H_2O_{(s)} rightleftharpoons H_2O_{(ell)}Whic | vedclass

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(C) $1$. At $273 \ K$ and $1 \ atm$, the system $H_2O_{(s)} \rightleftharpoons H_2O_{(\ell)}$ is in equilibrium, so $\Delta G = 0$. Since melting is a

Vedclass · Accepted Answer

$A, B, C$ only

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(C) $1$. At $273 \ K$ and $1 \ atm$, the system $H_2O_{(s)} \rightleftharpoons H_2O_{(\ell)}$ is in equilibrium, so $\Delta G = 0$. Since melting is an endothermic process involving an increase in disorder, $\Delta S > 0$. Thus, statement $(A)$ is correct.$2$. According to Le Chatelier's principle, for the process $H_2O_{(s)} \rightleftharpoons H_2O_{(\ell)}$, the density of liquid water is greater than that of ice. Increasing pressure favors the side with higher density (liquid phase). Therefore, more ice will melt. Statement $(B)$ is correct.$3$. Since increasing pressure shifts the equilibrium towards the liquid phase, the melting point of ice decreases. Statement $(C)$ is correct and $(D)$ is incorrect.

List-$I$ (Equilibrium)	List-$II$ (Conditions for the process)
$P. A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$ (Endothermic)	$1. \text{High temperature}$
$Q. 2AB_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{3(g)}$ (Exothermic)	$2. \text{Low temperature}$
$R. 2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$ (Endothermic)	$3. \text{High temperature}$
	$4. \text{Low temperature}$
	$5. \text{Independent of pressure}$

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