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Question:
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Calculate the work done in the following reaction at $300 \mathrm{~K}$ and at constant pressure.
$$
\begin{aligned}
& \left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \\
& 4 \mathrm{HCl}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}
\end{aligned}
$$
Options:
$$
\begin{aligned}
& \left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \\
& 4 \mathrm{HCl}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}
\end{aligned}
$$
Solution:
2886 Upvotes
Verified Answer
The correct answer is:
$2494 \mathrm{~J}$
$\begin{aligned} \mathrm{W} & =-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\ & =-(4-5) \times 8.314 \times 300 \\ & =1 \times 8.314 \times 300 \\ & =2494.2 \mathrm{~J}\end{aligned}$
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