The bond dissociation enthalpies of $H_{2} (g), {Cl}_{2} (g)$ and $HCl (g)$ are $435, 243$ and $431 kJ {mol}^{- 1}$ respectively. The enthalpy of formation of $HCl (g)$ will be

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$770 kJ {mol}^{- 1}$

$1109 kJ {mol}^{- 1}$

$247 kJ {mol}^{- 1}$

$- 92 kJ {mol}^{- 1}$

answer is D.

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Detailed Solution

$\begin{array}{l} H_{2} (g) + {Cl}_{2} (g) \to 2 HCl (g) \\ Δ_{f} H (HCl, g) = \frac{1}{2} [ε (H - H) + ε (Cl - Cl) - 2 ε (H - Cl)] = \frac{1}{2} [(435 + 243 - 2 \times 431) kJ {mol}^{- 1}] \\ = - 92 kJ {mol}^{- 1} \end{array}$

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