Question

Derive the relation between K_p and K_C.

Answer 1

Consider a general reaction in which all reactants and products are ideal gases.

xA + yB ⇌  lC + mD

The equilibrium constant K_C is

The ideal gas equation is 

PV = nRT or P = \(\frac{n}{V}\)RT

Since, 

Active mass = molar concentration = \(\frac{n}{V}\)

P = Active mass x RT 

Based on the above expression, the partial pressure of the reacants and products can be expressed as

On substitution in equation (2),

By comparing equation (1) and (4), we get

\(K_P=K_C(RT)_{∆n_g}\)

where ∆n_g is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase. 

(i) If ∆n_g = 0, K_b = K_C(RT)⁰

K_p = K_C

Example: H_2(g) + I₂ ⇌ 2HI_(g)

(ii) where,

∆n_g = +Ve 

K_p = K_C (RT)^+ve 

K_P = K_c 

Example: 2NH_3(g) ⇌ N_2(g) + 3H_2(g)

(iii) When,

∆n_g = -Ve 

K_P = K_C (RT)^-ve

K_P < K_C 

Example: 2SO_2(g) +O_2(g) ⇌ 2SO_3(g)

Answer 2

Let the gaseous reaction is a state of equilibrium is-

aA(g)​+bB(g)​⇌cC(g)​+dD(g)​

Let pA​,pB​,pC​ and pD​ be the partial pressure of A,B,C and D repectively.

Therefore,

Kc​=[A]a[B]b[C]c[D]d​.....(1)

Kp​=pA​apB​bpC​cpD​d​.....(2)

For an ideal gas-

PV=nRT

⇒P=Vn​RT=CRT

Whereas C is the concentration.

Therefore,

pA​=[A]RT

pB​=[B]RT

pC​=[C]RT

pD​=[D]RT

Substituting the values in equation (2), we have

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

⇒Kp​=[A]a[B]b[C]c[D]d​(RT)[(c+d)−(a+b)]

⇒Kp​=Kc​(RT)Δng​(From (1)]

Here,

Δng​= Total no. of moles of gaseous product − Total no. of moles of gaseous reactant

Hence the relation between Kp​ and Kc​ is-

Kp​=Kc​(RT)Δng​