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Consider the surface-catalyzed exchange reaction $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{D}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HD}(\mathrm{g}) $$ Experimental studies show that this reaction occurs by the Langmuir- Hinshelwood mechanism by which both \(\mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{D}_{2}(\mathrm{~g}\) ) first dissociatively chemisorb to the surface. The rate-determining step is the reaction between the adsorbed \(\mathrm{H}\) and \(\mathrm{D}\) atoms. Derive an expression for the rate law for this reaction in terms of the gas-phase pressures of \(\mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{D}_{2}(\mathrm{~g}\) ). (Assume ideal-gas behavior.)

Short Answer

Expert verified
The rate law is \( \text{Rate} = \frac{k_1 K_H K_D P_{H_2} P_{D_2}}{(1 + K_H P_{H_2} + K_D P_{D_2})^2} \).

Step by step solution

01

Write the Mechanism Stages

According to Langmuir-Hinshelwood kinetics, each gas molecule dissociates and adsorbs onto the surface. The steps can be written as follows:1. Hydrogen gas adsorbs: \( \text{H}_2(\text{g}) \rightarrow 2 \text{H*} \)2. Deuterium gas adsorbs: \( \text{D}_2(\text{g}) \rightarrow 2 \text{D*} \)3. Surface reaction (Rate-determining): \( \text{H*} + \text{D*} \rightarrow \text{HD(g)} \)Here, \(*\) denotes an atom adsorbed on the surface.
02

Express the Rate of Product Formation

In the rate-determining step of \( \text{H*} + \text{D*} \rightarrow \text{HD(g)} \), the rate of formation of HD is proportional to the surface concentrations of adsorbed H and D:\[\text{Rate} = k_1 [\text{H*}][\text{D*}]\]Here, \(k_1\) is a rate constant.
03

Use Langmuir Adsorption Isotherms

For dissociative adsorption, the coverage of H and D on the surface can be described using Langmuir adsorption isotherms given by:\[\theta_H = \frac{K_H P_{H_2}}{1 + K_H P_{H_2} + K_D P_{D_2}}, \quad \theta_D = \frac{K_D P_{D_2}}{1 + K_H P_{H_2} + K_D P_{D_2}}\]where \( \theta_H \) and \( \theta_D \) represent the surface coverage of H and D, respectively, and \( K_H \) and \( K_D \) are the adsorption coefficients for \( \text{H}_2 \) and \( \text{D}_2 \).
04

Substitute Surface Coverage into Rate Expression

Substituting the expressions for \( \theta_H \) and \( \theta_D \) into the rate expression, we get:\[\text{Rate} = k_1 \left( \frac{K_H P_{H_2}}{1 + K_H P_{H_2} + K_D P_{D_2}} \right) \left( \frac{K_D P_{D_2}}{1 + K_H P_{H_2} + K_D P_{D_2}} \right)\]This can be simplified to:\[\text{Rate} = \frac{k_1 K_H K_D P_{H_2} P_{D_2}}{\left(1 + K_H P_{H_2} + K_D P_{D_2}\right)^2}\]This is the rate law for the reaction under the Langmuir-Hinshelwood mechanism.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

surface-catalyzed reactions
Surface-catalyzed reactions are fascinating events where reactants interact with a surface, often leading to enhanced reaction rates or producing outcomes not achievable in the gas phase. In the case of the Langmuir-Hinshelwood mechanism, the reaction takes place on a catalytic surface. The surface often made of metals or metal oxides, provides sites where gas molecules can adsorb and react.
This is crucial for processes like ammonia synthesis or hydrogenation reactions in the chemical industry.
In a surface-catalyzed reaction, the surface temporarily holds reactant molecules. It can modify their chemical bonds, making it easier for them to transform into products. This leads to more efficient reactions compared to those occurring without a catalyst. Understanding this concept is key to designing better catalysts that reduce energy consumption and improve the selectivity of industrial processes.
dissociative chemisorption
Dissociative chemisorption is a process where a molecule splits into two or more parts upon adsorbing onto a surface. This transformation is an essential step in surface-catalyzed reactions like our \(\mathrm{H}_{2}(\mathrm{~g}) + \mathrm{D}_{2}(\mathrm{~g}) \\rightarrow 2 \mathrm{HD}(\mathrm{g})\). Here, both hydrogen and deuterium gases dissociatively chemisorb, meaning that they break apart into individual hydrogen and deuterium atoms and bind to the catalytic surface. The dissociation is critical because it primes the atoms for the subsequent reaction, making them more reactive.
It is this capability to break bonds that distinguishes chemisorption from mere physisorption, which only involves weak, non-reactive, van der Waals interactions. So, in dissociative chemisorption, we essentially break molecular bonds to facilitate further reaction steps.
rate-determining step
In a multi-step reaction mechanism, the rate-determining step is the slowest step that sets the pace for the reaction. Think of it like a bottleneck in traffic. For the reaction between the adsorbed hydrogen and deuterium atoms to form HD, this step determines how fast the product is formed.
In the Langmuir-Hinshelwood mechanism, this surface reaction step, where \(\text{H*} + \text{D*} \rightarrow \text{HD(g)}\), is crucial as it decides the overall reaction rate. Since reactions on surfaces are often slower due to the additional step of adsorption and interaction between adsorbed species, understanding which step is rate-determining helps in optimizing the reaction conditions.
It allows chemists to focus on accelerating the slowest, most critical part of the process to improve yields and efficiency.
Langmuir adsorption isotherm
The Langmuir adsorption isotherm is vital for understanding how molecules distribute over a catalytic surface at equilibrium. It describes how the fractional coverage of a surface, denoted as \( \theta \_H \) and \( \theta _D \) for hydrogen and deuterium, correlates with the partial pressures of these gases and their affinity for the surface.
The formulae are given by\[\theta_H = \frac{K_H P_{H_2}}{1 + K_H P_{H_2} + K_D P_{D_2}}, \quad \theta_D = \frac{K_D P_{D_2}}{1 + K_H P_{H_2} + K_D P_{D_2}}\]Here, \( K_H \\) and \( K_D \\) are adsorption coefficients expressing how strongly each molecule adheres to the surface.
This model assumes a single layer of adsorbed species, no interactions between adsorbed molecules, and that the adsorption sites are uniform. Such assumptions simplify the complex reality, guiding us to derive the rate law when evaluating catalytic systems like our ongoing example.
ideal-gas behavior
Ideal-gas behavior is an assumption used for simplifying calculations regarding gases. It presumes that gas molecules interact minimally and occupy negligible volume. This is a useful concept, especially in theoretical evaluations, like deriving a rate law for reactions such as the Langmuir-Hinshelwood mechanism.
When treating \( \mathrm{H}_2(\mathrm{~g}) \textrm{ and } \mathrm{D}_2(\mathrm{~g}) \) as ideal gases, we can relate their partial pressures directly to their concentrations. This is critical for applying adsorption models or evaluating nucleation rates.
While in the real world, especially under high pressure or low temperature, gases deviate from ideal behavior, assuming ideal-gas conditions simplifies the mathematical models without distorting useful insights into the system's dynamics. It acts as a baseline from which deviations can be understood and corrected.

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Most popular questions from this chapter

The hydrogenation of ethene on copper obeys the rate law \\[ v=\frac{k\left[\mathrm{H}_{2}\right]^{1 / 2}\left[\mathrm{C}_{2} \mathrm{H}_{4}\right]}{\left(1+K\left[\mathrm{C}_{2} \mathrm{H}_{4}\right]\right)^{2}} \\] where \(k\) and \(K\) are constants. Mechanistic studies show that the reaction occurs by the Langmuir-Hinshelwood mechanism. How are \(k\) and \(K\) related to the rate constants for the individual steps of the reaction mechanism? What can you conclude about the relative adsorption of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})\) to the copper surface from the form of the observed rate law?

If the density of surface sites is \(2.40 \times 10^{14} \mathrm{~cm}^{-2}\) and every molecule that strikes the surface adsorbs to one of these sites, determine the fraction of a monolayer created by the exposure of a \(1.00-\mathrm{cm}^{2}\) surface to \(1.00 \times 10^{-4}\) langmuir of \(\mathrm{N}_{2}(\mathrm{~g})\) at \(298.15 \mathrm{~K}\).

The density of silver at \(20^{\circ} \mathrm{C}\) is \(10.50 \mathrm{g} \cdot \mathrm{cm}^{-3},\) and its unit cell is cubic. Given that the first five observed Bragg diffraction angles are \(\theta=19.10^{\circ}, 22.17^{\circ}, 32.33^{\circ}, 38.82^{\circ},\) and \(40.88^{\circ},\) find the type of unit cell and its length. Take the wavelength of the X-radiation to be \(\lambda=154.433 \mathrm{pm}\).

For conducting surface experiments it is important to maintain a clean surface. Suppose that a \(1.50-\mathrm{cm}^{2}\) surface is placed inside a high-vacuum chamber at \(298.15 \mathrm{K}\) and the pressure inside the chamber is \(1.00 \times 10^{-12}\) torr. If the density of the surface sites is \(1.30 \times 10^{16} \mathrm{cm}^{-2}\) and we assume that the only gas in the chamber is \(\mathrm{H}_{2} \mathrm{O}\) and that each of the \(\mathrm{H}_{2} \mathrm{O}\) molecules that strike the surface adsorbs to one surface site, how long will it be until \(1.00 \%\) of the surface sites are occupied by water?

The first-order surface reaction \\[ \mathrm{A}(\mathrm{g}) \Longrightarrow \mathrm{A}(\mathrm{ads}) \Longrightarrow \mathrm{B}(\mathrm{g}) \\] has a rate of \(1.8 \times 10^{-4} \mathrm{mol} \cdot \mathrm{dm}^{-3} \cdot \mathrm{s}^{-1} .\) The surface has a dimension of \(1.00 \mathrm{cm}\) by \(3.50 \mathrm{cm}\) Calculate the rate of reaction if the dimensions of the two sides of the surface were each doubled. [Assume that \(A(g)\) is in excess.

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