From the following data at $25^{\circ} \mathrm{C}$, calculate the $\Delta_{\mathrm{r}} \mathrm{H}^0$ for…
From the following data at $25^{\circ} \mathrm{C}$, calculate the $\Delta_{\mathrm{r}} \mathrm{H}^0$ for $\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{H}(\mathrm{g})+\mathrm{O}(\mathrm{g})$
reaction: $\Delta_{\mathrm{r}} \mathrm{H}^0\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)$
$$
\begin{array}{ll}
\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{OH}(\mathrm{g}) & 42.09 \\
\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) & -242 \\
\mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) & 436 \\
\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{O}(\mathrm{g}) & 496
\end{array}
$$
- $1174 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- $742 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- $926 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- $690 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
$\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) ; \Delta \mathrm{H}=+242 \mathrm{~kJ} / \mathrm{mol}$
$
\begin{aligned}
& \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{H} ; \Delta \mathrm{H}=+436 \mathrm{~kJ} / \mathrm{mol} \\
& \frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{O} ; \Delta \mathrm{H}=+\frac{496}{2} \mathrm{~kJ} / \mathrm{mol}=248 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$
By adding up above three equations:
$
\Delta_{\mathrm{r}} \mathrm{H}=242+436+248=926 \mathrm{~kJ} / \mathrm{mol}
$
Asked in: AP EAMCET 2022 (06 Jul Shift 1)
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