Consider the following reactions: (i) $\mathrm{H}_{(a q)}^{+}+\mathrm{OH}_{(a q)}^{-} \rightarrow…

Consider the following reactions: (i) $\mathrm{H}_{(a q)}^{+}+\mathrm{OH}_{(a q)}^{-} \rightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{i})}, \Delta \mathrm{H}$ $=-\mathrm{X}_1 \mathrm{~kJ} \mathrm{~mol}^{-1}$ (ii) $\mathrm{H}_{2(a q)}+\frac{1}{2} \mathrm{O}_{2(a q)} \rightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{I})} \Delta \mathrm{H}$ $=-\mathrm{X}_2 \mathrm{~kJ} \mathrm{~mol}^{-1}$ (iii) $\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(g)} \rightarrow \mathrm{CO}_{(g)}+\mathrm{H}_2 \mathrm{O}, \Delta \mathrm{H}$ $=-\mathrm{X}_3 \mathrm{~kJ} \mathrm{~mol}^{-1}$ (iv) $\mathrm{C}_2 \mathrm{H}_{2(g)}+\frac{5}{2} \mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_{2(g)}+\mathrm{H}_2 \mathrm{O}_{(i) \text {, }}$ $\Delta \mathrm{H}=+\mathrm{X}_4 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Enthalpy of formation of $\mathrm{H}_2 \mathrm{O}_{(l)}$ is:
  1. $+\mathrm{X}_3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  2. $-\mathrm{X}_4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  3. $+\mathrm{X}_1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  4. $-\mathrm{X}_2 \mathrm{~kJ}^{\mathrm{mol}}{ }^{-1}$

Solution

The given reaction represents the formation of $\mathrm{H}_2 \mathrm{O}$, and the $\mathrm{X}_2$ represents the enthalpy of formation of $\mathrm{H}_2 \mathrm{O}$. The enthalpy of formation is the heat evolved or absorbed when one mole of substance is formed from its constituent atoms. Related Theory A negative enthalpy change represents an exothermic change where energy is released from the reaction, a positive enthalpy change represents an endothermic reaction where energy is taken in from the surroundings.

Asked in: NEET 2007

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