Question

The correct statement(s) for a surface catalyzed unimolecular gaseous reaction is(are)

• A. It follows Freundlich adsorption isotherm
• B. Zero order kinetics is followed at sufficiently high pressure of the reacting molecules
• C. First order kinetics is followed at very low pressure of the reacting molecules
• D. Rate of the reaction is proportional to the fraction of surface covered by the reacting molecules

Hint:

Use the Langmuir isotherm \(\theta=Kp/(1+Kp)\) with rate \(=k\theta\), and take the high- and low-pressure limits.

Answer 1

The Correct Option is B, C, D

Step 1: Set up the Langmuir-Hinshelwood picture.
For a unimolecular reaction catalyzed on a solid surface, reactant \(A\) must first adsorb before it reacts, and the rate of the surface reaction is proportional to the fraction of surface covered by \(A\), \(\theta\):
\[ \text{rate} = k\theta \]
The coverage \(\theta\) itself follows the Langmuir adsorption isotherm (not Freundlich),
\[ \theta = \frac{Kp}{1+Kp} \]
where \(p\) is the pressure of the gas and \(K\) is the adsorption equilibrium constant.

Step 2: Check statement (A).
The Freundlich isotherm (\(\theta=kp^{1/n}\), an empirical power law) applies to heterogeneous surfaces with a spread of binding energies; surface-catalyzed unimolecular kinetics is instead derived from the Langmuir isotherm, which assumes a uniform surface with equivalent sites.
Statement (A) is wrong.

Step 3: Check statement (B), high pressure.
At sufficiently high pressure, \(Kp\gg1\), so \(\theta = \dfrac{Kp}{1+Kp}\approx 1\). The surface is essentially saturated, so \(\text{rate}=k\theta\approx k\), independent of pressure: zero order kinetics. Statement (B) is correct.

Step 4: Check statement (C), low pressure.
At very low pressure, \(Kp\ll1\), so \(\theta\approx Kp\). Then \(\text{rate}=k\theta\approx kKp\), directly proportional to pressure: first order kinetics. Statement (C) is correct.

Step 5: Check statement (D).
This is exactly the starting rate law, \(\text{rate}=k\theta\): the rate is proportional to the fraction of surface covered. Statement (D) is correct.

Final Answer:
The reaction follows a Langmuir (not Freundlich) isotherm, giving zero order kinetics at high pressure, first order kinetics at low pressure, with the rate always tracking the surface coverage \(\theta\).
\[ \boxed{\text{(B), (C), (D)}} \]