Question

Derive the relation between KP and KC for the equilibrium reaction

Answer 1

Derive the relationship between Kp and Kc

Relationship between Kp and Kc
Consider the following reversible reaction:
aA + bB ⇌ cC + dD
The equilibrium constant for the reaction expressed in terms of the concentration (mol / litre) may be expressed as:
K c = [C] c [D] d / [A] a [B] b

If the equilibrium involves gaseous species, then the concentrations may be expressed in terms of partial pressures of the gaseous substance. The equilibrium constant in terms of partial pressures may be given as:
K p = pcC pdD / paA pbB
Where pA, pB, pC and pD represents the partial pressures of the substance A, B, C and D respectively. If gases are assumed to be ideal, then according to ideal gas equation:
pV = nRT
p = nRT / V
Where p ———-> pressure in Pa
n ——————–> amount of gas in mol
V ——————–> Volume in m3
T ———————> temperature in Kelvin
n/V = concentration, C
or
p = CRT or [gas] RT

If C is in mol dm-3 and p is in bar, then R = 0.0831 bar dm3 mol-1 K-1

Therefore, at constant temperature, pressure of the gas P is proportional to its concentration C, i.e.

Let us suppose a general reaction:
aA + bB↔ cC + dD
The equilibrium constant will be given as:
Kp = (pC) c (pD) d /
(pA) a (pB) b ……. (1)
Now, p = CRT
Hence,
pA = [A] RT
where [A] is the molar concentration of A
Similarly,
pB = [B] RT
pC = [C] RT
pD = [D] RT
where [B], [C] and [D] are the molar concentration of B, C and D respectively
Substituting these values in expression for Kp i.e. in equation (1)

Kp = [([C] RT) c ([D] RT) d]/[([A] RT) a ([B] RT) b]

= [C] c [D] d (RT) c+d/[A] a [B] b (RT) a+b

= [C] c [D] d (RT) c+d – a+b/[A] a [B] b

= Kc (RT) c+d – a+b
= Kc (RT) ∆n
Where ∆n = (c + d) – (a + b) i.e. number of moles of gaseous products – number of moles of gaseous reactants in the balanced chemical reaction.

Hence relation between Kp and Kc is given as:
Kp = Kc (RT) ∆n

Answer 2

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