Chemical Equilibrium
Terms
A. Irreversible reactions:
A complete reaction of all reactants = irreversible reaction
Characteristic of irreversible reaction:
1. reaction in which one of product is a gas,
e.g.
2 H+ (aq) + Mg (s)  Mg2+ (aq) + H2 (g)
2. reaction in which one of product is a precipitate,
e.g.
Ag+ (aq) + Cl (aq)  AgCl (s)
3. reaction in which one of the product is a weak electrolyte,
e.g.
H+ (aq) + OH (aq)  H2O (l)
4. reaction in which a large amount of energy is released,
e.g.
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
B. Reversible reactions:
in which some products may react to give back the original reactants.
e.g.
2 NO2 (g) ===== N2O4 (g)
===== represents a reversible reaction.
C. Chemical equilibrium is recognized by the constancy of macroscopic properties such as concentrations, pressure and surface area
in a closed system at an uniform temperature.
At equilibrium
1. Both reactants and products are present, the amount of each substance does not change. (Macroscopic definition)
2. Forward and backward reactions continue at an equal rate. It is a dynamic process. (Microscopic definition)
Reactant ===== Product
d[Pr oduct ]
d [Re ac tan t ]

dt
dt
At equilibrium:
D. Steady state is recognized by the constancy of macroscopic properties in an open system, e.g. a Bunsen burner flame.
1. Quantitative aspects of equilibrium:
Consider the following reversible reaction:
k1
A + B  C+ D
k 1
Rate of forward reaction = k1[A][B]
Rate of backward reaction = k1[C][D]
At equilibrium, rate of forward reaction = rate of backward reaction
k1[A][B] = k1[A][D]
k1 [C][ D]

k 1 [ A][ B]
K
The equilibrium constant
where
K
[ C ][ D]
[ A][ B ]
k1
k 1
Consider the following reaction:
2 NO2 + F2 ===== 2 NO2F
Its reaction mechanism:
k1
Step I: NO2  F2  NO2  F
k 1
k2
Step II:
F  NO2  NO2 F
k 2
At equilibrium, each elementary process must be balanced by its reversible reaction.
k1[NO2][F2] = k1[NO2F][F]
(1)
k2[F][NO2] = k2[NO2F]
(2)

(1) (2)
k1[NO2][F2]k2[F][NO2] = k1[NO2F][F]k2[NO2F]
k1 k 2
[ NO2 F ]2

k 1 k 2 [ NO2 ]2 [ F2 ]
[ NO2 F ]2
K
[ NO2 ]2 [ F2 ]
In general:
For the reaction involving aqueous solution:
c
At equilibrium:
Kc 
a A (aq) + b B (aq) ===== c C (aq) + d D (aq)
d
[ C ] [ D]
[ A]a [ B]b
where Kc is the equilibrium constant expressed in concentrations.
For the reaction involving gaseous mixture:
a A (g) + b B (g) ===== c C (g) + d D (g)
At equilibrium: K p 
PCc PDd
PAa PBb
where Kp is the equilibrium constant expressed in concentrations.
Chemical equilibrium / page
1
For the general gaseous reaction:
a A (g) + b B (g) ===== c C (g) + d D (g)
Assume that these gases obey ideal gas law.
P
nRT
V
The partial pressure for gas A is PA = [A]RT
that for gas B is PB = [B]RT
that for gas C is PC = [C]RT
that for gas D is PD = [D]RT
the equilibrium constant expressed in terms of the partial pressure is
PCc PDd
PAa PBb
([C]RT ) c ([ D]RT ) d
KP 
([ A]RT ) a ([ D]RT ) d
[ C ] c [ D] d

 ( RT ) [( cd ) ( a b )]
[ A]a [ B]b
Kp 
= KC(RT)[(c+d)-(a+b)]
= KC(RT)n
n = number of mole of products - number of mole of reactant (for gaseous component only)
Degree of dissociation () is the extent to which a substance will dissociate at equilibrium.

dissociated .. amount
 100%
initial.. amount
Characteristic of equilibrium constant:
1. If a reaction is written in reverse, the equilibrium constant for the reverse reaction is the reciprocal of that for the original
reaction.
e.g.
N2 (g) + 3 H2 (g) ===== 2 NH3 (g)
K1
2 NH3 (g) ===== N2 (g) + 3 H2 (g)
K2
K1 
1
K2
2. If a reaction is multiplied by a certain factor n, its equilibrium constant be raised to nth power in order to obtain the
equilibrium constant for the new reaction equation.
e.g.
2 H2 (g) + O2 (g) ===== 2 H2O (g)
K1
H2 (g) +
1
O2 (g) ===== H2O (g)
2
K2
1
K 2  K12
3. The equilibrium constant for the overall reaction is the product of the equilibrium constants of the elementary reactions.
e.g. For the following mechanism:
2 NO (g) + O2 (g) ===== 2 NO2 (g)
K1
2 NO2 (g) ===== N2O4 (g) K2
Overall 2 NO (g) + O2 (g) ===== N2O4 (g) K3
K3  K1  K2
Ex. For the reaction
H2 (g) + I2 (g) ===== 2 HI (g)
When 40.64 g of I2 and 2.00 g of H2 are heated at a certain temperature, the equilibrium mixture contains 2.54 g of I2.
(a) How many moles of each species are present in the equilibrium mixture?
(b) Calculate the value of KC.
Ans. (a) H2 (g) + I2 (g) ===== 2 HI (g)
At equilibrium:
2.54
= 0.01 mol
254
2 40.64 2.54
No. of mole of H2 =  (

) = 0.85 mol
2
254 254
40.64 2.54
No. of mole of HI = 2  (

) = 0.30 mol
254 254
No. of mole of I2 =
(b) Let V be the volume of the reactor.
 0.3 
 
2
V 
[ HI ]
(0.3) 2
KC 


 10.59
[ H 2 ][ I 2 ]  0.85   0.01  0.85  0.01

 

 V   V 
2
Ex. At 400C, hydrogen iodide has a degree of dissociation of 20%.
Chemical equilibrium / page
2
2 HI (g) ===== H2 (g) + I2 (g)
(a) Calculate the value of KC.
(b) Calculate the composition of the equilibrium mixture produced if 1 mole of hydrogen and 2.1 moles of iodine react to
equilibrium at this temperature.
Ans. (a) Let the initial number of mole of HI be 1 mol dm3.
2 HI (g) ===== H2 (g) + I2 (g)
Concentration at initial:
1 mol dm3
Concentration at equilibrium:
1  0.2 M
0.1 M 0.1 M
KC 
01
.  01
.
1  0.2 2
 0.01563
(b) Let V be the volume of the reactor.
2 HI (g) ===== H2 (g) + I2 (g)
Concentration at initial:
0M
Concentration at equilibrium:
2x
M
V
Kc 
 1  x  2.1  x 



 V  V 
1
2.1
M
M
V
V
1 x
2.1  x
M
M
V
V
 2x 
 
V 
 1  x 2.1  x
0.01563 
4x 2
2
60 x  198.4 x  134.4  0
198.4  198.4 2  4  60  134.4
x
2  60
2
x1 = 2.356 (rejected, it is not reasonable for x > 1)
x2 = 0.9508
At equilibrium:
number of mole of HI = 1.902 mol
number of mole of H2 = 0.04917 mol
number of mole of I2 = 1.149 mol
Ex. For the equilibrium system at 298 K,
2 NOBr (g) ===== 2 NO (g) + Br2 (g)
34 % of NOBr is dissociated and a total pressure of 0.25 atmosphere is recorded. Determine the value of K P for the
dissociation at 298 K.
Ans. Let the initial number of mole of NOBr used be 1 mole.
2 NOBr (g) ===== 2 NO (g) + Br2 (g)
Number of mole at initial
1 mol
0 mol
0 mol
Number of mole at equilibrium
(1 0.34) mol
0.34 mol 0.17 mol
1  0.34
0.34
017
.
117
.
117
.
117
.
0.66
0.34
017
.
Partial pressure at equilibrium 0.25 
atm 0.25 
atm 0.25 
atm
117
.
117
.
117
.
2
0.34 
017
.

 0.25 
  0.25 

117
. 
117
.
KP 
 9.640  10 3 atm
2
0.66 

 0.25 


117
. 
Mole fraction at equilibrium
2. Measuring the extent of reaction:
Consider the following reaction:
a A (aq) + b B (aq) ===== c C (aq) + d D (aq)
At equilibrium:
[ C ] c [ D] d
Kc 
[ A]a [ B]b
A. K < 1
Backward reaction is favoured; concentration of reactant is dominant.
B. K > 1
Forward reaction is favoured; concentration of product is dominant.
3. Reaction quotient Q:
Consider the following reaction:
a A (aq) + b B (aq) ===== c C (aq) + d D (aq)
Chemical equilibrium / page
3
Q
[C]c [ D]d
[ A]a [ B]b
where Q is not an equilibrium constant;
and [A], [B], [C] and [D] are not necessary the concentration at equilibrium.
A. If Q = K, the reaction system is at equilibrium.
B. If Q < K, the reaction system will shift to the right hand side.
C. If Q > K, the reaction system will shift to the left hand side.
4. Characteristic of equilibrium:
At equilibrium:
A. Both reactants and products are present, the amount of each substance does not change. (Macroscopic definition)
B. Forward and backward reaction continue at an equal rate (a dynamic process). (Microscopic definition)
Reactant ===== Product
d[ product ]
d[reac tan t ]

dt
dt
At equilibrium:
C. An equilibrium may be approached and established from either side of a given system.
D. The equilibrium state is not always attained in chemical reactions.
e.g.
2 H2 (g) + O2 (g) ===== 2 H2O (g)
G < 0
The equilibrium is not attained in this system because the rate of the reaction between hydrogen and oxygen at room
temperature is too low.
From G, we know that the system tends towards equilibrium spontaneously. However, we cannot predict how rapidly a
system will approach the equilibrium state. The rate at which equilibrium is attained is determined by chemical kinetics.
E. Le Chatelier Principle:
If an equilibrium system is forced to change, the system moves in a direction that tends to remove the cause of the change.
5. Temperature effect on equilibrium constant:
Consider the following reversible reaction:
k1
A + B  C+ D
k2
Rate of forward reaction = k1[A][B]
Rate of backward reaction = k2[C][D]
At equilibrium, rate of forward reaction = rate of backward reaction
k1[A][B] = k2[A][D]
K
k1 [C][ D]

k2 [ A][ B]

Ea 1
RT
 Ea 2 )
H
k1 A1e
A1  ( Ea 1RT

K


e
 Ae RT
E
 a2
k2
A
2
A2 e RT
H
ln( K )  ln( A) 
RT
6. Effect of changes in concentration, pressure and temperature at equilibrium:
Factor
Forward Backward Position of
Equilibrium
reaction reaction
equilibrium
constant
rate
rate
Increase in
increase same
shift to product side
no change
[reactant]
Diagram
Chemical equilibrium / page
4
Decrease in
[reactant]
decrease same
shift to reactant side
no change
Increase in
[product]
same
increase
shift to reactant side
no change
Decrease in
[product]
same
decrease
shift to product side
no change
Increase in
temperature (H>0)
increase
more
increase
less
shift to product side
increase
(H<0)
increase
less
increase
more
shift to reactant side
decrease
Decrease in
temperature (H>0)
decrease
more
decrease
less
shift to reactant side
decrease
(H<0)
decrease
less
decrease
more
shift to product side
increase
Increase in pressure
(n>0)
increase
less
increase
more
shift to reactant side
no change
(n<0)
increase
more
increase
less
shift to product side
no change
(n=0)
no
change
no
change
no change
no change
Chemical equilibrium / page
5
Decrease in pressure
(n>0)
decrease
less
decrease
more
shift to product side
no change
(n<0)
decrease
more
decrease
less
shift to reactant side
no change
(n=0)
no
no
change
change
increase to the same
extennt
no change
no change
no change
no change
With positive
catalyst
7. Application of equilibrium principles: Haber process
Application of Le Chatelier Principle suggests that low temperature and high pressure will favour the production of ammonia.
N2 (g) + 3 H2 (g) ===== 2 NH3 (g)
H =  184 kJ mol1
However, it is expensive to build high pressure equipment and the reaction rate are slow at low temperatures.
The compromise used industrially involves an intermediate temperature around 500C and an intermediate pressure about 200
atmospheres in the presence of finely divided iron powder as catalyst to achieve a reasonable reaction rate. Under the above
conditions, only about 15 % yield of ammonia is obtained.
Chemical equilibrium / page
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