Thermodynamics And Thermochemistry Result 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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Answer:
Correct Answer: 29. $(-41.7$ $ \mathrm{kcal})$
Solution:
$\mathrm{C}_2 \mathrm{H}_2+ \mathrm{H}_2 \longrightarrow \mathrm{C}_2 \mathrm{H}_4 $
$\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}$
BETA
