In a reaction `DeltaH` and `DeltaS` both are more than zero. In which of the following cases, the reaction would not be spontaneous?

Options

A. `DeltaH gt T DeltaS`
B. `T DeltaS gt DeltaH`
C. `DeltaH = T DeltaS`
D. `DeltaG lt 0`

Answer

To determine the conditions under which a reaction with positive ΔH (enthalpy change) and positive ΔS (entropy change) is non-spontaneous, we can use the Gibbs free energy equation:

### Step-by-Step Solution:

1. **Understanding the Gibbs Free Energy Equation**:
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. **Condition for Non-Spontaneity**:
A reaction is non-spontaneous when ΔG is greater than 0:
\[ \Delta G > 0 \]

3. **Rearranging the Equation**:
For the reaction to be non-spontaneous, we can rearrange the Gibbs free energy equation:
\[ \Delta H - T \Delta S > 0 \]
This implies:
\[ \Delta H > T \Delta S \]

4. **Analyzing the Given Conditions**:
Since both ΔH and ΔS are positive, we need to find a condition where ΔH is greater than TΔS. This can happen at low temperatures because as T increases, TΔS increases, which could potentially make ΔG negative.

5. **Conclusion**:
The reaction will not be spontaneous if ΔH is greater than TΔS. Therefore, the correct option indicating non-spontaneity would be:
- **Option 1**: ΔH > TΔS