Question

What is the value of equilibrium for the following reaction at 400 K?

2NOCl (g) ----> 2NO (g) + $Cl_{2}$ (g)
DH° = 77.5 KJ$mol^{-1}$
R= 8.314J$K^{-1}$ $mol^{-1}$
DS = 135 J$K^{-1}$ $mol^{-1}$

Answer 1

Solution :

On putting given values in,

ΔG° = ΔH° - TΔS°

⇒ 77500 - 400 × 135

⇒ 23500 J

We know that, ΔG° = -2.303 RT log K

⇒ 23500 = -2.303 × 8.314 × 400 log K

⇒ log K = $-\frac{23500}{2.303 \times 8.314 \times 400}$

⇒ -3.0682 × 4.9314

∴ K = 8.544 × $10^{-4}$