Problem 47
Question
What is the rate law for each of the following elementary reactions? (a) \(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\) (b) \(\mathrm{Cl}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{H}(\mathrm{g})\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(\mathrm{aq}) \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\)
Step-by-Step Solution
Verified Answer
(a) Rate = k[NO][NO₃]; (b) Rate = k[Cl][H₂]; (c) Rate = k[(CH₃)₃CBr].
1Step 1: Understanding Elementary Reactions
An elementary reaction is a single-step reaction with only one transition state, and its rate law can be directly written using the stoichiometric coefficients of the reactants. The rate of such a reaction is proportional to the concentration of each reactant raised to the power of its coefficient in the balanced equation.
2Step 2: Rate Law for Reaction (a)
For the reaction \( \mathrm{NO} + \mathrm{NO}_3 \rightarrow 2 \mathrm{NO}_2 \), the stoichiometric coefficients for the reactants \( \mathrm{NO} \) and \( \mathrm{NO}_3 \) are both 1. Thus, the rate law is given by: \[ \text{Rate} = k[\mathrm{NO}] [\mathrm{NO}_3]\] where \( k \) is the rate constant.
3Step 3: Rate Law for Reaction (b)
For the reaction \( \mathrm{Cl} + \mathrm{H}_2 \rightarrow \mathrm{HCl} + \mathrm{H} \), the stoichiometric coefficients for the reactants \( \mathrm{Cl} \) and \( \mathrm{H}_2 \) are both 1. Thus, the rate law is: \[\text{Rate} = k[\mathrm{Cl}] [\mathrm{H}_2]\] where \( k \) is the rate constant.
4Step 4: Rate Law for Reaction (c)
For the reaction \( (\mathrm{CH}_3)_3 \mathrm{CBr} \rightarrow (\mathrm{CH}_3)_3 \mathrm{C}^+ + \mathrm{Br}^- \), the reactant \( (\mathrm{CH}_3)_3 \mathrm{CBr} \) has a stoichiometric coefficient of 1. Thus, the rate law is: \[ \text{Rate} = k[(\mathrm{CH}_3)_3 \mathrm{CBr}]\] where \( k \) is the rate constant.
Key Concepts
Rate LawReaction KineticsStoichiometric Coefficients
Rate Law
The rate law is a mathematical equation that describes the speed of a chemical reaction. For elementary reactions, this law is straightforward, as it can be directly determined from the stoichiometric coefficients of the reactants in the reaction equation. This is because each elementary reaction represents a single-step process following the specified molecular interactions.
To write a rate law for any elementary reaction, follow these steps:
To write a rate law for any elementary reaction, follow these steps:
- Identify the reactants involved in the step.
- Determine their stoichiometric coefficients from the balanced equation.
- Express the rate law using these coefficients as exponents for each reactant concentration.
Reaction Kinetics
Reaction kinetics is the study of the rates at which chemical processes occur and the factors that affect these rates. It looks at how quickly reactants are converted into products, and the sequence of steps that reactants go through, known as the reaction mechanism.
In elementary reactions, reaction kinetics can be quite simple. Since these processes occur in a single step, the rate-limiting factor typically conforms directly to the concentrations of the reactants as expressed in the rate law. By understanding the kinetics of a reaction, chemists can design processes to optimize production rates or develop strategies to control biological systems.
For example, in the reaction \( \mathrm{Cl} + \mathrm{H}_2 \rightarrow \mathrm{HCl} + \mathrm{H} \), kinetics tells us that the rate is proportional to the concentration of \( \mathrm{Cl} \) and \( \mathrm{H}_2 \), thus the reaction can be influenced by altering these concentrations or temperature.
In elementary reactions, reaction kinetics can be quite simple. Since these processes occur in a single step, the rate-limiting factor typically conforms directly to the concentrations of the reactants as expressed in the rate law. By understanding the kinetics of a reaction, chemists can design processes to optimize production rates or develop strategies to control biological systems.
For example, in the reaction \( \mathrm{Cl} + \mathrm{H}_2 \rightarrow \mathrm{HCl} + \mathrm{H} \), kinetics tells us that the rate is proportional to the concentration of \( \mathrm{Cl} \) and \( \mathrm{H}_2 \), thus the reaction can be influenced by altering these concentrations or temperature.
Stoichiometric Coefficients
Stoichiometric coefficients are numbers placed in front of compounds in a balanced chemical equation. They represent the ratio of moles of a reactant or product involved in the reaction. In the context of an elementary reaction, these coefficients not only help balance the equation but also dictate the form of the rate law.
In the reaction \( (\mathrm{CH}_3)_3 \mathrm{CBr} \rightarrow (\mathrm{CH}_3)_3 \mathrm{C}^+ + \mathrm{Br}^- \), the stoichiometric coefficient is 1 for the reactant \((\mathrm{CH}_3)_3 \mathrm{CBr}\). This informs us that for its rate law, the concentration term of the reactant is raised to the power of 1: \( \text{Rate} = k[(\mathrm{CH}_3)_3 \mathrm{CBr}] \).
Understanding these coefficients is crucial for predicting how changes in concentrations will affect the reaction rate, and thus these coefficients are fundamental for creating efficient reaction plans in both laboratory and industrial settings.
In the reaction \( (\mathrm{CH}_3)_3 \mathrm{CBr} \rightarrow (\mathrm{CH}_3)_3 \mathrm{C}^+ + \mathrm{Br}^- \), the stoichiometric coefficient is 1 for the reactant \((\mathrm{CH}_3)_3 \mathrm{CBr}\). This informs us that for its rate law, the concentration term of the reactant is raised to the power of 1: \( \text{Rate} = k[(\mathrm{CH}_3)_3 \mathrm{CBr}] \).
Understanding these coefficients is crucial for predicting how changes in concentrations will affect the reaction rate, and thus these coefficients are fundamental for creating efficient reaction plans in both laboratory and industrial settings.
Other exercises in this chapter
Problem 42
If the rate constant for a reaction triples when the temperature rises from \(3.00 \times 10^{2} \mathrm{K}\) to \(3.10 \times 10^{2} \mathrm{K},\) what is the
View solution Problem 44
When heated, cyclopropane is converted to propene (see Example 15.5 ). Rate constants for this reaction at \(470^{\circ} \mathrm{C}\) and \(510^{\circ} \mathrm{
View solution Problem 48
What is the rate law for each of the following elementary reactions? (a) \(\mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \longrightarrow \mathrm{I}(\mathrm{g
View solution Problem 49
Ozone, \(\mathrm{O}_{3},\) in the earth's upper atmosphere decomposes according to the equation $$2 \mathrm{O}_{3}(\mathrm{g}) \longrightarrow 3 \mathrm{O}_{2}(
View solution