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Determine the expression for fractional coverage as a function of pressure for the dissociative adsorption mechanism described in the text in which adsorption is accompanied by dissociation:$$R_{2}(g)+2 M(\text {surface}) \stackrel{k_{a}}{\rightleftharpoons_{k}} 2 R M(\text {surface})$$

Short Answer

Expert verified
The expression for the fractional coverage (θ) as a function of pressure (P) for the dissociative adsorption mechanism is: \[ θ = 1 - \frac{k}{k_a\cdot P + k} \]

Step by step solution

01

Write down the rate equations for adsorption and desorption

We can represent the rate of adsorption (Rate_ads) and desorption (Rate_des) as follows: Rate_ads = \(k_a\cdot P \cdot (1-θ)^2\) Rate_des = \(k\cdot θ^2\) Here, (1-θ) represents the fraction of empty sites, and θ represents the fraction of occupied sites. P is the pressure.
02

Write down the expressions for dynamic equilibrium

At equilibrium, the rate of adsorption is equal to the rate of desorption, and we can write: \(k_a\cdot P\cdot (1-θ)^2 = k\cdot θ^2\)
03

Solve for θ

Now, we will solve for θ, which is the fractional coverage: \(\frac{k_a\cdot P}{k} = \frac{θ^2}{(1-θ)^2}\) From this equation, we can rearrange to get: \( θ^2 = \frac{k_a\cdot P}{k} (1-θ)^2 \) Now, we will expand the equation: \[ θ^2 = \frac{k_a\cdot P}{k} (1 - 2θ + θ^2) \] Now, we can isolate θ of: \[ θ = 1 - \frac{k}{k_a\cdot P + k} \]
04

Write the final expression for fractional coverage

The expression for the fractional coverage (θ) as a function of pressure (P) for the dissociative adsorption mechanism is: \[ θ = 1 - \frac{k}{k_a\cdot P + k} \]

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Key Concepts

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

Adsorption and Desorption Rates
The phenomenon of adsorption involves the attachment of particles, typically atoms, molecules, or ions, from a gas, liquid, or dissolved solid onto a surface. This process is pivotal in many industrial and environmental systems, where surface reactions are a critical component. In the case of dissociative adsorption, molecules split and then adsorb onto the surface. Conversely, desorption is the release of these adsorbed particles back into the surrounding phase.

The rates at which these two processes occur have profound implications for the state of the system. The adsorption rate, commonly denoted as Rate_ads, depends on the adsorption coefficient (\( k_a \)), the pressure of the surrounding gas (P), and the availability of empty sites on the surface, expressed as (\(1 - \theta \))^2 where \theta is the fractional coverage. Hence the rate equation for adsorption becomes Rate_ads = \(k_a\text{·}P\text{·}(1-\theta)^2\). On the flip side, the desorption rate, or Rate_des, is influenced by the desorption coefficient (k) and the fraction of sites already occupied by adsorbed particles (\theta^2), leading to Rate_des = \(k\text{·}\theta^2\).
Dynamic Equilibrium
Dynamic equilibrium in the context of adsorption is the state where the rates of adsorption and desorption are equal, meaning that as many particles are attaching to the surface as are leaving it. With dissociative adsorption, the equilibrium is established between the gas-phase reactants and the species adsorbed on the surface.

When the system reaches this state of balance, no net change in concentration of reactants and products occurs, although individual molecules may still be in flux. Mathematically, at equilibrium, the equation \(k_a\text{·}P\text{·}(1-\theta)^2 = k\text{·}\theta^2\) holds true. This equality allows for the derivation of an expression for fractional coverage, which describes the proportion of surface sites occupied by adsorbed species.
Solving for Fractional Coverage
Fractional coverage (\theta) is a crucial parameter representing the fraction of the surface sites that are occupied by adsorbed species. Solving for \theta gives insight into how surface coverage changes with pressure and is fundamental for understanding the behavior of adsorbed species.

By equating the rate equations - Rate_ads and Rate_des - and rearranging, we obtain an equation that relates \theta to the pressure and the rate constants (\theta = 1 - \(\frac{k}{k_a\text{·}P + k}\)). This derivation follows from considering that at equilibrium, the amount of adsorbate on the surface remains constant over time, allowing this steady-state assumption to simplify the kinetic analysis. The derived expression allows predictions of the system behavior under varying pressures and is an essential tool for designing and characterizing adsorption systems.
Chemical Kinetics
Chemical kinetics is the study of the speed or rate at which chemical reactions proceed and the factors affecting these rates. In reactions involving adsorption, such as the dissociative adsorption of gas onto a surface, insights from kinetics are crucial for understanding both the rates of adsorption/desorption and the mechanisms controlling these processes.

The rate constants (\( k_a \) and k) in the rate equations reflect the kinetic properties of the system and are influenced by factors like temperature, surface area, and the nature of the adsorbate and surface. In essence, kinetics helps explain how quickly equilibrium is reached and how varying conditions affect the adsorption process. The accurate quantification and understanding of these rate constants are essential for controlling chemical reactions, optimizing industrial processes, and understanding natural phenomena.

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Most popular questions from this chapter

The enzyme fumarase catalyzes the hydrolysis of fumarate: Fumarate \((a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{L}\) -malate \((a q)\). The turnover number for this enzyme is \(2.5 \times 10^{3} \mathrm{s}^{-1},\) and the Michaelis constant is \(4.2 \times 10^{-6} \mathrm{M}\). What is the rate of fumarate conversion if the initial enzyme concentration is \(1 \times 10^{-6} \mathrm{M}\) and the fumarate concentration is \(2 \times 10^{-4} \mathrm{M} ?\)

In Marcus theory for electron transfer, the reorganization energy is partitioned into solvent and solute contributions. Modeling the solvent as a dielectric continuum, the solvent reorganization energy is given by \\[ \lambda_{s o l}=\frac{(\Delta e)^{2}}{4 \pi \varepsilon_{0}}\left(\frac{1}{d_{1}}+\frac{1}{d_{2}}-\frac{1}{r}\right)\left(\frac{1}{n^{2}}-\frac{1}{\varepsilon}\right) \\] where \(\Delta e\) is the amount of charge transferred, \(d_{1}\) and \(d_{2}\) are the ionic diameters of ionic products, \(r\) is the separation distance of the reactants, \(n^{2}\) is the index of refraction of the surrounding medium, and \(\varepsilon\) is the dielectric constant of the medium. In \\[ \text { addition, }\left(4 \pi \varepsilon_{0}\right)^{-1}=8.99 \times 10^{9} \mathrm{Jm} \mathrm{C}^{-2} \\] a. For an electron transfer in water \((n=1.33 \text { and } \varepsilon=80 .)\) where the ionic diameters of both species are \(6 \AA\) and the separation distance is \(15 \AA\), what is the expected solvent reorganization energy? b. Redo the earlier calculation for the same reaction occurring in a protein. The dielectric constant of a protein is dependent on sequence, structure, and the amount of included water; however, a dielectric constant of 4 is generally assumed consistent with a hydrophobic environment. Using light-scattering measurements the dielectric constant of proteins has been estimated to be \(\sim 1.5\)

Using the preequilibrium approximation, derive the predicted rate law expression for the following mechanism: $$\begin{array}{l} \mathrm{A}_{2} \stackrel{k_{1}}{=} 2 \mathrm{A} \\ \mathrm{A}+\mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{P} \end{array}$$

The reaction of nitric oxide \((\mathrm{NO}(g))\) with molecular hydrogen \(\left(\mathrm{H}_{2}(g)\right)\) results in the production of molecular nitrogen and water as follows: \\[ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \\] The experimentally-determined rate-law expression for this reaction is first order in \(\mathrm{H}_{2}(g)\) and second order in \(\mathrm{NO}(g)\) a. Is the reaction as written consistent with the experimental order dependence for this reaction? b. One potential mechanism for this reaction is as follows: \\[ \begin{array}{l} \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} \\] Is this mechanism consistent with the experimental rate law? c. An alternative mechanism for the reaction is \\[ \begin{array}{l} 2 \mathrm{NO}(g) \frac{k_{1}}{\sum_{k-1}} \mathrm{N}_{2} \mathrm{O}_{2}(g) \text { (fast) } \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{2}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} \\] Show that this mechanism is consistent with the experimental rate law.

The hydrogen-bromine reaction corresponds to the production of \(\operatorname{HBr}(g)\) from \(\mathrm{H}_{2}(g)\) and \(\mathrm{Br}_{2}(g)\) as follows: \(\mathrm{H}_{2}(g)+\operatorname{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) .\) This reaction is famous for its complex rate law, determined by Bodenstein and Lind in 1906: \\[ \frac{d[\mathrm{HBr}]}{d t}=\frac{k\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]^{1 / 2}}{1+\frac{m[\mathrm{HBr}]}{\left[\mathrm{Br}_{2}\right]}} \\] where \(k\) and \(m\) are constants. It took 13 years for a likely mechanism of this reaction to be proposed, and this feat was accomplished simultaneously by Christiansen, Herzfeld, and Polyani. The mechanism is as follows: \\[ \begin{array}{l} \operatorname{Br}_{2}(g) \stackrel{k_{1}}{\sum_{k_{-1}}} 2 \operatorname{Br} \cdot(g) \\ \text { Br' }(g)+\mathrm{H}_{2}(g) \stackrel{k_{2}}{\longrightarrow} \operatorname{HBr}(g)+\mathrm{H} \cdot(g) \\ \text { H\cdot }(g)+\operatorname{Br}_{2}(g) \stackrel{k_{3}}{\longrightarrow} \operatorname{HBr}(g)+\operatorname{Br} \cdot(g) \\ \operatorname{HBr}(g)+\mathrm{H} \cdot(g) \stackrel{k_{4}}{\longrightarrow} \mathrm{H}_{2}(g)+\operatorname{Br} \cdot(g) \end{array} \\] Construct the rate law expression for the hydrogen-bromine reaction by performing the following steps: a. Write down the differential rate expression for \([\mathrm{HBr}]\) b. Write down the differential rate expressions for \([\mathrm{Br} \cdot]\) and [H']. c. Because \(\mathrm{Br} \cdot(g)\) and \(\mathrm{H} \cdot(g)\) are reaction intermediates, apply the steady-state approximation to the result of part (b). d. Add the two equations from part (c) to determine [Br'] in terms of \(\left[\mathrm{Br}_{2}\right]\) e. Substitute the expression for \([\mathrm{Br} \cdot]\) back into the equation for \([\mathrm{H} \cdot]\) derived in part \((\mathrm{c})\) and solve for \([\mathrm{H} \cdot]\) f. Substitute the expressions for \([\mathrm{Br} \cdot]\) and \([\mathrm{H} \cdot]\) determined in part (e) into the differential rate expression for \([\mathrm{HBr}]\) to derive the rate law expression for the reaction.

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