The Question is from neet-chemistry surface-chemistry adsorption, and it was created by the author Mr. S M HUSSAIN

Question

A mixture of two gases, A (molecular weight 28) and B (molecular weight 44), is adsorbed on a solid surface.  If the partial pressures of A and B are equal, which statement is most likely true regarding their fractional surface coverages ($	heta_A$ and $	heta_B$) assuming Langmuir adsorption?

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Langmuir Adsorption Isotherm
	**Related Topics**: Surface Chemistry, Adsorption, Gas Laws
	**Prerequisites**: Partial Pressure, Equilibrium Constant
## SOLUTION PATHWAY
**Given Information**:
		$P_A = P_B$
		Molecular weight of A = 28
		Molecular weight of B = 44
**Method & Steps**:
	1. **Recall the Langmuir Adsorption Isotherm:**
		Reasoning: The Langmuir isotherm relates the fractional surface coverage ($	heta$) to the partial pressure ($P$) of the adsorbate and the Langmuir constant ($K$).
		Formula/Rule: $	heta = \frac{KP}{1 + KP}$
	2. **Consider the Langmuir constant (K):**
		Working: The Langmuir constant ($K$) is related to the enthalpy of adsorption.  A larger $K$ implies stronger adsorption.
		Key Point: Generally, molecules with higher molecular weights and stronger intermolecular forces tend to have higher enthalpies of adsorption and thus larger $K$ values.
	3. **Compare $	heta_A$ and $	heta_B$:**
		Result: Since $P_A = P_B$ and B has a higher molecular weight (44) than A (28), it's likely that $K_B > K_A$.  Therefore, according to the Langmuir isotherm, $	heta_B > 	heta_A$.
		Verification:  If $K_B > K_A$ and $P_A = P_B$, then plugging these into the Langmuir equation will yield $	heta_B > 	heta_A$.
 ## OPTION ANALYSIS
[A] :x: $	heta_A > 	heta_B$: Incorrect. This would be true if $K_A > K_B$, which is less likely given the higher molecular weight of B.
[B] ✓: $	heta_B > 	heta_A$: Correct.  B is likely to adsorb more strongly due to its higher molecular weight, leading to greater surface coverage.
[C] :x: $	heta_A = 	heta_B$: Incorrect. This would only be true if $K_A = K_B$, which is unlikely.
[D] :x: $	heta_A + 	heta_B = 1$: Incorrect. This is not necessarily true. The sum of fractional coverages can be less than or equal to 1, depending on the pressures and the Langmuir constants.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Assuming equal adsorption strengths:  Don't assume $K_A = K_B$ just because $P_A = P_B$. Molecular weight and intermolecular forces influence adsorption strength.
		2. Misinterpreting the Langmuir Isotherm:  Understand the relationship between $	heta$, $K$, and $P$.
	**Quick Tips**:
		1. Higher molecular weight often implies stronger adsorption.
		2. The Langmuir constant ($K$) reflects the strength of adsorption.

Mr. S M HUSSAIN · Answer

θ_A > θ_B

Mr. S M HUSSAIN · Answer

The correct Option is "θ_B > θ_A"

Mr. S M HUSSAIN · Answer

θ_A = θ_B

Mr. S M HUSSAIN · Answer

θ_A + θ_B = 1

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