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Determine the predicted rate law expression for the following radical-chain reaction: \\[ \begin{array}{l} \mathrm{A}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{A} \\ \mathrm{A} \cdot \stackrel{k_{2}}{\longrightarrow} \mathrm{B} \cdot+\mathrm{C} \end{array} \\] $$\begin{array}{l} \mathrm{A} \cdot+\mathrm{B} \cdot \stackrel{k_{3}}{\longrightarrow} \mathrm{P} \\\ \mathrm{A} \cdot+\mathrm{P} \stackrel{k_{4}}{\rightarrow} \mathrm{B} \end{array}$$

Short Answer

Expert verified
The predicted rate law expression for the given radical-chain reaction mechanism is given by Rate = \(k_1 [A_2]\). This is based on the identification of the first step in the reaction as the rate-determining step and considering the formation and consumption rates of A radicals in the overall reaction steps.

Step by step solution

01

Understand the given reaction mechanisms

The reaction occurs in the following steps: 1. \(A_2 \stackrel{k_{1}}{\longrightarrow} 2A\): The molecule A2 decomposes into two A radicals with rate constant \(k_1\). 2. \(A\cdot \stackrel{k_{2}}{\longrightarrow} B\cdot + C\): A radical reacts to form B radical and C with rate constant \(k_2\). 3. \(A\cdot + B\cdot \stackrel{k_{3}}{\longrightarrow} P\): An A radical reacts with a B radical to form the product P with rate constant \(k_3\). 4. \(A\cdot + P \stackrel{k_{4}}{\rightarrow} B\): An A radical reacts with the product P to form B with rate constant \(k_4\).
02

Identify the rate-determining step

A rate-determining step is the slowest step in a chemical reaction. In our case, the rate-determining step is the first step of the reaction: \(A_2 \stackrel{k_{1}}{\longrightarrow} 2A\). This is because the formation of A radicals is the initial and limiting step of the reaction.
03

Determine the rate law expression for the rate-determining step

The rate law expression can be determined using the following equation: Rate = \(k_{1} [A_2]\) Since the rate-determining step is the first step, the rate law expression will be governed by the concentration of A2 and the rate constant \(k_1\).
04

Identify the overall rate law expression for the reaction

To find the overall rate law expression, we first need to take into account all radicals formed during the reaction process. Notice that A radicals are formed in the first step and then consumed in the second, third, and fourth steps. Therefore, taking into account the formation and consumption of A radicals, we can write the overall rate law expression as: Rate = \(k_1 [A_2] = k_2 [A\cdot] = k_3 [A\cdot][B\cdot] = k_4 [A\cdot][P]\)
05

Finalize the predicted rate law expression

Since the rate-determining step is the first step and in the overall reaction, it is represented by \(k_1 [A_2]\), the rate law expression for this radical-chain reaction is: Rate = \(k_1 [A_2]\)

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Key Concepts

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

Radical-chain reaction
In a radical-chain reaction, molecules undergo a series of steps where radicals are generated and then react with stable molecules. Radicals are highly reactive species that have unpaired electrons. These reactions are common in processes like polymerization and combustion. Each radical-chain reaction typically consists of three main steps:
  • Initiation: This step involves the formation of radicals from non-radical species. In the exercise, \(A_2\) decomposes into two \(A\) radicals.
  • Propagation: Radicals react with stable molecules to generate new radicals. For example, \(A\cdot\) reacts to form \(B\cdot\) and \(C\), keeping the chain going.
  • Termination: Two radicals combine to produce a stable molecule, ending the chain reaction. This could involve the reaction of \(A\cdot\) radicals in combination with others in different ways.
This sequence of steps results in the amplification of reaction progress, as each radical formed can lead to more chain reactions. Understanding radical-chain reactions helps in modeling complex chemical processes and in designing ways to control them.
Rate-determining step
The rate-determining step (RDS) is the slowest step in a reaction mechanism and determines the overall rate of the reaction. In our exercise, the decomposition of \(A_2\) to form two \(A\) radicals is identified as the rate-determining step. This step is crucial because it sets the pace at which all subsequent steps can occur.
The significance of the rate-determining step lies in:
  • Governing the speed of the overall reaction.
  • Limiting the concentration of intermediates or products generated.
  • Determining the form of the rate law expression.
Because it controls the reaction rate, the rate law is often formulated based on the reactants involved in the rate-determining step. In this example, the rate law depends on the concentration of \([A_2]\) and the rate constant \(k_1\), highlighting the importance of understanding the slowest step in multi-step reactions.
Reaction mechanism
A reaction mechanism is a step-by-step description of how a chemical reaction occurs at the molecular level. It illustrates the sequence of steps that lead from reactants to products, detailing the formation and breaking of bonds over the course of the reaction. Each step in a mechanism is characterized by:
  • The specific intermediate species involved, like \(A\cdot\) and \(B\cdot\) radicals.
  • The rate constants that quantify the speed of each step, such as \(k_1\), \(k_2\), etc.
  • The transition states that represent the highest energy configurations the reactants must overcome.
The exercise outlines a mechanism that includes initiation, propagation, and termination stages, typical of radical-chain reactions. Understanding reaction mechanisms allows chemists to predict the behavior of the reaction under different conditions and can aid in the development of catalysts to increase efficiency.
Chemical kinetics
Chemical kinetics involves the study of reaction rates and the factors affecting them. Through kinetics, scientists learn how different variables, like concentration, temperature, and catalysts, influence the speed of reactions. Key components of chemical kinetics include:
  • Rate laws, which relate the rate of a reaction to the concentration of its reactants.
  • Rate constants (e.g., \(k_1, k_2\)), which indicate the intrinsic speed of reaction steps under given conditions.
  • Reaction order, which tells the power to which the reactant concentrations are raised in the rate law.
In the exercise, kinetic principles are used to deduce the rate law expression of the radical-chain reaction. We derived the equation \( \text{Rate} = k_1 [A_2] \) from the slowest step, showcasing how kinetics helps in predicting and controlling chemical reactions. By understanding the kinetics of a reaction, scientists can manipulate conditions to optimize reaction rates for industrial or laboratory processes.

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Most popular questions from this chapter

The overall reaction for the halogenation of a hydrocarbon (RH) using Br as the halogen is \(\mathrm{RH}(g)+\mathrm{Br}_{2}(g) \rightarrow\) \(\operatorname{RBr}(g)+\operatorname{HBr}(g) .\) The following mechanism has been proposed for this process: \\[ \begin{array}{l} \operatorname{Br}_{2}(g) \stackrel{k_{1}}{\longrightarrow} 2 \operatorname{Br} \cdot(g) \\ \operatorname{Br} \cdot(g)+\operatorname{RH}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{R} \cdot(g)+\operatorname{HBr}(g) \\\ \mathrm{R} \cdot(g)+\operatorname{Br}_{2}(g) \stackrel{k_{3}}{\longrightarrow} \operatorname{RBr}(g)+\operatorname{Br} \cdot(g) \\ \operatorname{Br} \cdot(g)+\mathrm{R} \cdot(g) \stackrel{k_{4}}{\longrightarrow} \operatorname{RBr}(g) \end{array} \\] Determine the rate law predicted by this mechanism.

Many surface reactions require the adsorption of two or more different gases. For the case of two gases, assuming that the adsorption of a gas simply limits the number of surface sites available for adsorption, derive expressions for the fractional coverage of each gas.

A proposed mechanism for the formation of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) from \(\mathrm{NO}_{2}(g)\) and \(\mathrm{O}_{3}(g)\) is \\[ \begin{array}{c} \mathrm{NO}_{2}(g)+\mathrm{O}_{3}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{NO}_{3}(g)+\mathrm{O}_{2}(g) \\ \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)+\mathrm{M}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{M}(g) \end{array} \\] Determine the rate law expression for the production of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) given this mechanism.

Consider the following mechanism for ozone thermal decomposition: \\[ \begin{array}{l} \mathrm{O}_{3}(g) \stackrel{k_{1}}{\rightleftharpoons} \mathrm{O}_{2}(g)+\mathrm{O}(g) \\ \mathrm{O}_{3}(g)+\mathrm{O}(g) \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{O}_{2}(g) \end{array} \\] a. Derive the rate law expression for the loss of \(\mathrm{O}_{3}(g)\) b. Under what conditions will the rate law expression for \(\mathrm{O}_{3}(g)\) decomposition be first order with respect to \(\mathrm{O}_{3}(g) ?\)

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.

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