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Bond dissociation enthalpy of $\mathrm{H}_2, \mathrm{Cl}_2$ and $\mathrm{HCl}$ are 434,242 and $431 \mathrm{kJmol}^{-}$ ${ }^1$ respectively. Enthalpy of formation of $\mathrm{HCl}$ is
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The correct answer is:
$-93 \mathrm{kJmol}^{-1}$
$$
\begin{aligned}
& 2[\mathrm{H}]+2[\mathrm{Cl}] \longrightarrow 2 \mathrm{HCl}, \Delta \mathrm{H}=-862 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{H}_2 \longrightarrow 2[\mathrm{H}], \Delta \mathrm{H}=434 \mathrm{~kJ} \\
& \mathrm{Cl}_2 \longrightarrow 2[\mathrm{Cl}], \Delta \mathrm{H}=242 \mathrm{~kJ} \\
& \mathrm{H}_2+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{HCl}, \Delta \mathrm{H}=-186 \mathrm{~kJ} \\
& \therefore \Delta \mathrm{H}_4 \text { of } \mathrm{HCl}=\frac{-186}{2}=-93 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
\begin{aligned}
& 2[\mathrm{H}]+2[\mathrm{Cl}] \longrightarrow 2 \mathrm{HCl}, \Delta \mathrm{H}=-862 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{H}_2 \longrightarrow 2[\mathrm{H}], \Delta \mathrm{H}=434 \mathrm{~kJ} \\
& \mathrm{Cl}_2 \longrightarrow 2[\mathrm{Cl}], \Delta \mathrm{H}=242 \mathrm{~kJ} \\
& \mathrm{H}_2+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{HCl}, \Delta \mathrm{H}=-186 \mathrm{~kJ} \\
& \therefore \Delta \mathrm{H}_4 \text { of } \mathrm{HCl}=\frac{-186}{2}=-93 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
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