Which of the following statements is correct for the spontaneous adsoption of a gas?

Options

A. (a) `DeltaS` is negative and therefore, `DeltaH` should be highly positive
B. (b) `DeltaS` is negative and therefore, `DeltaH` should be highly negative
C. (c) `DeltaS` is positive and therefore, `DeltaH` should be negative
D. (d) `DeltaS` is positive and therefore, `DeltaH` should also be highly positive

Explanation

To determine the correct statement regarding the spontaneous adsorption of a gas, we can analyze the thermodynamic principles involved, particularly focusing on the changes in entropy (ΔS) and enthalpy (ΔH) during the process.

### Step-by-Step Solution:

1. **Understanding Adsorption**:
- Adsorption is the process where gas molecules accumulate on the surface of a solid (adsorbent). This leads to a change in the system's entropy.

2. **Entropy Change (ΔS)**:
- During the adsorption of a gas, the gas molecules transition from a more disordered state (gas phase) to a more ordered state (adsorbed on a surface).
- This results in a decrease in randomness or disorder, which means that the entropy change (ΔS) is negative. Therefore, ΔS < 0.

3. **Gibbs Free Energy (ΔG)**:
- The spontaneity of a process is determined by the Gibbs free energy change, given by the equation:
\[ ΔG = ΔH - TΔS \]
- For a process to be spontaneous, ΔG must be negative (ΔG < 0).

4. **Analyzing the Equation**:
- Since we established that ΔS is negative, we can substitute this into the Gibbs free energy equation:
\[ ΔG = ΔH - T(ΔS) \]
- Here, T is the temperature (which is always positive). Thus, T(ΔS) will also be negative because ΔS is negative.

5. **Condition for Spontaneity**:
- For ΔG to be negative when ΔS is negative, ΔH must also be sufficiently negative to outweigh the negative contribution from T(ΔS).
- Therefore, for spontaneous adsorption, ΔH must be highly negative (ΔH < 0).

6. **Conclusion**:
- From the analysis, we conclude that the correct statement regarding the spontaneous adsorption of a gas is:
- ΔS is negative, and therefore ΔH should be highly negative.
- This corresponds to **Option B**.

### Final Answer:
**Option B: ΔS is negative and therefore ΔH should be highly negative.**