Chemistry › Thermodynamics (C) › Laws of Thermochemistry and Enthalpy Change
Heat of combustion $\Delta \mathrm{H}^{\circ}$ for $\mathrm{C}(s), \mathrm{H}_2(g)$ and $\mathrm{CH}_4(g)$…
Heat of combustion $\Delta \mathrm{H}^{\circ}$ for $\mathrm{C}(s), \mathrm{H}_2(g)$ and $\mathrm{CH}_4(g)$ are $-94,-68$ and $-213 \mathrm{kcal}$ $\mathrm{mol}^{-1}$, then, $\Delta \mathrm{H}^{\circ}$ for $\mathrm{C}(s)+2 \mathrm{H}_2(g) \rightarrow$ $\mathrm{CH}_4(g)$ is:
$-17 \mathrm{kcal}$ $-111 \mathrm{kcal}$ $-170 \mathrm{kcal}$ $-85 \mathrm{kcal}$
Solution
(i) $\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(g)}$;
$\Delta \mathrm{H}_i=-94 \mathrm{kcal} / \mathrm{mole}$
(ii) $2 \mathrm{H}_{2(g)}+\mathrm{O}_{2(g)} \rightarrow 2 \mathrm{H}_2 \mathrm{O}_{(l)}$;
$\Delta \mathrm{H}_{\mathrm{ii}}=-68 \times 2 \mathrm{kcal} / \mathrm{mole}$
(iii) $\mathrm{CH}_{4(g)}+2 \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}+$
$2 \mathrm{H}_2 \mathrm{O}_{(l)} ; \Delta \mathrm{H}_{\mathrm{iii}}=-213 \mathrm{kcal} / \mathrm{mol}$
(iv) $\mathrm{C}_{(s)}+2 \mathrm{H}_{2(g)} \rightarrow \mathrm{CH}_{4(\mathrm{~g})}$;
$\Delta \mathrm{H}_{\mathrm{IV}}=$ ?
By applying Hess's law we can compute $\Delta \mathrm{H}_{\mathrm{iv}}$.
$\begin{aligned}
& \therefore \Delta \mathrm{H}_{\mathrm{iv}}=\Delta \mathrm{H}_{\mathrm{i}}+\Delta \mathrm{H}_{\mathrm{ii}}-\Delta \mathrm{H}_{\mathrm{iii}} \\
& =(-94-68 \times 2+213) \mathrm{kcal} \\
& =-17 \mathrm{kcal}
\end{aligned}$
Asked in: NEET 2002
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