For the reaction H_2O_{(s)} rightleftharpoons | vedclass

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(B) For the phase transition $H_2O_{(s)} \rightleftharpoons H_2O_{(l)}$ at $0\,^\circ C$ $(273.15\,K)$ and $1\,atm$ pressure,the system is in a state

Vedclass · Accepted Answer

$\Delta H = T\Delta S$

Vedclass · Answer

(B) For the phase transition $H_2O_{(s)} \rightleftharpoons H_2O_{(l)}$ at $0\,^\circ C$ $(273.15\,K)$ and $1\,atm$ pressure,the system is in a state of equilibrium.At equilibrium,the change in Gibbs free energy is $\Delta G = 0$.From the thermodynamic relation $\Delta G = \Delta H - T\Delta S$,substituting $\Delta G = 0$ gives $0 = \Delta H - T\Delta S$.Therefore,$\Delta H = T\Delta S$.

For the reaction $H_2O_{(s)} \rightleftharpoons H_2O_{(l)}$ at $0\,^\circ C$ and normal pressure:

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