Question

question 7 of 30
what is the relationship between the enthalpy (δh) and entropy (δs) of a reaction that is always spontaneous?

a. -δh, -δs
b. +δh, +δs
c. -δh, +δs
d. +δh, -δs

Explanation:

Step1: Recall Gibbs Free Energy Formula

The spontaneity of a reaction is determined by the Gibbs free energy change ($\Delta G$) using the formula: $\Delta G = \Delta H - T\Delta S$, where $T$ is temperature in Kelvin (always positive, $T>0$). A reaction is spontaneous when $\Delta G < 0$.

Step2: Analyze Conditions for $\Delta G < 0$ Always

• For $\Delta G$ to be negative always (regardless of temperature), we need the terms $\Delta H$ and $-T\Delta S$ to both contribute to a negative $\Delta G$.
• Case 1: $\Delta H < 0$ (exothermic, negative enthalpy change) and $\Delta S > 0$ (positive entropy change, increase in disorder).
• If $\Delta H$ is negative, the $\Delta H$ term is negative.
• If $\Delta S$ is positive, then $-T\Delta S$ becomes negative (since $T>0$ and $\Delta S>0$, so $-T\Delta S < 0$).
• Adding two negative terms ($\Delta H$ and $-T\Delta S$) will always give a negative $\Delta G$, so the reaction is always spontaneous.
• Let's check other options:
• Option A: $\Delta H < 0$, $\Delta S < 0$: Then $-T\Delta S$ is positive (since $\Delta S < 0$, $-T\Delta S = -T\times(\text{negative}) = \text{positive}$). So $\Delta G = \text{negative} + \text{positive}$. The sign of $\Delta G$ depends on temperature (at low $T$, $\Delta G$ might be negative; at high $T$, positive). Not always spontaneous.
• Option B: $\Delta H > 0$, $\Delta S > 0$: $\Delta H$ is positive, $-T\Delta S$ is negative. Sign of $\Delta G$ depends on temperature (at high $T$, $\Delta G$ negative; at low $T$, positive). Not always spontaneous.
• Option D: $\Delta H > 0$, $\Delta S < 0$: $\Delta H$ positive, $-T\Delta S$ positive (since $\Delta S < 0$, $-T\Delta S = -T\times(\text{negative}) = \text{positive}$). So $\Delta G = \text{positive} + \text{positive} = \text{positive}$, reaction is non - spontaneous.

Answer:

C. $-\Delta H, +\Delta S$