First Year 2011-2012
Chemical equilibrium
By
Dr. Hisham Ezzat Abdellatef
Professor of Pharmaceutical Analytical
Chemistry
Classification of reaction
Homogeneous
Heterogeneous
only one phase
mixture is not uniform
e.g:
In the gas phase:
H2(g) + I2 → 2 HI(g)
In the liquid phase:
CH3COCl + CH3OH → CH3COOCH3 + HCI
Acetyl Chloride methyl alcohol methylacetate
→
irreversible reaction
or complete reaction
A + B → AB

reversible reaction
N2(g) + 3 H2(g)  2 NH3(g)
at equilibrium ----the rate of forward reaction equal the rate of
backward reaction
Factors influencing equilibrium:
1.
2.
3.
4.
forward and reverse rates
partial pressures
concentrations
temperature
Law of mass action
A2(g) + B2(g)  2 AB(g)
rate of the forward reaction
= Kf. [A2] [B2]
rate of the reverse reaction is = Kr [AB]2
At equilibrium
ratef = rater
Kf [A2] [B2] = Kr [AB] 2
K 
[AB]
2
[A 2 ][B 2 ]
K which is called the equilibrium constant
In general, for any reversible reaction
aA + bB  eE + fF
e
Kc 
[E] [F]
a
[A] [B]
eE +fF
f
 aA + bB
a
K 
\
b
K \ 
1
Kc
[A] [B]
e
[E] [F]
b
f
Mechanisms of more than one step 1
2 NO2CI(g)  2 NO2(g) + CI2(g)
2
Kc 
[NO 2 ] [Cl 2 ]
[NO 2 Cl]
2
mechanism consisting of two steps:
1- NO2CI  NO2 + Cl
K1 
[NO 2 ] [Cl]
2. NO2CI  NO2 + CI2
K2 
[NO 2 ] [Cl 2 ]
Kc = K1 K2 =
[NO 2 Cl]
[NO 2 ] [Cl]
[NO 2 ] [Cl] [NO 2 ] [Cl 2 ]
[NO 2 Cl]
[NO 2 ] [Cl]
2

[NO 2 ] [Cl 2 ]
[NO 2 Cl]
2
• For reactions involving gases
PαC
using partial pressures instead of concentration.
N2(g) + 3 H2(g)  2 NH3(g)
2
Kp 
p NH3
(Kc ≠ Kp)
Kc 
[NH 3 ]
2
[N 2 ] [H 2 ]
2
3
Kp 
p NH3
3
3
PN2 .P H2
PN2 .P H2
The Relationship Between Kp and KC:
aA + bB  eE + fF
e
f
a
b
e
PE . PF
Kp 
Kc 
PA . PB
Assuming ideal gas
concentration of a gas X
[X] =
nX
V

[E] [F]
f
a
b
[A] [B]
PV= nRT
PX
RT
Px is its partial pressure
 Px = [X] RT
e
Kp 
f
PE . PF
a
b
PA . PB
e

[E] (RT)
a
[A] (RT)
e
a
f
[F] (RT)
b
[B] (RT)
e
f
b
Kp =
[E] [F]
[A]
Kp = Kc. (RT)n(g)
a
[B]
f
b
(RT)
(e  f ) (a  b)
Example 1:
For the reaction
N2O4(g)  2 NO2(g)
The concentrations of the substances present in an equilibrium
mixture at 25°C are
[N2O4] = 4.27 x 10-2 mol/L
[NO2] = 1.41 x 10-2 mol/L
what is the value of Kc for this temperature.
Solution:
Kc =
[NO 2 ]
2
[N 2 O 4 ]

(1.41x10
(4.27X10
2
mol/l)
2
mol/l)
2
 4.66x10
3
mol/l
Example 2:
At 500 K. 1.0 mol of ONCI(g) is introduced into a one - liter container. At
equilibrium the ONCI(g) is 9.0% dissociated:
2 ONCI(g)  2 NO(g) + CI2(g)
Calculate the value of Kc for equilibrium at 500 K.
Solution:
[ONCI(g) ] = 1 mol/L since 9.0% dissociated,
Number of moles dissociated = [X] = 9 x1.0 mol
100
 0.09 mol ONCI
at equilibrium, [ONCI] = 1.0 mol/L - 0.09 mol/L = 0.91 mol/L
amounts of CI2 :
2 ONCI  2NO + CI2
2x
x
0.09 mol 0.045 mol

2ONCI
at start
2NO +
1.0 mol/L
Change
at equilibrium
----+ 0.09
0.09
- 0.09 mol/L
0.91
-----+ 0.045
0.045
2
Therefore, at Kc =
(0.09 mol/l)
=
2
[NO] [Cl 2 ]
[ONCl]
(0.045mol/
(0.9 mol/l)
2
l)
2
 4.4x10
CI2
4
mol/l
Example 3:
For the reaction
2 SO3(g)  2 SO2(g) + O2(g)
at 1100 K
Kc is 0.0271 mol/L. what is Kp at same temperature.
Solution:
n = 3-2 =1
Kp = Kc (RT)+1 = 0.0271 mol/L x (0.0821 L. atm / K.
mol) (1100 K) = 2.45 atm
Example 4:
What is Kc for the reaction?
N2(g) + 3 H2(g)  2 NH3(g)
At 500°C if Kp is 1.5 x 10-5 / atm-2 at this temperature.
Solution:
n = 2 - 4 = -2
n
Kp = Kc (RT)
T = 273 + 500 = 773 K
-2
Kp = Kc (RT)
Kc = Kp (RT)2 = x [0.0821.atm/K.mol x 773K]2
= (1.5 x I0-5/ atm2) (4.03 x 103 L2. atm2 / mol2)
= 6.04 x 10-2 L2 / mol2
Try
At 127oC, K = 2.6 x 10-5 mol2/L2 for the reaction
2NH3(g)  N2(g) + 2H2(g)
Calculate Kp at this temperature
Reaction quotient (Q).
Predicting the Direction of a Relation:
• For the reaction
PCl2(g)  PCl3(g) + Cl2(g) at 250oC
Kc =
[PCl 3 ][Cl 2 ]
[PCl 5 ]
 0.0415mol/
l
• Suppose that a mixture of 1.00 mol of PCI5(g)/
0.05 mol of PCI3(g)/ and 0.03 mol of CI2(g) is placed
in 1.0 L container.
Is this an equilibrium system,
or
will a net reaction occur in one direction or the
other?
1- Q < Kc
from left to right (the forward direction) to
approach equilibrium.
2- Q = Kc The system is in equilibrium.
3- Q > Kc
from right to left (the reversible direction) to
approach equilibrium.
Q=
[PCl 3 ][Cl 2 ]
[PCl 5 ]
Q (0.015 mol/L)

0 . 05 x 0 . 03
0.1
<
 0.015 mol/l
kc (0.0415 mol/L).
The system is not at equilibrium.
The reaction will proceed from left to right.
Example 4:
• For the reaction 2 SO2(g) + O2(g)  2 SO3(g)
at 827°C, kc is 36.9 L / mol. If 0.05 mol of SO2(g), 0.03
mol of O2(g), and 0.125 mol of SO3(g) are mixed in a
1.0 L container, in what direction will the reaction
proceed?
• Solution:
Q 
[SO 3 ]
2
2
[SO 2 ] [O 2 ]

[0.125mol/ l]
2
2
[0.05 mol/l] [0.03mol/l ]
 208 L/moL
Since Q (208 L/mol) > kc (36.9 L/mol), the reaction
will proceed from right to left (SO3 will dissociate).
Heterogeneous Equilibria:
• The concentration of a pure solid or a pure
liquid is constant and do not appear in the
expression for the equilibrium constant.
• For example
CaCO3(S)  CaO(S) + CO2(g)
Kc = [CO2]
3 Fe(s) + 4 H2O(g)  Fe3O4(s) + 4 H2(g)
Example 5:
Kc for the HI equilibrium at 425°C is 54.5:
H2(g) + I2(g)  2 HI(g)
A quantity of HI(g) is placed in a 1.01 container
and allowed to come to equilibrium at 425°C
What are the concentrations of H2(g) and I2(g) in
equilibrium with 0.5 mol/L of HI(g)
Solution:
at equilibrium [H2] = [I2]
[H2] = [I2] = x
[HI] = 0.5 mol/L
Kc 
[HI]
[H 2 ] [I 2 ]
[0.5 mol/l]
X
X 
2
(0.5)
54.5
 x = 0.068 mol/L
The equilibrium concentration is:
[HI] = 0.5 mol/ L
[H2] = [I2] = 0.068 mol/ L
2
2
4
 54.5
2
 54.5
2
 0.00456 mol /L
2
2
Example 6:
For the reaction
H2(g) + CO2(g)  H2O(g) + CO(g)
kc is 0.771 at 750°C. If 0.01 mol of H2 and 0.01
mol of CO2 are mixed in 1 liter container at
750°C, what are the concentrations of all
substances present at equilibrium?
• Solution:
If x mol of H2 reacts with x mol of CO2 out of the
total amount supplied, x mol H2O and x mol CO
will be produced. Hence
H2(g) + CO2(g)
 H2O(g) + CO(g)
At start
0.01 mol/L
0.01 mol/L
----
-----
Change
-x
-x
+x
+x
at equilibrium
0.01 -x
0.01 -x
X
X
Kc 
Kc 
[H 2 O] [CO]
 0.771
[H 2 ] [CO 2 ]
X
2
(0.01 - X)
2
square root
X
0.01 - X
 0 . 878
X = 0.0878 – 0.878 X
X= 0.00468 mol/l
At equilibrium, therefore
[H2] = [CO2] = 0.01 mol/L - 0.00468 mol/L = 0.0053 mol/L
[H2O] = [CO] = 0.00468 mol/L
Example 7:
• For the reaction
C(s) + CO2(g)  2 CO(g)
Kp is 167.5 atm at 1000°C. What is the partial pressure of
CO(g) in an equilibrium system in which the partial
pressure ofCO2(g) is 0.1atm?
• Solution:
2
Kp 
PCO
P CO2
 167.5 atm
2
PCO
0.1
• PCO2= 16.8
• PCO= 4.10 atm
 167.5 atm
Example 8:
Kp for the equilibrium:
FeO(s) + CO(g)  Fe(s) + CO2(g)
at 1000°C is 0.403. If CO(g) at a pressure of 1.0
atm, and excess FeO(s) are placed in a container
at 1000°C, what are the pressures of CO(g) and
CO2(g) when equilibrium is attained?
Solution:
Let x equal the partial pressure of CO2 when equilibrium is attained
FeO(s) + CO(g)  Fe(s) + CO2(g)
At start
1.0 atm
---
Change
-x
+x
1.0 - x atm
x
At equilibrium
Kp 
PCO2
PCO
 0 . 403
X atm
1.0 atm - X
 0 . 403
X = PCO2 = 0.287 atm
1.0 – X = Pco = .713
• If a change is made to an equilibrium, the
equilibrium shifts in the direction that
consumes the change
– Case 1: Changing the amounts of reactants /
products.
– Case 2: Changing the volume by changing
pressure.
– Case 3: Changing the temperature.
• If the concentration of substance is increased, the
equilibrium will shift in a way that will decrease
the concentration of the substance that was
added.
e.g:
H2(g) + I2(9)  2 HI(g)
•
Increase H2 or I2 → shift to to formation of HI
• Removal of H2 or I2 ← Reaction shift to
decomposition of HI.
• Increasing the pressure causes a shift in the direction
that will decrease the number of moles of gas.
2 SO2(g) + O2(g)  2 SO3(g)
3 moles
2 moles
• When the pressure on an equilibrium mixture is
increased (or the volume of the system decreased), the
position of equilibrium shifts to the right., and vice
versa.
• For reactions in which n = 0, pressure changes have
no effect on the position equilibrium.
e.g:
N2(g) + O2(g)  2 NO(g)
For the reaction
N2(g) + 3 H2(g)  2 NH3(g)
H = - 92.4 KJ
Since H is -ve, the reaction to the right evolves
heat
N2(g) + 3 H2(g)  2 NH3(g) + 92.4 KJ
The highest yields of NH3 will be obtained at the
lowest temperatures and high pressures.
Also consider the reaction
CO2(g) + H2(g)  CO(g) + H2O(g) H = + 41.2KJ
Since H is + ve , we can write the equation
41.2 KJ + CO2(g) + H2(g) —— CO(g) + H2O(g)
Increasing the temperature always favors the
endothermic change, and decreasing the
temperature always favors the exothermic change.
The addition of a catalyst
causes a system to achieve equilibrium faster
but does not alter the position of equilibrium.
If an inert gas is
introduced into a reaction vessel containing
other gases at equilibrium, it will not affect
the position of equilibrium because it will not
alter the partial pressures or the
concentrations of any of the substances
already present.