At what temperature is the equilibrium constant equal 1.0 x 103?

Question

The reaction of nitrogen with hydrogen to form ammonia is thermodynamically favorable.
N2(g) + 3H2(g) = 2NH3(g) DeltaHº = -92.2 kJ

The equilibrium constant for this reaction is 6.0 x 105 at 298 K. At what temperature is the equilibrium constant equal 1.0 x 103? (R = 8.31 J/mol·K)

Answer 1

Use the van't Hoff equation.

ln (K2/K1) = (delta H/R) (1/T1- 1/T2)
ln (1.67 x 10^-3) = (-92200/8.31) (1/298 - 1/T2)
-6.40 = -11095 (0.00336 - 1/T2)
-0.00057 = 0.00336 - 1/T2
1/T2 = 0.00393
T2 = 254.5K