The enthalpy of reaction for the reaction : 2H_2(g)+O_2(g) 2H_2O(… - Vidyadip

Question

The enthalpy of reaction for the reaction : $2\text{H}_2(\text{g})+\text{O}_2(\text{g})\rightarrow2\text{H}_2\text{O}(\text{l})\ \text{is}\ \Delta_\text{r}\text{H}^\ominus=-572\text{kJ}\ \text{mol}^{-1}$
What will be standard enthalpy of formation of $H_2O(l)$?

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Answer

According to the definition of standard enthalpy of formation, the enthalpy change for the following reaction will be standard enthalpy of formation of $H_2O (l)\text{H}_2(\text{g})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{H}_2\text{O}(\text{l}).$
or the standard enthalpy of formation of $H_2O(l)$ will be half of the enthalpy of the given equation i.e., $\Delta_\text{r}\text{H}^\ominus$ is also halved.$\Delta_\text{f}\text{H}^\ominus_{\text{H}_2\text{O}}(\text{l})=\frac{1}{2}\times\Delta_\text{r}\text{H}^\ominus=\frac{-572\text{kJ}\ \text{mol}^{-1}}{2}=-286\text{kJ}\ \text{mol}^{-1}$

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