For the reaction at $298 \, K$,$2 A + B \rightarrow C$. Given $\Delta H = 400 \, kJ \, mol^{-1}$ and $\Delta S = 0.2 \, kJ \, K^{-1} \, mol^{-1}$. At what temperature will the reaction become spontaneous,considering $\Delta H$ and $\Delta S$ to be constant over the temperature range?

(D) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative.
The relationship is given by $\Delta G = \Delta H - T \Delta S$.
At equilibrium,$\Delta G = 0$,so $T = \frac{\Delta H}{\Delta S}$.
Substituting the given values: $T = \frac{400 \, kJ \, mol^{-1}}{0.2 \, kJ \, K^{-1} \, mol^{-1}} = 2000 \, K$.
Since $\Delta H$ is positive (endothermic) and $\Delta S$ is positive,the reaction becomes spontaneous when $T \Delta S > \Delta H$.
Therefore,the reaction will be spontaneous at $T > 2000 \, K$.