Adsorption of a gas on solid metal surface is spontaneous and exothermic, then: 

Options

A. a. H increases
B. b. S increases
C. c. G increases
D. c. S decreases

Answer

To solve the question regarding the adsorption of a gas on a solid metal surface being spontaneous and exothermic, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Terms**:
- **Spontaneous Process**: A process that occurs without needing to be driven by an external force. For a spontaneous process, the change in Gibbs free energy (ΔG) is negative.
- **Exothermic Process**: A process that releases heat. For an exothermic reaction, the change in enthalpy (ΔH) is negative.

2. **Use the Gibbs Free Energy Equation**:
The relationship between Gibbs free energy (ΔG), enthalpy (ΔH), and entropy (ΔS) is given by the equation:
\[ ΔG = ΔH - TΔS \]
where:
- ΔG = change in Gibbs free energy
- ΔH = change in enthalpy
- T = temperature (in Kelvin)
- ΔS = change in entropy

3. **Analyze the Given Conditions**:
- Since the process is spontaneous, we have ΔG < 0.
- Since the process is exothermic, we have ΔH < 0.

4. **Determine the Sign of ΔS**:
- Rearranging the Gibbs free energy equation gives:
\[ ΔG = ΔH - TΔS \]
- Since ΔH is negative and ΔG is also negative, we can analyze the term \( -TΔS \):
- If ΔS is positive, then \( -TΔS \) would be negative, which could potentially make ΔG positive if ΔH is not sufficiently negative.
- If ΔS is negative, then \( -TΔS \) would be positive, which would help keep ΔG negative since ΔH is already negative.

5. **Conclusion**:
- For the process to remain spontaneous (ΔG < 0) with ΔH < 0, it implies that ΔS must be negative. Thus, the entropy of the system decreases during the adsorption process.

### Final Answer:
The correct conclusion is that the change in entropy (ΔS) decreases during the adsorption of a gas on a solid metal surface.