For the reaction at 298 K, 2A + B → C. ΔH = 400 kJ mol^-1 and ΔS = 0.2 kJ mol^-1 K^-1.

Question

For the reaction at \(298 \,K, 2A+B \rightarrow C.\) \(\Delta H = 400\, kJ\,mol ^{-1}\) and \(\Delta S = 0.2 \) kJ \(mol ^{-1} \, K ^{-1}.\) the reaction will become spontaneous above _____ K.

Answer 1

Correct answer is : 2000

\(\Delta G = \Delta H - T \Delta S\)

\(\Delta G = 0 \rightarrow \Delta H = T \Delta S\)

\(T = \frac {\Delta H}{\Delta S} = \frac {400}{0.2} = 2000K\)