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: $\Delta_{\mathrm{r}} \mathrm{H}^0\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ $$ \begin{array}{ll} \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{OH}(\mathrm{g}) & 42.09 \\ \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) & -242 \\ \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) & 436 \\ \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{O}(\mathrm{g}) & 496 \end{array} $$
  1. $1174 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  2. $742 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  3. $926 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  4. $690 \mathrm{~kJ} \mathrm{~mol}^{-1}$

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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