MHT CET · Chemistry · Thermodynamics (C)
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)
- A 340 kJ
- B 432 kJ
- C 370 kJ
- D 864 kJ
Answer & Solution
Correct Answer
(B) 432 kJ
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\left(\mathrm{B} \cdot \mathrm{E})_{\mathrm{in}}-(\mathrm{B} \cdot \mathrm{E})_{\text {out }}\right. \\ & \Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\mathrm{B} \cdot \mathrm{E}_{\mathrm{H}-\mathrm{H}}+\mathrm{B} \cdot \mathrm{E}_{\mathrm{Cl}-\mathrm{Cl}}-2 \times \mathrm{B} \cdot \mathrm{E}_{\mathrm{H}-\mathrm{Cl}} \\ & -185=435+244-2 x \\ & 2 x=435+244+185 \\ & x=432 \mathrm{KJ}\end{aligned}\)
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