Q11 / 100 · BITSAT 2023 (Memory Based Paper 2) · Chemistry
Given that bond energies of $\mathrm{H}-\mathrm{H}$ and $\mathrm{Cl}-\mathrm{Cl}$ are $430 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $240 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively and $\Delta \mathrm{H}_{\mathrm{f}}$ for $\mathrm{HCl}$ is $-90 \mathrm{~kJ} \mathrm{~mol}^{-1}$, bond enthalpy of $\mathrm{HCl}$ is
- A$380 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- Correct answerB$425 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- C$245 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- D$290 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
$\frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \longrightarrow \mathrm{HCl}$
$\begin{aligned}<br />& \Delta H_{\mathrm{HCl}}=\sum \text { B.E. of reactant }-\sum \text { B.E. of products } \\<br />&-90=\frac{1}{2} \times 430+\frac{1}{2} \times 240-\text { B.E. of } \mathrm{HCl} \\<br />& \therefore \quad \text { B.E. of } \mathrm{HCl}=215+120+90=425 \mathrm{~kJ} \mathrm{~mol}^{-1}<br />\end{aligned}$
$\begin{aligned}<br />& \Delta H_{\mathrm{HCl}}=\sum \text { B.E. of reactant }-\sum \text { B.E. of products } \\<br />&-90=\frac{1}{2} \times 430+\frac{1}{2} \times 240-\text { B.E. of } \mathrm{HCl} \\<br />& \therefore \quad \text { B.E. of } \mathrm{HCl}=215+120+90=425 \mathrm{~kJ} \mathrm{~mol}^{-1}<br />\end{aligned}$
Hence, the correct answer is option: 'B'.