Match the following processes with their corresponding entropy changes:
$a$. Liquid to vapor conversion $1$. $\Delta S = 0$ $b$. Reaction not spontaneous at any temperature,$\Delta H = (+)$ $2$. $\Delta S = (+)$ $c$. Reversible expansion of an ideal gas $3$. $\Delta S = (-)$
| $a$. Liquid to vapor conversion | $1$. $\Delta S = 0$ |
| $b$. Reaction not spontaneous at any temperature,$\Delta H = (+)$ | $2$. $\Delta S = (+)$ |
| $c$. Reversible expansion of an ideal gas | $3$. $\Delta S = (-)$ |
(A) The correct matches are:
$a-2$: Conversion of liquid to vapor increases disorder,so $\Delta S = (+)$.
$b-3$: For a reaction to be non-spontaneous at all temperatures where $\Delta H = (+)$,the entropy change $\Delta S$ must be negative $((-))$ because $\Delta G = \Delta H - T\Delta S$. If $\Delta S = (-)$,then $\Delta G$ will always be positive.
$c-1$: For a reversible process,the entropy change of the system and surroundings is zero,but specifically for the reversible expansion of an ideal gas in an isolated system,$\Delta S = 0$ is often considered in specific contexts,though generally,it refers to the equilibrium state.
Correct sequence: $a-2, b-3, c-1$.
$a-2$: Conversion of liquid to vapor increases disorder,so $\Delta S = (+)$.
$b-3$: For a reaction to be non-spontaneous at all temperatures where $\Delta H = (+)$,the entropy change $\Delta S$ must be negative $((-))$ because $\Delta G = \Delta H - T\Delta S$. If $\Delta S = (-)$,then $\Delta G$ will always be positive.
$c-1$: For a reversible process,the entropy change of the system and surroundings is zero,but specifically for the reversible expansion of an ideal gas in an isolated system,$\Delta S = 0$ is often considered in specific contexts,though generally,it refers to the equilibrium state.
Correct sequence: $a-2, b-3, c-1$.
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