stage is > decrease in internal energy in one stage. Hence two stage expansion will come to rest between the point 2 and 2” but nearly with point 2”.



Adiabatic compression using single-stage & infiete stage compression (Work done by surrounding)irrerverseble > (Work done by surroundings)reversible  (T2)irrev. > (T1)rev as internal energy would increase to a greater extent in irreversible path and therefore temp. would also increase more.

If the same pressure gradiant is created in single stage then the piston would come to rest at point 2’ on its own. If the piston is taken by single stage. Compression from volume V1 to V2 then the point will be 2”

PAGE _ 17

Adiabatic Irreversible process (calculation of state parameters) irrev State A   State B P1,V1,T1 P2,V2,T2

W=

P1V1 P2 V2 nR(T2  T1 ) P2 V2  P1V1  = – Pext. (V2 – V1) , T = T  1  1 1 2

(i) Free expansion – Always going to be irrerversible and since Pext = 0 so

due = – Pext . dV =0 If no. heat is supplied q = 0 then E = 0 Ex.

so

T = 0.

A sample of an ideal monoatomic gas initially at (1 L, 10 atm, 300 K) is expanded till its volume becomes 4 L (a) using reversible isothermal expansion (b) using by expanding against constant exteranl pressure of 1 atm isothermally. (c) using reversible adiabatic process (d) by expanding aganist constant external pressure of 1 atm adiabatically. (e) by expanding againest constant external pressure of 1 atm under a process in which final pressure of gas becomes 5 atm. Calculate work, heat and E in each of the above process.

LECTURE#5 st

Application of  Law U = Q + W U = Q –PV  (a)



W = –P V

Constant volume process du = dq + dw du = dq – pdv  du = (dq)v Heat given at constant volume = change in internal energy du = (dq)  du = (nCdT)v du = nCvdT Cv is Specific molar heat capacity at constant volume. CV =

( dq) 1 du 1 = x dT dT n n

Cv is defined as amount of heat given per degree rise is temperature per mole at constant volume, for an ideal gas 1 du Cv = . n dT

(b)



1 d ( fnRT ) / 2 f = R n dT 2

Constant pressure process : du = dQ + dW du = dQ . PdV  dQ = du PdV ............... (i) Defining a new thermodynamic function H  Enthalpy (state function and extensive property) H = u + Pv as dH = du + d (Pv) as P = constant dH = du + Pdv .............. (ii)

PAGE _ 18

from equation (i) & (ii)  dH = (dq)p only at constant pressure. Heat given at constant pressure = Change in enthalpy dH = (ncdT)p dH = nCp dT Heat capacity at constant pressure Cp = * * *

1 dH 1 (dq)p = n dT n dT

Cp is defined as the amount of heat given to increases the temperature of 1 mole of substance by 1º at constant pressure. This can always be used anywhere. Relation between Cp ad Cv for an ideal gas H = u + Pv  dH = du + d(Pv) for an ideal gas Pv = nRT d(Pv) = d(nRT) = nRdT nCp dT = nCvdT + nRdT  Cp – Cv = R only for ideal gas

Calculation of H, U, work, heat etc. Case - I

For an ideal gas undergoing a process. the formula to be used are du = nCvdT =

f nRdT 2

f  dH = nCpdT =   1 nRdT 2 

W=–

 Pext. dv

dH = dv + d (Pv) H = u + nR T du = dQ + dw Q.5

Sol.

Calculate the maximum work done when pressure on 10 g of hydrogen is reduced form 20 to 1 atm at a constant temperature of 273 K. The gas behaves ideally. will there be any change in internal energy? Also,calculate ‘q’. We have,

P1 W = – 2.303 nRT log P 2 n = number of moles of hydrogen =

wt. in grams 10 mol. wt. = 2 = 5 moles.

20 = – 8180 calories. 1 further, the change in state of the system is from a gas to a gas and therefore, at constant temperature, internal energy will not change i.e., E = 0. Again, q = E – W = 0 – (– 8180 ) = 8180 calories. Thus W = – 2.303 × 5 × 2 × 273 × log

PAGE _ 19

Q.

Calculate q, w. E & H for processes AB, BC & CD & for whole of the process (for a centain mass of a monoatomic gas) and also prove that only for isobaric process q = H and only for isochoric precess q =E P

P/2 P/4 V

2V

4V

V

Sol. Ex.

A sample of an ideal monoatomic gas initially at (1 L, 10 atm, 300 K) is expanded till its volume becomes 4 L (a) using reversible isothermal expansion (b) using by expanding against constant exteranl pressure of 1 atm isothermlly. (c) using reversible adiabatic process (d) by expanding aganist constant external pressure of 1 atm adiabatically. (e) by expanding againest constant external pressure of 1 atm under a process in which final pressure of gas becomes 5 atm. In All above processes calculate q us E & H.

Sol. Ex.

Write formula for calulation of q, W E and H. for following process on an ideal gas (a) Reversible isothermal expansions (b) Irreversible isothermal expansions (against constant external pressure of Pext ) (c) Reversible adiabatic expanstion (d) Imerersible adiabatic expansions (against constant external pressure of Pext ) (e) Reversible isocharic process (f) Reversible isobaric process (g) Free expansion from Vi  Vf

Case - II

For solids and liquid system : du = nCvdT 

f nRdT 2

f  dH = nCp dT    1 nRT 2 

W=–

 Pext. dv

H = u + (P2V2 – P1V1) du = dq + dw Ex.

Sol.

A liquid of volume of 100 L and at the external pressure of 10 atm it the is confined inside an adibatic bath. external pressure of the liquid is suddenly increased to 100 atm and the liquid gets compressed by 1 L against this pressure then find (i) work (ii) u (iii) H W=–1 work dom = –100 × – 1 = 100 lit atm q = 0 w = u  100 = u H = u + (P1V2 – P1V1 ) = 100 + (100 × 99 – 100 × 10) = 100 + 100 × 89 = 9000 lit atm. 1 lit atm = 101.3 Jule.

PAGE _ 20

Case III: Calculation of q, W, E and H during phase wanstions. Phase transitions generally take place at constant pressure (unless specified) and at constant temperature. so T = 0 But stile H and E are non zero (different from process on an ideal gas) Because during phase transitions thought the kinetic energy of molecules of substance remains same but the potential energy gets modified or changed and since E is summation of all type of energies so E  0. Also during this phase transition (PV)  0 Hence H  0. Now to calculate heat – generally latent heats of transitions are given and since process is taking place at constant pressure so q = H To calculate W use dW = – Pext dV and then calculate E using Ist law of thermodyanmics Ex.

Calculate q, W, E and H when 90 gm of water at 100ºC is vapawised into vapour at constant external pressure of 1 atm. Given that latent heat of vapavisation of water = 540 cal/gm and assume that steam behaves as are ideal gas. density of water = 1 gm/ml.

Sol. Ex.

Calculate q, W, E and H when 100 gm of calulate form of CaCO3 is comerted into its aragonite form given density of calculate = 2 gm/CC density of arangonite = 2.5 gm/CC and CaCO3 CaCO3 Calcite Arangonite H = 2 kJ/mole Generally for solid  Solid solid  Liquid a solid  Liquid transitions W << q so. E ~  q = H while for gaseous conversion for example Solid  gas Liquid  gas q = H  E. as W will be significant

Q.2

Sol.

Calculate W and E for the conversion of 1 mole of water into 1 mole of steam at a temperature of 100ºC and at a pressure of 1 atmosphere. Latent heat of vaporisation of water is 9720 cal/mole. P = 1 atm

V1 = volume of 1mole of water at 100ºC = 18 mL. V2 = volume of 1 mole of steam at 100ºC =

373 x 22400 = 30605 mL. 273

(Charles’s law )

Now we have, W = – p ( V2 – V1 ) = – 1 (30605 – 18) = – 30587 atm ml = –

30587 101 .3 × cal = – 741 calories 1000 4.2

Again we have, E = q + W = 9720 + (– 741) = 8979 calories.

PAGE _ 21

LECTURE # 6 IIND LAW OF THERMODYNAMICS : Requirement : Ist law insufficient to Explain Neutralization, Melting of ice > 0° Combustion, Freezing of water < 0° Fall of water , Boiling of water > 100° C Cooling of a hot body, Free expansion of a gas, mixing of gases. Spontaneous /Natural /Irreversible Processes : If a system moves along in a direaction on its known by its own system, sorrounding interactions without the help of any external agency then that direction is known as spontaneous direction and the process is knwon as a pontaneous process. Limitation : First law cannot predict the direction of any process. But in nature a process is found to have a specific direction at certain conditions. 1. Energy can be the factor : To achieve stability * All exothermic processes are spontaneous. * All endothermic processes are non spontaneous. Exceptions * melting of ice (above 0°C) although H > 0 , E > 0 * Vaporization of water (above 100°C) H > 0 , E > 0 * Spreading of ink in water H = 0 , E = 0 * Mixing of two gases * Endothermic solution of salt H > 0 , E > 0 * It is observed that nature tries to distribute energy and matter uniformaly (randomization)

Second law of thermodynamics : Entropy of the universe is constanlty inereasing. S universe = S system + S surrounding > 0

for a spontaneous process.

Where (S) = entropy is the measure of randomness or disorder. It is a state function and extensive property. Mathematically

dq rev dq rev or S = T T dq is state function only for reversible process ds =



1 T At Higher T - entropy is already high. Therefor heat addition will not introduce much change in the entropy. dS  dqrev , dS 

Entropy calulation : 1.

For calculating Ssurrounding : As the surrounding is an infinite heat reservoir hence no. amount of heat given to the sorrounding can cause turbulance in it, hence all heat exhanges with sorroundings are considered to be reversible. Ssurrounding = For

dq surroundin g



T

surrounding T is constant

q surroundin g 1 dq surr = T T Bu the law of conservation of energy qsurr = – qsystem



Ssurr =



Ssurr =



 q system T

PAGE _ 22

(b)

For Ssystem

As ‘S’ is a state function ‘S’ will not depend on path.  S irrev = S rev B

S =

 A

e.g.

dqrev T reversible path .

ENTROPY CALCULATION FOR SOLID OR LIQUID SYSTEM. ‘Cu’ block kept in open atmosphera

T2

Suniverse = Ssystem + Ssurr =





T1

dq surr m.s. dT  T T



q surr T2 Suniverse = m.s. n T + T 1 2 but q surr = – qsystem = – ms (T2 – T1) = ms (T1 – T2)



T2 ms (T1  T2 ) Suniverse = ms n T + >0 T2 1



qA qB Ssystem = T  T 1 2

if qA = q  qB = – q 

q q Ssystem = T  T  0 1 2

FOR PHASE TRANSFORMATIONS Sfusion=

dq

T



f T

Q

 dq  T

Hfus T

for constant pressure Sfus. = for constant volume Sfus. =

Svap =

dQ

T



Ufusion T

1 Q dQ  T T



for constant Pressure, Svap = for constant Volume, Svap =

H vap T

H vap T

PAGE _ 23

LECTURE # 7 ENTROPY CALCULATION FOR AN IDEAL GAS UNDERGOING A PROCESS. irr 

State A

State B

Sirr

P1, V1, T1 P2, V2, T2 As ‘s’ is a state function S will not depend upon path B

Sirrev = Srev, system =



A

dq rev T

from the first law du = dQ + dw dQ = du – dw for reversible path dq = dqrev dQrev = du + Pgas dv B

Srev, system =

rev path.

{Pext = Pgas reversible}

du  Pgas dv



T

A

for an ideal gas Pgas T



nR & du = ncV dT V T2





S

sys

=



T1

V

2 nc v dT nRdv  T V V



1

T2 V2 Ssystem = ncv ln T + nR ln V 1 1

ISOTHERMAL PROCESS : (a) Reversible Re v State A   State B P1,V1,T P2,V2,T

S system = nCVln

V T2 + nRln 2 V1 T1

 0 Ssorrounding =  FLOT

dq system

T du = dq + dw  dq = – dw

V2  q = –w as w = – nRTn V 1 V2 q = nRTn V 1

 Ssys. =

 nRTn T

V2 V1

 –nRn

V2 V1

PAGE _ 24

Suniverse= Ssystem + Ssurr = nRn (b)

V2 V2 –nRn =0 V1 V1

rreversible  irrev State A   State B P1,V1,T P2,V2,T

Ssystem = nRn

SSurrounding =

V2 V1

 qsystem T

qsystem is calculated using FLOT qsystem = –Pext(V2– V1) Ssurrounding =

p ext ( V2  V1 ) T

Suniverse = nRn

p ext ( V2  V1 ) V2 – >0 V1 T

=

1 T

  V2 )  Pext ( V2  V1 ) (nRT n V 1   ( wgas )rev ( wgas )irrev

As irrevensible work of gas is less then reversible work.

ADIABATIC PROCESS (a) Reversible Adiabatic rev State A   State B P1,V1, T1 P2,V2, T2

T2 V2 Ssystem = nCV n T + nR n V 1 1 TV–1 = constant

T2  V1    T1 =  V2 

 1

 V1   (S)system = nCVn   V2  S surr = –

(b)

q =0; T

 1

V2 + nRn V = 0 1

S universe = 0 + 0 = 0

Irreversible Adiabatic irre State A   State B P1,V1,T1 P2,V2,T2 Using irreversible adiabatic process it is not possible to reach same state B which was reached by reversible adiabatic.

T2 V2 S system = nCV n T + nr n V > 0 1 1

[(T2)irrev > (T2)rev]

PAGE _ 25

S

surr

=–

q sys T

=0

as q = 0



For irreversible exparsion the decrease in temprature will be lesser as work done is lesser hence decrease in entropy due to fall in temprature will be lesser in case of irrversible expansion. Hence net entropy would increase.

Q.

For an ideal gas Cp = 2.5 R, calculate the change in entropy suffered by 3 moles of gas on being heated from 300 K to 600 K at (a) constant pressure (b) constant volume.

Sol.

(a) S = nCpn

T2 T1

+ Rn

P1 P2

At constant pressure : S = n Cpn

T2 T1

= 3 × 2.5 Rn2

= 43.22 JK–1 mol–1. (b) At constant volume : S = nCVn

T2 T1

= 3 × 1.5 Rn2 = 25.93 JK–1 mol–1.

Q.

Sol.

(a) One mole of an idal gas expands isothermally and reversibly at 25ºC from a volume of 10 litres to a volume of 20 litres. (i) What is the change in entrogy of the gas? (ii) How much work is done by the gas? (iii) What is q (surroundings)? (iv) What is the change in the entropy of the surroundings? (V) What is the change in the entropy of the system plus the surroundings? (b) Also answer the questions opening a stopcock and allowing the gas to rush into an evacuated bulb of 10–L volume.

V2 20 (a) (i) S = 2.303 nR log V = 2.303 ×1 × 8.314 × log = 5.76 J/K. 10 1 V2 (ii) W rev = –2.303nRT log V 1 = –2.303 × 1 × 8.314 × 298 × log

20 = –1781 J. 10

(iii) For isothermal process, E = 0 and heat is absorbed by the gas, qrev = E – W = 0 –(–1718) = 1718 J.  qrev = 1718 J. ( process is reversible) 1718 (iv) DSsurr= – = – 5.76 J/K. 298 As entropy of the system increases by 5.76 J, the entropy of the surroundings decreases by 5.76J, since the process is carried out reversibly. (v) Ssys + Ssurr = 0..... for reversible process (b) (i) S = 5.76 J/K, which is the same as above because S is a state function (ii) W = 0 ( pext = 0) (iii) No heat is exchanged with the surroundings. (iv) Ssurr = 0. (v) The entrogy of the system plus surroundings increases by 5.76 J/K, as we expect entropy to increase in an irreversible process.

PAGE _ 26

LECTURE # 8 THIRD LAW OF THERMODYNAMICS : The entropy of perfect crystals of all pure elements & compounds is zero at the absolute zero of temperature . Thus absolute value of entropy (unlike the absolute value of enthalpy) for any pure substance can be calculated at any given temperature . In standard state (298 K, 1 atm), it is standard absolute entropy Sº . These values are tabulated : 

Sº =  Sºproducts   Sºreactants.

Entropy calculation For chemical reaction aA + bB  cC + dH 0 0 S0system = ( ni Sm )product – ( ni Sm )reactant 0 where Sm = standard molar entropy. It can calculated using third law of thermodynamics.

Ssurr =

 Q system T

=–

 Hsystem T

* For a perfactly crystline substance at 0 K, entropy = 0

Gibb’s free energy G) : A system parameter to prodict the spontitancity of chemical reaction was introduced by Gibb’s so that entropy calculation for the surrounding need not be carried out. for a spontaneous change Sunivense = Ssystem + Ssurrounding Ssurr =

 Q system

=–

 Hsystem

T T TS univ = TS system – Hsystem

Introducing a new thermodyamic function G = Gibb’s free energy (State function and an extensive property) Gsystem = H system – TS system  G = H – TS Gsystem = H system – TS system .................. (ii) Comparing equation (i) and (ii) Gsystem = – TS universe New critatria of spontancity : (i) If G system is (–ve) < 0  process is spontaneous (ii) If G system is > 0  process is non spontaneous (iii) If G system = 0  system is at equilibrium So At every temparature G  0 G system

S system

G system = H system – TS system

+ Ve

– Ve

+

– Ve

+ Ve

–

+ Ve

+ Ve

– Ve at high temprature

– Ve

– Ve

+ Ve at high temprature

PAGE _ 27

Gº = standrad free energy change : When the reactants under standrad conditions gets converted into products which is also under standered condition, then the free energy change is known as Gº (it is a contant) for 1 mole at 1 bar. At standered conditions : G° = H° – TS° = constant for a GIVEN REACTION AB  C +  D G° = (Gbb’s energy of product) – (Gibb’s energy of reactant) 0 0 0 0 0 0 G° =  Gm ,C + Gm,D – Gm,A –  G m,B , where Gm can not be calculated as Hm cannot be calculated.

hence, we can convert this relation in to Gibb’s energy of formation of substance. G° = G0f ,C + G0f ,D – G0f ,A –  G 0f ,B

G0f : standered Gibb’s energy of formation. G0f (elements in their standard states) = 0 G° = G0f ,product – G0f ,reac tan ts For a reaction in progress G = Gº + RT n Q Q is reaction quotient , G > 0 backward is feasible , G < 0 forward is feasible At equilibrium G = 0  Q = K  Gº = – RT n K at equilibrium

Physical interpretation of G. The free energy change of a process symbolises the maximum amount of non-expansional (compression) work which can be performed. du = dq + dw  du = dq + dwexpon + dwexp dwnon-exp = du – dq – dwexp 

dwnon-exp = du – dq + PdV



dwnon-exp = dH – dq

For maximum work of non expansion the process should be reversible isobaric. 

dq = dqrev = T.dS

dwnon-exp = dH – TdS = G Gibb’s -Helmholtz equations : It interrelates G snd H at constant pressure :

 ( G)  G = H + T  T   p

or

G  H  ( G  =  T   p T

or

 ( G)  –S =  dT   p

 E cell  S = nF  T  (G = – nEF)  p

Thus temperature coefficient of the emf of a cell is Q.

 E cell  S  T  =  p nF

At what temperature, will the following process would not be spontaneous ? A(g)



A ()

H = – 20 kJ (A) > 200 K

S = – 50 J/K (B) > 300 K

(C) 350 K

(D*) > 400 K

PAGE _ 28

Sol.

G > 0 H > TS 

H 

T <   S    or T > Q.

H S

[Since H & S both are negative] or T > 400 K

Find whether the given reaction is spontaneous at 1 atm & 298 K. Ag(s) +

1 Cl (g)  2 2

AgCl(s)

Given Hºf (AgCl) = – 127 KJ and Sº of AgCl(s), Ag(s) and Cl2(g) are 96, 42.7 & 223 JK–1mol–1 respectively. Sol.

Sº = 96 – 42.7 –

223 2

= – 58.2 JK–1

Gº = Hº – TSº = – 127 × 103 + 298 × 58.2 = – 109.65 × 103 So spontaneous. Q. Sol.

Ka for CH3COOH at 25ºC is 1.754 × 10-5 . At 50ºC, Ka is 1.633 × 10-5 What are Hº and Sº for the ionisation of CH3COOH? (Gº)298 = – 2.303RT log K = – 2.303 ×8.314 × 298 ×log (1.754 ×10-5) = 27194 J. (Gº)323 = 2.303 × 8.314 × 323 × log ( 1.633 × 10-5) = 29605 J. Gº = Hº – TSº 27194 = Hº – 298 Sº 29605 = Hº – 323 Sº Hº = – 1.55 kJ/mol  Sº = – 96.44J/mol.K

LECTURE#9 THERMOCHEMISTRY : Deals primarily with the transfer of heat between a chemical system and its surroundings when a charge of phase or a chemical reaction takes place with the system. In laboratory, the majority of chemical reactions takes place under the constant atmospheric pressure so heat transfer is equal to the change in the enthalpy of the system. However if any reaction is carried out at constant volume, then heat transfer will be equal to charge in internal energy of the system.

Enthalpy of a substance :

   

Every substance has a fixed value of enthalpy under any particular state. Through its exact value cannot be calculated but it has some finite fixed value. The enthalpy of one mole of a substance called molar enthalpy in any particular state can be assigned symbol Hm(substance). for example molar enthalpy of water vapours at 398 K and 1 atm pressure may be represented as Hm (H2O, g, 398 K, 1 atm). In very simple works enthalpy can be considered as heat constant (amoiunt) of substance, and during reaction this heat is being released or absorbed. Molar enthalpy of substance under standard conditions is called standard molar enthalpy of a substance. Standard state of any substance means. For a gas standard state means ideal gas at 1 bar partial pressure at any give temperature. For a liquid – pure liquid at one bar pressure at 1 bar pressure at any given temperature. For a pure crystalline solid – pure crystalline solid at 1 bar pressure and at any given temperature For any substance or ion in solution – the species should be in unit molality (can also be taken as 1M concentration), at one bar pressure and at any given temperature.

PAGE _ 29

Molar standard enthalpy of water vapours at 398 K will be represented as H°(H2O, g , 398 K) and molar standard enthalpy of liquid water at 398 K will be represented as Hm° (H2O, l , 398 K) (It is hypothetical but can be calculated) We cannot exactly calculate enthalpy content of a substance only the change in enthalpy can be calculated when substance is taken from one state to other. For example :

0 0 Let enthalpy contant initially be Hm ,1 & finally entalpy contant be Hm , 2 0 0 Then H° = Hm , 2 – Hm ,1

= heat added at constant pressure to change temperature from 25°C to 50°C. = CPT = (18 cal/mole °C) (25°C) = 450 cal Q.7 Ans.

Why does heat get released/absorbed during chemical reactions? because the reactants have a fixed enthalpy content before the reaction and when these are converted into the products which have a different enthalpy content heat gets released or absorbed. Even if temperature of reactions remains constant yet due to change in bonding energies E  0. If Hproducts > Hreactants Reaction should be endothermic as we have to give extra heat to reactants to get these converted into products and if Hproducts < Hreactants Reaction will be exothermic as extra heat content of reactants will be released during the reaction. Enthalpy change of a reaction : Hreaction = Hproducts – Hreactants H°reactions

Ex.

= H°products – H°reactants = positive

–

endothermic

= negative

–

exothermic

Calculate H° of the reaction 2NH3 (g)  N2 (g) + 3H2 (g) Given that the molar entalpy contents (hypothetical) are H°m (NH3,g) = 200 cal/mole H°m (H2,g) = 100 cal/mole H°m (N2,g) = 120 cal/mole

ENTHALPY OF FORMATION : It is not possible to determine the absolute value of the molar enthalpy of a substance. However, based on the following convention, the relative values of standard molar enthalpies of formation of various substances can be built. The standard enthalpy of formation of every element in its stable state of aggregation at one bar pressure and at specified temperature is assigned a zero value.The specified temperature is usually taken as 25 °C. A few exmaples are Hf° (O2 , g) = 0 Hf° (C, graphite) = 0 Hf° (C, diamond)  0 Hf° (Br2 , ) = 0 Hf° (S, rhombic) = 0 Hf° (S, monoclinic)  0 Hf° (P, white) = 0 Hf° (P, black)  0

PAGE _ 30

The standard enthalpy of formation of a compound is the change in the standard enthalpy when one mole of the compound is formed starting from the requisite amounts of elements in their stable state of aggregation. The chemical equations corresponding to enthalpy of formation of few substances are given below.

1 1 H2 (g) + Br (l)  HBr(g) 2 2 2

Enthalpy of formation of HBr(g) :

Hf°(HBr, g)

0 0 =  B Hm (B) = Hm (HBr, g) –

Enthalpy of formation of SO2 (g) : Hf° (SO2, g) =

0 Hm

(SO2, g) –

1 0 1 0 (Br2, 1) H (H , g) – Hm 2 m 2 2

...(1)

S (rhombic) + O2 (g)  SO2 (g) 0 (S, Hm

0 rhombic) – Hm (O2, g)

...(2)

But above equations cannot be for calculation of enthalpy of reaction as the molar enthalpies of different species can not be exactly known.

ENTHALPY OF REACTION FROM ENTHALPIES OF FORMATION: The enthalpy of reaction can be calculated by Hr° =  B Hf°,products –  B Hf°,reactants B is the stoichiometric coefficient above equation holds good for any reaction as the same reference state is used for reactants and products (shown in figure).

Ex.

Calculate the standard enthalpy of reaction ZnO(s) + CO(g)  Zn(s) + CO2 (g). Given, Hf° (ZnO, s) = – 350 KJ / mole–1 Hf° (CO2, g) = – 390 KJ / mole Hf° (CO, g) = – 110 KJ / mole

HESS’S LAW OF CONSTANT HEAT SUMMATION : The heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps. The chemical equation can be treated as ordinary algbraic expressions and can be added or subtracted to yield the required equation. The corresponding enthalpies of reactions are also manipulated in the same way so as to give the enthalpy of reaction for the desired chemical equation. Since rH stands for the change of enthalpy when reactants (substances on the left hand side of the arrow) are converted into products (substances on the right hand side of the arrow) at the same temperature and pressure, it follows that if the reaction is reversed (i.e., products are written on the left hand side and reactants on the right hand side), then the numerical value of rH remains the same, but its sign changes. The utility of Hess’s law is considerable. In almost all the thermichemical numericals, some way or the other, Hess’s law is used. One of the important applications of Hess’s law is to determine enthalpy of reaction which is difficult to determine experimentally. For example, the value rH for the reaction C(graphite) +

1 O (g)  CO(g) 2 2

PAGE _ 31

which is difficult to determine experimentally, can be estimated from the following two reactions for which rH can be determined experimentally. C(graphite) + O2(g)  CO2(g) CO(g) +

r H 1

1 O (g)  CO2(g) 2 2

r H 2

Substracting the latter from the former, we get C(graphite) + Consequently,

1 O (g)  CO(g) 2 2

 rH =  rH 1 –  rH 2

LECTURE#10 LATTICE ENERGY OF A CRYSTAL (BORN-HABER CYCLE) : The lattice energy is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous ions.The larger the lattice energy, the more stable the ionic compound and the more tightly the ions held.Consider the following sequence of steps for the formation of NaCl crystals from Na(s) and Cl2(g) (i) Vaporization of Na(s) Na(s)  Na (g) Hsub + – (ii) Ionization of Na(g) Na(g)  Na (g) + e Hion (iii)

Dissociation of chlorine

(iv) (v)

Formation of Cl– (g) Condensation of Na+(g) and Cl– (g)

1 Cl (g)  Cl (g) 2 2 Cl(g) + e–  Cl– (g)

Hdiss Helectron gain

Na+(g) + Cl–(g)  NaCl(s) Hlattice ––––––––––––––––––––– ––––– 1 Net change : Na(s) + Cl (g)  NaCl(s) Hf 2 2 Hess’s law, we can write Hf = Hsub + Hion + Hdiss + Helectron gain + Hlattice All other enthalpies can be determined experimentally.Hence, Hlattice can be determined from the above relation. The lattice energy is the negative of Hlattice value. Q.

Set up a Born-Haber cycle to find the lattice energy of NaCl crystal. Given Hf° (NaCl) = – 410 kJ mol–1 . Ionization enthalpy of Na = 495 kJ mol–1 , electron affinity of chlorine = 365 kJ mol–1 , sublimation enthalpy of Na = 315 kJ mol–1 and dissociation enthalpy of Cl2(g) = 240 kJ mol–1 .

Q.

from the following thermochemical equation, calculte the enthalpy of formation of cane sugar (C12H22O11) : (i) C12 H22 O11 (s) + 12 O2(g)  12 CO2(g) + 11H2O()

H = – 5650 kJ mol–1

(ii) C(s) + O2 (g)  CO2(g)

H = – 390 kJ mol–1

(iii) H2 (g) + Q.

1 O (g)  H2O() 2 2

H = – 285 kJ mol–1

From the datat at 25°C (i) Fe2O3(s) + 3C(graphite)  2Fe(s) + 3CO(g)

H = – 490 kJ mol–1

(ii) FeO(s) + C(graphite)  Fe(s) + CO(g)

H = – 155 kJ mol–1

(iii) C(graphite) + O2 (g)  CO2(g)

H = – 390 kJ mol–1

(iv) CO(g) +

1 O (g)  CO2(g) 2 2

H = – 280 kJ mol–1

PAGE _ 32

Q.8

Sol.

Calculate heat of the following reaction at constant pressure, F2O(g) + H2O(g)  O2 (g) + 2HF(g) The heats of formation of F2O (g), H2O(g) and HF (g) are 5.5 kcal, –57kcal and 64 kcal respectively. Given that (i)

F2 (g) +

1 O (g)  F2O (g); H = 5.5 kcal 2 2

(ii)

H2 (g) +

1 O (g)  H2O (g); H = – 57 kcal 2 2

(iii)

1 1 H2 (g) + F2 (g)  HF (g); H = – 64 kcal 2 2

F2O and H2O in eqns. (i) and (ii) and in the eqn. given in the problem are on the opposite sides, while HF in eqn. (iii) and in the eqn. given in the problem is on the same sides. Thus applying, [ – Eqn. (i) –Eqn. (ii) + 2 × Eqn. (iii) ], we get –F2 (g) –

1 1 O (g) – H2 (g) – O2(g) + H2 + F2 (g)  –F2O (g) – H2O(g) + 2HF(g); 2 2 2

H = – 5.5 – (– 57) + 2 × ( – 64) = or F2O (g) + H2O (g)  O2 (g) + 2HF(g); H = – 76.5 kcal.

ENTHALPY OF COMBUSTION : It is the enthalpy change when one mole of a compound combines with the requisite amount of oxygen to give products in their stable forms. For example, the standard enthalpy of combustion of methane at 298.15 K is – 890.36 kJ mol–1. This implies the following reaction : CH4 (g) + 2O2(g)  CO2(g) + 2H2O () H° = – 890.36 kJ mol–1 The standard enthalpy of combustion of methane at 298.15 K may be written as Hc° (CH4, g, 29815 K) = – 890.36 kJ mol–1 The data on the enthalpy of combustion can be determined experimentally. With the help of such data, we can determine the enthalpy of formation of a compound, which otherwise is difficult or impossible to determine experimentally. Consider for example, the enthalpy of formation of CH4(g): C(graphite) + 2H2(g)  CH4(g) First of all, the combination of carbon and hydrogen does not occur readily. Secondly, if the reaction is even completed, the end product would not be pure methane. Therefore, the enthalpy of formation of methane can be determined indirectly through the enthalpy of combustion of methane : CH4(g) + 2O2(g)  CO2(g) + 2H2O() HC°(CH4, g) = Hf°(CO2, g) + 2Hf°(H2O , ) – Hf° (CH4, g) therefore Hf°(CH4, g) = Hf°(CO2, g) + 2Hf°(H2O , ) – HC° (CH4, g) The enthalpies of formation of CO2 and H2O can be determined experimentally by the combustion of carbon (graphite) and hydrogen. Thus, knowing the mesured value of HC°(CH4, g), the enthalpy of formation of CH4 can be calculated. The value is Hf°(CH4, g) = Hf°(CO2, g) + 2 Hf°(H2O , ) – HC° (CH4, g) = [– 393.51 + 2 (–285.83) – (–890.36)] kJ mol–1 = – 74.81 kJ mol–1 or, equivalently, we may add the following three chemical equations. C(graphite) + O2 (g)  CO2 (g) CH° = – 393.51 kJ mol–1

1 O (g)  H2O()] 2 2 – [CH4(g) + 2O2 (g)  CO2 (g) + 2H2O()] ––––––––––––––––––––––––––––––––––––––– C(graphite) + 2H2(g)  CH4 (g) 2 [H2(g) +

rH° = 2(– 285.83) kJ mol–1 rH° = –(– 890.36) kJ mol–1 –––––––––––––––––––––– fH° = – 74.81 kJ mol–1

PAGE _ 33

Ex.

Calculate the entalpy change of the following reaction 2C2H2(g)  C6H6(g) Given Enthalpy of combustion of C2H2(g) = – 1.30 MJ mol–1 and that of C6H6(g) = – 3.30 MJ mol–1

Ex.

Using the combustion data given below, compute the enthalpy of formation of isoprene (g) and its resonance energy. Given Hf (from bond enthalpies) = 103 kJ mol–1 Data : Enthalpy of combustion of isoprene (g) HC = – 3120 kJ mol–1 Enthalpy of formation of CO2 (g) Hf = – 393 kJ mol–1 Enthalpy of formation of H2O() Hf = – 285 kJ mol–1

MEASUREMENT OF ENTHALPY OF COMBUSTION : Enthalpies of combustion are usually measured by placing a known mass of the compound in a closed steel container (known as bomb calorimeter) which is filled with oxygen at about 30 bar pressure. The calorimetre is surrounded by a known mass of water. The entire apparatus is kept in an insulated jacket to prevent heat entering into or leaving from the container, as shown in figure. The sample is ignited electrically to bring about the combustion reaction. The heat evolved is used in raising the temperature of water and the calorimetre.

If total heat capacity of calorimeter and all of its contents = C, rise in temperature = T then heat released = M q = CT of this heat is because of mass m if substance then due to 1 mole heat released =   q m

= EC° (constant volume reactions) Now HC° can be calculated by using HC° = EC° + ng RT. Where ng is the change in stoichiometric number of gaseous species in the balanced chemical equation representing the combustion process. Q.

A 0.12 g sample of solid magnesium( molar mass = 24 gmole–1) is burned in a constant volume bomb calorimeter that has a heat capacity of 1.5 kJ 0C–1.The calorimeter contains 300 mL of water ( density 1g mL– 1 ) and its temperature is raised by 1 0C. Calculate the enthalpy of combustion of magnesium at 300K.

Q.9

A gas mixture of 3.67 litres of ethylene and methane on complete combustion at 25ºC produces 6.11 litres of CO2 . find out the amount of heat evolved on burning one litre of the gas mixture. The heats of combustion of ethylene and methane are – 1423 and –891 kJ mol-1 at 25ºC. [ IIT 1991 ] 3.67 lit.

Sol. C2H4

or

+

CH4

+ O2  CO2 + H2O

x lit. (say) x moles

(3.67 – x)lit.

6.11 lit.

(3.67 – x) moles 6.11 moles

PAGE _ 34

Applying POAC for C atoms, 2 × x + 1 × ( 3.67 – x ) = 1 × 6.11; x = 244 lit. Thus, the volume of C2H4 = 2.44 lit., and volume of CH2 = 1.23 lit.

 volume of C2H4 in a 1–litre mixture =

2.44 = 0.665 lit. 3.67

and volume of CH4 in a 1– litre mixture = 1 –0.665 = 0.335 lit. Now, thermochemical reactions for C2H4 and CH4 are C2H4 + 3O2  2CO2 + 2H2O; H = – 1423 kJ CH4 + 2O2  CO2 + 2H2O; H = – 891 kJ As H values given are at 25ºC, let us first calculate the volume occupied by one mole of any gas at 25ºC (supposing pressure as 1atm) Volume per mole at 25ºC =

298 × 22.4 = 24.45 lit. 273

Thus, heat evolved in the combustion of 0.665 lit. of C2H4 = –

and heat evolved in the combustion of 0.335 lit.of CH4 = 

1423 × 0.665 = – 38.70 kJ 24.45

891 × 0.335 = –12.20kJ. 24.45

total heat evolved in the combustion of 1 litre of the mixture = –38.70 + (–12.20) = – 50.90kJ.

LECTURE#11 ENTHALPY OF SOLUTION : The integral enthalpy of solution at the given concentration is the enthalpy change when one mole of the solute is dissolved in a definite quantity of solvent to produce a solution of a desired concentration. While recording integral enthalpies of solution it is a general practice to state the amount of the solvent in which 1 mole of solute is dissolved ; Thus HCl(g) + 10H2O()  HCl (10H2O)

H1 = –69.5 kJ mol–1

indicates that when 1 mol of hydrogen chloride gas is dissolved in 10 mol of water, there is an evolution of 69.5 kJ of heat. Other values are (i)

HCl(g) + 25 H2O()  HCl (25H2O)

H2 = –72.3 kJ mol–1

(ii)

HCl(g) + 40 H2O()  HCl (40H2O)

H3 = –73.0 kJ mol–1

(iii)

HCl(g) + 200 H2O()  HCl (200H2O)

H4 = –74.2 kJ mol–1

(iv) HCl(g) + aq  HCl (aq) H5 = –75.0 kJ mol–1 Whenever amount of solvent is not specified then take its amount to be very large just like in equation no. (iv).

ENTHALPY OF DILUTION : The enthalpy of dilution is the change in enthalpy when a solution containing 1 mole of solute is diluted from one concentration to another for example for the reaction HCl . 40 H2O + aq  HCl (aq), the enthalpy change can be calculated as follows HCl (g) + aq  HCl (aq) H 1 = – 75 kJ mol–1 HCl(g) + 40 H2O  HCl . 40H2O H 2 = – 73 kJ mol–1 Subtracting, we have HCl . 40 HCl + aq  HCl (aq) H = – 2 kJ mol–1

PAGE _ 35

ENTHALPY OF HYDRATION : Enthalpy of hydration is used in following to ways. (i) Enthalpy of hydration of anhydrous or partially hydrated salts. (ii) Enthalpy of hydration of gaseous ions. (i) Enthalpy of hydration of a given anhydrous or partially hydrated salt is the enthalpy change when it combines with the requisite amount of water to form a new hydrated stable salt. For example, the hydration of anhydrous cupric sulphate is represented by CuSO4 (s) + 5H2O ()  CuSO4 . 5 H2O(s) There is a almost invariably a liberation of heat in such reactions, i.e. the value of H is negative. CuSO4(s) + 800 H2O ()  CuSO4 (800 H2O) Hr° = – 68 kJ mol–1 CuSO4 . 5H2O (s) + 795 H2O ()  CuSO4 (800 H2O) Hr° = + 10 kJ mol–1 by subtraction, we get CuSO4(s) + 5H2O ()  CuSO4 . 5 H2O(s)Hr° = – 78 kJ mol–1 (ii) Enthalpy of hydration of any gaseous ion is the enthalpy change when 1 mole of the gaseous ion is hydrated in large amount of water to form aqeous ion. By convention, the standard enthalpy of formation of H+(aq) is taken to be zero. Enthalpy of hydration of Cl¯ gaseous ions will be represented by : Cl¯(g) + aq.  Cl¯(aq)

H =

ENTHALPY OF NEUTRALIZATION : The amount of heat released when one gram equivalent of an acid is neutralised by one gram equivalent of a base. OR The amount of heat released in formation of one mole of water when an acid is neutralised by a base. OR Enthalpy of neutralization is defined as the enthalpy change when one mole of H+ in dilute solution combines with one mole of OH¯ to give rise to undissociated water, i.e. H+(aq) + OH¯(aq)  H2O()

H= –57.1 kJ/mole = –13.7 kcal/mol

for SA + SB this heat of neutralisation is always equal to –13.7 kcal/mole or –57.1 kJ/mole. for any other combination of acid and base this heat is less than –13.7 kcal/mole or –57.1 kJ/mole.

ENTHALPY OF IONIZATION : Whenever a weak acid (or base) reacts with a strong base (or acid), the release of heat is less than 57.1 kJ mol–1. It is because of the fact that these acids or bases are not completely ionized in solution. Some of the heat is consumed in ionizing there acids and bases this heat is known as enthalpy of ionization. Examples are : HCN + Na+ OH–  Na+ + CN– + H2O rH° = – 12 kJ mol–1 + – + – CH3COOH + Na OH  Na + CH3COO + H2O rH° = – 49 kJ mol–1 The enthalpy of ionization can be calculated as follows. The neutralization of a weak acid, say HCN, may be represented in two steps, namely, (i)

Ionization

HCN  H+ + CN¯

H°1 = x

(ii) Neutralization H+ + OH¯  H2O H°2 = – 57.1 kJ/mole The complete reaction is obtained by adding the above two steps. Thus HCN + OH¯  H2O + CN¯ H° = – 12 kJ/mole Obviously, H° = H°1 + H°2 H°1 = H°H°2 = [–12 – (–57.1)] = 45.1 kJ/mole Greater the enthalpy of ionization of any weak acid or weak base, weaker will be the acid or base.

PAGE _ 36

Ex. Sol.

Enthalpy of neutralization of HCl by NaOH is –57.1 kJ/mol and by NH4OH is –51.1 kJ/mol. Calculate the enthalpy of dissociation of NH4OH. Given that H+(aq) + NH4OH(aq)  NH4+(aq) + H2O() H = –51.1 kJ/mole We may consider neutralization in two steps. (i) Ionization NH4OH(aq)  NH4+(aq) + OH¯(aq) H1 = ? (ii) Neutralization H+(aq) + OH¯(aq)  H2O() H2 = – 57.1 kJ/mole Thus, H = H1 + H2 Therefore, H1 = H – H2 = – 51.1 kJ/mol + 57.1 kJ mol–1 = 6.0 kJ/mol

ENTHALPY OF TRANSITION : Enthalpy of transition is the enthalpy change when one mole of one allotropic form changes to another. For example : C(graphite)  C(diamond) Htrs0 = 1.90 kJ mol–1 so if C(graphite) + O2(g)  CO2(g) HC0 = – 393.51 kJ mol-1 and C(diamond) + O2(g)  CO2(g) HC0 = – 395.41 kJ mol–1 Subtracting, we have, C(graphite)  C(diamond) Htrs0 = 1.90 kJ mol–1

ENTHALPY OF PRECIPTION : Enthalpy of precipitation is the enthalpy change when one mole of a precipitate is formed. For example : BaCl2(aq.) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq) rH0=–24.27 kJ mol–1

ENTHALPY OF FORMATION OF IONS : The enthalpy change when one mole of hydrated ions is obtained from element in its standard state as. ½ Cl2 (g) + aq  Cl– (aq) Hr° = Hf° (Cl–, aq) By convention, the standard enthlpy of formation of H+(aq) is taken to be zero. We have seen that H+ (aq) + OH–(aq)  H2O(l) rH0 = – 57.1 kJ mol–1 0 0 0 + 0 For this reaction, Hr = Hf (H2O,l ) – {Hf (H ,aq) + Hf (OH–,aq)} 0 Hence at 25 C, we get Hf0 (H+, aq) + Hf0 (OH– ,aq) = Hf0 (H2O, l ) – Hr0 so Hf0 (OH– ,aq) = {– 286.1 – (– 57.1)} kJ mol–1 = – 229.00 kJ mol–1 With the enthalpies of formation of these two ions, the enthalpy of formation of any other ion can be found from the enthalpies of formation and solution of its pure compound with H+ or OH¯. for example, the enthalpy of formation of Na+ can be calculated from the enthalpy of formation and enthalpy of infinite dilute solution of NaOH. The two values are : The chemical equation for the formation of infinite dilute solution of NaOH(s) is NaOH(s) + nH2O()  Na+(aq) + OH¯(aq) aqH°(NaOH, s) = –44.50 kJ mol–1 Since there are equal amounts of water on both sides of the above equation, the two enthalpies give no net effect and thus aqH°(NaOH, s) = fH°(Na+, aq) + fH°(OH¯, aq) – fH°(NaOH, s) or fH° (Na+, aq) = aqH° (NaOH, s) – fH°(OH°, aq) + fH°(NaOH, s) = [–44.50 – (–229.99) + (–425.61)] kJ mol–1 = –240.12 kJ mol–1 Similarly, from NaCl(aq) or HCl(aq), the enthalpy of formation of Cl¯(aq) can be determined, and so on. These are recorded in Appendix-II. The changes in enthalpy of any ionic reaction can then be found from these ionic enthalpies of formation and the usual enthalpies of formation of compounds. Ex.1 Sol.

The enthalpy of formation of H2O(l) is –285.83 kJ mol–1 and enthalpy of neutralization of a strong acid and a strong base is –55.84 kJ mol–1. What is the enthalpy of formation of OH¯ ions ? Given that H+(aq) + OH¯(aq)  H2O() f H 0 –285.83 kJ mol–1 neutH = fH(H2O, ) – fH(OH¯, aq) Hence fH (OH¯, aq) = fH(H2O, ) – neutH = [–285.83 – (–55.84)] kJ mol–1 = –229.99 kJ mol–1

PAGE _ 37

Ex.2

Ex.3

Calculate H° for the reaction Ag+(aq) + Cl¯(aq)  AgCl(s) at 25°C. Given fH° (Ag+, aq) = 105.58 kJ mol–1, fH° (Cl°, aq) = –167.16 kJ mol–1 and fH°(AgCl, s) = – 127.07 kJ mol–1 For the reaction Ag+(aq) + Cl¯(aq)  AgCl(s) we have H° = fH° (AgCl, s) – fH°(Ag+, aq) – fH°(Cl¯, aq) = [–127.07 – 105.58 – (–167.16)] kJ mol–1 = –65.49 kJ mol–1 Calculate the enthalpy change when one mole of HCl(g) is dissolved in a very large amount of water at 25°C.

Sol.

The change in state is HCl(g) + aq  H+(aq) + Cl¯(aq) Given : fH¯(HCl, g) = – 92.31 kJ mol–1 and fH° (Cl¯, aq) = – 167.16 kJ mol–1 For the reaction

Sol.

HCl(g) + aq  H+(aq) + Cl¯(aq) we have H° H°

= H°(Cl¯,aq) – fH°(HCl, g) = [–167.16 – (–92.31)] kJ mol–1 = – 74.85 kJ mol–1

LECTURE#12 BOND ENTHALPIES : The bond enthalpy is the average of enthalpies required to dissociate the said bond present in different gaseous compounds into free atoms or radicals in the gaseous state. While bond dissociation enthalpy is the enthalpy required to dissociate a given bond of some specific compound.for example the enthalpy of dissociation of the O–H bond depends on the nature of molecular species from which the H atom is being separated. For example, in the water molecule. H2O(g)  H(g) + OH(g) Hr0 = 501.87 kJ mol–1 However, to break the O–H bond in the hydroxyl radical required a different quantity of heat : OH(g)  O(g) + H(g) Hr0 = 423.38 kJ mol–1 The bond enthalpy, OH, is defined as the average of these two values, that is :

501.87mol 1  423.38kJmol 1 = 462.625 kJ mol–1 2 In the case of diatomic molecules, such as H2, the bond enthalpy and bond dissociation enthalpy are identical because each refers to the reaction. H2(g)  2H(g) H – H = Hr0 = 435.93 kJ mol–1 Thus, the bond enthalpy given for any particular pair of atoms is the average value of the dissociation enthaplies of the bond for a number of molecules in which the pair of atoms appears. OH =

ESTIMATION OF ENTHALPY OF A REACTION FROM BOND ENTHALPIES : Let the enthalpy change for the gaseous reaction C2H4(g) + HCl(g)  C2H5Cl(g) (g) be required from the bond enthalpy data. This may be calculated as follows :  Enthalpy required to  H =  break reactants into  gasesous atoms

= =

  Enthalpy released to        form products from the    gasesous atoms 

[4C–H + C=C + H–Cl] + [–5C–H – C–C – C–Cl] (C=C + H–Cl) – (C–H + C–C + C–Cl)

RESONANCE ENERGY : H°resonance = H°f, experimental – H°f, calclulated = H°c, calclulated– H°c, experimental

PAGE _ 38

Q.10

Sol.

Ex.4

Sol.

Calculate the heat of formation of benzene from the following data, assuming no resonance. Bond energies : C – C = 83 kcal C C = 140 kcal C – H = 99 kcal Heat of atomisation of C = 170 .9 kcal Heat of atomisation of H = 52.1 kcal C C C We have to calculate H for the reaction H = ? 6C (s) + 3H2(g)  C6 H6 (g) C C C For reactants : Heat of atomisation of 6 moles of C = 6 ×170.9 kcal heat of atomisation of 6 moles of H = 6 × 52.1 kcal For products : Heat of formation of 6 moles of C – H bonds = – 6 × 99 Heat of formation of 3 moles of C–C = – 3 × 83 Heat of formation of 3 moles of C C bonds = –3 × 140 on adding, we get heat of formation of C6H6,i.e., H = 6 ×170.9 + 6 × 52.1– 6 × 99 – 3 × 83 × 140 = 75.0 kcal

Find the bond enthalpy of S – S bond from the following data : C2H5 – S – C2H5(g) C2H5 – S – S – C2H5(g) S(g) Given that

fH° = – 147.23 kJ mol–1 fH° = –201.92 kJ mol–1 fH° = 222.80 kJ mol–1

(i) 4C(s) + 5H2(g) + S(s)  C2H5 – S – C2H5(g)

fH° = –147.23 kJ mol–1

(ii) 4C(s) + 5H2(g) + 2(s)  C2H5 – S – S – C2H5(g) Subtracting Eq. (i) from Eq. (ii), we get

fH° = –201.92 kJ mol–1

C2H5 – S – C2H5(g) + S(s)  C2H5 – S – S – C2H5(g) Adding to this, the following equation

fH° = –54.69 kJ mol–1

S(g)  S(s)

fH° = –222.80 kJ mol–1

We get C2H5 – S – C2H5(g) + S(g)  C2H5 – S – S – C2H5(g) fH° = –277.49 kJ mol–1 In the last equation 277.49 kJ of heat id evolved because of the S–S bond formation. Hence, the bond enthalpy of S – S is 277.49 kJ mol–1. Diagrammatically, we may represent the above calculation as follows :

H

H | C | H

H | C | H

S

H | C | H

H | C | H

H(g) + S(s)

2C  C + 10C  H + 2C  S

4C(g)

+

10H(g) +

H

H

H | C | H

H | C | H

vapH(S)

S(g) + S(g)

S

S

H | C | H

H | C | H

H(g)

-2C  C - 10C  H -2C  S - S  S

H

H

H

C

C

H

H

S

S

H

H

C

C

H

H

H

PAGE _ 39

According to Hess’s law H° = Enthalpy involved in bond breaking – Enthalpy involved in bond making = [2C – C + 10C–H + 2C–S + Dvap H(S)] + [–2C–C – 10C–H – 2C–S – S–S] = vapH°(S) – S–S or S–S = vapH°(S) – H° = vapH°(S) – [fH°(C2H5 – S – S – C2H5] – fH°(C2H5 – S – C2H5)] = [222.80 – {201.92 – (–174.23)}] kJ mol–1 = 277.49 kJ mol–1 Ex.5

Using the bond enthalpy data given below, calculate the enthalpy change for the reaction C2H4(g) + H2(g)  C2H6(g) Data :

Sol.

Bond Bond enthalpy C–C 336.81 kJ mol–1 C=C 606.68 kJ mol–1 C–H 410.87 kJ mol–1 H–H 431.79 kJ mol–1 Diagrammatically, we any represent the given reaction as follows :

H

H H(g) H

(g) + H

C=C H

H

H Break bonds

2C(g) +

4H(g)

H | C | H

Break bonds

+

2H(g)

H | C | H

H

make bond

H

H

H

C

C

H

H

H

The heat required to dissociate C2H4(g) and H2(g) into the gaseous atoms is For breaking 1C=C 606.68 kJ mol–1 For breaking 4C–H 4 x 410.87 kJ mol–1 For breaking 1H–H 431.79 kJ mol–1 Total

Ex.6

Sol.

2681.95 kJ mol–1

Thus H = –2802.03 kJ mol–1 + 2681.95 kJ mol–1 = –120.08 kJ mol–1 From the following data : Enthalpy of formation of CH3CN = 87.86 kJ mol–1 Enthalpy of formation of C2H6 = – 83.68 kJ mol–1 Enthalpy of sublimation of graphite = 719.65 kJ mol–1 Enthalpy of dissociation of nitrogen = 945.58 kJ mol–1 Enthalpy of dissociation of H2 = 435.14 kJ mol–1 C – H bond enthalpy = 414.22 kJ mol–1 Calculate (i) C – C and (ii) C  N (i) C – C : We have C2H6(g)  2C(graphite) + 3H2(g)

H1 = 83.68 kJ mol–1

2C(graphite)  2C(g)

H2 = 2 x 719.65 kJ mol–1

3H2(g)  6H(g) Adding, we get C2H6(g)  2C(g) + 6H(g)

H3 = 3 x 435.14 kJ mol–1 H4 = 2828.4 kJ mol–1

PAGE _ 40

Now H4 = C–C + 6C–H Therefore C–C = [2828.4 – 6 x 414.22] kJ mol–1 = 343.08 kJ mol–1 Diagrammatically, the above calculations may be represented as follows :

2C(graphite) + 2H2(g)

H2

fH

H

H | C | H

H

-6C H - C C

H3

2C(g) +

H | C | H

6H(g)

H

H

H

C

C

H

H

H

Applying Hess’s law, we get fH = H2 + H3 – 6C–H – C–C –83.68 kJ mol–1 = (2 x 719.65 + 3 x 435.14 – 6 x 414.22) kJ mol–1 – C–C C–C = (1439.3 + 1305.42 – 2485.32 + 83.68) kJ mol–1 = 343.08 kJ mol–1 (ii) C  N : We have CH3CN(g)  2C(graphite) +

3 1 H (g) + N2(g) 2 2 2

H1 = –87.86 kJ mol–1

2C(graphite)  2C(g)

H2 = 2 x 719.65 kJ mol–1

3 H (g)  3H(g) 2 2

H3 =

3 x 435.14 kJ mol–1 2

H4 =

1 x 945.58 kJ mol–1 2

1 N (g)  N(g) 2 2 Adding, we get

CH3CN(g)  2C(g) + 3H(g) + N(g) H5 = 2476.94 kJ mol–1 Now H5 = C–C + 3C–H + C  N Therefore C  N = [2 476.94 – 3 x 414.22 – 343.08] kJ mol–1 = 891.2 kJ mol–1 Diagrammatically, the above calculations may be represented as follows :

2C(graphite) + 3/2H2(g) + 1/2 N2(g)

H2

H3

fH

H

H | C | H

H4

C

N(g)

–3CH – CC – C C H

2C(g)

+

3H(g)

+

N(g)

H

C

C

H

H Applying Hess’s law, we get fH = H2 + H3 + H4 – (3C–H + C–C + CN)

3 1 x 435.14 + x 945.58 – 3 x 414.22 – 343.08 – C  N 2 2 = (1439.30 + 652.71 + 472.79 – 1242.66 – 343.08 – 87.86) kJ mol–1 = 891.2 kJ mol–1

87.86 kJ mol–1 = 2 x 719.65 + C  N

PAGE _ 41

Ex.7

Calculate the bond enthalpy C – H from the following data at 298 K : Enthalpy of combustion of methane H = –890.36 kJ mol–1 Enthalpy of combusion of C(graphite) H = –393.51 kJ mol–1 H2(g) +

Sol.

1 O (g)  H2O() 2 2

H = –285.85 kJ mol–1

Enthalpy of dissociation of H2(g) Enthalpy of sublimation of C(graphite) We have

H = 435.93 kJ mol–1 H = 716.68 kJ mol–1

(i) CH4(g) + 2O2(g)  CO2(g) + 2H2O()

H298K = –890.36 kJ mol–1

(ii) C(graphite) + O2(g)  CO2(g)

H298K = –393.51 kJ mol–1

(iii) H2(g) +

1 O (g)  H2O() 2 2

H298K = –285.85 kJ mol–1

(iv) H2(g)  2H(g)

H298K = 435.93 kJ mol–1

(v) C(graphite)  C(g) If we do the following manipulations Eq. (i) – Eq. (ii) – 2Eq. (iii) + 2Eq. (iv) + Eq. (v)

H298K = 716.68 kJ mol–1

We get CH4(g)  C(g) + 4H(g) The corresponding H298K is given is given by H298K = [–890.36 + 393.51 + 2 x 285.85 + 2 x 435.93 + 716.68] kJ mol–1 = 1 663.39 kJ mol–1 that is, 1 663.39 kJ is required to dissociate one mole of CH4(g) into gaseous atoms in which 4 mol of C – H bonds are broken. Therefore C–H = Ex.8

1663.39 kJ mol 1 = 415.85 kJ mol–1 4

Using the bond enthalpy data given below, estimate the enthalpy of formation of gaseous isoprene CH2  C  CH  CH2 | CH3

Data

Sol.

Bond enthalpy of C – H bond = 413.38 kJ mol–1 Bond enthalpy of C – C bond = 347.69 kJ mol–1 Bond enthalpy of C = C bond = 615.05 kJ mol–1 Enthalpy of sublimation of carbon (graphite) = 718.39 kJ mol–1 Enthalpy of dissociation of H2(g) = 435.97 kJ mol–1 For isoprene, we have to form 2C – C bonds ; 2C = C bonds and 8C – H bonds For which energy released is [2(–347.69) + (–615.05) + 8(–413.38)] kJ mol–1 = 5 232.52 kJ mol–1 that is, H (from gaseous atoms) = –5 232.52 kJ mol–1 The reaction corresponding to this is 5C(g) + 8H(g)  C5H8(g) fH1 = – 5232.52 kJ mol–1 But we want fH corresponding to the following equation 5C(graphite) + 4H2(g)  C5H8(g) fH = ? This can be obtained by the following manipulations : 5C(g) + 8H(g) C5H8(g) rH2 = – 5232.52 kJ mol–1 5C(graphite)  5C(g) rH3 = 5 × 718.39 kJ mol–1 4H2(g)  8H(g) rH4 = 4 × 435.97 kJ mol–1 Adding, we get 5C(graphite) + 4H2(g)  C5H8(g) fH = 103.31 kJ mol–1

PAGE _ 42

Diagrammatically, the above calculations may be represented as follows.

5C(graphite) + 4 H2(g) 5fH3

 fH

CH2 = C – CH2(g)

4fH4 5C(g) + 8H(g)

CH3 –2C=C – 2CC– 8C

Applying Hess’s law, we get f H = 5H3 + 4H4 – 2C = C – 8C – H = (5 x 718.39 + 4 x 435.97 – 2 x 615.05 – 2 x 347.69 – 8 x 413.38) kJ mol–1 = 103.31 kJ mol–1

LECTURE#13 Ex.9

Sol.

Calculate the resonance energy of benzene compared with one Kekule structure. Given the following data fH°(methane, g) = –74.85 kJ mol–1 fH°(ethane, g) = –84.68 kJ mol–1 fH°(ethylene, g) = 52.3 kJ mol–1 fH°(benzene, g) = 82.93 kJ mol–1 Enthalpy of sublimation of carbon (graphite) = 718.39 kJ mol–1 Dissociation enthalpy of H2 = 435.89 kJ mol–1 In order to calculate the resonance energy of benzene, we need to compute fH° from the bond enthalpy data. For this, we need C – C, C = C and C – H bond enthalpies. These can be calculated as follows : (i) Bond enthalpy of C – H from fH° (methane) : We have CH4(g)  C(graphite) + 2H2(g)

H° = +74.85 kJ mol–1

C(graphite)  C(g)

H° = +718.39 kJ mol–1

2H2(g)  4H(g) ; Adding, we get,

H° = 2 x 435.89 kJ mol–1

CH4(g)  C(g) + 4H(g) Now H1° = 4C–H, therefore

H1° = 1 665.02 kJ mol–1

C–H =

1665.02 kJ mol 1 = 416.255 kJ mol–1 4

(ii) Bond enthalpy of C – C from fH°(ethane) : We have C2H6(g)  2C(graphite) + 3H2(g)

H° = 84.68 kJ mol–1

2C(graphite)  2C(g)

H° = 2 x 718.39 kJ mol–1

3H2(g)  6H(g) Adding, we get Now Thus

H° = 3 x 435.89 kJ mol–1

C2H6(g)  2C(g) 6H(g) H2° = 2 829.13 kJ mol–1 H2° = C–C + 6C–H C–C = [2 829.13 – 6 x 416.255] kJ mol–1 = 331.60 kJ mol–1

(iii) Bond enthalpy of C = C from fH° (ethylene) : we have C2H4(g)  2C(graphite) + 2H2(g)

H° = –52.3 kJ mol–1

2C(graphite)  2C(g)

H° = 2 x 718.39 kJ mol–1

2H2(g)  4H(g) Adding, we get Now

C2H4(g)  2C(g) + 4H(g) H3° = C=C + 4C–H

H° = 2 x 435.89 kJ mol–1 H2° = 2 256.26 kJ mol–1

PAGE _ 43

Therefore C–C = [2 256.26 – 4 x 416.255] kJ mol–1 = 591.24 kJ mol–1 (iv) fH° (benzene) from the bond enthalpy data : We have 6C(g) + 6H(g)  C6H6(g) H° = –(3C–C + 3C=C + 6C–H) = –5 266.05 kJ mol–1 6C(graphite)  6C(g) H° = 6 x 718.39 kJ mol–1 3H2(g)  6H(g) H° = 3 x 435.89 kJ mol–1 Adding, we get 6C(graphite) + 3H2(g)  C6H6(g) H2° = +315.96 kJ mol–1 (v) Resonance energy of benzen(g) : Actual value of fH° = 82.93 kJ mol–1 Calculated value of fH° = 351.96 kJ mol–1 Thus, benzene becomes more stable by +269.03 kJ mol–1. Therefore, its resonance energy is 269.03 kJ mol–1. Ex.10

Sol.

The enthalpy of formation of ethane, ethylene and benzene from the gaseous atoms are –2 839.2, –2 275.2 and –5 536 kJ mol–1, respectively. Calculate the resonance energy of benzene, compared with one Kekule structure. The bond enthalpy of C–H bond is given as equal to 410.87 kJ mol–1. Bond enthalpy of C – C bond = Enthalpy required to break C2H6 into gaseous atoms –6 x Bond enthalpy of C – H bond = 2 839.2 kJ mol–1 –6 x 410.87 kJ mol–1 = 373.98 kJ mol–1 Bond enthalpy of C = C bond = Enthalpy required to break C2H4 into gaseous atoms –4 x Bond enthalpy of C – H bond = 2 275.2 kJ mol–1 – 4 x 410.87 kJ mol–1 = 631.72 kJ mol–1 For the formation of benzene having Kekule structure, we have to form 3C – C bonds, 3C = C bonds and 6 C – H bonds for which enthalpy released is [3(–373.98 + 3 (–631.72) + 6 (–410.87) kJ mol–1 = –5 482.32 kJ mol–1 But the given value of fH is fH(actual) = –5 536 kJ mol–1 Hence the resonance energy of benzene compared with one Kekule structure = fH (actual) – fH (Kekule structure) = (–5 536 + 5 482.32) kJ mol–1 = –53.68 kJ mol–1

RELATION BETWEEN ENERGY AND ENTHALPY OF A REACTION : rH = rU + (vg)RT where vg is the change in the stoichiocmetric number of gaseous species in going from reactants to products. It should be noted that while computing vg of a reaction, only the stoichiometric numbers of gaseous is counted and those of liquids and solids are completely ignored. Case - III For chemical reactions : aA (s) + bB() + cC(g) a’A’(s) + b’B’ () + c’ C’ (g)

 Pext. dv

W =– =

P

ext.

v f  v i 

= – Pext. [(VA’(s) + (VB’() + (VC’(g) – (VA’(s) + (VB’() + (VC’(g)] =  W = – Pext. [VC’(g) – VC’(g)]

 n c 'RT n c RT  Pext. (nc '  nc )RT  =– = – Pext.  P Pext. P ext   ext W = – (nc’ – nc)RT

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

 

W = – ng RT du = dq + dw if at constant pressure dq = dH du = dH – pdv du = dH – ng RT dH = du + ng RT

KIRCHOFF’S EQUATION (Variation of Hº with temprature) Since the enthalpy (or standard enthalpy) of a substance is dependent on state of the substance, value of enthalpy of a substance changes with temperature and hence the enthalpy change of reaction is also dependent on temperaturte at which the reaction is being carried out. This change is enthalpy change (or E, of reaction is carried out at constant volume) is represented by Kirchoff’s Equations. P + Q  R + S at temperature T1 let the standard enthalpy of reaction be H°1 , then H1° = H°m (R,T1) + H°m (S, T1) – H°m (P, T1) – H°m (Q, T1) If the same reaction is carried out at temperature T2, then H2° = H°m (R,T2) + H°m (S, T2) – H°m (P, T2) – H°m (Q, T2) Then the change in enthalpy (or difference in enthalpy at these two temperatures) H° = H2° – H1° =  {Hm°} (R,T2) – Hm° (R,T1)} +  {Hm°} (,T2) – Hm° (,T1)} –  {Hm°} (P,T2) – Hm° (P,T1)} -  {Hm°} (Q,T2) – Hm° (Q,T1)} Hm° (R, T2) – Hm° (R,T1) = CP, R (T2 – T1) = Heat requirfed at constant pressure to increase temperature of one mole of R from T1 to T2 Similarly Hm° (S,T2) – Hm° (S,T1) = CP,S (T2 – T1) Hm° (P,T2) – Hm° (P,T1) = CP,P (T2 – T1) and Hm° (Q,T2) – Hm° (Q,T1) = CP,Q (T2 – T1) so H° = H2° – H1° =  CP,R (T2 – T1) + CP,S(T2 – T1) –  CP,P (T2 – T1) + CP,Q(T2 – T1) = [ CP,R + CP,S –  CP,P – CP,Q] (T2 – T1) = CP (T2 – T1) CP = CP, R + CP,S – CP,P – CP,Q = Difference in molar heat capacities of products and reactants. so

H2° = H1° +

 C

P .dT

for example for the reaction N2 (g) + 3H2 (g)  2NH3 (g) H2° = H1° + CP (T2 – T1) where

CP = 2CP, N3  CP, N2  3CP, H2

* for a constant volume reaction

E 02  E10  C V .dT



Ex.

0 Calculate H 400 K for the reaction

1 N (g) + O2 (g)  NO2 (g) 2 2

H0300 K = – 35 KJ mole–1

Given , CP (NO2, g) = 35 JK–1 mole–1 CP (O2, g) = 30 JK–1 mole–1 CP (N2, g) = 30 JK–1 mole–1 Ex.

If Hm° (H2O, , 298 K) = 200 kJ/mole, Hm° (H2 ,g,298 K) = 50 kJ/mole, Hm° (O2, gas, 298 K) = 80 kJ/mole (a) Calculate entalpy change of the reaction 2H2 (g) + O2(g)  2H2O() at 298 K (b) If CP (H2O, ) = 18 cal/mole K and assuming H2, O2 are behaving as ideal gases then calculate enthalpy change of above reaction at 348 K.

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