Which of the following conditions may lead to a non-spontaneous change?

Options

A. `DeltaH` and `DeltaS` both `+ve`
B. `DeltaH` is `-ve` and `DeltaS` is `+ve`
C. `DeltaH` and `DeltaS` both `-ve`
D. `DeltaH` is `+ve` and `DeltaS` is `-ve`

Explanation

To determine the conditions that may lead to a non-spontaneous change, we can analyze the relationship between the change in free energy (ΔG), change in enthalpy (ΔH), and change in entropy (ΔS).

### Step-by-Step Solution:

1. **Understanding Non-Spontaneous Change**:
- A non-spontaneous change is characterized by a positive change in free energy (ΔG > 0). This means that the process requires energy input to occur.

2. **Free Energy Equation**:
- The relationship between ΔG, ΔH, and ΔS is given by the equation:
\[ ΔG = ΔH - TΔS \]
- Here, T is the temperature in Kelvin.

3. **Conditions for Non-Spontaneity**:
- For ΔG to be positive (ΔG > 0), we can analyze the terms in the equation:
- If ΔH is positive (ΔH > 0), this indicates that the process is endothermic and absorbs heat.
- If ΔS is negative (ΔS < 0), this indicates that the disorder of the system is decreasing.

4. **Combining Conditions**:
- When both conditions are met (ΔH > 0 and ΔS < 0), the equation becomes:
\[ ΔG = ΔH - TΔS \]
- Since ΔH is positive and ΔS is negative, the term -TΔS becomes positive (as a negative multiplied by a negative is positive).
- Therefore, ΔG will be positive, indicating a non-spontaneous change.

5. **Conclusion**:
- Thus, the conditions that lead to a non-spontaneous change are:
- ΔH is positive (endothermic process).
- ΔS is negative (decrease in disorder).
- Therefore, the correct option is (D): ΔH is positive and ΔS is negative.