The bond dissociation enthalpies of H2(g),Cl2(g) and HCl(g) are 435, 243 and 431 kJ mol−1 respectively. The enthalpy of formation of HCl(g) will be

A
$247 kJ {mol}^{- 1}$
B
$770 kJ {mol}^{- 1}$
C
$1109 kJ {mol}^{- 1}$
D
$- 92 kJ {mol}^{- 1}$

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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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

Solution:

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