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Chapter 14: Problem 109

The gas-phase decomposition of ozone is thought to occur by the following two- step mechanism. Step \(1: \quad \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{O}_{2}(g)+\mathrm{O}(g)\) (fast) Step \(2: \quad \mathrm{O}(g)+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{O}_{2}(g)\) (slow) (a) Write the balanced equation for the overall reaction. (b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.) (c) Is \(\mathrm{O}\) a catalyst or an intermediate? (d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

Short Answer

Expert verified
The balanced equation for the overall reaction is \(2 O_{3}(g) \longrightarrow 3 O_{2}(g)\). The rate law for the given two-step mechanism is \(\textrm{rate} = \frac{k_2 K}{[O_2]} [O_3]^2\). O is an intermediate in this reaction mechanism. If the reaction occurred in a single step, the rate law would change to \(\textrm{rate} = k [O_3]^2\).

Step by step solution

01

(a) Balanced Equation for the Overall Reaction

To find the balanced equation for the overall reaction, we add both steps and find the net equation. \( Step 1: O_{3}(g) \rightleftharpoons O_{2}(g) + O(g) \) (fast) \( Step 2: O(g) + O_{3}(g) \longrightarrow 2 O_{2}(g) \) (slow) Now, add both steps and cancel out the common species on both sides of the equation. In this case, we have the O atom appearing in both steps. \( O_{3}(g) + O(g) + O_{3}(g) \longrightarrow O_{2}(g) + O(g) + 2 O_{2}(g) \) Now we cancel out the O atoms \( 2 O_{3}(g) \longrightarrow 3 O_{2}(g) \) This is the balanced equation for the overall reaction.
02

(b) Deriving the Rate Law

To derive the rate law for this mechanism, we start by finding the rate law for each step individually. Step 1 is a fast equilibrium reaction and can be described using the equilibrium constant K: \( K = \frac{[O_{2}][O]}{[O_{3}]} \) \( [O] = \frac{[O_{3}]K}{[O_{2}]} \) Now we need to find the rate law for Step 2. This is a slow bimolecular reaction, so its rate law is: \( \textrm{rate}_2 = k_2 [O][O_{3}] \) Since we are interested in the overall rate, we substitute the expression for [O] from Step 1: \( \textrm{rate} = k_2 \left( \frac{[O_{3}]K}{[O_{2}]} \right) [O_{3}] \) Then we obtain: \( \textrm{rate} = \frac{k_2 K}{[O_2]} [O_3]^2 \) This is the rate law consistent with the given mechanism.
03

(c) Determining if O is a Catalyst or an Intermediate

In this problem, we can conclude that O is not a catalyst since it does not appear in the overall balanced equation and does not speed up the reaction by being consumed and regenerated during the course of the reaction. Instead, O is an intermediate, as it is formed in Step 1 and then consumed in Step 2.
04

(d) Would the Rate Law Change if the Reaction Occurred in a Single Step?

If the reaction occurred in a single step, the reaction would directly connect ozone with the products. \( 2 O_3(g) \longrightarrow 3 O_2(g) \) This single-step reaction suggests a bimolecular reaction between two O3 molecules to form the products. In this case, the rate law would be different from that of the two-step mechanism: \( \textrm{rate} = k [O_3]^2 \) Thus, the rate law would change if the reaction occurred in a single step.

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Key Concepts

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

Rate Law
The rate law is a mathematical expression that represents the speed or rate of a chemical reaction based on the concentration of its reactants. In the decomposition of ozone, the reaction occurs through a two-step mechanism. The rate law is based on the slowest step, also known as the rate-determining step. This is like finding the bottleneck in a process.When deriving the rate law for the given mechanism, we consider both steps. However, because Step 2 is the slow step, it determines the rate. The expression for this rate is derived from the concentrations of the reactants involved: - Since Step 2 involves an intermediate, we substitute its concentration using the equilibrium expression from Step 1:The final rate law for the ozone decomposition is: \( \text{rate} = \frac{k_2 K}{[O_2]} [O_3]^2 \)This shows how crucial it is to identify which part of the reaction mechanism dictates the speed of the overall reaction.
Intermediate
Intermediates are species that appear in the reaction steps but not in the overall balanced equation. They are formed in one step and consumed in another. In our ozone decomposition mechanism, the oxygen atom (O) plays the role of an intermediate. Here's how intermediates work: - In Step 1, ozone decomposes to produce molecular oxygen and an oxygen atom (O). This oxygen atom does not show up in the overall reaction but is crucial for Step 2. - It is quickly consumed in Step 2 by reacting with another ozone molecule to form more oxygen. Unlike catalysts, which are regenerated, intermediates do not appear in the net chemical equation. They are vital for understanding how the reaction progresses through its mechanism, helping us to hypothesize the steps that connect reactants to products.
Balanced Chemical Equation
A balanced chemical equation represents a reaction鈥檚 stoichiometry鈥攓uantitatively showing the relationship between reactants and products. For the decomposition of ozone, it involves combining the individual steps of the mechanism to eliminate intermediates and show the net transformation.Steps to balance the equation:- Start by writing each step: Step 1 produces an oxygen atom, and Step 2 consumes it.- Combine these steps and remove the oxygen atom from both sides since it appears as both a reactant and product but doesn't appear in the net reaction.The result is: \(2 O_3(g) \longrightarrow 3 O_2(g) \)This balanced equation clearly shows that two ozone molecules decompose to form three oxygen molecules, striking a balance between the number of atoms of each element on both sides of the equation. Balancing ensures that we respect the law of conservation of mass in a chemical reaction, where no atoms are lost or gained, merely rearranged.

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Most popular questions from this chapter

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\) (b) (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)\)

The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{array}{l} \mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ \mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{array} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) If the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction?

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: \(\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-} .\) This rapid reaction gives the following rate data: $$ \begin{array}{ccc} \hline\left[\mathrm{OCI}^{-}\right](M) & {\left[\mathrm{I}^{-}\right](M)} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1.5 \times 10^{-3} & 1.5 \times 10^{-3} & 1.36 \times 10^{-4} \\ 3.0 \times 10^{-3} & 1.5 \times 10^{-3} & 2.72 \times 10^{-4} \\ 1.5 \times 10^{-3} & 3.0 \times 10^{-3} & 2.72 \times 10^{-4} \\ \hline \end{array} $$ (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{OCl}^{-}\right]=2.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=5.0 \times 10^{-4} \mathrm{M}\)

Consider the following hypothetical aqueous reaction: \(\mathrm{A}(a q) \rightarrow \mathrm{B}(a q)\). A flask is charged with \(0.065 \mathrm{~mol}\) of \(\mathrm{A}\) in a total volume of \(100.0 \mathrm{~mL}\). The following data are collected: $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 10 & 20 & 30 & 40 \\ \hline \text { Moles of A } & 0.065 & 0.051 & 0.042 & 0.036 & 0.031 \\ \hline \end{array} $$ (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that there are no molecules of \(\mathrm{B}\) at time zero and that A cleanly converts to B with no intermediates. (b) Calculate the average rate of disappearance of A for each 10 -min interval in units of \(M /\) s. (c) Between \(t=0 \mathrm{~min}\) and \(t=30 \mathrm{~min},\) what is the average rate of appearance of \(\mathrm{B}\) in units of \(\mathrm{M} / \mathrm{s}\) ? Assume that the volume of the solution is constant.

(a) What are the units usually used to express the rates of reactions occurring in solution? (b) As the temperature increases, does the reaction rate increase or decrease? (c) As a reaction proceeds, does the instantaneous reaction rate increase or decrease?
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