If a process is both endothermic and spontaneous then

Options

A. `Delta S gt 0`
B. `Delta S lt 0`
C. `Delta H lt 0`
D. `Delta G gt 0`

Answer

To analyze the situation where a process is both endothermic and spontaneous, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

### Step-by-Step Solution:

1. **Understanding the Terms**:
- **Endothermic Process**: This means that the process absorbs heat from the surroundings, which implies that \(\Delta H > 0\).
- **Spontaneous Process**: This means that the process occurs naturally without external intervention, which implies that \(\Delta G < 0\).

2. **Setting Up the Gibbs Free Energy Equation**:
- We know from the Gibbs free energy equation that:
\[ \Delta G = \Delta H - T \Delta S \]
- Here, \(\Delta H\) is positive (since the process is endothermic), and \(\Delta G\) must be negative (since the process is spontaneous).

3. **Rearranging the Equation**:
- To satisfy the condition of spontaneity (\(\Delta G < 0\)), we can rearrange the equation:
\[ \Delta G < 0 \implies \Delta H - T \Delta S < 0 \]
- This can be rewritten as:
\[ \Delta H < T \Delta S \]

4. **Interpreting the Conditions**:
- Since \(\Delta H\) is positive, for the inequality \(\Delta H < T \Delta S\) to hold true, \(\Delta S\) must also be positive. This means that the entropy of the system is increasing during the process.

5. **Conclusion**:
- Therefore, if a process is both endothermic and spontaneous, it must involve an increase in entropy (\(\Delta S > 0\)) that is sufficient to outweigh the positive enthalpy change (\(\Delta H > 0\)) at the given temperature \(T\).