An endothermic reaction has $\mathrm{D}_{\mathrm{r}} \mathrm{H}_{298}^{\circ}=+389.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ , $D_{1} S_{298}^{\circ}=+497.2 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ and $D_{1} G_{298}^{\circ}=+241.3 \mathrm{~kJ} m ol^{-1}$ . Find the temperature, T(K), at which the reaction becomes spontaneous.

Question

Quizplus · Accepted Answer

For an endothermic reaction, $\mathrm{D}_{\mathrm{r}} \mathrm{H}$ is positive.

At equilibrium, $\mathrm{D}_{\mathrm{r}} G=0$.

Using the equation 
$$\mathrm{D}_{\mathrm{r}} G=\mathrm{D}_{\mathrm{r}} H-T \mathrm{D}_{\mathrm{r}} S$$

we can find the temperature at which the reaction becomes spontaneous by setting $\mathrm{D}_{\mathrm{r}} G$ to zero and solving for T:

\begin{aligned} \mathrm{D}_{\mathrm{r}} G &=\mathrm{D}_{\mathrm{r}} H-T \mathrm{D}_{\mathrm{r}} S \ 0 &=\mathrm{D}_{\mathrm{r}} H-T \mathrm{D}_{\mathrm{r}} S \ T &=\frac{\mathrm{D}_{\mathrm{r}} H}{\mathrm{D}_{\mathrm{r}} S} = \frac{+389.5 \mathrm{~kJ} \mathrm{~mol}^{-1}}{+497.2 ~\mathrm{JK}^{-1} \mathrm{mol}^{-1}} \ &= 783~\mathrm{K} \end{aligned}

Therefore, the answer is (B) 783.