Search any question & find its solution
Question:
Answered & Verified by Expert
Calculate enthalpy change for the change $8 \mathrm{~S}(g) \longrightarrow \mathrm{S}_{8}(g)$, given that
$$
\begin{array}{l}
\mathrm{H}_{2} \mathrm{~S}_{2}(g) \longrightarrow 2 \mathrm{H}(g)+2 \mathrm{~S}(g), \\
\Delta H=239.0 \mathrm{k} \mathrm{cal} \mathrm{mol}^{-1} \\
\mathrm{H}_{2} \mathrm{~S}(g) \longrightarrow 2 \mathrm{H}(g)+\mathrm{S}(g), \\
\qquad \Delta H=175.0 \mathrm{kcal} \mathrm{mol}^{-1}
\end{array}
$$
Options:
$$
\begin{array}{l}
\mathrm{H}_{2} \mathrm{~S}_{2}(g) \longrightarrow 2 \mathrm{H}(g)+2 \mathrm{~S}(g), \\
\Delta H=239.0 \mathrm{k} \mathrm{cal} \mathrm{mol}^{-1} \\
\mathrm{H}_{2} \mathrm{~S}(g) \longrightarrow 2 \mathrm{H}(g)+\mathrm{S}(g), \\
\qquad \Delta H=175.0 \mathrm{kcal} \mathrm{mol}^{-1}
\end{array}
$$
Solution:
2012 Upvotes
Verified Answer
The correct answer is:
$-512.0 \mathrm{kcal}$
$$
\begin{array}{l}
\Delta H_{\mathrm{S}-\mathrm{S}}+2 \Delta H_{\mathrm{H}-\mathrm{S}}=239 \\
2 \Delta H_{\mathrm{H}-\mathrm{S}}=175
\end{array}
$$
Hence,
$\Delta H_{\mathrm{S}-\mathrm{S}}=239-175=64 \mathrm{kcal} \mathrm{mol}^{-1}$
Then, $\Delta H$ for $8 \mathrm{~S}_{(\mathrm{g})} \rightarrow \mathrm{S}_{8(\mathrm{~g})}$ is $8 \times(-64)=-512 \mathrm{kcal}$
\begin{array}{l}
\Delta H_{\mathrm{S}-\mathrm{S}}+2 \Delta H_{\mathrm{H}-\mathrm{S}}=239 \\
2 \Delta H_{\mathrm{H}-\mathrm{S}}=175
\end{array}
$$
Hence,
$\Delta H_{\mathrm{S}-\mathrm{S}}=239-175=64 \mathrm{kcal} \mathrm{mol}^{-1}$
Then, $\Delta H$ for $8 \mathrm{~S}_{(\mathrm{g})} \rightarrow \mathrm{S}_{8(\mathrm{~g})}$ is $8 \times(-64)=-512 \mathrm{kcal}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.