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Chapter 13: Problem 117

Nitryl bromide, \(\mathrm{NO}_{2} \mathrm{Br}\), decomposes into nitrogen dioxide and bromine. $$ 2 \mathrm{NO}_{2} \mathrm{Br}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)+\mathrm{Br}_{2}(g) $$ A proposed mechanism is $$ \begin{array}{r} \mathrm{NO}_{2} \mathrm{Br} \longrightarrow \mathrm{NO}_{2}+\mathrm{Br} \\ \mathrm{NO}_{2} \mathrm{Br}+\mathrm{Br} \longrightarrow \mathrm{NO}_{2}+\mathrm{Br}_{2} \end{array} $$ Write the rate law predicted by this mechanism.

Short Answer

Expert verified
The rate law predicted by the mechanism is \( \text{Rate} = k[\mathrm{NO}_{2} \mathrm{Br}] \).

Step by step solution

01

Identify the Rate-Determining Step

The rate-determining step is the slowest step in the reaction mechanism and dictates the speed of the overall reaction. In this case, it's the first reaction: \( \mathrm{NO}_{2} \mathrm{Br} \rightarrow \mathrm{NO}_{2} + \mathrm{Br} \).
02

Determine the Rate Expression

For the rate-determining step, the rate expression is based on the reactants involved. Here, the concentration of \( \mathrm{NO}_{2} \mathrm{Br} \) governs the rate of the reaction. Hence, the rate expression is: \( \text{Rate} = k[\mathrm{NO}_{2} \mathrm{Br}] \) where \( k \) is the rate constant.
03

Write the Rate Law

From the rate expression obtained in the previous step, the rate law for the reaction based on the proposed mechanism is \( \text{Rate} = k[\mathrm{NO}_{2} \mathrm{Br}] \). This suggests that the concentration of nitryl bromide is directly proportional to the rate of reaction.

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Key Concepts

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

Reaction Mechanism
Chemical reactions often occur through a sequence of simpler steps known as a reaction mechanism. These steps provide a detailed pathway for how reactants are converted into products. The mechanism can consist of one or more elementary reactions.
Understanding the mechanism is crucial as it helps to explain the kinetic behavior of the entire reaction.
  • In the decomposition of nitryl bromide, the mechanism involves two steps.
  • The first step, where nitryl bromide dissociates into nitrogen dioxide and a bromine atom, is typically slower and crucial for understanding the reaction rate.
  • The second step involves the bromine atom reacting with another molecule of nitryl bromide to form bromine gas and another molecule of nitrogen dioxide.
Each step in the mechanism contributes to the formation of the final products, but not all steps contribute equally to the rate of the reaction.
Rate Law
The rate law provides a mathematical expression that relates the rate of a chemical reaction to the concentration of its reactants. By developing a rate law from a proposed mechanism, chemists predict how changes in concentration can affect the speed of a reaction.
A rate law typically takes the form: \[ \text{Rate} = k[A]^m[B]^n \] where \(k\) is the rate constant, and \([A]\) and \([B]\) are the molar concentrations of the reactants raised to some power, \(m\) and \(n\), respectively.
  • For the decomposition of nitryl bromide, the rate law derived from the mechanism is \( \text{Rate} = k[\mathrm{NO}_2\mathrm{Br}] \).
  • This indicates the reaction rate depends solely on the concentration of nitryl bromide.
  • The rate constant \(k\) is unique to each reaction under specific conditions such as temperature.
By applying the rate law, one can predict the speed at which a reaction proceeds based on initial concentrations.
Rate-Determining Step
The rate-determining step in a chemical mechanism is the slowest step, akin to a bottleneck in a process. Understanding this step helps identify which part of the mechanism controls the overall reaction rate.
The decomposition of nitryl bromide provides a clear example:
  • The first step: \(\mathrm{NO}_2\mathrm{Br} \rightarrow \mathrm{NO}_2 + \mathrm{Br}\) is slower in comparison to the second step.
  • Because this first step sets the pace for the entire process, it is labeled the rate-determining step.
By analyzing the rate-determining step, it's possible to make informed predictions about the reaction's overall speed and its dependence on reactant concentrations.
This step is crucial for calculating the rate law accurately since the rate of the whole reaction is governed by it.
Decomposition Reaction
Decomposition reactions involve breaking down a compound into simpler substances or elements. This class of reactions is fundamental in chemistry for understanding how compounds interact and change.
In the context of nitryl bromide decomposition:
  • The compound \(\mathrm{NO}_2\mathrm{Br}\) breaks down into nitrogen dioxide \(\mathrm{NO}_2\) and bromine gas \(\mathrm{Br}_2\).
  • Such reactions are typically characterized by a single reactant that splits into two or more products.
  • These reactions can be driven by various factors, including heat, light, or the presence of a catalyst.
Decomposition reactions are ubiquitous in both nature and industry, playing roles in processes ranging from metabolic pathways to the production of essential chemicals.

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Most popular questions from this chapter

By means of an example, explain what is meant by the term reaction intermediate.

Consider the reaction \(3 \mathrm{~A} \longrightarrow 2 \mathrm{~B}+\mathrm{C}\). a. One rate expression for the reaction is Rate of formation of \(\mathrm{C}=+\frac{\Delta[\mathrm{C}]}{\Delta t}\) Write two other rate expressions for this reaction in this form. b. Using your two rate expressions, if you calculated the average rate of the reaction over the same time interval, would the rates be equal? c. If your answer to part b was no, write two rate expressions that would give an equal rate when calculated over the same time interval.

The rate constant for a certain reaction is \(1.4 \times\) \(10^{-5} M^{-1} \min ^{-1}\) at \(483 \mathrm{~K}\). The activation energy for the reaction is \(2.11 \times 10^{3} \mathrm{~J} / \mathrm{mol}\). What is the rate constant for the reaction at \(611 \mathrm{~K} ?\)

Sketch a potential-energy diagram for the reaction of nitric oxide with ozone. $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ The activation energy for the forward reaction is \(10 \mathrm{~kJ} ;\) the \(\Delta H^{\circ}\) is \(-200 \mathrm{~kJ}\). What is the activation energy for the reverse reaction? Label your diagram appropriately. {-}$

Iron(III) chloride is reduced by tin(II) chloride. $$ 2 \mathrm{FeCl}_{3}(a q)+\mathrm{SnCl}_{2}(a q) \longrightarrow 2 \mathrm{FeCl}_{2}(a q)+\mathrm{SnCl}_{4}(a q) $$ The concentration of \(\mathrm{Fe}^{3+}\) ion at the beginning of an experiment was \(0.03586 M\). After \(4.00 \mathrm{~min}\), it was \(0.02715 \mathrm{M}\). What is the average rate of reaction of \(\mathrm{FeCl}_{3}\) in this time interval?
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