Assertion : If both `DeltaH^(@) and DeltaS^(@)` are positive then reaction will be spontaneous at high temperature.
Reason : All processes with positive entropy change are spontaneous.

Options

A. If both assertion and reason are true and reason is the correct explanation of assertion.
B. If both assertion and reason are true but reason is not the correct explanation of assertion
C. If assertion is true but reason is false.
D. If both assertion and reason are false.

Explanation

To solve the assertion-reason question, we need to analyze both the assertion and the reason provided.

### Step-by-step Solution:

1. **Understanding the Assertion**:
- The assertion states: "If both ΔH and ΔS are positive, then the reaction will be spontaneous at high temperature."
- To determine if this is true, we need to recall the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
- For a reaction to be spontaneous, ΔG must be negative.

2. **Analyzing ΔH and ΔS**:
- Given that both ΔH (enthalpy change) and ΔS (entropy change) are positive:
- ΔH > 0 (endothermic reaction)
- ΔS > 0 (increase in disorder)
- As temperature (T) increases, the term \(T \Delta S\) will also increase since ΔS is positive.

3. **Behavior of ΔG at High Temperatures**:
- At high temperatures, the term \(T \Delta S\) can become large enough to outweigh ΔH:
\[ \Delta G = \Delta H - T \Delta S \]
- If \(T \Delta S\) becomes greater than ΔH, then ΔG will be negative, indicating that the reaction is spontaneous.

4. **Conclusion for Assertion**:
- Therefore, the assertion is correct: if both ΔH and ΔS are positive, the reaction can indeed be spontaneous at high temperatures.

5. **Understanding the Reason**:
- The reason states: "All processes with positive entropy change are spontaneous."
- While it is true that a positive ΔS indicates an increase in disorder, it does not guarantee spontaneity on its own.

6. **Counterexample for the Reason**:
- If ΔH is significantly positive and the temperature is low, the term \(T \Delta S\) may not be sufficient to make ΔG negative:
\[ \Delta G = \Delta H - T \Delta S > 0 \]
- This means that even with a positive ΔS, the reaction may not be spontaneous if ΔH dominates.

7. **Conclusion for Reason**:
- Thus, the reason is incorrect: not all processes with positive entropy change are spontaneous, especially when ΔH is also positive and significant.

### Final Answer:
- The assertion is correct, but the reason is incorrect. Therefore, the correct answer is option C.