Thermodynamics And Thermochemistry Result Question 30
30. The molar heats of combustion of $\mathrm{C} _2 \mathrm{H} _2(g), \mathrm{C}$ (graphite) and $\mathrm{H} _2(\mathrm{~g})$ are $310.62$ $ \mathrm{kcal}, 94.05 $ $\mathrm{kcal}$ and $68.32 $ $\mathrm{kcal}$ respectively. Calculate the standard heat of formation of $\mathrm{C} _2 \mathrm{H} _2(g)$.
(1983, 2M)
Show Answer
Answer:
Correct Answer: 30. $(54.2 $ $\mathrm{kcal})$
Solution:
- The standard state formation reaction of $\mathrm{C}_2 \mathrm{H}_2(g)$ is :
$ 2 \mathrm{C}(\mathrm{g})+\mathrm{H}_2(g) \longrightarrow \mathrm{C}_2 \mathrm{H}_2(g) ; \quad \Delta H_f^{\circ} $
$\Delta H_r^{\circ}= \Sigma \Delta H_{\text {comb }}^{\circ} \text { (reactants) }-\Sigma \Delta H_{\text {comb }}^{\circ} \text { (products) } $
$= -2 \times 94.05-68.32-(-310.62) $
$= 54.2 \mathrm{kcal}=\Delta H_f^{\circ}\left(\mathrm{C}_2 \mathrm{H}_2\right)$
BETA
