One mole of ice is converted into water at 27 | vedclass

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(B) At equilibrium,the process of melting ice to water occurs at $273 \, K$ where $\Delta G = 0$. Using the Gibbs-Helmholtz equation: $\Delta G = \Del

Vedclass · Accepted Answer

$5954.13$

Vedclass · Answer

(B) At equilibrium,the process of melting ice to water occurs at $273 \, K$ where $\Delta G = 0$. Using the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T \Delta S$. Since $\Delta G = 0$,we have $\Delta H = T \Delta S$. Here,$\Delta S = S_{(l)} - S_{(s)} = 60.01 - 38.20 = 21.81 \, J \, mol^{-1} \, K^{-1}$. Therefore,$\Delta H = 273 \, K 	imes 21.81 \, J \, mol^{-1} \, K^{-1} = 5954.13 \, J \, mol^{-1}$.

One mole of ice is converted into water at $273 \, K$. The entropy of $H_2O_{(s)}$ and $H_2O_{(l)}$ are $38.20 \, J \, mol^{-1} \, K^{-1}$ and $60.01 \, J \, mol^{-1} \, K^{-1}$ respectively. The enthalpy change for the conversion is $....... \, J \, mol^{-1}$.

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