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Assume each reaction is carried out in an open container. For which reaction will $\Delta H=\Delta E ?$
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Verified Answer
The correct answer is:
$\mathrm{H}_{2(g)}+\mathrm{Br}_{2(g)} \rightarrow 2 \mathrm{HBr}_{(\mathrm{g})}$
$\mathrm{For}_2(g)+\mathrm{Br}_2(g) \rightarrow 2 \mathrm{HBr}(g)$
$$
\begin{array}{ll}
\because & \Delta n_g=0 \\
\therefore & \Delta \mathrm{H}=\Delta \mathrm{E}+\Delta n_g \mathrm{RT} \\
\Rightarrow & \Delta \mathrm{H}=\Delta \mathrm{E}
\end{array}
$$
$$
\begin{array}{ll}
\because & \Delta n_g=0 \\
\therefore & \Delta \mathrm{H}=\Delta \mathrm{E}+\Delta n_g \mathrm{RT} \\
\Rightarrow & \Delta \mathrm{H}=\Delta \mathrm{E}
\end{array}
$$
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