Which of the following reaction is expected never to be spontaneous :-

Options

A. `2O_(3)rarr 3O_(2)`
`Delta H=-Ve, Delta S = +Ve`
B. `Mg+H_(2)rarr MgH_(2)`
`Delta H =-Ve, Delta S=-Ve`
C. `Br_(2)(I)rarr Br_(2)(g)`
`Delta H=-Ve, Delta S=-Ve`
D. `2Ag+3N_(2)rarr2Ag N_(3)`
`Delta H=+Ve, Delta S=-Ve`

Answer

To determine which reaction is expected never to be spontaneous, we need to analyze the relationship between the change in internal energy (ΔU), change in enthalpy (ΔH), and change in entropy (ΔS) using the formula:

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

### Step-by-Step Solution:

1. **Understand the Conditions for Spontaneity**:
- A reaction is spontaneous if the change in internal energy (ΔU) is negative.
- The formula for ΔU can be rearranged as:
\[ \Delta U = \Delta H - T \Delta S \]
- For ΔU to be negative, we need ΔH to be negative and TΔS to be positive, or ΔH to be sufficiently negative to outweigh a positive TΔS.

2. **Analyze Each Option**:
- **Option A**: ΔH is negative and ΔS is positive.
- Here, ΔU = ΔH - TΔS will be negative (since ΔH is negative and TΔS is positive). This reaction is spontaneous.
- **Option B**: ΔH is negative and ΔS is negative.
- Here, ΔU = ΔH - TΔS will still be negative if the magnitude of ΔH is greater than TΔS. This reaction can be spontaneous under certain conditions.
- **Option C**: ΔH is negative and ΔS is negative.
- Similar to Option B, ΔU can still be negative if ΔH is sufficiently negative. This reaction can also be spontaneous.
- **Option D**: ΔH is positive and ΔS is negative.
- Here, ΔU = ΔH - TΔS will always be positive because both ΔH is positive and TΔS is negative. This means that ΔU cannot be negative, making this reaction non-spontaneous.

3. **Conclusion**:
- The reaction that is expected never to be spontaneous is **Option D**, where ΔH is positive and ΔS is negative.

### Final Answer:
**Option D** is the reaction that is expected never to be spontaneous.

---