When ΔH is positive and small, and ΔS is positive and large, the reaction will be

Analysis

To solve this, we need to apply the Gibbs Free Energy equation, which determines the spontaneity of a chemical reaction.

Concept Explanation

The spontaneity of a reaction is governed by the Gibbs Free Energy equation:

Where:

• $\Delta G$ is the change in Gibbs Free Energy.
• $\Delta H$ is the change in Enthalpy (heat content).
• $T$ is the absolute temperature (in Kelvin).
• $\Delta S$ is the change in Entropy (disorder).

The rules for spontaneity are:

1. If $\Delta G < 0$ (negative), the reaction is spontaneous.
2. If $\Delta G > 0$ (positive), the reaction is non-spontaneous.
3. If $\Delta G = 0$, the reaction is at equilibrium.

Step-by-Step Solution

1. Identify the given values: • $\Delta H$ is positive ($+$). • $\Delta S$ is positive ($+$). • The term $T\Delta S$ is "large" because $\Delta S$ is large.

2. Substitute into the equation: $$\Delta G = (\text{positive small value}) - (T \times \text{positive large value})$$

3. Analyze the result: Since the $T\Delta S$ term is large and it is being subtracted from a small positive $\Delta H$, the resulting value of $\Delta G$ will be negative.

4. Conclusion: Because $\Delta G$ is negative, the reaction will proceed spontaneously.

Identification of the Answer

Based on the Gibbs Free Energy criteria, when the entropy factor ($T\Delta S$) outweighs the enthalpy factor ($\Delta H$), the reaction becomes spontaneous.

The correct option is D.

Exam Tip

Remember the "Spontaneity Table":

• $\Delta H (-)$ and $\Delta S (+)$: Always spontaneous.
• $\Delta H (+)$ and $\Delta S (-)$: Never spontaneous.
• $\Delta H (+)$ and $\Delta S (+)$: Spontaneous at high temperatures (because $T\Delta S$ becomes large).
• $\Delta H (-)$ and $\Delta S (-)$: Spontaneous at low temperatures.