Question

For a particular reaction, Δ H ∘ = 20.1 kJ/mol and Δ S ∘ = 45.9 J / (mol ⋅ K). Assuming these values change very little with temperature,
at what temperature does the reaction change from nonspontaneous to spontaneous in the forward direction? T = K

Answer 1

Answer: 438 K

Explanation:

According to Gibbs equation:

$\Delta G=\Delta H-T\Delta S$

$\Delta G$ = Gibb's free energy change

$\Delta H$ = enthalpy change = 20.1 kJ/mol = 20100 J/mol

T = temperature

$\Delta S$ = entropy change = 45.9 J/Kmol

A reaction is at equilibrium when $\Delta G$ = Gibb's free energy change is zero and becomes spontaneous when $\Delta G$ = Gibb's free energy change is negative.

$\Delta H=T\Delta S$

$20100=T\times 45.9J/Kmol$

T=437.9K

Thus the temperature at which the reaction change from nonspontaneous to spontaneous in the forward direction is 438 K