Problem 102
Question
Write the rate laws for the following elementary steps and identify them as uni-, bi-, or termolecular steps: a. \(\mathrm{Cl}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g)\) b. \(2 \mathrm{NO}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) \(^{\bullet} \mathrm{c} .^{14} \mathrm{C} \rightarrow_{7}^{14} \mathrm{N}+_{-1}^{0} \beta\)
Step-by-Step Solution
Verified Answer
Question: Identify the type of each elementary step and write the corresponding rate law.
a. Cl(g) + O3(g) -> ClO(g) + O2(g)
b. 2 NO2(g) -> N2O4(g)
c. \(^{14}_7 C \rightarrow^{14}_7 N +^{-1}_0 \beta\)
Answer:
a. Bimolecular step, Rate = k[Cl][O3]
b. Bimolecular step, Rate = k[NO2]^2
c. Unimolecular step, Rate = k[\(^{14}_7 C\)]
1Step 1: a. Cl(g) + O3(g) -> ClO(g) + O2(g)
In this elementary step, one molecule of Cl(g) and one molecule of O3(g) react with each other to produce one molecule of ClO(g) and one molecule of O2(g). So, the overall rate law will be represented by:
Rate = k[Cl][O3]
Since there are two reactant molecules involved in this step, we classify it as a bimolecular step.
2Step 2: b. 2 NO2(g) -> N2O4(g)
In this elementary step, two molecules of NO2(g) react with each other to produce one molecule of N2O4(g). So, the overall rate law will be represented by:
Rate = k[NO2]^2
Since there are two reactant molecules of the same species involved in this step, we classify it as a bimolecular step.
3Step 3: c. \(^{14}_7 C \rightarrow^{14}_7 N +^{-1}_0 \beta\)
In this elementary step, one atom of \(^{14}_7 C\) decays, producing one atom of \(^{14}_7 N\) and one beta particle (\(^{-1}_0 \beta\)). Since only one reactant particle is involved in this step, the overall rate law will be represented by:
Rate = k[\(^{14}_7 C\)]
As there is only one reactant particle involved in this step, we classify it as a unimolecular step.
Key Concepts
Elementary StepsMolecularityChemical Kinetics
Elementary Steps
Elementary steps are crucial components in chemical reaction mechanisms. Each elementary step represents a simple reaction, where reactants directly form products without any intermediates or complex stages. They're like building blocks of a reaction, depicting the actual collision events happening in the reaction at a molecular level.
In the context of rate laws, each elementary step has a clear relationship between the concentration of the reactants and the speed at which the reaction occurs. This relationship is expressed as the rate law, which is generally determined directly by the stoichiometry of the reactants in the step.
For instance, consider the elementary step \[Cl(g) + O_3(g) \rightarrow ClO(g) + O_2(g)\]Here, the rate law given by the stoichiometry is \[Rate = k[Cl][O_3]\]This expression implies that the rate of reaction is directly proportional to the concentrations of Cl and O_3. Thus, elementary steps provide essential insights into the detailed dynamics of how reactions progress.
In the context of rate laws, each elementary step has a clear relationship between the concentration of the reactants and the speed at which the reaction occurs. This relationship is expressed as the rate law, which is generally determined directly by the stoichiometry of the reactants in the step.
For instance, consider the elementary step \[Cl(g) + O_3(g) \rightarrow ClO(g) + O_2(g)\]Here, the rate law given by the stoichiometry is \[Rate = k[Cl][O_3]\]This expression implies that the rate of reaction is directly proportional to the concentrations of Cl and O_3. Thus, elementary steps provide essential insights into the detailed dynamics of how reactions progress.
Molecularity
Molecularity refers to the number of reactant particles involved in an elementary step. It's an important concept in classifying elementary reactions.
There are three main types of molecularity:
There are three main types of molecularity:
- **Unimolecular**: Involves a single reactant particle breaking down to form products. An example is \[^{14}_7 C \rightarrow^{14}_7 N +^{-1}_0 \beta\]This step has a molecularity of one, making it unimolecular.
- **Bimolecular**: Involves two reactant particles. These could be two of the same kind or different. For example, in \[2NO_2(g) \rightarrow N_2O_4(g)\]two NO2 molecules participate, classifying the step as bimolecular.
- **Termolecular**: Involves three reactant particles. This is less common due to the low probability of three particles colliding simultaneously.
Chemical Kinetics
Chemical kinetics is a branch of chemistry concerned with the rates of chemical reactions and the factors affecting these rates. Understanding kinetics is vital to comprehending how and why reactions occur at the speed they do.
This field covers several key areas:
This field covers several key areas:
- **Rate Laws**: These are mathematical expressions derived from experimental data that indicate the relationship between the concentration of reactants and the reaction rate. For elementary reactions, these laws can often be deduced from the stoichiometry of the reactants involved.
- **Reaction Mechanisms**: These are detailed descriptions of the steps through which reactants transform into products. Each step in a mechanism is represented as an elementary step.
- **Factors Affecting Reaction Rates**: Temperature, concentration, and catalysts are primary influences that can accelerate or decelerate a reaction.
Other exercises in this chapter
Problem 99
The hypothetical reaction \(\mathrm{A} \rightarrow \mathrm{B}\) has an activation energy of \(50.0 \mathrm{kJ} / \mathrm{mol} .\) Draw a reaction profile for ea
View solution Problem 101
Write the rate laws for the following elementary steps and identify them as uni-, bi-, or termolecular steps: a. \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightarro
View solution Problem 103
A common classroom demonstration of a reaction involves mixing \(30 \%\) hydrogen peroxide with a solution of potassium iodide. The following mechanism has been
View solution Problem 104
A proposed mechanism for the reaction of \(\mathrm{NO}_{2}(g)\) and \(\mathrm{CO}(g)\) is $$\begin{array}{ll}\text { Step 1 } & 2 \mathrm{NO}_{2}(g) \rightarrow
View solution