For a particular reversible reaction at temperature `T, DeltaH` and `DeltaS` were found to be both `+ve`. If `T_e` is the temperature at equilibrium, the reaction would be spontaneous when :

Options

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

Answer

To solve the problem, we need to analyze the conditions under which a reversible reaction is spontaneous given that both ΔH (enthalpy change) and ΔS (entropy change) are positive.

### Step 1: Understand the Gibbs Free Energy Equation
The Gibbs free energy change (ΔG) for a reaction 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

### Step 2: Analyze the Conditions for Spontaneity
A reaction is spontaneous when ΔG < 0. Therefore, we need to find conditions under which:
\[ \Delta H - T \Delta S < 0 \]
Rearranging this inequality gives:
\[ \Delta H < T \Delta S \]

### Step 3: Relate Temperature to Equilibrium Temperature
At equilibrium, the Gibbs free energy change is zero (ΔG = 0). This leads to:
\[ \Delta H = T_E \Delta S \]
Where \(T_E\) is the equilibrium temperature. Rearranging this gives:
\[ T_E = \frac{\Delta H}{\Delta S} \]

### Step 4: Substitute into the Spontaneity Condition
From the spontaneity condition derived in Step 2, we can substitute \(T_E\):
\[ \Delta H < T \Delta S \implies \frac{\Delta H}{\Delta S} < T \]
This implies:
\[ T_E < T \]

### Conclusion
Thus, for the reaction to be spontaneous, the temperature \(T\) must be greater than the equilibrium temperature \(T_E\):
\[ T > T_E \]

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

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