Question

The signs of $\Delta H$ and $\Delta S$ for a reaction to be spontaneous at all temperatures respectively are

• A. Positive , positive
• B. Positive , negative
• C. Negative , negative
• D. Negative , positive

Hint:

To remember this conceptually: nature favors lower energy (negative enthalpy, exothermicity) and higher disorder (positive entropy). When both driving forces favor the reaction, it will occur spontaneously at any temperature.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks to identify the required thermodynamic signs of enthalpy change ($\Delta H$) and entropy change ($\Delta S$) that guarantee a chemical reaction will be spontaneous under all temperature conditions.

Step 2: Key Formula or Approach:
The spontaneity of a reaction at constant temperature ($T$) and pressure is determined by the change in Gibbs Free Energy ($\Delta G$), as defined by the Gibbs-Helmholtz equation:
\[ \Delta G = \Delta H - T\Delta S \] For a reaction to be spontaneous, the change in Gibbs Free Energy must be negative ($\Delta G < 0$).

Step 3: Detailed Explanation:

• Let us analyze the mathematical sign of $\Delta G$ under different combinations of signs for $\Delta H$ and $\Delta S$:

• Case 1: $\Delta H$ is negative ($\Delta H < 0$) and $\Delta S$ is positive ($\Delta S > 0$):
- Since $\Delta H$ is negative, the first term in the equation is negative.
- Since $T$ (absolute temperature in Kelvin) is always positive ($T > 0$) and $\Delta S$ is positive, the term $-T\Delta S$ will always be negative.
- Adding two negative numbers always results in a negative value. Therefore, $\Delta G = (\text{Negative}) + (\text{Negative}) = \text{Negative}$ at all possible temperatures. This reaction is always spontaneous.

• Case 2: $\Delta H$ is positive ($\Delta H > 0$) and $\Delta S$ is negative ($\Delta S < 0$):
- The first term $\Delta H$ is positive, and $-T\Delta S$ becomes positive.
- Thus, $\Delta G$ will always be positive, making the reaction non-spontaneous at all temperatures.
- Case 3: $\Delta H$ is negative and $\Delta S$ is negative:
- $\Delta G$ is negative only at low temperatures (where the favorable enthalpy term dominates).
- Case 4: $\Delta H$ is positive and $\Delta S$ is positive:
- $\Delta G$ is negative only at high temperatures (where the favorable entropy term dominates).

• Therefore, only a combination of negative enthalpy change ($\Delta H < 0$, exothermic) and positive entropy change ($\Delta S > 0$, increasing randomness) guarantees spontaneity across all temperatures.

Step 4: Final Answer:
The signs of $\Delta H$ and $\Delta S$ for a spontaneous reaction at all temperatures are negative and positive, respectively.