Question

For the equilibrium, 2NOCl_(g) ⇌ 2NO_(g) + Cl_2(g) the value of the equilibrium constant, K_c is 3.75×10^–6 at 1069 K. Calculate the K_p for the reaction at this temperature?

Answer 1

K_P ​= K_C​.(RT)^Δng

⇒ K_C ​= 3.75 × 10⁻⁶

⇒ T = 1069K

Δ_ng = moles of product - moles of reactant

Δ_ng = 3 − 2 = 1

⇒ K_P ​= 3.75 × 10⁻⁶ × (0.0831 × 1069)

⇒ K_P​ = 0.033

Answer 2

We know that, K_p = K_c (RT)^Δn 

For the above reaction, 

Δn = (2+1) – 2 = 1 

Using, K_p = K_c (RT)^Δn 

K_p = 3.75 ×10^–6 (0.0831 × 1069) 

K_p = 0.033.