Chap. 6: Reaction Equilibrium in Ideal

Gas Mixtures

6.1: Chemical Potentials in an ideal Gas

Mixture

6.2: Ideal-Gas Reaction Equilibrium

6.3: Temperature Dependence of the

Equilibrium Constant

6.4: Ideal-Gas Equilibrium Calculations

6.6: Shifts in Ideal-Gas Reaction

Equilibria

6.1: Chemical Potentials in an Ideal Gas

Mixture

1. Chemical potential of a pure gas

μi(T, P) — Intensive property

dμ = dGm = -SmdT + VmdP

At constant T (dT = 0): dμ = VmdP

For an ideal gas, Vm = (RT/P) → dμ = (RT/P)dP

If the gas undergoes an isothermal change of

states from P1 to P2: 2

1

2

1

ln2

1

P

P

P

PPdRT

P

dPRTd

1

212 ln,,

P

PRTPTPT

Generally, we have μ(T, P ) values in the table,

thus

See Fig-6.1 on p. 175.

2. Chemical potentials in an ideal gas mixtures

For an ideal gas mixture:

(1) PV = ntotRT (2) Pi = χi·P

The standard state of component i of an ideal

gas at temperature T is defined to be pure ideal

gas i at T and pressure P = 1 bar.

bar 1 where, ln

PP

PRTT

The condition for phase equilibrium between

the mixture and pure i is: μi =μi*, where

μi : The chemical potential of the ith

component in the gas mixture at T.

μi *: The chemical potential of pure i at (T, Pi*).

For an ideal gas mixture:

μi(T, P, χ1, χ2, ··· ) =μi*(T, χiP) = μi*(T, Pi)

The chemical potential of pure gas i at T, Pi:

Equ-6.4 is a way to use the chemical potential

to of pure gas i to calculate the mixture one.

*6.4-Equ ln*

P

PRTT i

ii

6.2: Ideal Gas Reaction Equilibrium

1. The Reaction Equilibrium:

Example: for the ideal gas reaction

aA + bB ↔ cC + dD

In equilibrium: aμA + bμB = cμC + dμD

i

ii 0

P

Pd

P

Pc

P

Pb

P

PaRTdcba

P

PdRTd

P

PcRTc

P

PbRTb

P

PaRTa

A, B, C, DiP

PRTT

DCBADCBA

DD

CC

BB

AA

iii

lnlnlnln

0lnln

lnln Thus,

where, ln Because

P

dcba

b

eqB

a

eqA

d

eqD

c

eqC

P

P

ib

P

Pa

P

P

d

P

Pc

P

P

d

D

c

C

b

B

a

A

DCB

i

Aii

i

iTmiT

m

KRTΔGPPP

PPK

K

PRTΔG

P

P

P

P

P

P

P

PRTΔG

dcbaTGG

G

BA

DC

ln then,,

:reactions phase gasfor constant mequilibriu theDefine

mequilibriuat is :Note , ln

lnlnlnln

:becomes *6.5-Equcondition mequilibriu theTherefore,

substance pure afor Because

,,

,,

,,

3

,H,N

2

,NH

,

NHHN322

,,

,,

22

3

3222 ,3 ,1 )(NH2)(H3)(N :Example

ess)dimensionl is ( and ln

:Define ln

lnln

:number" tricstoichiome" heremember t still weIf

P

P

P

P

P

P

P

PK

ggg

KeKKRTΔG

P

PK

P

PRTΔG

P

PRT

P

PRTTΔG

eqeq

eq

i

eqi

P

PRT

G

PPT

i

eqi

P

i

eqi

T

i

eqi

i

eqi

i

i

iiT

i

T

ii

i

Example 6.1: see p. 179

2. Concentration and Mole Fraction

Equilibrium Constant

Define the “molar concentration”, Ci, of species

i in the mixture as Ci = ni /V .

Note: for an ideal gas, Ci = Pi /RT

dfba

b

eqB

a

eqA

d

eqD

f

eqF

b

eqB

a

eqA

d

eqD

f

eqF

PP

RTC

C

C

C

C

C

C

C

C

P

RTC

P

RTC

P

RTC

P

RTC

K

dDfFbBaA

,,

,,

,,

,,

then, If

n

P

i

eqi

n

CP

i

eqi

C

C

dfba

b

eqB

a

eqA

d

eqD

f

eqF

P

P

PKKK

K

P

RTCKKΔn

C

CK

K

C

P

RTC

C

C

C

C

C

C

C

C

K

i

i

where,

.Constant mEquilibriuFraction -Mole" define Similarly,

, let weif ,

.Constant mEquilibriuion Concentrat Standard" theDefine

Lmol 1L 1

mol 1 Where

,

i

i

,

,,

,,

3. Qualitative discussion of chemical

equilibrium

0 < KP° < ∞

1) KP° >> 1: Reactant → Product

2) KP° << 1: Reactant ← Product

3) Moderate KP°: incomplete reactions, for

example, N2 + 3H2 ↔ 2NH3

6.3: Temperature dependence of the

equilibrium constant

222

,,

,

,

,

,2

ln

:Equation Gibbs

because ,1ln

ln ln

RT

H

RT

STG

RT

S

RT

G

dT

Kd

SSdT

GdS

dT

dG

dTSdGdPVdTSdG

dT

dGG

dT

dG

dt

d

GGdT

Gd

RTRT

G

dT

Kd

RT

GKKRTG

P

i

imiim

im

mmmmm

i

im

i

imi

i

imiP

PPT

211

2

12

2

1

2

212

2

2

12

2

11ln

: thenconstant, ausually is small, is If

this)evaluate to (Need ln

lnln

ln ln

Equation Hofft Van' ln

:6.36*-Equ

2

1

2

1

2

1

TTR

H

TK

TK

THTTT

THdTRT

TH

TK

TK

dTRT

THTKTK

dTRT

HKddT

RT

HKd

RT

H

dT

Kd

P

P

T

TP

P

T

TPP

T

TPP

P

Example 6.2: Change of KP° with T, p. 183.

Find KP° at 600 K for N2O4(g) ↔ 2NO2(g)

(a) Using the approximation that ΔH° is

independent of T.

(b)Using the approximation that ΔCP° is

independent of T.

(c) Using the NIST-JANAF tables (see 5.9)

Solution:

Read p. 183 – 184.

6.4: Ideal gas equilibrium calculations

Measure ΔGT° of a reaction, then use

ΔGT° = -RTlnKP° to determine KP°

ΔGT° → KP° by thermodynamic calculations

The equilibrium composition of an ideal gas

reaction is a function of T and P (or T and V)

and the initial composition (mole numbers)

n1,0 , n2,0 , ··· , of the mixture.

The equilibrium extent of reaction ξeq

Δni ≡ ni,eq - ni,0 = νiξeq

ni = ni,0 + νiξeq

The specific steps to find the equilibrium

composition of an ideal gas reaction mixture:

1) Calculate ΔGT of the reaction:

Using a table of ΔGT or ΔHT and ΔST values.

2) Calculate KP using ΔG = -RTlnKP

3) Use the stoichiometry of the reaction to express

the equilibrium mole numbers ni = ni,0 + νiξeq

4) (a) If the reaction is running at fixed T and P,

use Pi = χiP and ni = ni,0 + νiξeq to express Pi in

terms of ξeq .

iiTfiT GΔΔG

,

4) (b) If the reaction is running at fixed T and V,

use Pi = niRT/V and ni = ni,0 + νiξeq to express Pi

in terms of ξeq .

5) Substitute the Pi’s (expressed as a function of

ξeq) into the equilibrium constant expression

and solve for ξeq :

6) Calculate the equilibrium mole numbers from

ξeq and the expressions for ni in step 3).

Example: see p. 187 about the “ICE” table.

i

i

iP

P

PK

"." , if ;m"Equilibriu" , if ;"" , If

:QuotientReaction

25.3

0.1 fixed, is if or,

,25.3

0.1 fixed, is if Thus,

25.3

250.0 ,

25.3

30.2 ,

25.3

0.1

mol25.3250.030201 m,equilibriuIn

20.50 3- 2.0 - 1.0 :(mol) mEquilibriu

2 3- - :(mol) Change

0.50 2.0 1.0 :(mol) Initial

)(NH2 )(H3 )(N

3

HN

2

NH

N

N

NN

NH

NH

H

H

N

N

322

22

3

2

2

22

3

3

2

2

2

2

PPPPPP

P

tottottot

tot

KQKQKQ

PP

PQ

V

RT

z

z

V

RTnPV

Pz

zPPP

z

z

n

n

z

z

n

n

z

z

n

n

zzz.z.n

zzz

zzz

ggg

Example 6.4: see p. 187.

Suppose that a system initially contains 0.300

mol of N2O4(g) and 0.500 mol of NO2(g) and the

equilibrium is attained at 25 °C and 2.00 atm.

Find the equilibrium composition.

Px

xPPP

x

xPP

x

x

n

n

x

x

n

n

xnxn

xxxn

xx

xx

gg

eeKKRTΔG

GΔΔG

tottot

tot

PP

iifi

RT

ΔG

0.800

20.500 ,

0.800

-0.300

0.800

20.500 ,

0.800

-0.300

:pressures partial and fractions Mole (4)

mol 2500.0 mol, 300.0

mol 800.02500.02300.0 Thus,

2 0.500 - 0.300 (mol) mEquilibriu

2 - (mol) Change

0.500 0.300 (mol) Initial

)(2NO )(ON : tableICE"" )3(

148.0 ln (2)

kJ/mol 73.431.51289.971 )1(

224242

2

2

42

42

242

298

NONOONON

NO

NO

ON

ON

NOON

242

298.15KKJ/mol 31447.8

J/mol 4730

298

,298298

mol 148.0176.02500.02500.0

mol 476.0176.0300.0300.0 )6(

!not , need wemol, 176.0 ,mol 324.0

00.2325 2.03654.0730

torr750

torr1520 where,

500.0240.0

42250.0148.0

800.0

0.300

800.0

20.5000.148

/

/ )5(

2

42

42

2

NO

ON

1221

2

2

2

2

2

2

ON

2

NO

xn

xn

xxxx

xx

P

P

P

P

xx

xx

P

P

x

x

P

P

x

x

PP

PPKP

Example 6.5: Self-reading, see p. 189.

KP° = 6.51 at 800 K for the ideal gas reaction:

2A + B ↔ C + D. If 3.000 mol of A, 1.000 mol of

B, and 4.000 mol of C are placed in an 8000-cm3

vessel at 800 K, find the equilibrium amounts of

all species.

6.6: Shifts in ideal gas reaction equilibrium

1. Reaction quotient, QP:

1) QP < KP : reaction shifts to right

2) QP > KP : reaction shifts to left

3) QP = KP : reaction in equilibrium

2. Isobaric temperature change

.by determined

is ofsign The positive. always is and Because

1

ln

2

222

H

dT

dKRTK

RT

HK

dT

dK

RT

H

dT

dK

KRT

H

dT

Kd

PP

PPP

P

P

i

iPiPQ

If ΔH° > 0 (endothermic reaction), dT > 0

will lead to dKP° > 0. The reaction will “ →”

because KP°(current) > KP°(original) = QP°

If ΔH° < 0 (exothermic reaction), dT > 0 will

lead to dKP° < 0. The reaction will “ ←”

because KP°(current) < KP°(original) = QP°

An increase in T at constant P in a closed

system shifts the equilibrium in the direction

in which the system absorbs heat from the

surroundings.

3. Isothermal pressure change:

An increase in P at constant T in a closed

system shifts the equilibrium in the direction

in which the system volume decreases.

4. Isochoric addition of inert gas:

Reaction not shifts because QP = KP due to

the no change of partial pressures of

reactants and products.

5. Addition of a reactant gas: See p. 196 – 197.

1) At const-T, const-V: reaction shift to right.

2) At const-T, const-P: Tricky!!

Le Châtelier’s Principle:

In a system at equilibrium, a change in one of

the variables that determines the equilibrium

will shift the equilibrium in the direction

counteracting the change in that variable.

6. Shifts in system with more than one reaction

Determined by experiments!