Question

The equilibrium constant Kp for the reaction,

H_2(g) + I_2(g) → 2HI_(g) is 130 at 510 K. Calculate ΔG^o for the following reaction at the same temperature: 2HI_(g) → H_2(g) + I_2(g) [Given: R = 8.314 J K^-1 mol^-1 ]

Answer 1

Given: H_2(g) + I_2(g) ⇌ 2HI_(g); Kp = 130, T = 510 K, R = 8.314 J K^-1 mol^-1

To find: ΔGº for reaction, 2HI_(g) ⇌ H_2(g) + I_2(g)

Formula: ΔGº = -2.303 RT log₁₀K_p

Calculation:

H_2(g) + I_2(g) ⇌ 2HI_(g); Kp = 130,

2HI_(g)⇌H_2(g) + I_2(g) ; Kp = 1/130

ΔGº= - 2.303 RT log₁₀K_p

₌-2.303 x 8.314 x 510 x log₁₀ (1/130)

= 20642.7 J mol^-1
=20.64 kJ mol^-1