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Chapter 36: Problem 10

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

Short Answer

Expert verified
The final rate law expression for the given mechanism using the pre-equilibrium approximation is: \[R = k_2\left(\frac{ -[A_2]鈧� + \sqrt{[A_2]鈧�^2 + 4K[A_2]鈧�}}{ 4 }\right)[B]\] Here, R represents the rate of product (P) formation, [A2]鈧� and [B] denote the initial concentrations of A2 and B, respectively, K is the equilibrium constant for the first elementary reaction, and k2 is the rate constant for the second elementary reaction.

Step by step solution

01

Identify the reactions

The given mechanism has two elementary reactions: 1. \(A_2 \stackrel{k_1}{\rightleftharpoons} 2A\) 2. \(A + B \stackrel{k_2}{\longrightarrow} P\) Step 1 tells us that the given mechanism includes two elementary reactions. The first elementary reaction goes in both directions, and its rate constants are k1, as shown in the forward direction. The second elementary reaction goes in one direction, and its rate constant is k2.
02

Apply pre-equilibrium approximation

Using the assumption that equilibrium is reached for the first reaction, we can write the equilibrium constant, K, as: \[K = \frac{ [A]^2 }{ [A_2] }\]
03

Find the concentration of A

Let's denote the initial concentration of A2 as [A2]鈧� (since initially there's no A). When some amount x of A2 dissociates, the concentrations become [A2] = [A2]鈧� - x and [A] = 2x. Substitute these values in the equilibrium constant expression: \[K = \frac{ (2x)^2 }{ [A_2]鈧� - x }\] Solving for x, we get the concentration of A as a function of K and [A2]鈧�: \[x = \frac{ [A] }{ 2 }=\frac{ -[A_2]鈧� + \sqrt{[A_2]鈧�^2 + 4K[A_2]鈧�}}{ 4 }\]
04

Determine the rate of formation of P

The rate of formation of P, R, can be expressed as follows using the second elementary reaction: \[R = k_2[A][B]\] Now, substitute the expression for [A] from Step 3: \[R = k_2\left(\frac{ -[A_2]鈧� + \sqrt{[A_2]鈧�^2 + 4K[A_2]鈧�}}{ 4 }\right)[B]\]
05

Write the final rate law expression

The final rate law expression for the given mechanism using the pre-equilibrium approximation is: \[R = k_2\left(\frac{ -[A_2]鈧� + \sqrt{[A_2]鈧�^2 + 4K[A_2]鈧�}}{ 4 }\right)[B]\] In this expression, R is the rate of formation of P, [A2]鈧� is the initial concentration of A2, [B] is the concentration of B (which is also constant throughout the reaction), K is the equilibrium constant of the first elementary reaction, and k2 is the rate constant of the second elementary reaction.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the branch of physical chemistry that is concerned with understanding the rates of chemical reactions. It's essential for predicting how a reaction progresses over time. Basic principles of kinetics indicate that reaction rates depend on the concentration of reactants and the rate constant, a parameter that is influenced by factors such as temperature and catalyst presence.

The rate law expression, as seen in the given exercise, demonstrates this relationship by indicating how the rate of product formation (R) changes in response to various concentrations of reactants. In our scenario, the final rate law was derived using both the pre-equilibrium approximation鈥攚hich assumes a fast initial step reaching equilibrium鈥攁nd the stoichiometry of the elementary reactions involved.
Reaction Mechanisms
A reaction mechanism is a step-by-step sequence of elementary reactions by which overall chemical change occurs. Understanding the mechanism is crucial for developing the rate law, which is the mathematical relationship between the rate of a reaction and the concentration of reactants.

In the exercise, the mechanism is a two-step process involving the dissociation of a molecule of A2 into two A atoms, followed by the reaction of A with B to form product P. The rate law, therefore, is a reflection of these steps, and the pre-equilibrium approximation is utilized to simplify the expression, by treating the fast initial step as if it has reached equilibrium.
Equilibrium Constant
The equilibrium constant (K) is a numerical value that represents the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to the power of its coefficient in the balanced equation. This is only applicable for reversible reactions and is a direct measure of the extent of the reaction at equilibrium.

For the given exercise, the equilibrium constant is used to relate the concentrations of the reactant A2 with the intermediate A. By applying the preequilibrium approximation, we can use K to find out the concentration of A available for the second step of the reaction. To solve for the concentration of A, we apply algebraic manipulation as shown in step 3 of the solution.
Elementary Reactions
Elementary reactions are the simplest reaction steps in a mechanism, each involving a single event or collision. They can be unimolecular, involving the rearrangement or dissociation of a single molecule, or bimolecular, involving two reacting particles. The overall reaction rate can be influenced by both the rate constants of elementary steps and their molecularity.

In the textbook problem, we have two elementary reactions, with the first being a reversible dissociation of A2 into A, and the second being the combination of A and B to form P. The rate law for elementary reactions is typically derived from their molecularity and is a critical step in understanding the full reaction mechanism.

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Most popular questions from this chapter

The chlorination of vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}+\) \(\mathrm{Cl}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3},\) is believed to proceed by the following mechanism: \\[ \begin{array}{l} \mathrm{Cl}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{Cl} \\ \mathrm{Cl} \cdot+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{2} \\ \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{2} \cdot+\mathrm{Cl}_{2} \stackrel{k_{3}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3}+\mathrm{Cl} \\ \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{2} \cdot+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{2} \cdot \stackrel{k_{4}}{\longrightarrow} \text { stable species } \end{array} \\] Derive the rate law expression for the chlorination of vinyl chloride based on this mechanism.

Consider the following mechanism, which results in the formation of product \(P:\) \\[ \begin{array}{l} \mathrm{A} \stackrel{k_{1}}{\rightleftharpoons_{k-1}} \mathrm{B} \frac{k_{2}}{\rightleftharpoons_{-2}} \mathrm{C} \\ \mathrm{B} \stackrel{k_{3}}{\rightarrow} \mathrm{P} \end{array} \\] If only the species \(A\) is present at \(t=0,\) what is the expression for the concentration of \(\mathrm{P}\) as a function of time? You can apply the preequilibrium approximation in deriving your answer.

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:$$R_{2}(g)+2 M(\text {surface}) \stackrel{k_{a}}{\rightleftharpoons_{k}} 2 R M(\text {surface})$$

Consider the formation of double-stranded (DS) DNA from two complementary single strands (S and S') through the following mechanism involving an intermediate helix (IH): \\[ \begin{array}{l} \mathrm{S}+\mathrm{S}^{\prime} \frac{k_{1}}{\mathrm{k}_{-1}} \mathrm{IH} \\ \mathrm{IH} \stackrel{k_{2}}{\longrightarrow} \mathrm{DS} \end{array} \\] a. Derive the rate law expression for this reaction employing the preequilibrium approximation. b. What is the corresponding rate-law expression for the reaction employing the steady state approximation for the intermediate IH?

a. For the hydrogen-bromine reaction presented in Problem P36.7 imagine initiating the reaction with only \(\mathrm{Br}_{2}\) and \(\mathrm{H}_{2}\) present. Demonstrate that the rate law expression at \(t=0\) reduces to \\[ \left(\frac{d[\mathrm{HBr}]}{d t}\right)_{t=0}=2 k_{2}\left(\frac{k_{1}}{k_{-1}}\right)^{1 / 2}\left[\mathrm{H}_{2}\right]_{0}\left[\mathrm{Br}_{2}\right]^{1 / 2} \\] b. The activation energies for the rate constants are as follows: $$\begin{array}{cc} \text { Rate Constant } & \Delta \boldsymbol{E}_{a}(\mathbf{k J} / \mathbf{m o l}) \\ \hline k_{1} & 192 \\ k_{2} & 0 \\ k_{-1} & 74 \end{array}$$ c. How much will the rate of the reaction change if the temperature is increased to \(400 .\) K from \(298 \mathrm{K} ?\)
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