Calculate bond enthalpy of $\mathrm{H}-\mathrm{Cl}$ bond from following reaction.…

Calculate bond enthalpy of $\mathrm{H}-\mathrm{Cl}$ bond from following reaction. $\mathrm{H}_2(g)+\mathrm{Cl}_2(g) \rightarrow 2 \mathrm{HCl}(g), \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-185 \mathrm{~kJ}$ (Given bond enthalpies of $\mathrm{H}-\mathrm{H}, \mathrm{Cl}-\mathrm{Cl}$ bonds are $435.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $244 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively)
  1. 340 kJ
  2. 432 kJ
  3. 370 kJ
  4. 864 kJ

Solution

$\begin{aligned} & \Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\left(\mathrm{B} \cdot \mathrm{E})_{\mathrm{in}}-(\mathrm{B} \cdot \mathrm{E})_{\text {out }}\right. \\ & \Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\mathrm{B} \cdot \mathrm{E}_{\mathrm{H}-\mathrm{H}}+\mathrm{B} \cdot \mathrm{E}_{\mathrm{Cl}-\mathrm{Cl}}-2 \times \mathrm{B} \cdot \mathrm{E}_{\mathrm{H}-\mathrm{Cl}} \\ & -185=435+244-2 x \\ & 2 x=435+244+185 \\ & x=432 \mathrm{KJ}\end{aligned}$

Asked in: MHT CET 2022 (08 Aug Shift 2)

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