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