MCQ
The bond enthalpies of $H_2, X_2$ and $HX$ are in the ratio of $2 : 1 : 2.$ If the standard enthalpy for formation of $HX$ is $-50\, kJ \,mol^{-1} ,$ the bond enthalpy of $H_2$ is....$kJ\, mol^{-1}$
- ✓$200$
- B$400$
- C$100$
- D$300$
✓
Answer
Correct option: A.
$200$
a
$\mathrm{B.E}$ of $\mathrm{H}_{2}, \mathrm{X}_{2}\, \& \,\mathrm{Hx}$ is $2 \mathrm{x}, \mathrm{x}, 2 \mathrm{x}$
$\mathrm{B.E}$ of $\mathrm{H}_{2}, \mathrm{X}_{2}\, \& \,\mathrm{Hx}$ is $2 \mathrm{x}, \mathrm{x}, 2 \mathrm{x}$
$\frac{1}{2} \mathrm{H}_{2}+\frac{1}{2} \mathrm{X}_{2} \rightarrow \mathrm{HX}$
$\Delta \mathrm{H}_{\mathrm{r}}=-50=\Sigma \mathrm{B} \cdot \mathrm{E}(\mathrm{R})-\Sigma \mathrm{B} \cdot \mathrm{E}(\mathrm{P})$
$=\left[\frac{1}{2} \times \mathrm{BE}\left(\mathrm{H}_{2}\right)+\frac{1}{2} \times \mathrm{B} \mathrm{E}\left(\mathrm{X}_{2}\right)\right]-[\mathrm{B} \cdot \mathrm{E}(\mathrm{HX})]$
$-50=\frac{1}{2} \times 2 \mathrm{x}+\frac{\mathrm{x}}{2}-2 \mathrm{x}=\frac{-\mathrm{x}}{2}$
$\mathrm{x}=100\, \frac{\mathrm{kJ}}{\mathrm{mol}}$
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