For a reaction $\mathrm{A}+\mathrm{B} \rightarrow$ products $\Delta \mathrm{H}$ is $-84.2 \mathrm{~kJ}$ and…
For a reaction $\mathrm{A}+\mathrm{B} \rightarrow$ products $\Delta \mathrm{H}$ is $-84.2 \mathrm{~kJ}$ and $\Delta \mathrm{S}$ is $-200 \mathrm{~J} \mathrm{~K}^{-1}$. Calculate the highest value of temperature so that the reaction will proceed in forward direction.
$421 \mathrm{~K}$
$237 \mathrm{~K}$
$168 \mathrm{~K}$
$273 \mathrm{~K}$
Solution
$\begin{aligned}
\mathrm{T} & =\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}} \\
\therefore \quad \mathrm{T} & =\frac{-84200 \mathrm{~J}}{-200 \mathrm{~J} \mathrm{~K}^{-1}}=421 \mathrm{~K}
\end{aligned}$
Since $\Delta H$ and $\Delta S$ are negative, the reaction is spontaneous at low temperatures. Therefore, the highest temperature is $421 \mathrm{~K}$ and the reaction will proceed in forward direction spontaneously below $421 \mathrm{~K}$.