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The bond dissociation energy of gaseous $H_{2},$ ${Cl}_{2}$ and $HCl$ are 104, 58 and $103 kcal {mol}^{- 1}$ respectively. The enthalpy of formation for HCl gas will be:

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-44.0 kcal

44.0 kcal

22.0 kcal

-22.0 kcal

answer is B.

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Detailed Solution

In the reaction we have,

$\begin{array}{l} \frac{1}{2} H_{2} + \frac{1}{2} {Cl}_{2} ⟶ HCl \\ Δ H_{f} = - [103 - \frac{1}{2} (104 + 58)] = - 22 kcal \end{array}$

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