A reaction with ΔH° = -50 kJ/mol and ΔS° = -100 J/mol·K is spontaneous at 298 K. What is the temperature range (in K) where the reaction becomes non-spontaneous?
Correct: B
Using ΔG° = ΔH° - TΔS°, for ΔG° ≥ 0 (non-spontaneous), T ≥ ΔH°/ΔS° = (-50 x 10³ J/mole)/(-100 J/mol·K) = 500 K. Thus, reaction is non-spontaneous above 500 K.