In a reaction `P+Q rarr R+S`, there is no change in entropy, Enthalpy change for the reaction `(DeltaH)` is `"12 kJ mol"^(-1)`. Under what conditions, reaction will have negative value of free energy change?

Options

A. If `DeltaH` is positive
B. If `DeltaH` is negative.
C. If `DeltaH` is `"24 kJ mol"^(-1)`.
D. If temperature of reaction is high

Explanation

To determine the conditions under which the reaction \( P + Q \rightarrow R + S \) will have a negative value of free energy change (\( \Delta G \)), we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

### Step-by-Step Solution:

1. **Identify Given Values**:
- The change in entropy (\( \Delta S \)) is given as 0.
- The change in enthalpy (\( \Delta H \)) is given as \( 12 \, \text{kJ mol}^{-1} \).

2. **Substitute the Values into the Gibbs Equation**:
Since \( \Delta S = 0 \), the equation simplifies to:
\[ \Delta G = \Delta H - T \cdot 0 \]
This further simplifies to:
\[ \Delta G = \Delta H \]

3. **Analyze the Condition for Negative Free Energy Change**:
We want \( \Delta G < 0 \). Given that \( \Delta G = \Delta H \):
\[ \Delta H < 0 \]
However, since \( \Delta H \) is given as \( 12 \, \text{kJ mol}^{-1} \) (which is positive), we see that:
\[ \Delta G = 12 \, \text{kJ mol}^{-1} > 0 \]
This indicates that under the current conditions, the reaction will not have a negative free energy change.

4. **Conclusion**:
Since \( \Delta H \) is positive, and \( \Delta S \) is zero, the reaction cannot have a negative free energy change under any temperature conditions. Therefore, the reaction will not be spontaneous.

### Final Answer:
The reaction will not have a negative value of free energy change under any conditions given that \( \Delta H \) is positive and \( \Delta S \) is zero.