Chapter 6 Chemical equilibrium

6 Chemical equilibrium6.1 Equilibrium condition & affinity of chemical reaction6.2 Isothermal of chemical reaction6.3 Equilibrium constant and chemical equation6.4 Chemical equilibrium of heterogeneous reaction6.5 Calculation of equilibrium constant6.6 Standard forming Gibbs free energy6.7 Calculation of equilibrium const form Q function6.8 Effect of T, P, indifferent gas on chemical equilibrium6.9 Simultaneous equilibrium6.10 Coupling reaction

6.1 Equilibrium condition and affinity of chemical reaction

Chemical reaction system

Extent of reaction

Equation of thermodynamics

Direction and limit of chemical reaction

Affinity of chemical reaction

6.1.1 Chemical reaction system

D E F GD E F G

0,0 t D,0n E,0n F,0n G,0n ,tt Dn En Fn Gn

Dekonder defined extent of reaction ξ

B B,0

B

n n

B

B

dd

n

vB: negative for reactant; positive for resultant.

6.1.1.1 Extent of reactionAt any moment of the reaction, we

can use any reactant/resultant to stands for the extent of reaction, the values are all the same.

GD E F

D E F G

dd d dd

nn n n

The changing value of every substance must satisfy:

BB

0 B

6.1.1.2 Equation of thermodynamics

When T, P certain,

B

BBdddd npVTSG

, B B B BB B

d d dT pG n （ ） B B(d d )n

, B BB

( ) (a) T p

G

r m , B BB

(b) T pG （ ） When ξ=1 mol:

6.1.1.3 Conditions of equation (1) The chemical reaction of isothermal,

isobaric and it do not export work; (2) Chemical potential μB of every

substances keep unchangeable. Formula (a): the micro change happens in

the limit system; Formula (b): extent of reaction is 1 mol

in the abundant system which. The concentration/ μB is also unchangeable.

6.1.2 Direction and limit of chemical reaction

can be used to judge chemical direction. (ΔrGm)T,P <0, reaction process to the

right spontaneously; (ΔrGm)T,P >0, reaction process to the

left spontaneously; (ΔrGm)T,P=0, reaction equilibrium.

PT

G

,

BB

Bv PTmrG ,and

6.1.2.2 Judgment The slope of the curve.

. Because change is small, extent of reaction is in 0~1 mol.

PT

G

,

right spontaneously

left spontaneously

equilibrium

0)( ,

pT

G

0)( ,

pT

G

0)( ,

pT

G

6.1.2.3

6.1.2.4 Mixing Gibbs free energy D+E=2F R, G maximum, D and

E no mixing; P, G maximum, after D

and E mixing;T, G maximum of all

the substances in equilibrium, containing mixing G;

S, G of pure resultant F.

6.1.2.5 van’t Hoff equilibrium box

In order to make the reaction process to the end, process must be in the van’t Hoff equilibrium box (point S from point R directly), prevent any mixture between the reactants or between the reactant and the resultant.

6.1.2.6

6.1.3 Affinity of chemical reaction In 1922, Belgium expert, De donder first b

rought in the concept of chemical reaction affinity, A, a state function.

def

, B BB

( )T p

GA

mr- GA or

A has the properties of “ potential ”:

A>0 reaction process towards the positive direction

A<0 reaction process towards the converse direction

A=0 reaction gets equilibrium

6.1.3 A used as judgment

6.2 Isothermal of chemical reaction

Chemical potential of gas (B)

Isothermal of chemical reaction

Equilibrium constant

Judgment of reaction direction

6.2.1 Chemical potential of gas (B)

If the gas is ideal gas, fB=pB .

Put chemical potential expression into the calculation formula of (ΔrGm)T,p :

BB B( , ) ( ) ln

fT p T RT

p $

$

B

BB,mr )( pTG BB B B

B B

( ) lnf

T RTp

$$

Br m , r m B

B

( ) ( ) lnT p

fG G T RT

p $

$ r m B BB

( ) ( )G T T $ $

Go

Make:

6.2.2.2 Chemical potential of gas (B)

6.2.3 Isothermal of chemical reaction For a reaction, D E G Hd e g h

g hG H

r m d eD

rE

m

( / ) ( / )( ) ln

( / ) ( / )

f p f pG T RT

f p f pG

$ $$

$ $

r m ( ) ln fG T RT Q $

Qf is called fagasity quotient. ΔrG$m(T)

is also can be worked out. thereby, ΔrGm can be calculated.

6.2.4 Equilibrium constant

In equilibrium, ΔrGm =0, therefore:g h

G Hr m d e

D E

( / ) ( / )ln

( / ) ))

( /(

f p f pRT

f p fT

pG

$ $

$$

$

ln fRT K $

K$f , thermodynamics equilibrium

constant, or standard equilibrium constant. It is only temperature function.

6.2.5 Judgment of reaction direction

For ideal gas

r m ln lnf fG RT K RT Q $

r m ln lnp pG RT K RT Q $

Chemical reaction isothermal formula is also can be denote by:

K$p>Qp ΔrGm <0 reaction process to the right

spontaneously K$

p<Qp ΔrGm >0 reaction process to the left spontaneously K$

p=Qp ΔrGm =0 reaction gets to equilibrium

The changing value of standard Gibbs free energy;Obviously, in the chemical reaction equation, the comput

ation coefficient assumes the multiple relation, the value of ΔrG$

m (T) is also assumes multiple relation, while the value of K$

f assumes exponential relation.

HI(g)2g)(Ig)(H 22

HI(g)g)(Ig)(H 221

221 （ 1 ）

（ 2 ）r m,2 r m,12G G $ $

2,2 ,1( )f fK K$ $

6.3 Equilibrium constant and chemical equation

r m ( ) ln fG T RT K $ $

Kp by pressure BHG

Bd BD E

g h

p e

p pK p

p p

6.3.1 Kp

When ΣvB =0, the units of Kp is 1.

Experimental equilibrium constant:Use the real pressure, mole fraction or concentration of reactant/resultant.

6.3.2 Kx

For the ideal gas, it accord with th

e Dalton partial pressure law, BB pxp

BHGBB

D E

g h

x d e

x xK x

x x

BB

pKK px

About mole fraction equilibrium constant.

6.3.3 Kc

For the ideal gas,

BHGBB

D E

g h

c d e

c cK c

c c

BB

)(

RTKK pc

cRTp

About mole concentration equilibrium constant

6.3.4 Ka

Liquid reaction use activity to stands for the equilibrium constant Ka

B

BB

aK a

B BB

ca

c

$

B

B( )a c rK K K c $

therefore

Because

6.4 Chemical equilibrium of Heterogenous reaction

The reaction covering a gas phase and a liquid/solid phase is called heterogenous chemical reaction. If the agglomerate phase is pure state, the chemical potential of pure state is its standard state chemical potential.

therefore, Kp=exp(pjp/pi

r).

There is a reaction, gas can be seen as ideal gas:

CaCO3=CaO(s)+CO2(g)

K$p=p(CO2)/p$

p(CO2) is called the dissociation pressure of CaCO3 (s).

Example

6.4.2 Dissociation pressureWhen certain substance dissociates at some gas pressure it produces, this gas pressure is called dissociation pressure.This pressure has fixed value in certain TIf the gas it produces is more than one kind, therefore, the summation of all the gas pressure is called dissociation pressure.

NH4HS(s)=NH3(g)+H2S(g)

Dissociation pressure p=p(NH3)+p(H2S)

Therefore, the thermodynamics equilibrium constant:

3 2(NH ) (H S)p

p p

pK

p $

$$

214 ( / )p p $

example

6.5 Calculation of equilibrium constant

6.5.1 Equilibrium conversion rate It is called theory conversion rate, is the

percentage of reactant turning into the resultant, after getting equilibrium,

%100devotion of quantum material the

mequilibriu gettingafter resultant theinto gets material of quantum

rateion translatmEquilibriu

6.5.2 Industrial conversion rate Because the reaction is not in equilibrium , the real conversion rate is smaller than equilibrium conversion rate.

6.5.3 Measuring of equilibrium constant6.5.3.1 Physical method

6.5.3.2 Chemical method

6.6 Standard forming Gibbs free energy

standard mole reaction Gibbs free energy changing value

Mole standard forming Gibbs free energy

The ionic mole standard forming Gibbs free energy

The use of numerical value of ΔfG$

m

6.6.1 Standard forming Gibbs free energy

Under the standard pressure, the Gibbs free energy changing value cause by the steady single substance forming 1 mol compound

ΔrG$m ( compound, substance state, tempe

rature ) It is usually can be check out in the table

at 298.15K.

6.6.2.1 Calculate the thermodynamics equilibrium constant

r m ln aG RT K $ $

r mexp( / )aK G RT $ $

6.6.2 The application of ΔrG$m

6.6.2.2 Calculation of Special K Some of equilibrium constant can not be measured easily.

1

22C(s) O (g) CO(g)

2 2 r m(1) C(s) O (g) CO (g) (1)G $

12 2 r m2(2) CO(g) O (g) CO (g) (2)G $

12 r m2(3) C(s) O (g) CO(g) (3)G $

r m r m r m(3) (1) (2)G G G $ $ $ (1)(3)

(2)p

pp

KK

K

$$

$

(1)-(2) we can get (3)

6.6.2.3 Estimating reaction feasibility

r m r m ln pG G RT Q $

can be used to judge the direction of the reaction approximately.

0,,mr f)( wpTG

6.7 Calculating equilibrium constant from partition function

Sharing zero-point energy value of chemical equilibrium

Free energy function

Heat content function

6.7.1 Sharing zero-point energy value of chemical equilibriu

m The energy zero selection of

the same substance particle will not effect the calculation of its energy changing value.

When both of the quantum number of transition and vibrate are zero (J=0,v=0), their energy level is fixed as zero.

6.7.1.2 Sharing energy expression of chemical equilibrium

As for many substances, the energy zero point of every substance is different. The lowest energy level at 0K is chosen, the difference of energy from energy zero to sharing zero is ε0.

The partition function expression of F, G, H, U has an item U0 (U0=Nε0 ) while S, Cv and p are unchangeable.

6.7.1.2

6.7.2 Free energy functionFree energy function

0lnq

G NkT UN

0( )ln

G T U qNk

T N

0( )G T U

T

0( )G T H

T

is also free energy function

At 0K, U0=H0

Therefore ,

is called free energy function

Therefore

Because

When N=1 mol, NK=R, suppose at the condition of standard state

m m( ) (0)ln

G T H qR

T L

$ $

Free energy function can be worked out from q. The free energy function of every substance at different temperature can be checked out in the table.

6.7.2.2 Free energy function

6.7.2.3 Calculation of equil. Const. D + E = G + H

r m ( ) lnG T RT K $ $

r m ( )ln

G TR K

T

$$

r m r m r m( ) (0) (0)G T U U

T T T

$ $ $m m r m

BB

( ) (0) (0){ }G T U U

T T

$ $ $

the second item is the change value of reaction thermodynamics energy at 0K.

6.7.2.4 Calculation of ΔrU$

m (0) value

6.7.2.4.1. From K$ value and the value of every substance free energy function, inversely.

r m

m m r mB

B

( )ln

( ) (0) (0) { }

G TR K

T

G T U U

T T

$$

$ $ $

6.7.2.4.2 From G definition

r m r m r m( ) ( ) ( )

G H TS

G T H T T S T

$ $ $

m mr m r m

r m

( ) (0){ [ ] ( )} ( )

(0)

G T UT S T H T

T

U

$ $$ $

$

Both Plus one, minus one ΔrU$m (0), then we can

get:

6.7.2.4.3 From Kirchhoff formula

r m r m 0( ) (0) d

T

pH T H C T $ $

r m r m r m 0(0) (0) ( ) d

T

pU H H T C T $ $ $

6.7.2.4 From molecule dissociation energy D

E F G H

r m E F G H(0) ( ) ( )U D D D D $

6.7.2.5 From heat content function

The heat content change of reaction and the value of heat content function are already known, so we can get the value of ΔrU$

m (0).

m m r mr m

( ) (0) (0)( ) [ { } ]

H T U UH T T

T T

$ $ $$

6.7.3 Heat content function

The value of heat content function can be worked out by q. When T is 298.15K, the value of H$

m(298.15K)-U$

m(0) can be checked out in the table.

m m,

ln( )

( ) (0)V N

qRT R

T

H T U

T

$ $

Make use of the value of heat content function to calculate the heat content change of the chemical reaction:

m m r mr m

( ) (0) (0)( ) [ { } ]

H T U UH T T

T T

$ $ $$

6.7.3.2 Calculation of Chem. Heat

6.7.3.3 Calculation of K

KN is the equilibrium constant which is denoted by the molecule number, q is the total partition function after disparting the zero energy.

' *G G

' ' * *D E D E

N

q NK

q q N N 0' exp( )q q

kT

Suppose D+E=G

If further disparting the V in the transition partition function, therefore, use f to denote the partition function:

0' exp( )q V fkT

*G G 0

* *D E D E

exp( )c

c fK

c c f f kT

6.7.3.3

Work out the f of every partition function, we can get the equilibrium constant value Kc.

6.8 Effect of T, p and indifferent gas on chemical equilibriumEffect of T

Effect of p

Effect of indifferent gas

6.7.2.3 Calculation of equil. Const. D + E = G + H

r m ( ) lnG T RT K $ $

r m ( )ln

G TR K

T

$$

r m r m r m( ) (0) (0)G T U U

T T T

$ $ $m m r m

BB

( ) (0) (0){ }G T U U

T T

$ $ $

the second item is the change value of reaction thermodynamics energy at 0K.

6.7.2.4 Calculation of ΔrU$

m (0) value

6.7.2.4.1. From K$ value and the value of every substance free energy function, inversely.

r m

m m r mB

B

( )ln

( ) (0) (0) { }

G TR K

T

G T U U

T T

$$

$ $ $

6.7.2.4.2 From G definition

r m r m r m( ) ( ) ( )

G H TS

G T H T T S T

$ $ $

m mr m r m

r m

( ) (0){ [ ] ( )} ( )

(0)

G T UT S T H T

T

U

$ $$ $

$

Both Plus one, minus one ΔrU$m (0), then we can

get:

6.7.2.4.3 From Kirchhoff formula

r m r m 0( ) (0) d

T

pH T H C T $ $

r m r m r m 0(0) (0) ( ) d

T

pU H H T C T $ $ $

6.7.2.4 From molecule dissociation energy D

E F G H

r m E F G H(0) ( ) ( )U D D D D $

6.7.2.5 From heat content function

The heat content change of reaction and the value of heat content function are already known, so we can get the value of ΔrU$

m (0).

m m r mr m

( ) (0) (0)( ) [ { } ]

H T U UH T T

T T

$ $ $$

6.7.3 Heat content function

The value of heat content function can be worked out by q. When T is 298.15K, the value of H$

m(298.15K)-U$

m(0) can be checked out in the table.

m m,

ln( )

( ) (0)V N

qRT R

T

H T U

T

$ $

Make use of the value of heat content function to calculate the heat content change of the chemical reaction:

m m r mr m

( ) (0) (0)( ) [ { } ]

H T U UH T T

T T

$ $ $$

6.7.3.2 Calculation of Chem. Heat

Example 1/ 3227 计算甲醇反应 CO2 ＋ 2H2 ＝ CH3OH(g) 在 1000 K 时平衡常数 Kp 。 已知下列数据（ T=1000 K 时） ： -{[G (T) － U(0K)]/T}/JK-1mol-1 ΔfU (0 K)/kJmol-1 CO 204.054 -113.813 H2 136.984 0 CH3OH(g) 257.651 -190.246

Answer-1

Δ{[G (T)-U (0 K)]/T} =220.371 J·K-1·mol-1

ΔU (0 K)=Δ[ΔfU (0 K)] =-76.433 kJ·mol-1

-RlnK =Δ{[G (T)-U (0 K)]/T} +ΔrU (0 K)/T

=143.938

K =3.03×10-8

Exampl-2/3205试由下面数据计算反应 N2(g) + 3H2(g) ＝ 2NH3(g) 在 1000 K 时的平衡常

数 K 。 已知 : 1000 K N2(g) H2(g) NH3(g)

{(G (T) -H)/T}/JKmol-1 -197.9 -137.0 -203.5 298 K(H (T)- H )/kJmol-1 8.669 8.468 9.920 ΔfH /kJmol-1 -46.10

- RlnK =ΔrG /T =Δ[(G (T)-H )/T]1000 K +ΔrH /T Δ[(G (T)- H )/T]1000 K= 201.9 J·K-1·mol-1 ΔrH (298K) = 2[ΔH (NH3)]= -92.2 kJ·mol-1 Δ[H (T)- H ]298 K = -14.23 kJ·mol-1 ΔrH = -Δ[H (T)- H ]298 K+ΔrH (298 K) = - 77.97 kJ·mol-1 -RlnK=ΔrG /T =Δ[(G (T)-H )/T]1000 K+ΔrH /T = 123.9 K = 3.381×10-7

Example 3/ 3152 已知 298 K 时下列数据： H2O(g) H2(g) O2(g) ΔfH/kJmol-1 -241.83 0 0 S/JK-1mol-1 188.74 130.58 205.03 试求 H2O(g) H2(g) ＋ (1/2)O2(g) 在 25℃时的 K 。

Answer 3ΔrH =-ΔfH (H2O) =241.83 kJ·mol-1

ΔrS = S (H2)+(1/2)S (O2)- S (H2O)= 44.355 J·K-1·mol-1

ΔrG =ΔrH -TΔrS =228 612 J·mol-1

K =exp(-ΔrG /RT) =0.84×10-40

Example 4/ 3165 在 448 ～ 688 K 的温度区间内，用分光光度法研究了下面的气相反应： I2 ＋环戊烯 2HI ＋环戊二烯 得到 K 与温度的关系为： lnK = 17.39 － (51 034/4.575)×(1/T) (1) 计算在 573 K 时，反应的 ΔrG ， ΔrS ， ΔrH (2) 若开始时用等量的 I2 和环戊烯混合，温度为 573 K ，起始总压为 101.325 kPa ， 试求平衡后 I2 的分压；

AnswerΔrG =-RTlnK =(92 743-144.58T/K)-- (1) 代以 T=573 K 得 ΔrG =9870 J·mol-1 将 ΔrG =ΔrH -TΔrS 与 (1) 式比较 , 得： ΔrH =92 743 J·mol-1 , ΔrS =144.6 J·K-1·mol-1

(2)K =exp(-ΔrG /RT) =0.1260 I2(g) + 环戊烯 (g) = 2HI(g) + 环戊二烯 (g) 平衡： (p/2)-p (p/2)-p 2p p K={[p2(HI)p]/[p(I2)p( 环 )]}(1/p) =0.1260 解此三次方程得： p=0.1552p

6.8.1 Effect of T

For endothermic reaction, ΔrHm$ >0 , T incr

eases, K$p increase too, it is good for the p

ositive reaction. For exothermic reaction, ΔrHm

$ <0 , T increases the temperature, K$

p decreases. it is bad for the positive reaction.

r m2

d ln

dpK H

T RT

$ $

Gibbs-Helmholtz equation:2

( )(2) [ ]p

GHT

T T

Then

6.8.1.2 Effect of T If the T cover is not so big, and ΔrH$

m can be considered as constant, so we can get the fixed integral formula is:

2 r m

1 21

( ) 1 1ln ( )

( )p

p

K T H

R T TK T

$ $

$

If ΔrH$m is relative with T, therefore, put

the relative formula into the differential coefficient formula to integral, and use the table to work out the integral constant.

6.8.1.3 For ideal gas reaction When ideal gas is denoted by concentrati

on, because p=cRT, we can get

r m2

d ln

dcK U

T RT

$ $

2 r m

1 21

( ) 1 1ln ( )

( )c

c

K T U

R T TK T

$ $

$

This formula is useful in the gas reaction thermodynamics.

6.8.2 Effect of p

According to the principle of Le chatelier, increasing pressure, the reaction process to the direction of volume decreasing. Here, we can use the effect on the equilibrium constant by pressure to explain the principle from essential.

K$p is only temperature function

B B , f pp c RT K K $ $

r m B BB

( ) ln pG T RT K $ $ $

ln( ) 0p

T

K

p

$

6.8.2.1 For the ideal gas

6.8.2.1.2

therefore

BB

B( )p

pK

p$

$BB

B( )c RT

p

$

BB

B( )c

cK

c$

$

BB( )p c

c RTK K

p

$$ $

$

ln( ) 0c

T

K

p

$

Because

K$c is only temperature function

6.8.2.1.3B B m B

B

/ , x p p p V RT BB

B( )p

pK

p$

$

B BBB

B B( )x

pK x

p

BB( )p x

pK K

p

$

$

6.8.2.1.4

Kx is relative with P,The gas molecule numer decreases, adding pressure,

the reaction process towards positive direction, it is true in opposite.

BB

0

BB

ln ln( ) ( ) / 0p x

T T

K Kp

p p

$

mB

B

ln( ) /x

T

K V

p RTp

6.8.2.2 For the agglomerate reaction

r m B B

B

( ) ln aG T RT K $ $ $

*BB

( )( )T

TV

p

$*B

B

( )( )T

TV

p

$

*Bln

( )aT

K V

p RT

$

ΔV*B>0, increasing P, K$

a decreases, it is bad for the reaction, it is true in opposite. When P is not too large, because ΔV*

B

is not so big, the effect by P can be ignored.

6.8.3 Effect of indifferent gasB

B( )p xpK Kp

$

$B

BB BB

( )pxp

$

BB

B BBB

B

( )pnp n

$

Increases the indifferent gas, n increases, the item in the bracket decreases. Because K$

p is fixed value, therefore, will increase, the content of resultant will increase.

BB

0

B BB

n

6.9 Simultaneous equilibrium

In one reaction system, if several reactions happen at the same time, when getting equilibrium state, this condition is called simultaneous equilibrium. We need to consider the change of every substance amount in every reaction. In every equilibrium equation, the amount of the same substance must keep identical.

Example

At 600K, CH3Cl(g) and H2O(g) form CHOH, then forming (CH3) 2O, two equilibrium coexist:

(1) CH3Cl(g)+H2O(g)=CH3OH(g)+HCl(g)

(2) 2CH3OH(g)=(CH3) 2O(g)+H2O(g)

K$p,1=0.00154, K$

p,2=10.6. The amount of CH3OH and H2O is the same in equilibrium, respectively. What is the equilibrium translation rate of CH3Cl.

6.9.3 Example 2 Suppose the mole fraction of CH3Cl and H2

O is 1.0, when getting equilibrium, the forming mole fraction of HCl is x, the forming (CH3)2O is y.

(1) CH3Cl(g)+H2O(g)=CH3OH(g)+HCl(g) 1-x 1-x+y x-2y x (2) 2CH3OH(g)=(CH3) 2O(g)+H2O(g) X-2y y 1-x+y

6.9.4 Example 2 Because both of the reactions are zero. The

refore K$p=Kx

BB

,1

,2 2

( 2 ) 0.00154(1 )(1 )

(1 ) 10.6( 2 )

p

p

x y xKx x y

y x yKx y

$

$

Unite these two equations, then we can get x=0.048, y=0.009. The translation rate of CH3Cl is 0.148 or 4.8%.

6.10 Coupling reaction

Suppose two chemical reaction happen in the system, if the resultant of one reaction in another reaction is one reactant, therefore these two reactions are called coupling reaction.

Use the reaction whose ΔrG$m value

is very negative to bring along the reaction whose ΔrG$

m is larger than zero.

(1) A B C D

(2) C E F H

6.10 example

6.10.2 example 2

Therefore:(3) TiO2+C(s)+2Cl2 (g)=TiCl4 (l)+CO2(g) ΔrG$

m,3=-232.44kJ.mol-1

After couple reaction (1) and (2), reaction (3) can take place smoothly.

-12 2 4 2 r m,1 (1) TiO (s) 2Cl (g) TiCl (l) O (g) 161.94 kJ mol G $

-12 2 r m,2(2) C(s) O (g) CO (g) 394.38 kJ molG $

at 298.15K:

PTBn

G,)(

mPTB

PTTB

T Vn

V

n

G

PP

,, ][])([)(

)( VdPSdTdG

P

P

P

P m

P

PdP

P

RTdPVd

P

PRTLnPTPT ),(),(

P

PRTLnPTPT ),(),(

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ExercisesP446/1,2,5P447/9, 10P447/1P448/9

P449/14P450/16P451/23P453/33P453/34