Relation between Kc and Kp.

Question

Explain :

Relation between K_c and K_p.

Answer 1

Consider a general reversible reaction :

aA_(g)+ bB_(g) ⇌ cC_(g) + dD_(g)

The equilibrium constant (K_p) in terms of partial pressure is given by equation:

K_p = \(\frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}\) …………(1)

For a mixture of ideal gases, 

The partial pressure of each component is directly proportional to its concentration at constant temperature.

For component A,

P_AV = n_ART

P_A = \(\frac{n_A}{V}\) × RT

\(\frac{n_A}{V}\) is molar concentration of A in mol dm^-3 V

∴ P_A = [A]RT where, [A] = \(\frac{n_A}{V}\)

Similarly, for other components, 

P_B = [B]RT, 

P_C = [C]RT, 

P_D = [D]RT

Now substituting equations for P_A, P_B, P_C, P_D in equation (1), we get

Where Δn = (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced chemical equation.

R = 0.08206 L atm K^-1 mol^-1

[Note : While calculating the value of K_p, pressure should be expressed in bar, because standard state of pressure is 1 bar. 1 pascal (Pa) = 1 N m^-2 and 1 bar = 10⁵Pa]