Question

Calculate the equilibrium constant Keq for the following reaction at 25°C. (S° values in J/mol•K: N2(g), 191.5 ; O2(g), 205.0 ; Cl2(g), 223.0 ; NOCl(g), 261.6) N2(g) + O2(g) + Cl2(g) ↔ 2NOCl(g) ΔH° = 103.4 kJ

a. 2.43 x 10−20

b. 8.82 x 10−29

c. 5.91 x 10−17

d. 2.01 x 10−10

e. 6.99 x 10−24

Answer 1

The reaction is

N2 + O2 + Cl2 ---------> 2 NOCl : delta H = 103.4 kJ

delta S of reaction = sum of entropies of products - sum of entropie s of reactants

= (2x261.6) - [191.5 + 205.0+223.0]

= -96.3 J /K

We know Delta G⁰ = delta H⁰ - T dleta S^o  

= 103.4x1000J - 298 K x (-96.3J/K)

= 132097.4 J

The relation between delta G0 and Keq is

Delta G⁰ = -2.303 RT log Keq

132097.4 J = -2.303 x 8.314J/mol.K x 298 K log K

Thus K = 7.04 x10^-24

Thus the correct option is e