7.5 HETEROGENEOUS EQUILIBRIA
Equilibrium in a system having more than one phase is called heterogeneous equilibrium. The equilibrium between water vapour and liquid water in a closed container is an example of heterogeneous equilibrium.
H2O(l) 
H2O(g)
In this example, there is a gas phase and a liquid phase. In the same way, equilibrium between a solid and its saturated solution,
Ca(OH)2 (s) + (aq) Ca2+ (aq) + 2OH–(aq)
is a heterogeneous equilibrium.
NEETprep Audio Note:
CaCO3 (s) 
CaO (s) + CO2 (g) (7.16)
On the basis of the stoichiometric equation, we can write,
Since [CaCO3(s)] and [CaO(s)] are both constant, therefore modified equilibrium constant for the thermal decomposition of calcium carbonate will be
This shows that at a particular temperature, there is a constant concentration or pressure of CO2 in equilibrium with CaO(s) and CaCO3(s). Experimentally it has been found that at 1100 K, the pressure of CO2 in equilibrium with CaCO3(s) and CaO(s), is
2.0 ×105 Pa. Therefore, equilibrium constant at 1100K for the above reaction is:
NEETprep Audio Note:
The value of equilibrium constant Kc can be calculated by substituting the concentration terms in mol/L and for Kp partial pressure is substituted in Pa, kPa, bar or atm. This results in units of equilibrium constant based on molarity or pressure, unless the exponents of both the numerator and denominator are same.
For the reactions,
H2(g) + I2(g) 2HI, Kc and Kp have no unit.
N2O4(g) 2NO2 (g), Kc has unit mol/L and Kp has unit bar
Equilibrium constants can also be expressed as dimensionless quantities if the standard state of reactants and products are specified. For a pure gas, the standard state is 1bar. Therefore a pressure of 4 bar in standard state can be expressed as 4 bar/1 bar = 4, which is a dimensionless number. Standard state (c0) for a solute is 1 molar solution and all concentrations can be measured with respect to it. The numerical value of equilibrium constant depends on the standard state chosen. Thus, in this system both Kp and Kc are dimensionless quantities but have different numerical values due to different standard states.
Kp = PCO2 = 2 × 105 Pa/105 Pa = 2.00
Similarly, in the equilibrium between nickel, carbon monoxide and nickel carbonyl (used in the purification of nickel),
Ni (s) + 4 CO (g) Ni(CO)4 (g),
the equilibrium constant is written as
It must be remembered that for the existence of heterogeneous equilibrium pure solids or liquids must also be present (however small the amount may be) at equilibrium, but their concentrations or partial pressures do not appear in the expression of the equilibrium constant. In the reaction,
Ag2O(s) + 2HNO3(aq) 2AgNO3(aq) +H2O(l)
Problem 7.6
The value of Kp for the reaction,
CO2 (g) + C (s) 2CO (g)
is 3.0 at 1000 K. If initially PCO2 = 0.48 bar and PCO = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO2. NEETprep Audio Note:
Solution
For the reaction,
let ‘x’ be the decrease in pressure of CO2, then
CO2(g) + C(s) 2CO(g)
Initial
pressure: 0.48 bar 0
At equilibrium:
(0.48 – x)bar 2x bar
Kp = (2x)2/(0.48 – x) = 3
4x2 = 3(0.48 – x)
4x2 = 1.44 – x
4x2 + 3x – 1.44 = 0
a = 4, b = 3, c = –1.44
= [–3 ± √(3)2– 4(4)(–1.44)]/2 × 4
= (–3 ± 5.66)/8
= (–3 + 5.66)/ 8 (as value of x cannot be negative hence we neglect that value)
x = 2.66/8 = 0.33
The equilibrium partial pressures are,
pCO2= 2x = 2 × 0.33 = 0.66 bar
pCO2= 0.48 – x = 0.48 – 0.33 = 0.15 bar