For a particular reversible reaction at temperature T, `DeltaHandDeltaS` were found to be both +ve. If `T_(e)` is the temperature at equilibrium, the reaction would be spontaneous when

Options

A. `T=T_(e)`
B. `T_(e)gtT`
C. `TgtT_(e)`
D. `T_(e)` is 5 times T

Answer

To determine when the reaction is spontaneous, we need to analyze the Gibbs free energy change (ΔG) for the reaction. The spontaneity of a reaction is indicated by the sign of ΔG, where a negative ΔG implies that the reaction is spontaneous.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- We are given that both ΔH (enthalpy change) and ΔS (entropy change) are positive for the reaction.
- We need to find the conditions under which the reaction is spontaneous at a temperature T.

2. **Recall the Gibbs Free Energy Equation**:
- The Gibbs free energy change is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]

3. **Substitute the Given Values**:
- Since both ΔH and ΔS are positive, we can express this in the equation:
\[ \Delta G = \text{(positive)} - T \times \text{(positive)} \]

4. **Determine the Condition for Spontaneity**:
- For the reaction to be spontaneous, we need ΔG to be less than 0:
\[ \Delta H - T \Delta S < 0 \]
- Rearranging this inequality gives:
\[ \Delta H < T \Delta S \]
- This implies:
\[ T > \frac{\Delta H}{\Delta S} \]

5. **Identify the Equilibrium Temperature**:
- The temperature at equilibrium (Te) can be defined as:
\[ T_e = \frac{\Delta H}{\Delta S} \]
- Therefore, for the reaction to be spontaneous, the temperature T must be greater than the equilibrium temperature Te:
\[ T > T_e \]

6. **Conclusion**:
- The reaction will be spontaneous when the temperature T is greater than the equilibrium temperature Te.

### Final Answer:
The reaction would be spontaneous when \( T > T_e \).