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In the reaction $\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{s})$ at $0^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$, the internal energy change is $-41 \mathrm{~kJ} / \mathrm{mol}$. What will be the value of molar enthalpy change?
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The correct answer is:
$-41 \mathrm{~kJ} / \mathrm{mol}$
$\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{s})$ at $0^{\circ} \mathrm{C}, 1 \mathrm{~atm}$
This process of phase change from liquid to solid happens at constant temperature and pressure. So the change in internal energy will be the molar enthalpy change.
$$
\Delta \mathrm{H}=\Delta \mathrm{U}=-41 \mathrm{~kJ} / \mathrm{mol}
$$
This process of phase change from liquid to solid happens at constant temperature and pressure. So the change in internal energy will be the molar enthalpy change.
$$
\Delta \mathrm{H}=\Delta \mathrm{U}=-41 \mathrm{~kJ} / \mathrm{mol}
$$
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