Question

For the reaction at 298 K: \[\ce{2A + B -> C}\]

ΔH = 400 KJ mol⁻¹; ΔS = 0.2 KJ K⁻¹ mol⁻¹ Determine the temperature at which the reaction would be spontaneous.

Answer 1

Given, T = 298 K

ΔH = 400 KJ mol⁻¹

ΔS = 0.2 KJ K⁻¹ mol⁻¹

ΔG = ΔH − TΔS

if T = 2000 K

ΔG = 400 − (0.2 × 2000) = 0

ΔH = 400 KJ mol⁻¹ if T > 2000 K

ΔG will be negative.

The reaction would be spontaneous only beyond 2000 K