Consider the following reaction The equilibrium constant is

Consider the following reaction.

The equilibrium constant is 0.708 at 910. K and 3.112 at 1250. K. What is the enthalpy of the reaction?

Solution

Given:
T1 = 910 K
T2 = 1250 K
K2/K1 = 3.112/0.708

use:
ln(K2/K1) = (delta H/R)*(1/T1 - 1/T2)
ln(3.112/0.708) = ( delta H/8.314)*(1/910.0 - 1/1250.0)
1.4806 = (delta H/8.314)*(2.989*10^-4)
delta H = 41183 J/mol
delta H = 41.1826 KJ/mol
Answer: 41.2 KJ/mol