Thermodynamics And Thermochemistry Result Question 28
28. The bond dissociation energies of gaseous $\mathrm{H} _2, \mathrm{Cl} _2$ and $\mathrm{HCl}$ are $104, 58$ and $103$ $ \mathrm{kcal} / \mathrm{mol}$ respectively. Calculate the enthalpy of formation of $\mathrm{HCl}$ gas.
$(1985,2 M)$
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Answer:
Correct Answer: 28. $(-22$ $ \mathrm{kcal} / \mathrm{mol})$
Solution:
$ \dfrac{1}{2} \mathrm{H}_2(g)+\dfrac{1}{2} \mathrm{Cl}_2(g) \longrightarrow \mathrm{HCl}(g) ; \quad \Delta H_f^{\circ} $
$\Delta H_f^{\circ} =\Sigma \mathrm{BE} \text { (reactants) }-\Sigma \mathrm{BE} \text { (products) } $
$\quad= \dfrac{1}{2}(104+58)-103=-22 $ $\mathrm{kcal} / \mathrm{mol}$
BETA
