Thermodynamics and Thermochemistry 2 Question 29
29. Given the following standard heats of reactions
(i) heat of formation of water $=-68.3 \mathrm{kcal}$
(ii) heat of combustion of acetylene $=-310.6 \mathrm{kcal}$
(iii) heat of combustion of ethylene $=-337.2 \mathrm{kcal}$
Calculate the heat of reaction for the hydrogenation of acetylene at constant volume $\left(25^{\circ} \mathrm{C}\right)$.
(1984, 4M)
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Solution:
- $\mathrm{C}{2} \mathrm{H}{2}+\mathrm{H}{2} \longrightarrow \mathrm{C}{2} \mathrm{H}_{4}$
$$ \begin{aligned} \Delta H^{\circ} & =\Sigma \Delta H_{\text {comb }}^{\circ}(\text { reactants })-\Sigma \Delta H_{\text {comb }}^{\circ} \text { (products) } \ & =-310.6-68.3-(-337.2) \ & =-41.7 \mathrm{kcal} \end{aligned} $$
