Chapter 6: Chemical Equilibrium

• The fundamental expression governing spontaneity is the Clausius inequality, written in the form

TdS ≥ ᵭq

• The equality is satisfied only for a reversible process. Because ᵭq = dU – ᵭw,

TdS ≥ dU - ᵭw or, equivalently, -dU +ᵭw + TdS ≥ 0

• It is useful to distinguish between expansion work, in which the work arises from a volume change in the system, and nonexpansion work. The above Equation can be written as:

-dU – PexternaldV + ᵭwnonexpansion + TdS ≥ 0

• This equation expresses the condition of spontaneity for an arbitrary process in terms of the changes in the state functions U, V, S, and T as well as the path-dependent functions PexternaldV and wnonexpansion.

• For an isolated system, w = 0 and dU = 0. Therefore dS ≥ 0.

Chapter 6: Chemical Equilibrium

1

The Gibbs Energy and Helmholtz Energy• It is useful to consider transformations at constant temperature and either constant

volume or constant pressure.

• Note that constant T and P (or V) does not imply that these variables are constant throughout the process, but rather that they are the same for the initial and final states of the process.

• For the isothermal process, TdS = d(TS), and the inequality can be written in the form:

• The combination of state functions U – TS, which has the units of energy, defines a new state function that we call Helmholtz energy, abbreviated A.

• Using this definition, the general condition of spontaneity for isothermal processes becomes

dA ≤ ᵭwexpansion +ᵭwnonexpansion

• Because the equality applies for a reversible transformation, the above Equation provides a way to calculate the maximum work that a system can do on the surroundings in an isothermal process.

onnonexpansiexpansion

onnonexpansiexpansion

dwdwTS)-d(UdwdwTdSdU

lyequivalentor,

2

The Gibbs Energy and Helmholtz Energy• In discussing the Helmoltz energy, dT = 0 was the only constraint applied. We now

apply the additional constraint for a constant volume process, dV = 0.

• If only expansion work is possible in the transformation, then dwexpansion = 0, because dV = 0.

• The condition for spontaneity at constant T and V takes on a simple form using the Helmholtz energy rather than the entropy, if nonexpansion work is also not possible.

• Since chemical reactions are more commonly studies under constant pressure conditions, we next consider the conditions for spontaneity for isothermal transformations that take place at constant pressure, P = Pexternal.

• At constant pressure and temperature, PdV = d(PV) and TdS = d(TS). In this case the inequality can be written in the form

• The combination of state functions H -TS, which has the units of energy, defines a new state function called the Gibbs energy, abbreviated G.

• Using the Gibbs energy, the condition for spontaneity and equilibrium for an isothermal process at constant pressure becomes dG ≤ ᵭwnonexpansion.

onnonexpansidwTSHdTSPVUd )()(

0dA

3

The Gibbs Energy and Helmholtz Energy• For a reversible process, the equality holds, and the change in the Gibbs energy is a

measure of the maximum nonexpansion work that can be produced in the transformation.

• If the transformation is carried in such a way that nonexpansion work is not possible, the condition for spontaneity and equilibrium is

• What is the advantage of using the state functions G and A as criteria for spontaneityrather than entropy?

• The Clausius inequality can be written in the form

• dSsurroundings = - ᵭq/T, the Clausius inequality is equivalent to the spontaneity condition: dS + dSsurroundings ≥ 0

• ∆Ssurroundings = -dH/T for a constant T, P process.

• ∆G = ∆H -T∆S -∆G/T= -∆H/T + ∆S = ∆Ssurroundings + ∆Ssystem

• By introducing G and A, it is no longer necessary to consider the surroundings explicitly. Knowledge of G and A for the system alone is sufficient to predict the direction of natural change.

0dG

0TdqdS

4

• The following conclusions can be drawn based on the Equation G = H -TS- the entropic contribution to G is greater for higher temperatures- a chemical transformation is always spontaneous if H < 0 (an exothermic

reaction and S > 0- a chemical transformation is never spontaneous if H > 0 (an endothermic

reaction and S < 0- for all other cases, the relative magnitudes of H and TS determine if the

chemical transformation is spontaneous.

• Similarly, for macroscopic changes at constant V and T in which no nonexpansion work is possible, the condition for spontaneity is A < 0, where

• Two contributions determine if an isothermal chemical transformation is spontaneous- U is an energetic contribution, and TS is an entropic contribution to A- the same conclusions can be drawn from this equation as for those listed

above for the case of G with U substituted for H.

The Gibbs Energy and Helmholtz Energy

STUA

5

The Differential Forms of U, H, A, and G

Maxwell relations:

VTV

PS

TP

VS

SV

PT

SP

VT

PT

VT

PS

VS

VdPSdTSdTTdSVdPTdSdGPdVSdTSdTTdSPdVTdSdAVdPTdSVdPPdVPdVTdSdH

PdVTdSdU

VPGS

TG

PVAS

TA

VPHT

SH

PVUT

SU

TP

TV

SP

SV

and

and

and

and

6

T, V, P Dependence of S

VTV

PS

TP

VS

PT

VT

RelationsMaxwell

TC

TH

TTSVdPTdSdH

TC

TU

TTSPdVTdSdU

P

PP

V

VV

1

1

dPPSdT

TSdS

TP

dVVSdT

TSdS

TV

Chapter 5

)gasidealfor(lnln, fT

iT

fV

iV i

f

i

fmV

V

VVnR

TTnCdVdT

TCS

)gasidealfor(lnln, fT

iT

Pf

iP i

f

i

fmP

P

PPnR

TTnCdPVdT

TCS

PPV

VTTV

V TP

11and11

(for ideal gas) 7

V, P Dependence of U and H

PTVT

PTT

VTT

TV

PS

TP

VS

TVTVV

PST

PHVdPTdSdH

PTPTP

VST

VUPdVTdSdU

and : RelationsMaxwell

dPPHdT

THdH

TP

dVVUdT

TUdU

TV

dPTVVdTCdH

PP )(

dVPTPTdTCdU

VV )(

Chapter 3

T8

The Dependence of G and A on P, V, and T

• We know how A changes with T and V

• S and P always take on positive values. Therefore, the Helmholtz energy of a pure substance decreases as either the temperature or the volume increases.

• In an analogous way for the Gibbs energy

• Whereas the Gibbs energy decreases with increasing temperature, it increases with increasing pressure.

PVAS

TA

TV

and

VPGS

TG

TP

and

9


• For a macroscopic change in P at constant T:

• In the above integral the initial pressure is chosen to be the standard state pressure Po = 1 bar.

• For liquids and solids, the volume is to a good approximation independent of Pover a limited range in P and

P

P

P

P

VdPPTGPTGdGo o

oo '),(),(

)(),('),(),( oo

o

oo oPPVPTGVdPPTGPTGP

P

10


• For ideal gas behavior:

oPPnRTTGdP

PnRTTGVdPTGPTG

P

P

P

P

ln)(''

)(')(),( o

o

o

o

o

Figure 6.1

11


• The dependence of G on T (actually the temperature dependence of G/T, since it is related to the equilibrium constant K). The dependence of G/T on temperature is given by

This results is known as the Gibbs-Helmholtz equation.

• Because

• The above equation also applies to the change in G and H associated with a process such as a chemical reaction, in which G becomes G.

2222

1TH

TTSG

TG

TS

TG

TG

TTTG

P

P

2

1/1TdT

Td

HTTH

Td

dTTTG

T

TG

P

2

211

12


• Integrating the previous equation at constant P,

121

1

1

2

2

2

1

2

1

11)()()(

1

TTTH

TTG

TTG

TdH

TGd

T

T

T

T

13

14

15

The Gibbs Energy of a Reaction Mixture• So far the discussion has been limited to systems at a fixed composition. The

state functions such as U, H, S, A, and G need to be revised to account for the fact that reactants are consumed and products are generated in chemical reactions.

• For a reaction mixture containing species 1, 2, 3…G is no longer a function of the variables T and P only. Because it depends on the number of moles of each species, G is written in the form G = G(T, P, n1, n2, n3,….). The total differential dG is

• Note that if the concentration of all the species is constant, all the dni = 0, and the above equation reduces to the form presented earlier.

• The above Equation can be simplified by defining the chemical potential, i, as

• It is important to realize that although i is defined mathematically in terms of an infinitesimal change in the amount dni of species i, the chemical potential i is the change in the Gibbs energy per mole of the substance i added at constant concentration.

....2...,,2

1...,,1...,,...,,

122121

dnnGdn

nGdP

PGdT

TGdG

nPTnPTnnTnnP

injnTPii n

G

,,

16

The Gibbs Energy of a Reaction Mixture• The two requirements are not contradictory. To keep the concentration constant,

one adds a mole of substance i to a huge vat containing many moles of the various species.

• In this case, the slope of a plot of G versus ni is the same if the differential (∂G/∂ni)P,T,nj ≠ ni is formed, where dni → 0, or the ratio (∆G/∆ni)P,T,nj ≠ ni is formed, where ∆ni is 1 mol.

• Based on the definition of i, the total differential dG can be written as

• Integrating the above equation at constant composition and at a constant T and Pfrom an infinitesimal size of the system where ni → 0 and therefore G → 0 to a macroscopic size where the Gibbs energy has the value G (because T and P are constant, the first two terms do not contribute; because the composition is constant, i is constant)

• Note that because i depends on the number of moles of each species present, it is a function of concentration.

iii

iiinnTnnP

dnVdPSdTdndPPGdT

TGdG

...,,...,, 2121

i

ii

G

i

in

ii nGdndG or,''0 0

17

The Gibbs Energy of a Reaction Mixture

• If the system consists of a single pure substance A, G = nAGm.A because G is an extensive property. Applying the definition of A

• This shows that A is equal to the molar Gibbs energy of A for a pure substance. However, this is not true for mixtures (as shown later).

• Why is i called the chemical potential of species i?

• This can be understood by assuming that the chemical potential for species i has the value i

I in region I, and iII in region II of a given mixture with i

I > iII.

• If dni moles of species i are transported from region I to region II, at constant Tand P, the change in G is given by

dG = - iI dni + i

II dni = (iII - i

I )dni < 0

Because dG < 0, this process is spontaneous.

Am

TPA

AmA

TPAA G

nGn

nG

,,

,

,

18

The Gibbs Energy of a Reaction Mixture

• For a given species, transport will occur spontaneously from a region of high chemical potential to one of low chemical potential.

• The flow of material will continue until the chemical potential has the same value in all regions of the mixture.

• Note the analogy between this process and the flow of mass in a gravitational process or the flow of charge in an electrostatic potential. Therefore, the term chemical potential is appropriate.

• An additional criterion for equilibrium in a multi-component mixture has been defined: At equilibrium, the chemical potential of each individual species is the same throughout a mixture.

19

The Gibbs Energy of a Gas in a Mixture• The conditions for equilibrium in a mixture of ideal gases are next derived in

terms of the i of the chemical constituents.

• It will be shown that the partial pressures of all constituents of the mixture are related to the thermodynamic equilibrium constant, KP.

• Consider first the simple system consisting of two volumes separated by a semi permeable membrane as shown in the left Figure (Fig. 6.2).

• On the left side, the gas consists solely of pure H2. On the right side, H2 is present as one constituent of a mixture. The membrane allows only H2 to pass in both directions.

• Once equilibrium has been reached with respect to the concentration of H2throughout the system, the hydrogen pressure (but not the total pressure) is the same on both sides of the membrane and

H2 pure = H2

mixture

• Recall that the molar Gibbs energy of a pure ideal gas depends on its pressure as G (T,P) = Go(T) + nRT ln(P/Po).

Figure 6.2

20

The Gibbs Energy of a Gas in a Mixture• For a pure substance, i, i = Gm,i. Therefore, the chemical potential equation can

be written in the form

• The chemical potential of a gas in a mixture depends logarithmically on its partial pressure. The above Equation applies to any mixture, not just to those for which an appropriate semipermeable membrane exists. The discussion can thus be generalized by referring to a component of the mixture as A.

• The partial pressure of species A in the gas mixture, PA, can be expressed in terms of xA, its mole fraction in the mixture, and the total pressure P:

PA = xAP

• The Equation for the chemical potential can thus be written as:

• Because lnxA < 0, the chemical potential of a gas in a mixture is less than that of the pure gas if P is the same for the pure sample and the mixture. Because matter flows from a region of high to one of low chemical potential, a pure gas initially separated by a barrier from a mixture at the same pressure flows into the mixture as the barrier is removed (spontaneous process).

),(ln)(),( 22o2

2o

22 HmixtureH

HHH

pureH PT

PPRTTPT

AApureA

mixtureAA

pureA

mixtureA

AAmixtureAAA

mixtureA

xRTnPTGPTGxRTPTPT

xRTPPRTTPTxRT

PPRTTPT

ln),(),(orln),(),(

lnln)(),(or lnln)(),( oo

oo

21

Calculating the Gibbs Energy of Mixing for Ideal Gases• We next obtain a quantitative relationship between ∆Gmixing and

the mole fraction of the individual components of the mixture.

• Consider the system shown in the left Figure (Fig. 6.3). The four compartments contain the gases He, Ne, Ar, and Xe at the same temperature and pressure (1 bar). The volume of the four compartments differ. (Same answer obtained for a pressure P)

• To calculate ∆Gmixing, we must compare G for the initial state shown in the Figure and the final state.

• For a initial state, in which we have four pure separated substances,Gi = GHe + GNe + GAr + GXe = nHeGo

m,He + nNeGom,Ne + nArGo

m,Ar + nXeGom,Xe

• For the final state in which all four components are dispersed in the mixture,Gf = nHe(Go

m,He + RT lnxHe)+ nNe(Gom,Ne + RT lnxNe) + nAr(Go

m,Ar + RT lnxAr) + nXe (Go

m,Xe + RT lnxXe)

• The Gibbs energy of mixing is Gf - Gi or

• ∆Gmixing = RTnHe ln xHe + RTnNe ln xNe + RTnAr ln xAr + RTnXe ln xXe

i

iii

ii xxnRTxnRT lnln

Figure 6.3

22

Calculating the Gibbs Energy of Mixing for Ideal Gases• Note that because each term in the last expression is negative,

∆Gmixing < 0. Therefore, mixing is a spontaneous process at constant T and P.

• The Equation can be used to calculate the value of ∆Gmixing for any given set of the mole fractions xi. To visualize the results for a binary mixture of species A and B, we set xA = x, so that xB = 1-x. It follows that

∆Gmixing = nRT [x lnx + (1-x) ln(1-x)]

• A plot of ∆Gmixing versus x is shown in Figure 6.4 for a binary mixture. Note that ∆Gmixing is zero for xA = 0 and xA = 1 because they are only pure substances in these limits. Also, ∆Gmixinghas a minimum for xA = 0.5, because there is a maximal decrease in G that arises from the dilution of both A and B when A and B are present in equal amounts.

• The entropy of mixing can be calculated from

i

iiP

mixingmixing xxnR

TGS ln

• As seen in the above Figure 6.5, the entropy of mixing is greatest for xA = 0.5.

Figure 6.4

Figure 6.5

23

• What lies behind the increase in entropy? Each of the two components of the mixture expands from its initial volume to the same final volume. Therefore, ∆Smixingarises purely from the dependence of S on V at constant T, if both components and the mixture are at the same pressure

• What is the driving force for the mixing of gases?

• As seen in the previous section, there are two contributions to ∆G, an enthalpic contribution ∆H and an entropic contribution T∆S.

• By calculating ∆Hmixing from ∆Hmixing = ∆Gmixing + T∆Smixing, using the Equations for ∆Gmixing and ∆Smixing, one sees that for the mixing of ideal gases, ∆Hmixing = 0.

• Because the molecules in an ideal gas do not interact, there is no enthalpic change associated with mixing. We conclude that the mixing of ideal gases is driven entirely by ∆Smixing.

• Although the mixing of gases is always spontaneous, the same is not true of liquids. Liquids can be either miscible or immiscible.

• For gases or liquids, ∆Smixing > 0. Therefore, if two liquids are immiscible, ∆Gmixing > 0 because ∆Hmixing > 0 and ∆Hmixing > T ∆Smixing. If two liquids mix, it is generally energetically favorable for one species to be surrounded by the other species. In this case, ∆Hmixing < T ∆Smixing, and ∆Gmixing < 0.

Calculating the Gibbs Energy of Mixing for Ideal Gases

)lnln(1ln1ln(lnln BBAAB

BA

AiB

fB

iA

fA xxxxnR

xnx

xnxR

VVn

VVnRS

24

• Consider the balanced chemical reaction A + B + C + ….→ M + N + O + …

in which Greek letters represent stoichiometric coefficients and the uppercase Roman letters represent the reactants and products.

• We can write an abbreviated expression for this reaction in the form the stoichiometric coefficients of the products are positive, and those of the reactants are negative.

• What determines the equilibrium partial pressures of the reactants and products?

In this equation, the individual dni are not independent because they are linked by the stoichiometric equation.

• It is convenient to introduce a parameter, , called the extent of reaction. If the reaction advances by moles, the number of moles of each species i changes according to

• 2NO2 (g) ↔ N2O4 (g), (2 - 2) moles of NO2 (g) and moles of N2O4 (g)

• An advancing reaction can be described in terms of the partial derivative of G with

Expressing Chemical Equilibrium in an Ideal Gas Mixture in Terms of the i

,0 i

iiX

i

iidndG

ddnnn iiiinitialii or

ireactionii

PTreaction

iii GGdGddG

, 25

• The direction of spontaneous change is that in which ∆Greaction is negative. This direction corresponds to (∂G/∂)T,P < 0.

• At a given composition of the reaction mixture, the partial pressures of the reactants and products can be determined, and the i can be calculated.

• Because i = i (T, P, nA, nB….), must be evaluated at specific values of T, P, nA, nB….

• Based on the value of ∆Greaction, the following conclusions can be drawn:- if (∂G/∂)T,P < 0, the reaction proceeds spontaneously as written- if (∂G/∂)T,P > 0, the reaction proceeds spontaneously in the opposite direction- if (∂G/∂)T,P = 0, the reaction system is at equilibrium, and there is no direction of spontaneous change.

• The most important value of is eq, How can this value be found?

• Let us consider the reaction system 2NO2 (g) ↔ N2O4 (g), with (2 - 2) moles of NO2(g) and moles of N2O4 (g) present in a vessel at a constant pressure of 1 bar and 298 K. The parameter could in principle take on all values between zero and one, corresponding to pure NO2 (g) and pure N2O4 (g), respectively.

• The Gibbs energy of the pure unmixed reagents and products, Gpure, is given by Gpure = (2 - 2 )Go

m(NO2, g) + Gom(N2O4, g),

where Gom is the conventional molar Gibbs energy.


i

reactionii G

26

• Note that Gpure varies linearly with as shown in Figure 6.6

• Because the reactants and products are mixed throughout the range accessible to , Gpure is not equal to Gmixture, which is given by Gmixture = Gpure + ∆Gmixing

• If Gpure alone determined eq, eq= 0 or eq = 1, depending on whether Gpure is lower for reagents or products. For the reaction under consideration ∆Go

f (N2O4, g) < ∆Gof (NO2, g) so that eq = 1

in this limit. In other words, if Gpure alone determined eq, every chemical reaction would go to completion.


• How does mixing influence the equilibrium position for the reaction system? The value of ∆Gmixing can be calculated (end-of-chapter problem, │∆Gmixing│ for this reaction system has a maximum at eq= 0.55).

• If Gpure alone determined eq then eq = 1, and if ∆Gmixing alone determined eq, then eq = 0.55. However, the minimum in ∆Greaction is determined by the minimum in Gpure + ∆Gmixing rather than the minimum of the individual components.

• For our specific case, eq = 0.72 at 298 K. The decrease in G on mixing plays a critical role in determining the position of equilibrium in a chemical reaction in that eq is shifted from 1.0 to 0.72.

Figure 6.6

27

• In summary, for reactions in which Gpure for reactants and products are very similar, eq will be largely determined by ∆Gmixing. For reactions in which Gpure for reactants and products are very different, eq will not be greatly influenced by ∆Gmixing and the equilibrium mixture will essentially consist of pure reactant or pure products.

• The direction of spontaneous change is determined by ∆Greaction. How can ∆Greaction be calculated?

• We distinguish between calculating ∆Greaction for the standard state in which Pi = Po

= 1 bar and including the pressure dependence of ∆Greaction.

• Tabulated values of ∆Gof for compounds are used to calculate ∆Go

reaction at Po = 1 bar and T = 298.15 K (Example Problem 6.6).

• Just as for enthalpy, ∆Gof for a pure element in its standard state at the reference

temperature is equal to zero because the reactants and products in the formation reaction are identical.

• Just as for ∆Horeaction,

• In the next section, the pressure dependence of G is used to calculate ∆Greaction for arbitrary values of the pressure.


i

ifireaction GG ,oo

28

• We introduce the pressure dependence of i in order to calculate ∆Greaction for reaction mixtures in which Pi is different from 1 bar.

• Consider the following reaction that takes place between ideal gas species A, B, C, and D:

A + B ↔ C + D

• Because all four species are present in the reaction mixture, the reaction Gibbs energy is given by

• The terms in the previous equation can be separated into those at the standard condition of Po = 1 bar and the remaining terms:

• In the above Equation

Calculating ∆Greaction and Introducing the Equilibrium Constant for a Mixture of Ideal Gases

oo

oo

oo

oo

, lnlnlnlnPPRT

PPRT

PPRT

PPRTGG B

Bi

AA

DD

CCifireaction

oo

ooo

ooooo

ln

lnlnlnln

PP

PP

PP

PP

RTG

PPRT

PPRT

PPRT

PPRTGG

BA

DC

reaction

BADCreactionreaction

i

ifiBADCreaction GTTTTG ,oooooo )()()()( 29

• The combination of the partial pressures of reactants and products is called the reaction quotient of pressures, which is abbreviated QP and defined as follows:

• With the definitions of ∆Goreaction and QP

• Note that ∆Greaction can be separated into two terms, only one of which depends on the partial pressures of the reactants and products.

• The above Equation can be used to predict the direction of spontaneous change for given partial pressures of the reactants and products.

• If the partial pressures of the products C and D are large, and those of the reactants A and B are small compared to their values at equilibrium, QP will be large.

• As a result, RTlnQP will be large and positive, and ∆Greaction=∆Goreaction+ RTlnQP > 0.

In this case the reaction as written from left to right is not spontaneous.


oo

oo

PP

PP

PP

PP

QBA

DC

P

Preactionreaction QRTGG lno

30

• Consider the opposite extreme. If the partial pressures of the reactants A and B are large, and those of the products C and D are small compared to their values at equilibrium, QP will be small. As a result, RT lnQP will be large and negative.

• In this case, ∆Greaction=∆Goreaction+ RT lnQP < 0 and the reaction will be spontaneous

as written from left to right, as a portion of the reactants combines to form products.

• A more interesting case is ‘equilibrium’ for which ∆Greaction= 0. At equilibrium, ∆Go

reaction= - RT lnQP.

• We denote this special system configuration by adding a superscript ‘eq’ to each partial pressure, and renaming QP as KP. The quantity KP is called the thermodynamic equilibrium constant:

• Because ∆Goreaction is a function of T only, KP is also a function of T only. The

thermodynamic equilibrium constant KP does not depend on the pressure. Note that KP is a dimensionless number.


RTGK

KRTG

PP

PP

PP

PP

RTG

reactionP

PreactionBeq

Aeq

Deq

Ceq

reaction

o

o

oo

ooo

lnor

lnly,equivalentor,ln0

31

• As shown in the previous section, the partial pressures of the reactants and products in a mixture of gases at equilibrium cannot take on arbitrary values because they are related through KP .

• In this section, by way of Example Problem 6.9, the calculation of equilibrium partial pressures for ideal gases is shown.

• In Example in Problem 6.9, we consider the dissociation of chlorine:Cl2 (g) → 2Cl (g)

• In this example, n0 moles of chlorine gas are placed in a reaction vessel, whose temperature can be varied over a wide range, so that molecular chlorine can partially dissociate to atomic chlorine- derive an expression for KP in terms of n0, , and P- define the degree of dissociation as = eq / n0 , where 2eq is the number of

moles of Cl (g) present at equilibrium, and n0 represents the number of moles of Cl2 (g) that would be present in the system if no dissociation occurred. Derive an expression for as a function of KP and P.

• From the solution it is clear that whereas KP(T) is independent of P, as calculated does depend on P. In this case, decreases as P increases for constant T.

Calculating the Equilibrium Partial Pressures in a Mixture of Ideal Gases

32

Calculating the Equilibrium Partial Pressures in a Mixture of Ideal Gasesa. We set up the following table:

Cl2 (g) 2Cl (g)

Initial number of moles no 0

Moles present at equilibrium no - eq 2eq

Mole fraction present at equilibrium, xi

Partial pressure at equilibrium, Pi =xiP

eq

eq

nn

o

o

eq

eq

n o

2

Pnn

eq

eq

o

o Pn eq

eq

o

2

o22

o

2

ooo

2

oo

o

2

oo

o

2

2

o 442

)(PP

nPP

nnPP

nn

PP

n

PPPP

TKeq

eq

eqeq

eq

eq

eq

eq

eq

eqCl

eqCl

P

o

2

o

oo

2

2

o22

o

2

4)(

)(or)(4

/;144)(b.

PPTK

TKTKPP(T)K

nPP

PP

nTK

P

PPP

eqeq

eqP

Because KP(T) depends

strongly on temperature, will also be a strong function of temperature. Note that depends on both KP and P. 33

• Based on our chemical intuition, we expect the degree of dissociation to be high at high temperatures and low at low temperatures (as shown in Example Problem 6.9, the degree of dissociation of Cl2 is dependent on the temperature).

• For this to be true, KP (T) must increase with temperature for this reaction. How can we understand the temperature dependence for this reaction and for reactions in general?

• The Equation for the equilibrium constant can be written as

• Using the Gibbs-Helmholtz equation, the preceding equation reduces to

• Because tabulated values of ∆Gof are available at 298.15 K, we can calculate

∆Goreaction and KP at this temperature; KP can be calculated at the temperature Tf by

integrating the above Equation between the appropriate limits:

The Variation of KP with Temperature

dTTGd

RdTRTGd

dTKd reactionreactionP )/(1)/(ln oo

2

oo )/(1lnRTH

dTTGd

RdTKd reactionreactionP

dTTH

RKd

fT

K

reactionTK

KP

fP

P

15.298

2

o)(

)K15.298(

1ln34

• If the temperature Tf is not much different from 298.15 K, it can be assumed that ∆Ho

reaction is constant over the temperature interval.

• This assumption is better than it might appear at first glance. Although H is strongly dependent on temperature, the temperature dependence of ∆Ho

reaction is governed by the difference in heat capacities ∆CP between reactants and products.

• If the heat capacities of reactants and products are nearly the same, ∆Horeaction is

nearly independent of temperature.

• With the assumption that ∆Horeaction is nearly independent of temperature, the

integral can be simplified to:

• Example Problem 6.10 Using the results of Example Problem 6.9 and the data tables, consider the dissociation reaction Cl2 (g) → 2 Cl (g)- calculate KP at 800, 1500, and 2000 K for P = 0.010 bar- calculate the degree of dissociation, , at 300, 1500, and 2000 K.

The Variation of KP with Temperature

K15.29811)K15.298(ln)(ln

o

f

reactionPfP TR

HKTK

35

• Many chemical reactions involve a gas phase in equilibrium with a solid or liquid phase. An example is the thermal decomposition of CaCO3(s):

CaCO3 (s) → CaO (s) + CO2 (g)• In this case, a pure gas is in equilibrium with two solid phases. At equilibrium,

• Because the equilibrium pressure is P ≠ Po, the pressure dependence of each species must be taken into account. However, from the previous discussion we know that the pressure dependence of G for a solid or liquid is very small

• Using the dependence of on P for an ideal gas, the above Equation becomes

• Rewriting this equation in terms of KP, we obtain

Equilibria Involving Ideal Gases and Solid or Liquid Phases

),,CaCO(),,CO(),,CaO(0

0

32 PsPgPs

nG

eqeqeq

iiitioncrea

),CaCO(),,CaCO(and),CaO(),,CaO( 3o

3o sPssPs eqeq

o2

3o

2ooo

o2

3o

2oo

ln),CaCO(),CO(),CaO(

orln),CaCO(),CO(),CaO(0

PPRTsgsG

PPRTsgs

COreaction

CO

RTG

PPK reactionCO

P

o

o2lnln

36

• Chemists often find it useful to express the concentrations of reactants and products in units other than partial pressures. Two commonly used units are mole fraction and molarity.

• We express the equilibrium constant in terms of mole fractions (for a mixture of ideal gases). The mole fraction, xi, and the partial pressure, Pi, are related by Pi = xiP. Therefore,

• Note that just as for KP, Kx is a dimensionless number.

• Because the molarity, ci, is defined as ci = ni /V = Pi /RT, we can write Pi/Po = (RT/Po)ci. To work with dimensionless quantities, we introduce ci / co, which is related to Pi /Po by

Expressing the Equilibrium Constant in Terms of Mole Fraction or Molarilty

oo

o

oo

oo

oo

oo

orPPKK

PPK

PP

xxxx

PPx

PPx

PPx

PPx

PP

PP

PP

PP

K

Pxx

bacd

bBeqa

Aeq

dDeqc

Ceq

bBeqa

Aeq

dDeqc

Ceq

bBeqa

Aeq

dDeqc

Ceq

P

oo

o

o cc

PRTc

PP ii

37

• Using this notation, we can express Kc in terms of KP:

• Recall that ∆ is the difference in the stoichiometric coefficients of products and reactants.

• Kx and Kc are in general different from KP. They are only equal in the special case that ∆ = 0.

Expressing the Equilibrium Constant in Terms of Mole Fraction or Molarilty

o

o

o

o

o

o

oo

oo

oo

oo

PRTcKK

PRTcK

PRTc

cc

cc

cc

cc

PP

PP

PP

PP

K

Pc

c

bacd

bBeqa

Aeq

dDeqc

Ceq

bBeqa

Aeq

dDeqc

Ceq

P

38

Assume that a sealed vessel at constant pressure of 1 bar initially contains 2.00 mol of NO2(g). The system is allowed to equilibrate with respect to the reaction 2 NO2(g) ↔ N2O4(g). The number of moles of NO2(g) and N2O4(g) at equilibrium is 2.00 – 2ξ and ξ, respectively, where ξ is the extent of reaction.

a. Derive an expression for the entropy of mixing as a function of ξ.b. Graphically determine the value of ξ for which ∆Smixing has its maximum value.

c. Write an expression for Gpure as a function of ξ.d. Plot Δ Greaction = Gpure + Δ Gmixing as a function of x for T = 298 K and graphically determine the value of ξ for which Δ Greaction has its minimum value. Is this value the same as for part (b)?

a) Derive an expression for the entropy of mixing as a function of x.

2 2 2 4 2 4ln ln

2.00 2 2.00 22.00 ln ln2.00 2.00 2.00 2.00

mixing NO NO N O N OS nR x x x x

R

39

b) A plot and a detail plot locating the maximum are shown below.

The value of x at the maximum is ξ = 0.55.

Similarly, ∆Gmixing = nRT [x lnx + (1-x) ln(1-x)]

40

c) Write an expression for Gpure as a function of ξ.

2 2 42.00 2 NO , N O ,pure f fG G g G g

d) Plot ∆Greaction = Gpure + ∆Gmixing as a function of ξ for T = 298 K and graphically determine the value of ξ for which ∆Greaction has its minimum value. Is this value the same as for part b)?

The minimum is at ξ = 0.72. The minimum is shifted to higher values of ξ relative to the ∆S vs ξ curve for mixing because of the contribution of Gpure to ∆Greaction. 41

• Suppose that we have a mixture of reactive gases at equilibrium. Does the equilibrium shift toward reactant or products as T or P is changed?

• To answer this question we first consider the dependence of KP on T. Because

• Note that KP will change differently with temperature for an exothermic reaction than an endothermic reaction.

• For an exothermic reaction d ln KP / dT < 0, and eq will shift toward the reactants as T increases. For an endothermic reaction, d ln KP / dT > 0, and eq will shift toward the products as T increases.

• The dependence of eq on pressure can be ascertained from the relationship between KP and Kx:

• Because KP is independent of pressure, the pressure dependence of Kx arises solely from (P/Po)-∆ .

• If the number of moles of gaseous products increases as the reaction proceeds, Kxdecreases as P increases, and eq shifts back toward the reactants.

The Dependence of eq on T and P

2

olnRTH

dTKd reactionP

oPPKK Px

42

• If the number of moles of gaseous products decreases as the reaction proceeds, Kxincreases as P increases, and eq shifts back toward the products.

• A combined change in T and P leads to a superposition of the effects just described.

• According to the French chemist Le Chatelier, reaction systems at chemical equilibrium respond to an outside stress, such that a change in T and P, by countering the stress.

• Consider the Cl2 (g) → 2Cl (g) reaction discussed in Example Problem 6.9 and 6.10 for which ∆Hreaction > 0.

• In this case, eq responds to an increase in T in such a way that heat is taken up by the system; in other words Cl2 (g) dissociates. This counters the stress imposed on the system by an increase in T.

• Similarly, eq responds to an increase in P in such a way that the volume of the system decreases.

• Specifically, eq changes in the direction that ∆ < 0, and for the reaction under consideration, Cl (g) is converted to Cl2 (g). This shift in the position of equilibrium counters the stress brought about by an increase in P.

The Dependence of eq on T and P

43

Chapter 4: Thermochemistry

½ N2 (g) + 3/2 H2 (g) → NH3 (g)

• In 2000, more than 2 million tons of ammonia were produced per weekusing the Haber-Bosch process

- More than 98% of the inorganic input to soil as fertilizer

44

The Haber process

½ N2(g, T, P) + 3/2 H2(g, T, P) = NH3(g, T, P)

Ho 298 K 46.21 kJ/mol

Go 298 K 16.74 kJ/mol

For P = 1 bar this is pretty good, lots of product. However, the reaction at room T is slow (kinetics, not thermodynamics).

Raising T to 800 K can speed it up. But since ∆Ho(T) < 0

(exothermic), Le Chatelier tells us that the equilibrium will shift toward the reactants.

45

Indeed: Kp (800 K) = 0.007

What is the option? Note that KX = P x Kp

Use Le Chatelier principle but with higher pressureEq. Shift towards products

Run reaction at high T and high P

For P = 1 bar, T = 800 K, Kp = 0.007

But at P = 100 bar, Kx = (100) Kp = 0.7 (much better!)

46

Q6.1) Under what conditions ∆A ≤ 0 is a condition that defines the spontaneity of a process?

Q6.2) Under what conditions is ∆G ≤ 0 a condition that defines the spontaneity of a process?

Q6.3) Which thermodynamic state function gives a measure of the maximum electric work that can be carried out in a fuel cell?

Q6.4) By invoking the pressure dependence of the chemical potential, show that if a valve separating a vessel of pure A from a vessel containing a mixture of A and B is opened, mixing will occur. Both A and B are ideal gases, and the initial pressure in both vessels is 1 bar.

Q6.5) Under what condition is KP = Kx?

Q6.6) It is found that KP is independent of T for a particular chemical reaction. What does this tell you about the reaction?

Q6.7) The reaction A + B → C + D is at equilibrium for ξ = 0.1. What does this tell you about the variation of Gpure with x ?

Questions on Concepts

47

Q6.8) The reaction A + B → C + D is at equilibrium for x = 0.5. What does this tell you about the variation of Gpure with ξ ?

Q6.9) Why is it reasonable to set the chemical potential of a pure liquid or solid substance equal to its standard state chemical potential at that temperature independent of the pressure in considering chemical equilibrium?

Q6.10) Is the equation valid for liquids, solids, and gases?

Q6.11) What is the relationship between the KP for the two reactions 3/2H2 + 1/2N2 → NH3 and 3H2 + N2 → 2NH3?

Questions on Concepts

T

U T PV

48

• N2 + 3H2 2NH3

How to improve yield: temperature and/or pressure? (ammonia yield in equilibrium will be higher if the temperature is decreased and the pressure is increased; Chapter 6).

• CH3OH (l) + 3/2O2 (g) CO2 (g) + 2H2O (l)Internal combustion engine versus electrochemical fuel cell, what are the relative efficiencies of the two propulsion systems? (electrochemical fuel cell is more efficient; Chapter 5). • Will the use of a catalyst promote to increase the yield of a reaction?(the maximum yield expected under equilibrium conditions must be calculated first. If the yield is insufficient, a catalyst is useless; Chapter 6).

Examples of Applications of Thermodynamics

49

Documents

Chapter 6: Chemical Equilibrium