In the reaction H_2O_{(l)} rightarrow H_2O_{( | vedclass

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(A) The reaction is $H_2O_{(l)} \rightarrow H_2O_{(s)}$ at $0^{\circ}C$ and $1 \ atm$. Since the change in volume $(\Delta V)$ for the phase transitio

Vedclass · Accepted Answer

$-41 \ kJ / mol$

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(A) The reaction is $H_2O_{(l)} \rightarrow H_2O_{(s)}$ at $0^{\circ}C$ and $1 \ atm$. Since the change in volume $(\Delta V)$ for the phase transition of water at $0^{\circ}C$ is negligible, the work done $(P\Delta V)$ is approximately zero. According to the first law of thermodynamics, $\Delta H = \Delta U   P\Delta V$. Since $P\Delta V \approx 0$, we have $\Delta H \approx \Delta U$. Given $\Delta U = -41 \ kJ / mol$, therefore $\Delta H = -41 \ kJ / mol$.

In the reaction $H_2O_{(l)} \rightarrow H_2O_{(s)}$ at $0^{\circ}C$ and $1 \ atm$, the internal energy change is $-41 \ kJ / mol$. What will be the value of molar enthalpy change?

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