Consider the reaction scheme A(g)+S k1⟹ A-S k2⟶ P(g) for which the rate law is v=k2 θA where θA

Consider the reaction scheme A(g)+S k1⟹ A-S k2⟶ P(g) for...

Consider the reaction scheme $$ \mathrm{A}(\mathrm{g})+\mathrm{S} \stackrel{k_{1}}{\Longrightarrow} \mathrm{A}-\mathrm{S} \stackrel{k_{2}}{\longrightarrow} \mathrm{P}(\mathrm{g}) $$ for which the rate law is $$ v=k_{2} \theta_{\mathrm{A}} $$ where $\theta_{\mathrm{A}}$ is the fraction of surface sites occupied by A molecules. Use the Langmuir adsorption isotherm (Equation 29.35) to obtain an expression for the reaction rate in terms of $K_{c}$ and $[\mathrm{A}]$. Under what conditions will the reaction be first order in the concentration of A?

Consider the reaction scheme
$$
\mathrm{A}(\mathrm{g})+\mathrm{S} \stackrel{k_{1}}{\Longrightarrow} \mathrm{A}-\mathrm{S} \stackrel{k_{2}}{\longrightarrow} \mathrm{P}(\mathrm{g})
$$
for which the rate law is
$$
v=k_{2} \theta_{\mathrm{A}}
$$
where $\theta_{\mathrm{A}}$ is the fraction of surface sites occupied by A molecules. Use the Langmuir adsorption isotherm (Equation 29.35) to obtain an expression for the reaction rate in terms of $K_{c}$ and $[\mathrm{A}]$. Under what conditions will the reaction be first order in the concentration of A?

Key Concepts

Langmuir Adsorption Isotherm

The Langmuir adsorption isotherm describes how the fraction of a surface covered by a substance depends on its concentration in the contacting phase and its adsorption equilibrium constant. It is typically expressed as ? = (Kc [A]) / (1 + Kc [A]), where ? is the fraction of occupied sites, [A] is the concentration of the adsorbing species, and Kc is the adsorption equilibrium constant. This relation results from assuming that adsorption occurs on a homogeneous surface with a fixed number of sites and that the adsorbed molecules do not interact with each other.

Adsorption Equilibrium Constant

The adsorption equilibrium constant, often denoted as Kc, quantifies the affinity between the adsorbate and the surface. It is defined by the ratio of the rate constants for adsorption and desorption and plays a critical role in determining the extent of surface coverage at a given adsorbate concentration. A larger Kc indicates stronger binding and a higher degree of surface coverage.

Surface Reaction Kinetics

Surface reaction kinetics involves the study of chemical reactions that occur on solid surfaces, such as in heterogeneous catalysis. In these reactions, the rate is often expressed in terms of the fraction of surface sites occupied by a reactant. Utilizing the Langmuir isotherm to express surface coverage enables the integration of adsorption equilibrium with kinetic rate expressions, linking the macroscopic reaction rate to molecular-level interactions.

Reaction Order

The reaction order describes how the rate of a reaction depends on the concentration of one or more reactants. In the context of a surface reaction governed by the Langmuir isotherm, the overall rate expression transitions from first order to fractional order depending on the adsorbate concentration. At low concentrations, where Kc [A] is much less than one, the rate expression simplifies to a form that is first order in [A], meaning the reaction rate is directly proportional to [A].

Transcript

00:01 All right so in this question we have tabulated data showing the initial concentration of a reactant and the rate of the reaction at different concentrations.

00:12 So the question says we should determine what the order of the reaction is, we should write the rate law of the reaction, and of course determine the rate constant and include that in the rate law expression.

00:27 So generally the rate law of that reaction will be r equals k, a raised to the power of x, where x is the order of the reaction, of the reaction with respect to a, and k is the rate constant, k is a rate constant, it's a rate constant, and r is the instantaneous rate of reaction.

01:03 We're going to use the values that we have been provided with, just enter each value.

01:10 When the rate of the reaction, when the rate of the reaction, let's say r1, was 0 .0078, the concentration of a, let's call that a1, was 0 .12 molar.

01:28 And by the way you can see that the rate is given to us in molar per second, molar per second.

01:36 Okay and then when the concentration a2 was 0 .16, the rate of the reaction, the rate of the reaction r2 was 0 .0104.

01:51 So we're going to put this into the expression that is 0 .0078 is equal to k into 0 .12 raised to the power of x, and 0 .0104 is equal to k into 0 .0104 raised to the power of x.

02:19 Divide both equations so that this and this will cancel and you have 0 .0078 divided by 0 .0104 is equal to 0 .12 divided by 0 .16.

02:39 It should be 0 .16, 0 .16 raised to the power of x.

02:50 So 0 .16, all of it raised to the power of x.

02:53 Okay so when you divide that you will have 0 .75 and this will be 0 .75 also.

03:11 Right so 0 .75 raised to the power of x.

03:14 Now this is raised to the power of one and since the bases are the same it means that x is equal to one.

03:21 So we can say that the order of the reaction is first order with respect to reactant a.

03:28 So that is the order of the reaction...

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