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At low pressure, the fraction of the surface covered follows:
(A) Zero-order kinetics
(B) First-order kinetics
(C) Second-order kinetics
(D) Fractional order kinetics.

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Last updated date: 14th Apr 2024
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MVSAT 2024
Answer
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Hint: The chemical kinetics of catalyzed reactions in which rate expressions are indispensable during the application of catalyzed reactions. These reactions are also known as heterogeneous catalyzed reactions which vary with the partial pressure of the reactant gases above the catalyst surface.

Complete step by step solution:
To predict the kinetics of heterogeneous catalyzed reactions by Langmuir isotherm expression for equilibrium surface coverages.
Consider the surface decomposition of a molecule,
${{A}_{(g)}}\rightleftharpoons {{A}_{(ads)}}\to products$
This process is called a unimolecular decomposition reaction, which occurs uniformly across the surface sites in molecule A may be adsorbed or limited to a number of special sites. The product which is formed in this reaction is weakly bonded to the surface. The surface decomposition step is the rate-determining step.
Examples of unimolecular decomposition are the decomposition of $N{{H}_{3}}$ to ${{N}_{2}}\And {{H}_{2}}$ on metal surfaces and decomposition of phosphine on gas.
According to Langmuir adsorption isotherm, the rate of surface decomposition is \[rate=k\theta \]
Substitute the surface coverage $\theta $, the required rate in terms of the pressure of the gas on the surface,
\[rate=\dfrac{kbP}{1+bP}\]
Consider the lower pressure condition, the steady-state coverage $\theta $, is very small for the reactant molecule and proportional to pressure.
bP<<1, then $1+bP\approx 1$
 then the rate becomes first order, rate=kb, if substance absorbs on surfaces, it is a first-order reaction.

Hence, the correct answer is option B.

Note: If the high-pressure conditions, the fraction of the surface covered follows the zeroth order of kinetics reaction. Because under these conditions the reactant molecules of steady surface coverage almost unity and follows a zero-order reaction.