Which of the following statement is correct for the spontaneous adsorption of a gas?

Options

A. `Delta S ` is nagative and, therefore , `Delta H` should be highly positive
B. `Delta S ` is nagative and therefore , `Delta H` should be highly nagative
C. `Delta S` is positive and , therefore, `Delta H` should be nagative
D. `Delta S` is positive and, therefore, `Delta H` should also be highly positive

Answer

To determine the correct statement regarding the spontaneous adsorption of a gas, we need to analyze the thermodynamic conditions that govern this process. Here’s a step-by-step breakdown of the reasoning:

### Step 1: Understanding Adsorption
Adsorption is a surface phenomenon where gas molecules (adsorbate) adhere to the surface of a solid (adsorbent). This process can be either physisorption (weak Van der Waals forces) or chemisorption (strong chemical bonds).

**Hint:** Remember that adsorption involves the interaction between the adsorbate and the adsorbent's surface.

### Step 2: Energy Changes in Adsorption
When gas molecules are adsorbed onto a surface, energy is released. This release of energy is associated with a change in enthalpy (ΔH). In spontaneous processes, this change in enthalpy is typically negative (ΔH < 0), indicating that energy is released.

**Hint:** Focus on the concept of enthalpy and its sign during spontaneous processes.

### Step 3: Entropy Considerations
Entropy (ΔS) measures the randomness or disorder of a system. When gas molecules are adsorbed onto a surface, their freedom of movement is restricted, leading to a decrease in randomness. Thus, the change in entropy for spontaneous adsorption is negative (ΔS < 0).

**Hint:** Consider how the movement of gas molecules is affected by adsorption.

### Step 4: Gibbs Free Energy Equation
The spontaneity of a process is determined by the Gibbs free energy change (ΔG), which is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
For a process to be spontaneous, ΔG must be negative (ΔG < 0).

**Hint:** Recall the Gibbs free energy equation and the conditions for spontaneity.

### Step 5: Analyzing the Conditions for Spontaneity
Given that ΔH is negative and ΔS is also negative, we need to analyze how these two terms interact in the Gibbs free energy equation. The term \( T \Delta S \) will be positive since T (temperature) is always positive. Therefore, for ΔG to be negative, the magnitude of ΔH must be greater than the magnitude of \( T \Delta S \).

**Hint:** Think about how the magnitudes of ΔH and \( T \Delta S \) affect the overall sign of ΔG.

### Step 6: Conclusion
From the analysis, we conclude that for the spontaneous adsorption of a gas:
- ΔH must be negative (indicating energy release).
- ΔS must be negative (indicating a decrease in randomness).
- The absolute value of ΔH must be sufficiently large to outweigh the positive \( T \Delta S \) term.

Thus, the correct statement regarding spontaneous adsorption of a gas is that ΔH should be negative.

**Final Answer:** The correct statement for the spontaneous adsorption of a gas is that ΔH should be negative (Option B).