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Heat of Formation

The bond diss...

Question

The bond dissociation energies of gaseous $H_{2}, {C l}_{2}$ and $H C l$ are $104, 58$ and $103 k c a l$ respectively. The enthalpy of formation of $H C l$ gas would be

- 44 k c a l

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44 k c a l

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- 22 k c a l

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22 k c a l

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Solution

The correct option is C $- 22 k c a l$
$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} ⟶ H C l$

$Δ H = \sum {B E}_{r e a c t a n t s} - \sum {B E}_{p r o d u c t s}$

$= [\frac{1}{2} B E (H_{2}) + \frac{1}{2} B E (C l_{2})] - B E (H C l) = [(\frac{1}{2} \times 104) + (\frac{1}{2} \times 58)] - 103$

$= (52 + 29) - 103 = - 22 k c a l$

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