MCQ
The bond dissociation energies of gaseous ${H_2},\,C{l_2}$ and $HCl$ are $104, 58$ and $ 103\, kcal $ respectively. The enthalpy of formation of $HCl$ gas would be.....$kcal$
- A$-44$
- B$44 $
- ✓$-22$
- D$22$
✓
Answer
Correct option: C.
$-22$
(c)Aim: $\frac{1}{2}{H_2} + \frac{1}{2}C{l_2} \to HCl$
$\Delta H = \sum B.E{._{{\rm{(Products)}}}} - \sum B.E{._{{\rm{(Reactants)}}}}$
$ = B.E.(HCl\,) - \left[ {\frac{1}{2}B.E.({H_2}) + \frac{1}{2}B.E.(C{l_2})} \right]$
$ = - 103 - \left[ {\frac{1}{2}(\, - 104\,) + \frac{1}{2}(\, - 58)} \right]$
$ = - 103 - (\, - 52 - 29\,) = - 22\,\,kcal$.
$\Delta H = \sum B.E{._{{\rm{(Products)}}}} - \sum B.E{._{{\rm{(Reactants)}}}}$
$ = B.E.(HCl\,) - \left[ {\frac{1}{2}B.E.({H_2}) + \frac{1}{2}B.E.(C{l_2})} \right]$
$ = - 103 - \left[ {\frac{1}{2}(\, - 104\,) + \frac{1}{2}(\, - 58)} \right]$
$ = - 103 - (\, - 52 - 29\,) = - 22\,\,kcal$.
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