1
Subtopics
1.Chemical Potential in an Ideal Gas
Mixture.
2.Ideal-Gas Reaction Equilibrium
3.Temperature Dependence of the
Equilibrium Constant
4.Ideal-Gas Equilibrium Calculations
2
1.1 Chemical Potential of a Pure Ideal Gas
Expression for μ of a pure gas
 dG=-S dT + V dP
 Division by the no of moles gives:
 dGm = dμ = -Sm dT + Vm dP
 At constant T,
 dμ = Vm dP = (RT/P) dP
 If the gas undergoes an isothermal change from P1 to P2:
2

P2

1
dP
P
1
P1
 μ (T, P2) - μ (T, P1) = RT ln (P2/P1)
 Let P1 be the standard pressure P˚
 μ (T, P2) – μ˚(T) = RT ln (P2/ P˚)
 μ = μ˚(T) + RT ln (P/ P˚)
pure ideal gas
 . d  RT
3
1.2 Chemical Potential in an Ideal Gas Mixture
 An ideal gas mixture is a gas mixture having the
following properties:
1) The equation of state PV=ntotRT
obeyed for all T, P & compositions.
(ntot = total no. moles of gas).
2) If the mixture is separated from
pure gas i by a thermally conducting
rigid membrane permeable to
gas i only, at equilibrium the partial
pressure of gas i in the mixture
is equal to the pure-gas-i system.
Pi  xi P
Mole
fraction of
i(ni/ntot)
At equilibrium, P*i = P i
4
1.2 Chemical Potential in an Ideal Gas Mixture
 Let μi – the chemical potential of gas i in the






mixture
Let μ*i – the chemical potential of the pure gas in
equilibrium with the mixture through the
membrane.
*
The condition for phase equilibrium:  i   i
The mixture is at T & P, has mole fractions x1, x2,….xi
The pure gas i is at temp, T & pressure, P*i.
P*i at equilibrium equals to the partial pressure of i,
Pi in the mixture: Pi  xi P
Phase equilibrium condition becomes:
i T , P, x1, x2 ,....  i* T , xi P  i* T , Pi 
gas in the mixture
(ideal gas mixture)
At equilibrium, P*i = P i
pure gas
5
1.2 Chemical Potential in an Ideal Gas Mixture
 The chemical potential of a pure gas, i:
*
i
T , Pi  
0
i
T   RT ln Pi
P
0

(for standard state, P  1bar )
0
 The chemical potential of ideal gas mixture:
i   (T )  RT ln Pi P
0
i
0

(for standard state, P  1bar )
0
6
2. Ideal-Gas Reaction Equilibrium
 All the reactants and products are ideal gases
 For the ideal gas reaction:
aA  bB  cC  dD
 the equilibrium condition:
 
i i
i
0
a A  b B  cC  d D




0
0
 Substituting i  i  RT ln Pi P


into μA , μB , μC and μD :

 dRT ln P

P 
a A0  aRT ln PA P0  b B0  bRT ln PB P0  cC0  cRT ln PC P0 
d D0
D
0
7
2. Ideal-Gas Reaction Equilibrium





 dRT ln P

P 
a A0  aRT ln PA P0  b B0  bRT ln PB P0  cC0  cRT ln PC P0 
d D0
 The equilibrium condition becomes:
D
GT 
GT0
0
v G
i

m,T ,i

v 
i

i
i
i


cC0  d D0  a A0  b B0   RT c ln PC P 0  d ln PD P 0  a ln PA P 0  b ln PB P 0
G
 
0

P
  RT ln
P
C ,eq
A,eq


 P
P  P
P
0 c
D ,eq
0 a
B ,eq


P 
P



0 d
0 b

KP
 where eq – emphasize that these are partial pressure at
equilibrium.
8

2. Ideal-Gas Reaction Equilibrium
 Defining the standard equilibrium constant (K P ) for
the ideal gas reaction: aA + bB
K
0
P

P

P
C ,eq
A,eq
 P
P  P
P
0 c
D ,eq
0 a
B ,eq
cC + dD

P 
P
0 d
P 0  1bar
0 b
 Thus,
G   RT ln K
0
0
P
9
2. Ideal-Gas Reaction Equilibrium
 For the general ideal-gas reaction: 0  i i Ai
 Repeat the derivation above,



G   RT  i ln Pi ,eq P   RT  ln Pi ,eq P
0
T
0
i

0 vi 
 Then, G   RT ln  Pi ,eq P

 i


0
T
 Define:

K   Pi ,eq P
0
P


0 i
i
n
a
i
 a1a2 .....an
i 1

0 vi
i
 Then,
G   RT ln K
0
 Standard equilibrium constant:
0
P
K e
0
P
 G 0 RT
(Standard pressure equilibrium constant)
10
Example 1
 A mixture of 11.02 mmol of H2S & 5.48mmol of CH4 was placed in an
empty container along with a Pt catalyst & the equilibrium
2H 2 S ( g )  CH4 ( g )  4H 2 ( g )  CS2 ( g )
was established at 7000C & 762 torr.
 The reaction mixture was removed from the catalyst & rapidly cooled to
room temperature, where the rates of the forward & reverse reactions
are negligible.
 Analysis of the equilibrium mixture found 0.711 mmol of CS2.
0
0
 Find K P & G
for the reaction at 7000C.
Pi  xi P
K
0
P

P

P
C ,eq
A,eq
 P
P  P
P
0 c
D ,eq
0 a
B ,eq

P 
P
0 d
0 b
G 0   RT ln K P0
1bar =750torr
11
Answer (Example 1)

2H 2 S ( g )



11.02mmol 2 ( 0.711mmol)
9.60mmol
CH 4 ( g )



5.48mmol 0.711mmol
 4.77 mmol
 4 H 2 ( g )  CS2 ( g )


 
4 ( 0.711mmol)
 2.84 mmol
 Mole fraction: xH S  (9.60 17.92) mmol  0.536
2
xCH 4  (4.77 17.92) mmol  0.266
 P = 762 torr,
 Partial pressure:
PH 2 S  0.536(762torr )  408torr
PCH 4  0.266(762torr )  203torr
0.711mmol
xH 2  0.158
xCS 2  0.0397
PH 2  120torr
PCS 2  30.3torr
 Standard pressure, P0 = 1bar =750torr.
K P0

P

P
H2
H 2S
 P
P  P
P
0 4
CS 2
0 2
CH 4
  120 750 30.3 750  0.000331
P  408 750 203 750
P
0 1
0 1
4
2
12
Answer (Example 1)
 Use
G   RT ln K
0
0
P
 At 7000C (973K),
G 0  [8.314 J / molK ][973 K ] ln[0.000331 ]
 64 .8kJ / mol
13
3. Temperature Dependence of the
Equilibrium Constant
ln K   G RT
0
P
0
 The ideal-gas equilibrium constant (Kp0) is a function of
temperature only.
 Differentiation with respect to T:
d ln K p0
dT
 From
dG 0
 S 0
dT
d ln K 0p
dT
G
1 d G


2
RT
RT dT
0
0

G 0
S 0 G 0  TS 0



2
RT
RT
RT 2
14
3. Temperature Dependence of the Equilibrium Constant
d ln K 0p
dT

G 0  TS 0
RT 2
0
0
d
ln
K

H
P
 Since G  H  TS ,

dT
RT 2
0
0
0
 This is the Van’t Hoff equation.
 The greater the | ΔH0 |, the faster K P0 changes with
temperature.
 Integration:
K T2 
H 0 T 
ln

dT
2
K T1  T1 RT
 Neglect the temperature dependence of ΔH0,
0
P
0
P
T2
K P0 T2  H 0  1 1 
  
ln 0

K P T1 
R  T1 T2 
15
Example 2
N2O4 ( g )  2NO2 ( g )
 Find K P0 at 600K for the reaction by using the
approximation that ΔH0 is independent of T;
K P0 T2  H 0  1 1 
  
ln 0

K P T1 
R  T1 T2 
Note:
Substance
0
kJ/mol
 f H 298
H T0 

 i  f H T0,i
i
ln K P0   G 0 RT
0
kJ/mol
 f G298
NO2 (g)
33.18
51.31
N2O4 (g)
9.16
97.89
16
Answer (Example 2)
 If ΔH0 is independent of T, then the van’t Hoff equation
K T2  H  1 1 
  
ln

K T1 
R  T1 T2 
gives
H T0 
 
i
0
P
0
P
f
H T0,i
0
0
H 298
 [2(33.18)  9.16]kJ / mol  57.20kJ / mol
i
0
G298
 [2(51.31)  97.89]kJ / mol  4730 J / mol
 From
ln K P0   G 0 RT
K P0  eG
0
K P0 , 298  e4730 8.314298  0.148
RT
 From
ln
K
K P0 ,600
0.148
0
P , 600

57200 J / mol 
1
1 


  11.609
8.314 J / mol.K  298.15K 600 K 
 1.63x10
4
17
3. Temperature Dependence of the Equilibrium
Constant
1
2
 Since d (T )  T dT , the van’t Hoff equation can be written
as:
d ln K
H

dT
2
dT
RT
0
P
0
d ln K P0
H

d 1 T 
R
 The slope of a graph of ln Kp0 vs 1/T at a particular
temperature equals –ΔH0/R at that temperature.
 If ΔH0 is essentially constant over the temperature range, the
graph of lnKp0 vs 1/T is a straight line.
 The graph is useful to find ΔH0 if ΔfH0 of all the species are
not known.
18
Example 3
 Use the plot ln Kp0 vs 1/T for N 2 ( g )  3H 2 ( g )  2 NH 3 ( g )
for temperature in the range of
300 to 500K
25
0
lnK
 Estimate the ΔH .
p
0
20
15
d ln
H

d 1 T 
R
K P0
10
5
0
1
R  1.987cal mol K
1
-5
0
0.001
0.002
0.003
0.004
0.005
T -1 /K -1
-10
-15
-20
Plot of lnKp0 vs 1/T
19
Answer (Example 3)
 T-1 = 0.0040K-1, lnKp0 = 20.0.
 T-1 = 0.0022K-1, lnKp0 = 0.0.
 The slope:
20.0  0.0
0.0040  0.0022K 1
 From
 So,
 1.11x10 4 K
d ln K P0
H

 1.11x10 4 K
d 1 T 
R
H 0  (1.987 cal mol 1 K 1 )(1.11x10 4 K )
 22 kcal / mol
20
4. Ideal-Gas Equilibrium Calculations
 Thermodynamics enables us to find the Kp0 for a
reaction without making any measurements on an
equilibrium mixture.
 Kp0 - obvious value in finding the maximum yield of
product in a chemical reaction.
 If ΔGT0 is highly positive for a reaction, this reaction
will not be useful for producing the desired product.
 If ΔGT0 is negative or only slightly positive, the
reaction may be useful.
 A reaction with a negative ΔGT0 is found to proceed
extremely slow - + catalyst
21
4. Ideal-Gas Equilibrium Calculations
 The equilibrium composition of an ideal gas reaction
mixture is a function of :
 T and P (or T and V).
 the initial composition (mole numbers) n1,0,n2,0….. Of
the mixture.
 The equilibrium composition is related to the initial
composition by the equilibrium extent of reaction (ξeq).
ni  ni ,eq  ni ,0   i eq
 Our aim is to find ξeq.
22
4. Ideal-Gas Equilibrium Calculations
Specific steps to find the equilibrium composition of an
ideal-gas reaction mixture:
0
0
1) Calculate ΔGT0 of the reaction using GT  i i  f GT ,i
and a table of ΔfGT0 values.
2) Calculate Kp0 using G 0   RT ln K P0
[If ΔfGT0 data at T of the reaction are unavailable,
Kp0 at T can be estimated using
K P0 T2  H 0  1 1 
  
ln 0

K P T1 
R  T1 T2 
which assume ΔH0 is constant]
23
4. Ideal-Gas Equilibrium Calculations
3) Use the stoichiometry of the reaction to express the
equilibrium mole numbers (ni) in terms of the initial
mole number (ni,0) & the equilibrium extent of reaction
(ξeq), according to ni=n0+νi ξeq.
4) (a) If the reaction is run at fixed T & P, use
Pi  xi P  ni i ni P (if P is known)
& the expression for ni from ni=n0+νi ξeq to express
each equilibrium partial pressure Pi in term of ξeq.
(b) If the reaction is run at fixed T & V, use
Pi=niRT/V
(if V is known)
to express each Pi in terms of ξeq


24
Ideal-Gas Equilibrium Calculations
5) Substitute the Pi’s (as function of ξeq) into the

equilibrium constant expression
0
0
K P  i Pi P
& solve ξeq.
6) Calculate the equilibrium mole numbers from ξeq
and the expressions for ni in step 3.

vi
25
Example 4
 Suppose that a system initially contains 0.300 mol of
N2O4 (g) and 0.500 mol of NO2 (g) & the equilibrium
is attained at 250C and 2.00atm (1520 torr).
N 2O4 ( g )  2 NO2 ( g )
 Find the equilibrium composition.
 Note:
Substance
0
 f G298
kJ/mol
1.
GT0  i i  f GT0,i
2.
G 0   RT ln K P0
NO2 (g)
51.31
3. ni=n0+νi ξeq.
N2O4 (g)
97.89
4. P  x P  n
i
i
i
  n P
5.
i
i
K P0  i Pi P 0 
vi
6. Get 𝜉 and find n
26
Answer (Example 4)
 Get:
G
 From
G 0   RT ln K P0
0
298
 2(51.31)  97.89  4.73 kJ / mol
4730 J / mol  8.314 J / mol.K 298.1K ln K P0
ln K P0  1.908
K P0  0.148
 By the stoichiometry,
 2 NO2 ( g )


N 2O4 ( g )



Let x moles react to reach equilibrium
nN 2O4  0.300  x  mol
27
2x
nNO2  0.500  2 x  mol
Answer (Example 4)
 Since T & P are fixed:
PNO2
 Use
0.500  2 x
 x NO2 P 
P
0.800  x
PN 2O4  x N 2O4 P 
K P0  i Pi P 0 
vi

K  PNO2 P
0
P
 P
0 2
N 2O4

0.500  2 x  P P
0.148 
0.800  x 2
2
P

0 2

0 1
0.800  x
0.300  x  P P 0
0.250  2 x  4 x 2 P
0.148 
0.240  0.500 x  x 2 P 0
 P0 
0.250  2 x  4 x 2
0.148  
2
P
0
.
240

0
.
500
x

x
 
28
0.300  x
P
0.800  x


Answer (Example 4)
 The reaction occurs at: P=2.00atm=1520 torr & P0=1bar=750torr.
 Clearing the fractions:
4.0730 x 2  2.0365 x  0.2325  0
 Use quadratic formula:
 b  b 2  4ac 


x 
2a
 So, x = -0.324 @ -0.176
 Number of moles of each substance present at equilibrium must be
positive.
 Thus, nN O  0.300  x  mol  0 x  0.300
nNO  0.500  2 x  mol  0 x  0.250
2
4
2
 So,
 0.250  x  0.300
 As a result,
29
nN 2O4  0.476 mol
 x  0.176
nNO2  0.148 mol
Example 5
 Kp0 =6.51 at 800K 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 800K.
 Find the equilibrium amounts of all
species.
1 bar=750.06 torr,
1 atm = 760 torr
R=82.06 cm3 atm mol-1 K-1
1.
GT0  i i  f GT0,i
2.
G 0   RT ln K P0
3. ni=n0+νi ξeq.
4. Pi=niRT/V
5.
K P0  i Pi P 0 
vi
6. Get 𝜉 and find n
30
Answer (Example 5)
 Let x moles of B react to reach equilibrium, at the equilibrium:

2
A
n A  ( 3 2 x ) mol
B

nB  1 x  mol

C

nc   4  x  mol
 The reaction is run at constant T and V.


 Using Pi=niRT/V & substituting into K P0   Pi P 0
i



 
D

nD  x mol

vi

 We get: K P0  PC P 0  1 PD P 0  1 PA P 0  2 PB P 0  1
0
0



n
RT
V
n
RT
V
P
n
n
VP
D
K P0  C
 C2 D
2
n A RT V  nB RT V  n A nB RT
 Substitute P0=1bar=750.06 torr, R=82.06 cm3 atm mol-1 K-1,

4  x x
mol 2
8000 cm3 bar
6.51 
3  2 x 2 1  x  mol 3 83.14 cm3 bar mol 1K 1 800K 
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Answer (Example 5)
 We get,
x 3  3.995 x 2  5.269 x  2.250  0
 By using trial and error approach, solve the cubic equation.
 The requirements: nB>0 & nD>0, Hence, 0 < x <1.
 Guess if x=0, the left hand side = -2.250
 Guess if x =1, the left hand side = 0.024
 Guess if x=0.9, the left hand side = -0.015
 Therefore, 0.9 < x < 1.0.
 For x=0.94, the left hand side = 0.003
 For x=0.93, the left hand side=-0.001
 As a result,
nA=1.14 mol, nB=0.07mol, nC=4.93mol, nD=0.93mol.
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