If `Delta H gt 0` and `Delta S gt 0`, the reaction proceeds spontaneously when :-

Options

A. `Delta H gt 0`
B. `Delta H lt T Delta S`
C. `Delta H = T Delta S`
D. None

Answer

To determine when a reaction with positive enthalpy change (ΔH > 0) and positive entropy change (ΔS > 0) will proceed spontaneously, we can use the Gibbs free energy equation:

### Step-by-Step Solution:

1. **Understand the Gibbs Free Energy Equation**:
The Gibbs free energy change (ΔG) is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
where:
- ΔG = change in Gibbs free energy
- ΔH = change in enthalpy
- T = absolute temperature (in Kelvin)
- ΔS = change in entropy

2. **Identify the Conditions**:
We know from the problem that:
- ΔH > 0 (the reaction is endothermic)
- ΔS > 0 (the disorder of the system increases)

3. **Determine Spontaneity**:
For a reaction to be spontaneous, we need ΔG to be negative:
\[ \Delta G < 0 \]
This can be rearranged to:
\[ \Delta H < T \Delta S \]
This means that for the reaction to be spontaneous, the temperature (T) must be high enough such that the term \(T \Delta S\) is greater than ΔH.

4. **Conclusion**:
Therefore, the reaction will proceed spontaneously when:
\[ T > \frac{\Delta H}{\Delta S} \]
In simpler terms, as the temperature increases, the \(T \Delta S\) term increases, making it possible for ΔG to become negative despite ΔH being positive.

### Final Answer:
The reaction proceeds spontaneously when the temperature (T) is greater than the ratio of ΔH to ΔS:
\[ T > \frac{\Delta H}{\Delta S} \]