Question

Calculate K_c and K_p for the reaction at 295 K, N₂O₄ ⇌ 2NO_2(g) if the equilibrium

concentrations are [N₂O₄] = 0.75 M and [NO₂] = 0.062 M, R = 0.08206 L atm K^-1 mol^-1

Answer 1

Given: R = 0.08206 L atm K^-1 mol^-1 , T = 295 K

At equilibrium , [N₂O₂] = 0.75 M, [NO₂] = 0.062 M

To find: Equilibrium constants, K_p and K_c

Formulae: i. K_c = \(\frac{[C]^c [D]^d}{[A]^a [B]^b}\)

ii. K_p = K_c (RT)^Δn

Calculation : The equilibrium reaction is given as N₂O_4(g) ⇌ 2NO_2(g)

The expression of K_c is

K is related to K by expression: K_p = K_c (RT)^Δn where, Δn = numbers of moles of gaseous products – number of moles of gaseous reactants

= 2 – 1 = 1

∴ K_p = K_c (RT)¹

∴ K_p = 5.13 × 10^-3 × 0.08206 × 295

∴ K_p = 123.9 × 10^-3 = 0.124

K_c and K_p for the reaction at 295 K are 5.13 × 10^-3  and 0.124 respectively.