Question

The reaction: 2 Cl2(g) + 2 H2O(g) ➝ O2(g) + 4 HCl(g)

has ΔG° = +11.2 kJ/mol and ΔH° = +114.4 kJ/mol at 800 K.

a) What is the numerical value of the equilibrium constant for this reaction at 800 K?

b) What is the numerical value of ΔG for the reaction in a vessel containing 3.6 bar Cl2, 1.4 bar H2O, 0.20 bar O2, and 0.50 bar HCl at 800 K?

Answer 1

2 Cl2 (g) + 2 H2O (g) $\rightarrow$ O2 (g) + 4 HCl (g)

Given: ∆G° = 11.2 KJ/mol

∆H° = 114.4 KJ/mol

T = 800 K

(a) To calculate equilibrium constant

∆G° = -R T ln K

11200 J/mol = -( 8.314 J k-1 mol-1) x 800 x 2.303 log K

log K = - 11200/(8.314 x 800 x 2.303)

= - 0.7311

K = 10-0.7311

K = 0.1857

b) Given: Cl2 = 3.6 bar, H2O = 1.4bar, O2 = 0.20 bar, HCl= 0.50 bar

∆G = ∆G° + RT lnQ ------(1)

Reaction quotient (Q) = product/ reactant

According to above balanced reaction

Q = (0.20) x (0.50)4 /(3.6)2 x (1.4)2

Q = 4.9 x 10-4 ​​​​​

ln Q = ln 4.9 x10-4

= -7.6211

Putting values in (1)

∆G = (11200 J/mol) + (8.314 x 800 x -7.6211)

= - 39.48 KJ/mol