For the equilibrium,H_{2}O_{(l)} rightlefthar | vedclass

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(D) For the given equilibrium $H_{2}O_{(l)} \rightleftharpoons H_{2}O_{(g)}:$$(I)$ At equilibrium,the Gibbs free energy change is $\Delta G = 0$.$(II)

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$\Delta G = 0, \Delta H > 0, \Delta S > 0$

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(D) For the given equilibrium $H_{2}O_{(l)} ightleftharpoons H_{2}O_{(g)}:$$(I)$ At equilibrium,the Gibbs free energy change is $\Delta G = 0$.$(II)$ Entropy change,$\Delta S$,measures the randomness of the system. Since the vapor state is more disordered than the liquid state,the entropy increases,meaning $\Delta S > 0$.$(III)$ The conversion of liquid $H_{2}O$ to vapor $H_{2}O$ is an endothermic process,which requires the absorption of heat energy. Therefore,the enthalpy change is $\Delta H > 0$.Thus,the correct set of conditions is $\Delta G = 0, \Delta H > 0, 	ext{ and } \Delta S > 0$.Therefore,option $(D)$ is the correct answer.

For the equilibrium,$H_{2}O_{(l)} \rightleftharpoons H_{2}O_{(g)},$ which of the following is correct?

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