Q68 / 200 · NEET 2008 (Screening) · Chemistry
Bond dissociation enthalpy of $\mathrm{H}_2, \mathrm{Cl}_2$ and $\mathrm{HCl}$ Enthalpy of formation of $\mathrm{HCl}$ is
- A$93 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- B$-245 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- Correct answerC$-93 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- D$245 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
Key Idea:
$\Delta H_{\text {reaction }}=\Sigma$ Bond energy of reactant $-\Sigma$ Bond energy of product
$\begin{array}{ll}
\text {Here, } & \Delta \mathrm{H}_{\mathrm{H}-\mathrm{H}}=434 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{Cl}-\mathrm{Cl}}=242 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{H}-\mathrm{Cl}}=431 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
\because & \frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \longrightarrow \mathrm{HCl}
\end{array}$
$\begin{aligned}
\Delta H_{\text {reaction }} & =\frac{1}{2} \Delta H_{\mathrm{H}-\mathrm{H}}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{Cl}-\mathrm{Cl}}-\Delta \mathrm{H}_{\mathrm{H}-\mathrm{Cl}} \\
& =\frac{1}{2} \times 434+\frac{1}{2} \times 242-431 \\
& =217+121-431 \\
& =-93 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$
Hence, the correct answer is option: 'C'.