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DEPARTMENT OF APPLIED CHEMISTRY

LECTURE NOTES

6151- ENGINEERING CHEMISTRY-II

UNIT II

CHEMICAL THERMODYNAMICS

Unit syllabus: Terminology of thermodynamics - Second law: Entropy - entropy change for an ideal gas, reversible and irreversible processes; entropy of phase transitions; Clausius inequality. Free energy and work function: Helmholtz and Gibbs free energy functions (problems); Criteria of spontaneity; Gibbs-Helmholtz equation (problems); Clausius-Clapeyron equation; Maxwell relations – Van’t Hoff isotherm and isochore(problems).

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SESSION : Terminology of thermodynamics.

The term thermodynamics means flow of heat. In general it deals with the inter conversion of various kinds of energy in physical and chemical systems.

Thermodynamics

1. Predicts the feasibility of a physical process or chemical reaction under given condition of temperature and pressure.

2. Predicts whether a chemical reaction would occur spontaneously or not under a given set of conditions

3. Helps to determine the extent to which a reaction would takes place.

Limitations of thermodynamics.

1. It predicts the extent to which a reaction can take place .however it does not say anything about the rate.

2. It applies only to matter in bulk and not to individual atoms or molecules.

Terms used in thermodynamics.

System and Surroundings: Any part of the universe which is selected for thermodynamic study is called system and the rest of the Universe in its neighborhood is known as surroundings.The system is separated from the surroundings by a real or imaginary boundary through which exchange of energy or matter may take place.

Types of System

Homogeneous system: A system is said to be “homogeneous” if it consists of only one phase and uniform thought. Ex. a solution of sugar in water.Heterogeneous system: When a system consists of two or more phases and is not uniform throughout it is called heterogeneous. Ex .Ice in water, chloroform in water.Open system: A system which can exchange both energy and matter with the surroundings.Ex. beaker of water.Closed system: A system which can exchange energy but not matter with the surroundings. Ex Water in as stoppered bottle.Isolated System: In system in which neither matter nor energy can exchange with its surroundings.

Properties of a system

The properties associated with a macroscopic system are called thermodynamic properties or variables.Extensive properties: These are thermodynamic properties which depend on the quantity of matter specified in the system e.g. mass, volume energy etc.Intensive properties: These are thermodynamic properties which depend on characteristics of matter but independent of its amount e.g. pressure, temperature, viscosity, density m.p, bp etc.

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Thermodynamic process:

The method by which the state of a system is changed is called a “Process”. It can be effected by changing any one of the state variables viz P,T,C etc.

Isothermal process: It is a process carried out at const temperature. Exchange of heat takes place between system and surroundings dT = 0

Adiabatic process: It is a process in which no exchange of heat takes place .The temperature of the system may increase or decrease during adiabatic process. dQ =0

Isobaric process : Carried out at constant pressure

Isochoric proces: Carried out at constant volume.

Reversible process: It is a process which takes place infinitesimally slowly so that the system is in thermodynamic equilibrium at any instant of the change. Since the process is carried out extremely slowly the properties of the system remain virtually unchanged and the direction may be reversed by small change in a variable like temperature, pressure etc. The driving force is greater than the opposing force only by a infinitesimal quantity and hence the process wouldrequire infinite time for its completion.

Irreversible process: It is a process which takes place rapidly or spontaneously so that it is equilibrium with its surroundings. The driving force differs from the opposing force by a large amount and hence it cannot be reversed unless some external force is applied. All the natural process re irreversible.

Reversible Process Irreversible process1 Driving force and opposing force

differ by small amount.Driving force and opposing force differ by a large amount

2 It is a slow process It is a rapid process3 The work obtained is more The work obtained is less4 It is am imaginary process It is a real process5 It consists of many steps It has only two steps i.e initial and

final6 It occurs in both the directions It occurs in only one direction7 It can be reversed by changing

thermodynamic variablesIt can be reversed.

Internal Energy: U or E

The energy stored in a substance by virtue of its constituent atoms and molecules is called Internal energy.It is the sum of vibrational energy, rotational energy, electronic energy etc.Internal Energy change (∆ )It is the difference in the internal energies of initial and final states of the system.∆ = +But for the chemical reactions ∆ = − ∆ = −

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Enthalpy or Heat content of a system (H)

It is “the heat content of the system: or “The sum of internal energy and pressure-Volume changework done) of s system , under a particular set of conditions”Mathematically = +Unit: KJ mol-1

Enthalpy change : It is the difference in the enthalpy of initial and final stages of the system. ∆ = −For the chemical reaction ∆ = −We know that = +Then ∆ = 2 + 2 − ( 1 + 1) = ( 2 − 1) + ( 2 − 1) ∆ = ∆ + ∆At constant volume ∆ = 0 ∆ = ∆

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SESSION : Second law of thermodynamics – various statements.

Second law of thermodynamics – various statements

There are three laws of thermodynamics : zeroth law, first law and second law.

Zeroth law of thermodynamics discusses about the thermal equilibrium between three bodies.

If two systems A and B are in thermal equilibrium with the system C, then A and B are in thermal equilibrium with each other..

First law of thermodynamics:

The law of conservation of energy.

Energy can be neither created nor destroyed, but it can be converted from one form to another.

The mathematical form of First law of thermodynamics is

ΔE = q – w

where ΔE, q and w represent respectively the change in internal energy, quantity of heat supplied and work done. For a small change,

dE = dq – dw ----- (1)

Work done (dw) can be represented as PdV in terms of pressure-volume changes for the gas. i.e dw = PdV ----- (2)

Second law of thermodynamics :

Second law of thermodynamics predicts the direction of of heat flow

It explains the reason for why complete conversion of energy into work is not possible.

It predicts the feasibility of a process

It introduces two new thermo dynamic functions entropy and free energy to explain spontaneity.

Clausius statement : It states that heat cannot flow itself from a cold body to a hot body spontaneously without the intervention of an external energy.

Kelvin statement: It is impossible to take heat from a hot body and convert it completely into work by a cyclic process without transferring a part of heat to cold body.

II law in terms of entropy: A spontaneous process is always accompanied by an increase in entropy of the universe.

Other statements:

All spontaneous process are irreversible

It is not possible to construct a machine functioning in a cycle which can convert heat completely into equivalent amount of work without produces changes elsewhere.

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SESSION : Entropy concept - entropy change for an ideal gas.

ENTROPY: Clausius introduced a new thermodynamic function called entropy. It is a measure of degree of disorder or randomness in a molecular system. It is also considered as a measure of unavailable form of energy.

The change in entropy ∆ of system is equal to the ratio of heat change to the temperature (T) of the reversible cyclic process.

∆ =

Unit: Cal /deg or JK-1 mol-1

Significance of entropy:

Measure of disorder of the system: All spontaneous process are accompanied by increase in entropy as well as increases in the disorder .Increase in entropy implies increase in disorder.Measure of probability: An irreversible process tend to proceed from less probable state to more probable state. Since entropy increases in a spontaneous process, entropy may be defined as a function of probability of thermodynamic state.

Entropy and unavailable energy: When heat is supplied to the system, some portion of heat is used to do some work. This portion of heat is available energy. The remaining portion is called unavailable energy. Hence entropy is defined as unavailable energy per unit temperature.

Entropy = Temperature

Entropy change of an ideal gas at constant temperature

From First law

= − -------------------------(1) From Second law

dS = -------------------------(2)

Using 1 and 2

= -------------------------(3)

But = and = ∴ = + -------------------------(4)

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At constant temperature (isothermal) process = 0 ∴ = 0 + -------------------------(5)

Using ideal gas equation = : = ∴ = = -------------------------(6)

ℎ ∫ = ∫ 2

( 2 − 1) = ∆ = -------------------------(7)

For n mole

∆ = . ------------------------(8)

1 1 = 2 2 =Equation 8 becomes ∆ = .

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SESSION : Entropy change for reversible and irreversible processes; entropy of phase transitions

Entropy change for a reversible ( non spontaneous)process:

Consider an isothermal and reversible expansion of an ideal gas.If the system absorbs q amount of heat from the surroundings at temperature T , the increase in entropy of the system is given by

∆ =

But the entropy of the surroundings decrease , because the surroundings loose the same of heat q i.e ∆ =Hence, the net change in the entropy is given by

∆ = ∆ + ∆ ∆ = +

∆ = 0

i.e in a reversible isothermal process, there is no net change entropy.

Entropy change for a irreversible ( spontaneous) process:

Consider a system maintained at higher temperature T1 and its surrounding maintained at a lower temperature T2.If q amount of heat passes irreversibly from the system to surroundings. then,

Decrease in entropy of the system , ∆ =Increase in entropy of the surroundings. ∆ =Net change in entropy is given by ∆ = ∆ + ∆ ∆ = +

= q

Since T1 > 2 : 1 − 2 is positive

∆ > 0In an irreversible process the entropy of the system increases

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Entropy change accompanying change of phase.

When there is a change of state from solid to liquid or liquid to vapour or solid to vapour there is change in entropy.

Solid to liquid:

Let us consider the process of melting of 1 mole of substance being carried out reversibvly .It would absorb molar heat of fusion at te4mperature equal to its melting point.

∆ = ∆HWhere ∆ - molar heat of fusion - fusion temperature.

Liquid to vapour

One mole of a substance changes from liquid to vapour state reversibly at its boiling point Tb

Under constant pressure .If ∆ , molar heat of vaporization then the entropy change accompanying the process

∆ = ∆H

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SESSION : Clausius inequality. Free energy and work function –concepts.

CLAUSIUS INEQUALITY OR THEOREM

Claussius theorem is a mathematical explanation for II law of thermodynamics.

It states that the cyclic integral of is always less than or equal to zero

≤Where = Differential heat transfer at the system boundary during a cycle.

T= Absolute temperature at the boundary. ∮ = Integration over the entire cycle.The Clausius inequality is valid for all cyclic, reversible or irreversible process.Consider two heat engines, one reversible and other irreversible. We assume the both the engines absorb same amount of heat QH from the heat source having the temperature of TH. Both the engines reject the heat QL to a heat sink at a temperature TL . Applying the first law of thermodynamics to both the engines.

W rev = QH – QL.rev

Wirrev = QH – QL.irrev

Since the reversible engine is more efficiemt then the irreversible engine it must must reject less heat( QL.rev) to the thermal sink than thatof rejected by the irreversible engine QL.irrev).So the reversible heat engine produces more work than the irreversible heat engine for the same heat input QH

W rev = QH – QL.rev > Wirrev = QH – QL.irrev

For reversible heat engine (Carnot)Consider the first reversible heat engine.The reversible heat transfer can only occur isothermally ( at constant T) ,so the cylic integral of the heat transfer dived by the temperature can be written as

∮ = - .. =

∮ = 0

For irreversible heat engine.The two heat engines (reversible and irreversible ) have the same value of heat transfer from the thermal source QH .But the heat rejection QL is more in irreversible engine than the reversible one.QL.irrev> QL.rev

∮ = - . < 0Thus for any reversible or irreversible heat engines we obtain the Clausius inequality

≤

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Free energy and work function –concepts.

In order to find out the spontaneity of a process ,we have to see the change in entropy of the system as well as surroundings .It is difficult to find out entropy of the surrounding everytime. So new thermodynamic functions are introduced , which can be determine more conveniently.

1.Helmholtz free energy A or Helmholtz work function

2.Gibbs free energy G or free energy

Helmholtz free energy or work function A

The work function A is defined as = −E- Energy content of the system

T- Absolute temperature

S- Entropy

For small change

∆ = ∆ − T∆SBy definition ∆ =

∆ = ∆ − (1)

According to I law of thermodynamics

∆ = - w (2)

Using 1 and 2 we get

∆ = − wrev

- ∆ =

Thus the decrease in function A gives the maximum work that can be obtained during an isothermal and reversible change

W represents the total work i.e expansion, electrical work etc. so it is also called work function.

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Gibbs free energy:

It is isothermally available energy present in the system

It is defined as = −G – available energy

H – total energy

TS – unavailable energy

For a small change.

∆ = ∆ − ∆But ∆ = ∆ + ∆

∴ ∆ = ∆ + ∆By definition ∆ = ∆ − ∆

∆ = ∆ − ∆ + ∆ ∆ = ∆ + ∆

For reversible isothermal change.

∆ = − ∆ = − + ∆ Or -∆ = − ∆

Where ∆ is work of expansion

−∆ = − − = net or useful work

−∆ =W useful

The decrease of free energy of a process at constant temperature and pressure is useful work obtainable from the system.

Standard freee energy change. ( ∆ )

The free energy change for a process at 25 oC in which the reactants are converted into products in their standard states

Thus. ∆ = ∑ ( ) − ∑ ( )

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SESSION : Criteria of spontaneity – Gibbs-Helmholtz equation .

Gibbs –Helmholtz equation (in terms of free energy and enthalpy)

(Relation between∆ ∆ )

Free energy (G) is related with enthalpy (H) as

= − --------------------------------------- (1)

Enthalpy (H) is related with internal energy (E) as

H = E + PV -------------------------(2)

∴ = + − ---------------(3)

Upon differentiation.

= + + − − --------------(4)

The first law of thermodynamics equation for an infinitesimal change may be written as

= + -----------------(5)

If work done ,dW is only due to expansion than

= + -----------------(6)

For reversible process = or = = = = -----------------(7)

Combining (4) and (7) we get

= + + − ( + ) − = − -----------------(8)

At constant pressure dP = 0 and the above equation 8 becomes

= −or

= -S -----------------(9)

Substituting (9) in (1)

= − = + -------------------------- (10)

This is one form of Gibbs –Helmholtz equation.

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Let G1 initial free energy of a system at T and G1 and dG1 = initial free energy of the system at T + dT

where dT is infinitesimally small and pressure is constant.

1 = − 1 where S1 is the entropy of the system in the initial state.

Now suppose that the free energy of the system in final state is G2 at T. Let G2 + d G2 is the free energy

of the system at T + dT in the final state then.

2 = − 2 Where S2 is the entropy of the system in the final state.

Subtracting (11) from (12)

2 − 1 = −( 2 − 1) ( ∆ ) = −∆ At constant pressure

(∆ )= - ∆

We know∆ = ∆ − ∆ − ∆ =

∆ ∆ ∆ ∆

= (∆ )

∆ = ∆ + (∆ )

This is called Gibbs –Helmoltz equation in terms of free energy and enthalpy at constant pressure.

Another form of Gibbs – Helmholtz equation is obtained by the differentiation of the above equation with

respect to temperature, and given by

Ә (∆G/T) / ӘT = - ∆H/RT2, which represents the variation of ∆G with the temperature

Application of Gibbs –Helmoltz equation.

1. Calculation of enthalpy change (∆ ) for the cell reaction .

If a Galvanic cell produces nF coulombs of electricity in a reversible manner, it must be equal to the

decrease in the free energy ∆ −∆ = JWhere n- no of electrons involved in the process. F- Faraday 96500 coulombs. E = EMF in V

Gibbs –Helmoltz equation

∆ = ∆ + (∆ )

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- =∆ + (∆ )

- =∆ + ( )

- =∆ − Or

∆ = − +∆ = − −

2.Calculation of emf of the cell

- =∆ −=

∆ +3.Calculation of entropy change (∆ )

∆ = ∆ − ∆ ∆ = ∆ + (∆ )

Comparing the above two equations

−∆ = (∆ )-----------A

−∆ =Substituting in A we get

∆ = ( )or ∆ =

Spontaneous and Non spontaneous process:

The physical or chemical changes which proceed by themselves without the intervention of any external agents are known as spontaneous process. All natural process are spontaneous. All spontaneous process proceed in one direction and are thermodynamically irreversible.

Examples:1. Heat flow from a hotter to a colder body till they attain thermal equilibrium.2. Water flows by itself from a higher level to a lower level.3. The expansion of gas into an evacuated space.

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The process which proceed in both directions are called spontaneous or reversible process.Free energy and spontaneity

∆ = ∆ − ∆The sign of ∆ depends on magnitude of ∆ ∆ .For a spontaneous process.

> or >I law states that = +∴ > +

Or − ( + ) > 0 + — < 0 ( + − ) < 0 ( − ) < 0 dG <0 = −

Thus spontaneous process involve a decrease in decrease in free energy.The free energy change (∆ ) is the criterion for predicting spontaneity or feasibility of a reaction. ∆ = ∆ − ∆The sign of ∆ depends on the sign and numerical value of ∆ ∆If ∆ − the process is feasible∆ + the process is not feasible∆ the process is reversible i.e exists in equilibrium.

Conditions for spontaneity

∆ ∆ ∆ = ∆ − ∆ Nature of process

(− ) exothermic + ve -ve spontaneous(− ) exothermic -ve - ve (low T) spontaneous(− ) exothermic -ve + ve (high)T Non spontaneous(+ )endothermic + + ve (at low)T Non spontaneous(+ ) endothermic + −ve (high)T spontaneous(+ ) endothermic + + ve Non spontaneous

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SESSION : Clausius-Clapeyron equation

Clausius – Claypeyron equation:

Consider a system consisting of only 1 mole of substance in two phases A and B. The free energies of the substance in two phases A and B be GA and GB

Let the temperature and pressure of the system be T and P respectively.The system is in equilibrium ,so there is no change in free energy GA = GB

If the temperature of the system rised to T + dT and Presure becomes P + dPGA + d GA = GB + d GB

G = H – TSG= E + PV –TSDiffentiating the above equation.dG = dE + PdV + Vdp-TdS –SdTdG = VdP-SdTdGA = VAdP –SA dT ------------ (a)dGB = VBdP –SB dT ------------(b)Where VAand VB are the molar volume of phases A and B respectively.SA and SB are molar entropies.Since GA = GB

dGA = dGB

Substituting in equation (a) and (b) VAdP –SA dT = VBdP –SB dT

= or

∆∆ = ∆ = = ∆This is Clayperon equation.

The above equation was modified by Claussius and called Claussius –Clayperon equation.Let us consider the following equilibrium

Solid ↔ vapour Vv ›› Vs

Liquid↔ vapour Vv ›› Vl

= ∆

= ∆

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PV = RT ; V= RT /P

= ∆ 2 P

= ∆2

Integrating between the limits. P1 and P2 coressponding to T1 to T2

ln = ∆ 21

2

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= ∆.

This is Claussius –Clapeyron equation.

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SESSION : Maxwell relations – statement, explanation & derivation

MAXWELL RELATIONSHIPdE = q − PdVG = H − TSH = E + PV A = E − TSUsing the above fundamental equations any thermodynamic relationships can be derived.Note: E,S,H,A,G,T,P,V are all state functions .

= −dE = TdS − Pdv (dE) = TdSDifferentiating with respect to V at constant S∂2E∂S ∂V = ∂T∂V − − − − − ( )

(dE) = −PdVDifferentiating with respect to S at constant V∂2E∂V ∂S = − ∂P∂S − − − − − ( )From equations and

= −= +dH = TdS + VdP ∂H∂S = T

Differentiating with respect to P at constant S∂2H∂S ∂P = ∂T∂P − − − − − (3) ∂H∂P = V

Differentiating with respect to S at constant P∂2H∂P ∂S = ∂V∂S − − − − − (4)Since H is a state function and dH is an exact differential then from equation 3 and 4

== −

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G = E + PV − TSdG = TdS − PdV + PdV + VdP − TdS − SdTdG = VdP − SdT ∂G∂T = −S

Differentiating with respect to P at constant T∂2G∂T ∂P = − ∂S∂P − − − − − (5) ∂G∂P = V

Differentiating with respect to T at constant P∂2G∂P ∂T = ∂V∂T − − − − − (6)− =

= −dA = −PdV − SdT ∂A∂T = −S

Differentiating with respect toV at constant T∂2A∂T ∂V = − ∂S∂V − − − − − (7) ∂A∂V = −P

Differentiating with respect toT at constant V∂2A∂V ∂T = − ∂P∂T − − − − − (8)=

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SESSION : Van’t Hoff isotherm and isochore

= − = −= + = = − = −= − =

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Van’t Hoff isotherm

It gives the quantitative relationship between the free energy ∆ G and equilibrium constant K

Consider a general reaction aA + bB � cC +dD

We know G = H –TS = E+ PV –TS

Or dG = dE +PdV +VdP-TdS-SdT

But dq = dE + PdV and dS = dq/T

� dG = VdP-SdT

At Constant temperature dG = VdP

PV = RT or V = RT /P for one mole of a gas.

( ) = ---------(1)

Where dG is the free energy change for 1 mole of any gas at constant temperature.

Integrating (1) we get

∫ = ∫

G = G0 + RT ln P _________(2) G0 - standard free energy G = G0 When P= 1 atm

Let the free energies of A,B, C and D at their respective pressure PA ,PB ,PC and PD are GA ,GB,GC and GD respectively .

Then the free energy change for the above reaction is given by

∆G = G Products – G reactants.

∆G = (cGC + dGD ) –(aGA + bGB)

From (2)

aGA = aG0 A+ RT ln PA

bGB = bG0 B+ RT ln Pb

cGA = CG0 C+ RT ln PC

dGA = dG D+ RT ln PD

∆G = (cGC + dGD ) –(aGA + bGB )

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= ( CG0 C+ RT ln PC + dG D+ RT ln PD ) –( aG0

A+ RT ln PA + bG0 B+ RT ln Pb )

∆ = ∆ + ln ( ) ( )( ) ( )

Where ∆ G0 = standard free energy change of the reaction. At equilibrium ∆ G =0

0 = ∆ + ln ( ) ( )( ) ( )

∆ = − ln ( ) ( )( ) ( )

∆ = − ∆ = − + ( ) ( )

( ) ( )The above relationship is called Vant’t Hoff isotherm.

Vant’t Hoff Isochore.

The effect of variation of equilibrium constant with temperature is given Vant’t Hoff Isochore

∆G 0 = - RT ln Keq

ln Keq = -∆G0 / RT = -(∆H0-T∆S0) /RT = -∆H0/RT +∆S0/R

= − ∆1 + ∆

= − ∆2 + ∆

− = ∆ 0 −ln = ∆ 0 −

= ∆. −The above equation iscalled the Vant’t Hoff isochore

Problem 1.Calculate the change in entropy accompanying the isothermal expansion of 4 moles of an ideal gas at 300K until its volume has increased three times.

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∆S = 2.3030 nR log (V2/V1) JK -1

= 2.303 Xx 8.314 log 3

= 36.54 JK-1

Problem 2: Calculate the entropy change when 100 g of ice converted into water at 0oC. Latent heat of fusion of ice is 80Cal/g.

∆ = ∆

For 1 g of ice ∆S= 0.293 cal K-1 g-1

For 100 g of ice = 29.3 cal K-1 g-1

Problem 3: Gibbs free energy of a raction at 300K and 310 K are -29 k Cal and 29.5 k.cal respectively.Determine its ∆H and ∆S at 300 K

∆ = ∆ + (∆ )(∆ ) = ∆ − ∆ = - 29.5-(-29) = -0.5

=T2-T1 = 10 K

∆H = - 14kcal

∆ = − (∆ ) = 0.05 k.cal K-1

Problem 4: The equilibrium constant Kp for a reaction is 3.0 at 673 K and 4.0 at 773 K. Calculate the value of ∆H0 for the reaction.(R=8.314 JK-1)

= ∆. −K1= 3.0 K2= 4.0 T1 = 673 T2 =773

∆H= 12.490 kJ

Review Questions : (Section A)

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1. What are the differences between reversible process and irreversible process?2. What are intensive and extensive properties?3. What is meant by state function?4. What is internal energy? Mention its unit.5. What is enthalpy? How is it related with internal energy?6. Write the mathematical form of I law of thermodynamics.7. State II law of thermodynamics.8. Define entropy.9. Write the mathematical expression and unit for entropy.10. Define work function.11. Define standard free energy.12. What is the significance of decrease in free energy?13. What is meant by spontaneous process.?14. Mention any two applications of Gibb’s Helmholtz equation.15. Write any two Maxwell’s relationships.16. What is significance of Maxwell’s relationship?17. Write the integral form of Claussius –Clapeyron equation.18. Mention any two applications of Claussius –Clapeyron equation.19. Write the significance of Van’t Hoff equation.20. What is Claussius inequality?

Review Questions:(Section B)

1. Compare the the reversible process with the irreversible process.(May 2014)

2. Derive an expression for entropy change of an ideal gas at constant temperature .(Nov2002)

3. Explain Clausius inequality.4. Explain the significance of free energy.(May 2014,May 2015)

5. Derive Gibb’s Helmoltz equation and discuss its application. .(May 2005,Jan 2005 ,Jan 2014 ,May 2014, Dec

2014,May 2015)

6. Discuss the criteria for chemical reaction to be spontaneous .(Nov 2002, Jan 2014)

7. What is meant by VantHoff reaction isotherm? Derive the expression for a reaction isotherm of general reaction aA +bB →cC + dD. (Jan 2014,Dec 2014,May 2015) .

8. Derive the Claussius- Claperon equation. Discuss its applications.(Dec 2012,Dec 2014)

9. Derive all the four Maxwell’s relations. (May 2014,May 2015)

10. By combing Van’t Hoff isotherm and Gibb’s Helmholtz equation. Illustrate the effect of temperature on equilibrium Constant.(May 2005)

Problems for Practice:

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1. Calculate the entropy change for the reversible isothermal expansion of 10 moles of an ideal gas to 50 times its original volume at 298 K

2. Compute free energy change when 5 moles of an ideal gas expands reversibly and isothermally at 300K from an initial volume of 50 L to 100L.

3. Check whether the following reaction is spontaneous at 25 C and 11000 C C(S) + H2O(l) → CO(S) + H2 (g)Given that ∆ ∆ are 31400 Cal/mol and 32 Cal/deg at 25 C.

4. Gibbs free energy of a reaction at 300K and 310K are -29 K Cal and -29.5 K Cal respectively. Determine ∆ ∆ at 300 K

5. The emf of standard cadmium cell at 298 K is 1.01832 V and the temperature coefficient of emf of the cell is 5X 10 -5 VK-1Calculate the values of ∆ , ∆ ∆ for the cell reaction

6. The vapour pressure of water at 95oC and 100OCare 634 nm and 760 nm respectively. Calculate the molar heat of vapourisation of water.

7. The value of equilibrium constant for a reaction is found at 10,000 at 25oC calculate ∆ 0for the reaction

8. The equilibrium constant KP for the reaction N 2 + 3H2 ↔ 2NH3 is 1.64 X 10-4 atm at 400 o C What will be the equilibrium constant at 5000C if heat of reaction in the temperature range is (∆ ) = − 05 86 and R = 8.3014JK-1 mol-1

9. ∆G for a reaction at 300K is -16K.Cal, ∆H for the reaction is -10K.Cal. What is the entropy (∆S) of the reaction? What will be ∆G at 303 K?

10. At what temperature will water boil when the atmospheric pressure is 528 mm Hg? Latent heat of water is 545.5 cal/g.

11. Calculate the standard entropy change for the reaction A�B, if the value of ∆H0 = 28.4 kJ/mol and equilibrium constant is 1.8 X 10-7 at 298K.