Question

What is the equilibrium constant K_eq for the following reaction at 400 K?

\[\ce{2NOCl_{(g)} ⇌ 2NO_{(g)} + Cl2_{(g)}}\], given that H⁰ = 77.2 kJ mol⁻¹ and ∆S⁰ = 122 JK⁻¹ mol⁻¹

Answer 1

T = 400 K; ∆H⁰ = 77.2 kJ mol⁻¹ = 77200 J mol⁻¹;

∆S⁰ = 122 JK⁻¹ mol⁻¹
∆G⁰ = −2.303 RT log K_eq

log K_eq = `(-∆"G"^0)/(2.303  "RT")`

log K_eq = `((-∆"H"^0 - "T"∆"S"^0))/(2.303  "RT")`

log K_eq = `-((77200 - 400  xx 122)/(2.303 xx 8.314 xx 400))`

log K_eq = `-((28400)/7659)`

log K_eq = − 3.7080

K_eq = antilog (−3.7080)

K_eq = 1.95 × 10⁻⁴