Chemical Equilibrium
※ What’s Chemical Equilibrium:
☆ Thermodynamics:
Spontaneous
  S 
 G 
Equilibrium
  S 
0
total
0
s y s t e m
 0 or  G 
total
system
0
☆ Kinetics:
 Concept introduced by A. W. Williamson in 1851 during
study esterifications
 Forward reaction rate = Reverse reaction rate

Equilibrium (at a specified temperature)
 Reaction rates are not ceased when equilibrium is
reached.
 Dynamical equilibrium
 Chemical equilibrium
※ Expression of Equilibrium Constant:
Considering a stoichiometric reaction: aA  bB  yY  zZ
Forward reaction rate: r  k  A  B
a
1
b
1
Reverse reaction rate: r  k Y   Z 
y
-1
-1
At equilibrium, r  r
1
1
1
z
k1 Y   Z 
 a b  Kc
k1  A  B
y
z
(A) Types of equilibrium constant:
Kc;
Kp;
Ksp
(B) The factors that affect equilibrium constant:
Temperature;
Reacting species
※ Chemical Potential:
 G 

 n 
 
i
i
 U 

 n 

T , P , ni  n j
i
 H 

 n 

S , V , ni  n j
i
 A 

 n 

S , P , ni  n j
i
T , V , ni  n j
At constant P and T, the direction of reaction is determined by the
chemical potential, i .
(A) Chemical potential for ideal solution:
 ( T , P )   ( T , P )  RT n x
*
i
i
i
 ideal solution
where  ( T , P ) is the chemical potential of the pure substance
*
i
i when it is at the temperature T and pressure P.
xi is the mole fraction of substance i in the solution.
(B) Chemical potential for non-ideal solution:
For non-ideal solution, the chemical potential of substance i
is defined by:
 ( T , P )   ( T , P ) RT n a
*
i
i
i
(definition of ai)
where  is the chemical potential of the pure substance i.
*
i
2
ai is the activity.
※ Extent of reaction:
 The degree of reaction progress
 The ratio of the amount of any species produced/formed to
the its stoichiometric coefficient
 A  B Y  Z
e.g.
A
B
Y
for reactants:

for products:

Z
n A
A
nY
Y


n B
B
n Z
Z
※ The general equilibrium expression:
Consider a closed system in which a single reaction occurs in a
single phase.
If an infinitesimal amount of reaction takes place
at constant T and P, the change in the Gibbs energy is given by:
dG    dn
N
i 1
i
i
where N is the number of species involved in the reaction;
dni is the differential change in the amount of the i-th
species.
The differential change in the amount of a species I is given
by dn   d .
i
i
3
 G 
Thus, we can identify  
  
  i i
N
i 1
T ,P
The quantity  G  is the rate of change of Gibbs energy per
 
T ,P
mole of reaction.
  G   0 (forward reaction spontaneous)
   
 

  G    0 (backward reaction spontaneous)

 G   0 (equilibrium)
  
T ,P
T ,P
T ,P
The chemical potential of a species is expressed by,
 ( T , P )   ( T , P ) RT n a
*
i
i
i
At equilibrium,     RT   n a  0
N
i
i 1

N
o
i
i 1
i
i,eq
N

    -RT n  a
N
o
i
i 1
i
i 1
i
i,eq
Define the equilibrium constant K of the reaction as
N
K  a
i 1
i
i,eq
and the quantity    is the standard reaction Gibbs energy,
N
i 1
o
i
i
G.
o
r
Therefore, we can obtain
 G   RT n K
o
r
(A) Equilibrium in pure gases:
4
a P
f
( ideal gas ) or a 
i
i
P
o
i
f
i
o
( real gas)
i
i
 f 
P
K P    (ideal gas) or K P    (real gas)
P 
f 
N
N
i
i
o
i 1
o
i 1
eq
eq
(B) Equilibrium in pure condensed phases:
Substances that are pure liquids or solids contribute a factor nearly
equal to unity to the equilibrium constant expression.
To a good
approximation, they can be omitted from it, i.e.
a  1 (pure liquid or solid)
i
(C) Equilibrium in solution:
a  x ( ideal solution ) or a   m ( real solution)
i
i
i
K   x
N
a
i 1
i
i

i
i
(ideal solution) or K   m  (real solution)
N
eq
a
i 1
i
i
i
eq
※Equilibrium constant in concentration units:
For a general reaction: A A  B B  YY  Z Z

∵ PV  nRT
 n
P    RT = CRT
V 
P P 
C C 
K 


 C C   RT 
P
P




uY
Y
P
uZ
Z
uA
A
uB
B
eq
∴ K  K  RT 
P
uY
uZ
Y
Z
uA
uB
A
B
uY  uZ  uA  uB
 K  RT 
C
eq
u
C
where   = the stoichiometric sum (係數和)

KC is dependent on T only.
5
u
※Equilibrium constant in mole fraction:
From Dalton’s law: P  y P
i
i

 P P 
 y y 
K            P
P P 
y y 
Y
Z
Y
P
Y
Z
A
Y
B
A
A
B
A
eq
∴ K  K  P
P

Z
Y  Z  A  B
Z
y
B
B
 K  P

eq

y
Ky is dependent on T and P.
 Units:
Thermodynamic equilibrium constant is dimensionless, but
practical equilibrium constant is not.
A B Z
Ex.1:
K
C
Z ;

 A B
K 
C

mole
liter

1
Ex. 2: A  B  Y  Z
K 
C

Y  Z  ;
 A B
In general,
Kc is dimensionless
K  ( unit )  u
 K depends on how the stoichiometric equation is written:
Z
K 
 A
Z
K 
 A
2
Ex.: A  2Z
1
2
AZ
C
C
1
G   RT nK
0
0
1
C
G   RT nK
2
6
0
0
2
C
∴ K   K
C
Ex.:
C

1
2
G 
1
0
;
2
G1
0
2
 BC
 A
 BY 

C
A BC
K 
C  BY
K
1C
2C
  Y   K
 A
2
K 
A  2B  Y
C
B
1C
K
2C
※ Determination of equilibrium constant:
(A) From Gibbs energy: G   RT n K
o
The Gibbs energy can be determined by:
 From standard Gibbs energy of formation:
G    G
o
i
i
0
f ,i
 From reaction enthalpy: G  H  TS
0
0
0
 From electromotive force: G   zFE
0
(B) From the reaction extent:
Procedures:
 Given the equilibrium extent of reaction,ξ
 To express the equilibrium pressure of various
components in terms ofξ
 To determine the equilibrium constant in terms ofξ
7
2NO
Ex.:
NO
initial
1 mole
final
(1-ξ) mole
4( g )
2
1
Pequilibrium
1
K 
P
2
PN O
2
0 mole

the reaction extent =ξ
2  mole
2
P
2
PNO
the total equilibrium pressure = P
2( g )
1
 2 
P

1   
1
4
1
P
2

P
4
2
1
2
P
Ex. The Gibbs energies of formation of NO2(g) and N2O4(g) are 51.3
kJ/mole and 102.0 kJ/mole, respectively, in the standard state,
(a) Calculate Kp and Kc for the reaction N O
2
4( g )
 2NO if
2( g )
the gas is ideal.
(b) Calculate Kx at 1 bar pressure.
(c) At what pressure is N2O4 50% dissociated?
(d) What is G if the standard state is 1 mole
0
Sol:
(a) G  2  51. 3  102 . 0  0 . 6 kJ
liter
?
0
n K  
0
P
G
0

RT
600
8 . 3145  298 . 15
K  0 .785 bar
0
P
 K  K  RT 
P
C
u
 K  RT 
C
8
mole
 0 . 242 ;
0
KP
K 
0
C
0 . 785  10

RT
8 . 3145  298 . 15
(b) K  K  P
P
y
KP
K 
y
(c) K 
P
u
2
 3 . 17  10 mole
liter
K P
y
 0 .785
P
4
5
2
1
P ; ξ=0.5; Kp= 0.785 bar
2
 0 .785 
4  0.5
2
1  0.5
2
P
P = 0.589 bar
(d) G   RT nK  8 . 3145  298 . 15  n3 . 17  10
0
2
C
= 8555 joule
mole
Ex. Write down the expression of equilibrium constant for
CaCO
3( S )
 CaO  CO
(S)
2( g )
Assumed CO2(g) is ideal gas.
Sol:
∵a
CaCO3
 1 and a
CaO
∴K  P
1
P
CO2
Ex. At 1393K,
Fe O
2
3( S )
 3CO  2Fe  3CO
( g)
(S)
2( g )
2CO 2CO O
2( g )
( g)
K1=0.0467
K2 = 1. 4  10
2( g )
 12
What is the equilibrium pressure of O2(g) in a vessel containing
9

Fe2O3(g) and Fe(s) at equilibrium at 1393 K?
Sol:
The total reaction is
 3CO  2Fe  3CO
Fe O
3( S )
2
( g)
(S)
2
2( g )
2CO 2CO O
2( g )
( g)
2 F eO
2
3( S )
3
2( g )
 4 F e  3O
(S)
KK K
2( g )
 K  0 . 0467   1. 4  10
2
K  P
12

3
 6 . 01  10
2
3
1
2
39
3
O2
 P  6 . 01  10
O2
 39

1
3
 1. 82  10 bar
13
※ Effect of pressure on equilibrium:
For an ideal mixture of ideal gases, the equilibrium constant can
be expressed in terms of the individual mole fraction and the total
pressure of the system; i.e. K  K  P
P

y
 if    0 ; K isindependent of pressure
y

equilibrium doesn’t shift
 if    0 ; K islower when pressureis higher
y

equilibrium toward reactants
 if    0 ; K is higher when pressureis higher
y

equilibrium toward products
※ Effect of volume on equilibrium:
10
V 
nRT
P
 if    0 ; K isindependent of volume
y

equilibrium doesn’t shift
 if    0 ; K islower when volumeislower
y

equilibrium toward reactants
 if    0 ; K is higher when volumeislower
y

equilibrium toward products
※ Effect of inert gas on equilibrium:
i
∵ K    P     y P
i
P
i
i
i
i
i
 n 
 P 

  n 
  
P  

n 
n 
i
i
i
i
 P
 K  
n
i

i
P
equilibrium doesn’t shift
 if    0 ; adding inert gas is to increase  n
i

i
Kn increasing; equilibrium toward products
 if    0 ; adding inert gas is to increase  n
i

i
n
 if    0 ; K isequal to K
n
i

P
i

 K



i
Kn decreasing; equilibrium toward reactants
※ Effect of temperature on equilibrium:
11
i
i
   G
From the Gibbs-Helmholtz equation:  
 T  T
 G   RT n K
o
   G
 
 T  T
0
H





T
0
2
P
0
P
H
  nK 



R



 T 

T
0
0
P
2
P
P
H
  nK 

 T 
RT

0
0
0
called the van’t Hoff equation.
P
2
P
 if H  0 (exothermic reaction), K decreases when T is
0
0
P
increasing

equilibrium toward reactants
 if H  0 (endothermic reaction), K increases when T is
0
0
P
increasing

equilibrium toward products
 if H is independent of temperature, then
0
n K  
0
H
P
0
RT
(∴slope = 

H
R
S
0
R
0
and intercept =
S
0
)
R
 if H is a function of temperature, i.e. C  d  eT  fT
0
2
P
H ( T )  H 25 C  d T  298 . 15 
0
m
m
T
1 e
2
2
1 
1
 288 . 15   f  

 T 298 . 15 
2
then, the equation can be integrated to be:
1  H d d  298 . 15 e 298 . 15  e f
f

nK P    2 m  




dT  C
2
2
3
2 
R T
T
T
2
2T
T
298 . 15  T 
0
2
0
If H , d , e , and f are known , KP can be calculated at any
0
m
12
other temperature.
 if Ky instead of Kp: K  K  P
P
  nK 
H
∴ 


RT
 T 
0

y
0
y
2
P
 if Kc instead of Kp: K  K  RT 
P
U
  nK 

∴ 
RT
 T 
0
u
C
0
C
2
P
★ Le Chatelier Principle:
When an independent variable of a system at equilibrium is
changed, the equilibrium shifts in the direction that tends to
reduce the effect of the change.
 When total pressure is increased, the equilibrium shifts in
the direction to reduce the number of molecules.
 An increase in temperature at constant pressure in a closed
system shifts the equilibrium in the direction in which the
system absorbs heat from the surroundings.
Ex.: Consider that a dissociation reaction of water becomes O2 and
H2:
HO H
2
( g)
2( g )

1
2
H  57 . 8 Kcal
0
O
2( g )
r
mole
(a) will the dissociation increase or decrease if the pressure is
13
reduced?
(b) will the dissociation increase or decrease if argon gas is
added, holding the total pressure equal to 1 bar?
(c) will the dissociation increase or decrease if the pressure is
raised by addition of argon at constant volume to the close
system?
(d) will the dissociation increase or decrease if oxygen gas is
added?
(e) will the dissociation increase or decrease if temperature is
raised?
Sol: (a) increase; (b) increase; (c) no change; (d) decrease;
(e) increase.
※ Coupling of reactions:
If two reactions exist in a chemical system, one of reaction
product is the reactant of another reaction.
This is called the
coupling of reactions.
e.g.
A+B=L+M
K1 < 1
L+C=N+Q
K2 > 1
A+B+C=M+N+Q
K3 = K1K2 > 1
The importance of the coupling of reaction is to increase the
production rate of product M.
14