Question

If $\Delta_r H^{\ominus}$ and $\Delta_r S^{\ominus}$ are standard enthalpy change and standard entropy change respectively for a reaction, the incorrect option is

• A. $\Delta_r H^{\ominus}$ = negative; $\Delta_r S^{\ominus}$ = positive; spontaneous at all temperatures
• B. $\Delta_r H^{\ominus}$ = negative; $\Delta_r S^{\ominus}$ = negative; non-spontaneous at low temperatures
• C. $\Delta_r H^{\ominus}$ = positive; $\Delta_r S^{\ominus}$ = positive; non-spontaneous at low temperatures
• D. $\Delta_r H^{\ominus}$ = negative; $\Delta_r S^{\ominus}$ = negative; spontaneous at low temperatures

Hint:

Spontaneity: $\Delta G = \Delta H - T\Delta S$. $\Delta G<0$ for spontaneous processes.

Answer 1

The Correct Option is B

Gibbs free energy determines spontaneity: $\Delta G = \Delta H - T\Delta S$. (1) $\Delta H<0$, $\Delta S>0$: $\Delta G$ will always be negative, so spontaneous at all temperatures.
(2) $\Delta H<0$, $\Delta S<0$: At low temperatures, the $T\Delta S$ term will be small, so $\Delta G$ can be negative (spontaneous). As temperature increases, the $T\Delta S$ term becomes larger, and $\Delta G$ may become positive (non-spontaneous). The option says "non-spontaneous" at low T, so this is incorrect.
(3) $\Delta H>0$, $\Delta S>0$: At low temperatures, the $T\Delta S$ term is small, and $\Delta G$ will be positive (non-spontaneous). At high temperatures, $T\Delta S$ can become larger than $\Delta H$, making $\Delta G$ negative (spontaneous).
(4) $\Delta H<0$, $\Delta S<0$: At low temperatures, the reaction can be spontaneous. This statement in itself isn't necessarily incorrect, but it can be depending on the magnitude of $\Delta H$ and $\Delta S$ compared to the temperature. Since we are looking for the "incorrect" option, (2) is definitely incorrect. (4) can be incorrect as well depending on the context. (4) can be considered a better answer if it means "always spontaneous" at low T, which is false.