Certain gas-phase reactions on a heterogeneous catalyst are first order at low gas pressures and zero order at high pressures. Can you suggest a reason for this?

Short Answer

Expert verified
The change from first order to zero order reaction with increasing gas pressure can be attributed to the increasing occupancy of the catalyst's active sites. At low pressures, the concentration of gases is low and hence the reaction rate is governed by the frequency of collision between gas molecules and the catalyst's surface (first order). At high pressures, all the active sites are occupied and the reaction rate becomes independent of gas concentration (zero order).

Step by step solution

01

Understanding Reaction Orders

To start with, an understanding of reaction orders is necessary. The order of a reaction is a concept in chemical kinetics that describes the relationship between the rate of a reaction and the concentration of its reactants. A zero-order reaction has a rate that is independent of the concentration of the reactant(s) while a first-order reaction has a rate that is directly proportional to the concentration of a single reactant.
02

Low Pressure Condition

Under low pressure conditions, the concentration of the gas molecules is relatively low. Therefore, the rate of the reaction is mainly determined by the frequency of collision between gas molecules and the catalyst's surface. Since the number of active sites available on the catalyst is huge, most of the gas molecules will find an active site to react with as soon as they collide with the catalyst's surface. Thus, the reaction is first order with respect to gas concentration.
03

High Pressure Condition

Under high pressure conditions, the concentration of gas molecules is significantly high. In this scenario, practically all the active sites on the catalyst’s surface are occupied by gas molecules. Subsequently, the arrival of more gas molecules does not increase the reaction rate as there are no free active sites for additional gas molecules to react with. Hence, the reaction becomes independent of the concentration of the gas and the reaction is zero order.

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Most popular questions from this chapter

A first-order reaction, \(\mathrm{A} \longrightarrow\) products, has a halflife of \(75 \mathrm{s},\) from which we can draw two conclusions. Which of the following are those two (a) the reaction goes to completion in 150 s; (b) the quantity of \(A\) remaining after 150 s is half of what remains after 75 s; (c) the same quantity of A is consumed for every 75 s of the reaction; (d) one- quarter of the original quantity of A is consumed in the first 37.5 s of the reaction; (e) twice as much A is consumed in 75 s when the initial amount of \(\mathrm{A}\) is doubled; (f) the amount of \(\mathrm{A}\) consumed in 150 s is twice as much as is consumed in 75 s.

The rate of a chemical reaction generally increases rapidly, even for small increases in temperature, because of a rapid increase in (a) collision frequency; (b) fraction of reactant molecules with very high kinetic energies; (c) activation energy; (d) average kinetic energy of the reactant molecules.

A simplified rate law for the reaction \(2 \mathrm{O}_{3}(\mathrm{g}) \longrightarrow\) \(3 \mathrm{O}_{2}(\mathrm{g})\) is $$\text { rate }=k=\frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]}$$ For this reaction, propose a two-step mechanism that consists of a fast, reversible first step, followed by a slow second step.

In the reaction \(A(g) \longrightarrow 2 B(g)+C(g),\) the total pressure increases while the partial pressure of \(\mathrm{A}(\mathrm{g})\) decreases. If the initial pressure of \(\mathrm{A}(\mathrm{g})\) in a vessel of constant volume is \(1.000 \times 10^{3} \mathrm{mmHg}\) (a) What will be the total pressure when the reaction has gone to completion? (b) What will be the total gas pressure when the partial pressure of \(\mathrm{A}(\mathrm{g})\) has fallen to \(8.00 \times 10^{2} \mathrm{mmHg} ?\)

In three different experiments, the following results were obtained for the reaction \(A \longrightarrow\) products: \([\mathrm{A}]_{0}=1.00 \mathrm{M}, t_{1 / 2}=50 \mathrm{min} ;[\mathrm{A}]_{0}=200 \mathrm{M}, t_{1 / 2}=\) \(25 \min ;[\mathrm{A}]_{0}=0.50 \mathrm{M}, t_{1 / 2}=100 \mathrm{min} .\) Write the rate equation for this reaction, and indicate the value of \(k.\)

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