A particular reaction at `27^(@)C` for which `DeltaH gt 0` and `DeltaS gt 0` is found to be non-spontaneous. The reaction may proceed spontaneously if

Options

A. The temperature is decreased
B. The temperature is increased
C. The temperature is kept constant
D. It is carried in open vessel at `27^(@)C`

Answer

To solve the problem, we need to analyze the conditions under which a reaction with positive enthalpy change (ΔH > 0) and positive entropy change (ΔS > 0) can become spontaneous. We will use the Gibbs free energy equation:

### Step-by-Step Solution:

1. **Understanding Gibbs Free Energy**:
The Gibbs free energy (G) is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
where:
- ΔG = change in Gibbs free energy
- ΔH = change in enthalpy
- T = temperature in Kelvin
- ΔS = change in entropy

2. **Conditions for Non-Spontaneity**:
For a reaction to be non-spontaneous, ΔG must be greater than zero:
\[ \Delta G > 0 \]
Given that ΔH > 0 and ΔS > 0, we can rewrite the equation:
\[ \Delta G = \Delta H - T \Delta S > 0 \]

3. **Rearranging the Inequality**:
Rearranging the inequality gives:
\[ \Delta H > T \Delta S \]
This means that the enthalpy change is greater than the temperature multiplied by the entropy change.

4. **Effect of Temperature**:
To make the reaction spontaneous (i.e., ΔG < 0), we need:
\[ \Delta H < T \Delta S \]
This implies that we can achieve spontaneity by increasing the temperature (T). As T increases, the term \(T \Delta S\) will also increase, potentially surpassing ΔH.

5. **Conclusion**:
Therefore, the reaction may proceed spontaneously if the temperature is increased.

### Final Answer:
The reaction may proceed spontaneously if the temperature is increased.

---