Chapter 6
Chemical equilibrium
Classification of reaction
Chemical
reactions
are
either
homogeneous
or
heterogeneous. In the former case only one phase is
present
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
In heterogeneous reactions the mixture is not uniform
throughout.
e.g:
Alumina
C2H5OH(l) ⇔
ethanol
Chemical equilibrium
C2H4(g) + H2O
ethane
Page 148
On the other hand, when a chemical reaction proceeds in
one direction, i.e: from reactant to products, the reaction is
known as irreversible reaction or complete reaction.
A + B → AB
On the contrary, reactions that proceed either in the
forward or in the backward (reverse) directions are known
as reversible reaction and the equation for the reaction can
be written.
N2(g) + 3 H2(g)  2 NH3(g)
The double arrow () indicates that the equation can be
read in either direction.
All reversible processes tend to attain a state of
equilibrium. For
a reversible
chemical
reaction,
an
equilibrium state is attained when the rate at which the
direct reaction is proceeding equals the rate at which the
reverse reaction is proceeding.
Chemical equilibrium
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• The Law of Mass Action:
Consider a hypothetical reversible reaction
A2(g) + B2(g)  2 AB(g)
The rate of the forward reaction is
rate f = Kf. [A2] [B2]
and the rate of the reverse reaction is
rate r = Kr [AB]2
At equilibrium
ratef = rater
Kf [A2] [B2] = Kf [AB] 2
Kf
[AB] 2

K r [A 2 ][B2 ]
Kf
= K which is called the equilibrium constant
Kr
[AB] 2
Therefore, K =
[A 2 ][B2 ]
The numerical value of K varies with temperature. It does
not vary with changes in the amounts of substances used
to establish the equilibrium, changes in pressure, or the
presence of catalyst.
Chemical equilibrium
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In general, for any reversible reaction
aA + bB  eE + fF
at constant temperature at equilibrium,
[E]e [F]f
Kc =
[A] a [B]b
where the quantities written within brackets denote
equilibrium molar concentrations (mol/liter).
 If the general equation were written in reverse form:
eE +fF

aA + bB
the expression for the equilibrium constant (K\) would be:
K\ =
[A] a [B]b
[E] e [F]f
Notice that K' is the reciprocal of Kc i.e K\ =
1
Kc
* If the law of chemical equilibrium hold for reactions that
occur by mechanisms of more than one step 1 :
2 NO2CI(g)  2 NO2(g) + CI2(g)
[NO 2 ]2 [Cl2 ]
Kc =
[NO 2Cl] 2
The reaction is believed to occur by means of a
mechanism consisting of two steps:
Chemical equilibrium
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1. NO2CI  NO2 + Cl
2. NO2CI  NO2 + CI2
therefore;
and
K1 =
K2 =
KC
[NO 2 ] [Cl]
[NO 2Cl]
[NO 2 ] [Cl2 ]
[NO 2 ] [Cl]
[NO 2 ] [Cl] [NO 2 ] [Cl2 ]
[NO 2 ]2 [Cl2 ]

= K1 K2 =
[NO 2Cl] [NO 2 ] [Cl]
[NO 2Cl] 2
In this case, the equilibrium constant for the overall
change is the product of the equilibrium constants of each
the steps:
KC = K1 K2
 For reactions involving gases, the partial pressures of
the reactants and products are proportional to their
molar
concentrations.
The
equilibrium
constant
expression for these reactions can therefore be
written
using
partial
pressures
instead
of
concentration. For example,
N2(g) + 3 H2(g)

2 NH3(g)
Chemical equilibrium
Page 152
2
pNH3
Kp =
PN2 .PH23
We will use the symbol, Kp, to donate equilibrium
constants derived from partial pressure and kc to indicate
equilibrium constants having molar concentrations in the
mass action expression (Kc ≠ Kp)
[NH 3 ]2
Kc =
[N2 ] [H2 ]3
2
pNH3
Kp =
PN2 .PH23
If Kc is a larger number, the forward reaction is fairly
complete. If Kc is a small number, the reverse reaction is
fairly complete.
The Relationship Between Kp and KC:
For the general equation,
aA + bB  eE + fF
PEe . PFf
Kp = a
PA . PBb
[E]e [F]f
Kc =
[A] a [B]b
 n 
Concentration has the units mol/L 

 V.
Assuming ideal gas behavior, we can use the gas law,
PV= nRT
Chemical equilibrium
Page 153
to obtain the concentration of a gas X
[X] =
nX
P
 X
V RT
Where Px is its partial pressure
 Px = [X] RT
[E] e (RT) e [F]f (RT) f
PEe . PFf

Kp = a
PA . PBb [A] a (RT) a [B]b (RT) b
[E] e [F]f
(RT) (e f )(ab)
Kp =
a
b
[A] [B]
Kp = Kc. (RT)n(g)
Where n(g) is the change in the number of moles of gas
when going from reactants to products.
ng = (number of moles of gaseous products) -(number
of moles of gaseous reactants)
Example 6.1: 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
Chemical equilibrium
Page 154
what is the value of Kc for this temperature.
Solution:
[NO 2 ]2 (1.41x10 2 mol/l) 2

 4.66x103 mol/l
Kc =
2
[N2O4 ] (4.27X10 mol/l)
Example 6.2: 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: The concentration of ONCI(g) is 1 mol/L since
ONCl is 9.0% dissociated,
Number of moles dissociated =
9
[X] =
x1.0 mol  0.09 mol ONCI
100
The concentration of ONCI at equilibrium, therefore, is
[ONCI] = 1.0 mol/L - 0.09 mol/L = 0.91 mol/L
We can derive the amounts of CI2 from the coefficient of
the chemical equation:
Chemical equilibrium
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2 ONCI 
2NO +
CI2
2x
0.09 mol
x
0.045 mol
2ONCI

2NO +
CI2
at start
1.0 mol/L
-----
------
Change
- 0.09 mol/L
+ 0.09
+ 0.045
0.91
0.09
0.045
at equilibrium
[NO] 2 [Cl2 ]
Therefore, at Kc =
[ONCl] 2
(0.09 mol/l) 2 (0.045mol/ l)
 4.4x10 4 mol/l
2
(0.9 mol/l)
=
Example 6.3: 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
Chemical equilibrium
Page 156
Example 6.4: 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
T = 273 + 500 = 773 K
Therefore,
Kp = Kc (RT)n
Kp = Kc (RT)-2
1.5 x 10 -5
Kc = Kp (RT) =
x [0.0821.atm/K.mol x 773K]2
2
atm
2
= (1.5 x I0-5/ atm2) (4.03 x 103 L2. atm2 / mol2)
= 6.04 x 10-2 L2 / mol2
Some facts dealing with expressions for equilibrium
constants may be summarized as follows:
1- The concentration terms for substances that appear
on the right side of the chemical equation are written
in the numerator of the expression for kc. The
concentration terms for substances that appear on the
left side are written in the denominator.
Chemical equilibrium
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2- No concentration terms are included for pure solids or
pure liquids.
3- The value of kc for a given equilibrium is constant at
constant temperature. At a different temperature the
value of kc is different.
Predicting the Direction of a Relation:
For the reaction
PCl2(g)  PCl3(g) + Cl2 at 250oC
Kc =
[PCl3 ][Cl2 ]
 0.0415mol/ l
[PCl5 ]
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?
If the mixture is as equilibrium, the concentrations of the
three substances, when substituted into the expression for
kc, will equal kc. A value obtained by substituting initial
concentration into the expression for the equilibrium
constant is called a reaction quotient (Q). For the
example at hand,
Q=
[PCl 3 ][Cl2 ] 0.05 x 0.03

 0.015 mol/l
[PCl5 ]
0.1
Chemical equilibrium
Page 158
In this case Q (0.015 mol/L) is smaller than kc (0.0415
mol/L). The system is not at equilibrium. The reaction
will proceed from left to right.
In general, the values of a Q and the corresponding kc
may be related in one of three ways:
1- Q < Kc In this instance, the reaction will move from left
to right (the forward direction) to approach equilibrium.
2- Q = Kc The system is in equilibrium.
3- Q > Kc In this instance, the reaction will move from
right to left (the reversible direction) to approach
equilibrium.
Example 6.5: 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 at 827°C, in what direction will the reaction
proceed?
Solution:
[SO3 ]2
[0.125mol/ l] 2

 208 L/moL
Q=
2
2
[SO2 ] [O2 ] [0.05 mol/l] [0.03mol/l ]
Chemical equilibrium
Page 159
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
1- CaCO3(S)  CaO(S) + CO2(g)
Kc = [CO2]
3 Fe(s) + 4 H2O(g)  Fe3O4(s) + 4 H2(g)
2-
[H2 ]4
Kc 
[H2O] 4
Equilibrium Concentrations Using K
Example 6.6: 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:
Chemical equilibrium
Page 160
The [H2] = [I2] at equilibrium since they are produced in
equal amounts by decomposition of HI(g)
[H2] = [I2] = x
[HI] = 0.5 mol/L
[HI] 2
Kc 
 54.5
4
[H2 ] [I2 ]
[0.5 mol/l] 2
 54.5
X2
(0.5)2
 0.00456 mol 2 /L2
x =
54.5
2
 x = 0.068 mol/L
The equilibrium concentration is:
[HI] = 0.5 mol/ L
[H2] = [I2] = 0.068 mol/ L
Example 6.7: 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:
Chemical equilibrium
Page 161
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
0.01 mol/L
----
-----
Change
mol/L
-x
-x
+x
+x
0.01 -x
0.01 -x
X
X
at equilibrium
Kc 
[H2O] [CO]
 0.771
[H2 ] [CO2 ]
X2
Kc 
(0.01 - X)2
If we extract the square root of both sides of this equation:

X
 0.878
0.01 - X
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
Chemical equilibrium
Page 162
[H2O] = [CO] = 0.00468 mol/L
Example 6.8: 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
PCO
Kp 
 167.5 atm
PCO2
2
PCO
 167.5 atm
0.1
PCO2= 16.8
PCO= 4.10 atm
Example 6.9: 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?
Chemical equilibrium
Page 163
Solution: Let x equal the partial pressure of CO2 when
equilibrium is attained
FeO(s) + CO(g)  Fe(s) + CO2(g)
At start
Change
At equilibrium
Kp 
1.0 atm
---
-x
1.0 - x atm
+x
x
PCO2
 0.403
PCO
X atm
 0.403
1.0 atm - X
X = PCO2 = 0.287 atm
1.0 – X = Pco = 0.713
Le Chatelier's Principle:
Le chatelier's principle predicts how a
system in equilibrium will respond to
changes in experimental conditions
(such as temperature or pressure),
which
states
Chemical equilibrium
that
a
system
in
Page 164
equilibrium reacts to a stress in a way that counteracts the
stress and establishes a new equilibrium state:
1- Concentration changes: 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.
H2(g) + I2(9)  2 HI(g)
e.g:
Increase H2 or I2 → shift to to formation of HI
Removal of H2 or I2 ← Reaction shift to decomposition of
HI.
2- Pressure changes: 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. In this way the
system counteracts the change, and vice versa.
Chemical equilibrium
Page 165
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 heterogeneous equilibrium the effect of pressure is
predicated by counting the number of moles of gas
indicated on each side of the equation.
3- Temperature changes: 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.
Chemical equilibrium
Page 166
The numerical value of the equilibrium constant charges
when the temperature is changed.
4- Effect of catalyst: The addition of a catalyst causes a
system to achieve equilibrium faster but does not alter
the position of equilibrium.
5- Addition of an inert gas: 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.
Chemical equilibrium
Page 167