1
Subtopics
Chemical Potential in an Ideal Gas
Mixture.
Ideal-Gas Reaction Equilibrium
Temperature Dependence of the
Equilibrium Constant
Ideal-Gas Equilibrium Calculations
2
The reaction Gibbs energy
 The reaction Gibbs energy:
 G 

 r G  
   p ,T
 the slope of the graph
of the Gibbs energy (G)
plotted against the
extent of reaction (ξ).
 ΔrG – the difference
between the chemical
potentials of the
reactants & products at
composition of the
reaction mixture.
the
3
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
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.
In an ideal gas mixture, P*i at equilibrium equals to the
partial pressure of i in the mixture. Pi  xi P
For ideal gas mixture,
i T , P, x1 , x2 ,....   T , xi P    T , Pi 
*
i
*
i
5
Chemical Potential in an Ideal Gas Mixture
 From chemical potential of a pure ideal gas:
 T , Pi    T   RT ln Pi P
*
i
0
i
0

(for standard state, P  1bar )
0
 For a ideal gas mixture:
i   (T )  RT ln Pi P
0
i
0

(for standard state, P 0  1bar )
6
Ideal-Gas Reaction Equilibrium
 For the reaction 0  i i Ai & specialize to the case of
all 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
 Substitute
    RT ln Pi P ,

0
i
0


a A0  RT ln PA P 0  b B0  RT ln PB P 0




 cC0  RT ln PC P 0  d D0  RT ln PD P 0

7
Ideal-Gas Reaction Equilibrium
 The equilibrium condition becomes:
GT0



cC0  d D0  a A0  b B0   RT
c ln P
C
G
0



P  P
P  P



P 0  d ln PD P 0  a ln PA P 0  b ln PB P 0

P
  RT ln
P
C ,eq
A,eq
0 c
D ,eq
0 a
B ,eq


P 
P
0 d
0 b
 where eq – emphasize that there are partial pressure at
equilibrium.
8
Ideal-Gas Reaction Equilibrium
 Defining the standard equilibrium constant (K0p) for
the ideal gas reaction: aA + bB
K
0
P

P

P
C ,eq
A,eq
 Then,
 P
P  P
P
0 c
D ,eq
0 a
B ,eq

P 
P
cC + dD
0 d
0 b
P 0  1bar
G   RT ln K
0
0
P
9
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:
 Then,

K   Pi ,eq P
0
P
i

 Standard equilibrium constant:
i
n
a
i
 a1a2 .....an
i 1

0 vi
G   RT ln K
0

0 i
0
P
K e
0
P
 G 0 RT
(Standard pressure equilibrium constant)
10
Ideal-Gas Reaction Equilibrium
 A chemical system which has reached the reversible
thermodynamic state shows no further net reaction.
Since G 0  0
 This does not mean that nothing is occurring.
 Actually, the chemical reactions in both directions
continue, but at the same rate.
 The result is no net change in concentrations, but
there is still a lot going on chemically.
11
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 was established at
7000C & 762 torr.
2H 2 S ( g )  CH4 ( g )  4H 2 ( g )  CS2 ( g )
 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.
 Find K P0 & G 0 for the reaction at 7000C.
12
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.
 It is independent of P, V, # of the reaction species
 Differentiation with respect to T:
d ln K
 From dG  S 0
0
dT
dT
d ln K p0
dT
0
p

G 0
1 d G 0


2
RT
RT dT

G 0 S 0 G 0  TS 0



2
RT
RT
RT
13
Temperature Dependence of the Equilibrium Constant
0
0
d
ln
K

H
 Since G  H  TS ,
P

dT
RT 2
0
0
0
 This is the Van’t Hoff equation.
 The greater the | ΔH0 |, the faster Kp0 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 
14
Example 2
N2O4 ( g )  2NO2 ( g )
 Find Kp0 at 600K for the reaction by using the
approximation that ΔH0 is independent of T;
Note:
Substance
0
 f H 298
kJ/mol
0
 f G298
kJ/mol
NO2 (g)
33.18
51.31
N2O4 (g)
9.16
97.89
15
Temperature Dependence of the Equilibrium Constant
1
2
 Since d (T )  T dT , the van’t Hoff equation can be:
d ln K P0
H

d 1 T 
R
 The slope of a graph of lnKp0 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.
16
Example 3
 Use the plot lnKp0 vs 1/T for N 2 ( g )  3H 2 ( g )  2 NH 3 ( g )
for temperature in the range of
200 to 1000K
25
0
lnK
 Estimate the ΔH .
p
0
20
15
10
5
0
-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
17
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
18
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.
19
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]
20
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


21
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
22
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.
N2O4 ( g )  2NO2 ( g )
 Find the equilibrium composition.
 Note:
Substance
0
 f H 298
kJ/mol
0
 f G298
kJ/mol
NO2 (g)
33.18
51.31
N2O4 (g)
9.16
97.89
23
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.
24
Le Chatelier’s Principle
 The statement of Le Chatelier’s principle is generally if
a system at equilibrium is stressed, the reaction will
shift to relieve that stress.
 Chemical reactions can be exposed to stresses
including a change in temperature, a change in
pressure, a change in concentration of one or more
of the participants and others.
(1) Effect of changes in temperature:
 The effect of an increase in temperature of a system
at equilibrium is a shift in the direction which
absorbs heat.
25
Le Chatelier’s Principle
(2) Effect of changes in pressure:
 When the pressure of a system at equilibrium increases,
the reaction occurs in the direction that lowers the
pressure by reducing the volume of gas.
(3) Effect of varying the concentration:
 Increasing the concentration of any component of a system
at equilibrium will cause a shift resulting in using up some
of the added substance.
(4) Effect of catalysts:
 Catalysts accelerate both forward & reverse reaction rates
equally.
 Catalysts can be used to shorten the amount of time it
takes to reach equilibrium when the original concentrations
do not match equilibrium concentration.
26
Answer (Example 1)

2H 2 S ( g )



11.02mmol 2 ( 0.711mmol)
9.60mmol
 Mole fraction:
CH 4 ( g )



5.48mmol0.711mmol
 4.77 mmol
 4 H 2 ( g )  CS2 ( g )


 
4 ( 0.711mmol)
 2.84 mmol
xH 2 S  (9.60 17.92) mmol  0.536
xCH 4  (4.77 17.92) mmol  0.266
0.711mmol
xH 2  0.158
xCS 2  0.0397
 P = 762 torr,
PH 2  120torr
 Partial pressure: PH S  0.536(762torr )  408torr
2
PCS 2  30.3torr
PCH 4  0.266(762torr )  203torr
 Standard pressure, P0 = 1bar =750torr.
K P0

P

P
H2
H 2S
27
 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
Answer (Example 1)
 Use
G   RT ln K
0
0
P
 At 7000C (973K),
G  [8.314 J / molK ][973K ] ln[ 0.000331]
0
 64.8kJ / mol
28
Answer (Example 2)
 If ΔH0 is independent of T, then the van’t Hoff equation gives
K T2  H  1 1 
  
ln

K T1 
R  T1 T2 
0
P
0
P
 From
0
H T0   i H m0 ,T ,i
0
H 298
 [2(33.18)  9.16]kJ / mol  57.20kJ / mol
i
 From K  e
0
P
0
G298
 [2(51.31)  97.89]kJ / mol  4730 J / mol
 G 0 RT
K
0
P , 298
e
4730 8.314298
K P0 ,600
 0.148
57200 J / mol 
1
1 
ln



  11.609
0.148 8.314 J / mol.K  298.15K 600 K 
K
29
0
P , 600
 1.63x10
4
Answer (Example 3)
 T-1 = 0.0040K-1, lnKp0 = 20.0.
 T-1 = 0.0022K-1, lnKp0 = 0.0.
 The slope:
 From
20.0  0.0
0.0040  0.0022K 1
 1.11x10 4 K
d ln K P0
H

 1.11x10 4 K
d 1 T 
R
 So, H 0  (1.987cal mol 1 K 1 )(1.11x10 4 K )
 22kcal / mol
30
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
31
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

K P0  i Pi P 0

K  PNO2 P
0
P
PN 2O4

0.300  x
 x N 2O4 P 
P
0.800  x
vi
 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 P0
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
 
32


Answer (Example 4)
 The reaction occurs at: P=2.00atm=1520 torr & P0=1bar=750torr.
 Clearing the fractions:
 Use quadratic formula:
4.0730 x 2  2.0365 x  0.2325  0

 b  b  4ac  
x
2
1/ 2
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,
33
nN 2O4  0.476 mol
 x  0.176
nNO2  0.148 mol
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

 We get: K 0  P P 0
P
C
 P
1
D
P
 P
0 1
0



n
RT
V
n
RT
V
P
D
K P0  C
nA RT V 2 nB RT V 
A
P
0
 P
2
B
D

nD  x mol

vi
P
0

1
nC nD VP 0
 2
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 
34
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.
35