In a reversible process,
`Delta S_(sys) + Delta S_(surr)` is

Options

A. `gt 0`
B. `lt 0`
C. `ge 0`
D. `= 0`

Explanation

To solve the question regarding the sum of the change in entropy of the system and the surroundings in a reversible process, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Second Law of Thermodynamics**:
The second law of thermodynamics states that for any process, the total change in entropy of the universe (system + surroundings) must be greater than or equal to zero. For reversible processes, this change is exactly zero.

2. **Defining the Entropy Changes**:
Let \( \Delta S_{\text{sys}} \) be the change in entropy of the system and \( \Delta S_{\text{surr}} \) be the change in entropy of the surroundings.

3. **Relationship in Reversible Processes**:
For a reversible process, the change in entropy of the surroundings is related to the change in entropy of the system by the equation:
\[ \Delta S_{\text{surr}} = -\Delta S_{\text{sys}} \]
This relationship holds because, in a reversible process, the system is in equilibrium with its surroundings at every instant.

4. **Combining the Entropy Changes**:
Now, we can add the changes in entropy of the system and the surroundings:
\[ \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = \Delta S_{\text{sys}} + (-\Delta S_{\text{sys}}) = 0 \]

5. **Conclusion**:
Therefore, for a reversible process, we conclude that:
\[ \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = 0 \]

### Final Answer:
In a reversible process, \( \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = 0 \).