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Chapter 14: Problem 70

A simplified rate law for the reaction \(2 \mathrm{O}_{3}(\mathrm{g}) \longrightarrow\) \(3 \mathrm{O}_{2}(\mathrm{g})\) is $$\text { rate }=k=\frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]}$$ For this reaction, propose a two-step mechanism that consists of a fast, reversible first step, followed by a slow second step.

Short Answer

Expert verified
The proposed two-step mechanism consists of a fast, reversible first step: \(O_{3} \longleftrightarrow O_{2} + O\), and a slow second step \(O_{3} + O -> 2 O_{2}\)

Step by step solution

01

Identify Rate Law Expression and Given Reaction

The given rate law expression is rate = \(k \left[\mathrm{O}_{3}\right]^{2}/\left[\mathrm{O}_{2}\right]\). And the overall reaction is \(2 O_{3}(g) → 3 O_{2}(g)\).
02

Propose the First Step

Propose a fast reversible step that matches with the rate law expression. Thus, the first step could be \(O_{3} \longleftrightarrow O_{2} + O\). This step is fast and reversible, meeting the requirements in the question.
03

Propose the Second Step

After the first step, the O radical should combine with another O3 to form O2; this must be the second step and is slow because the radical needs to collide with another O3 molecule. Thus, the second step could be \(O_{3} + O -> 2 O_{2}\). This step is slow, so it acts as the rate-determining step in this reaction mechanism.

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Most popular questions from this chapter

If even a tiny spark is introduced into a mixture of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g}),\) a highly exothermic explosive reaction occurs. Without the spark, the mixture remains unreacted indefinitely. (a) Explain this difference in behavior. (b) Why is the nature of the reaction independent of the size of the spark?

The following three-step mechanism has been proposed for the reaction of chlorine and chloroform. $$\begin{aligned} & \text { (1) } \quad \mathrm{Cl}_{2}(\mathrm{g}) \stackrel{k_{1}}{\rightleftharpoons_{k-1}} 2 \mathrm{Cl}(\mathrm{g})\\\ & \text { (2) } \quad \mathrm{Cl}(\mathrm{g})+\mathrm{CHCl}_{3}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{HCl}(\mathrm{g})+\mathrm{CCl}_{3}(\mathrm{g})\\\ &\text { (3) } \quad \mathrm{CCl}_{3}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) \stackrel{k_{3}}{\longrightarrow} \mathrm{CCl}_{4}(\mathrm{g}) \end{aligned}$$ The numerical values of the rate constants for these steps are \(k_{1}=4.8 \times 10^{3} ; \quad k_{-1}=3.6 \times 10^{3} ; \quad k_{2}=1.3 \times 10^{-2} ; k_{3}=2.7 \times 10^{2} .\) Derive the rate law and the magnitude of \(k\) for the overall reaction.

In the reaction \(A(g) \longrightarrow 2 B(g)+C(g),\) the total pressure increases while the partial pressure of \(\mathrm{A}(\mathrm{g})\) decreases. If the initial pressure of \(\mathrm{A}(\mathrm{g})\) in a vessel of constant volume is \(1.000 \times 10^{3} \mathrm{mmHg}\) (a) What will be the total pressure when the reaction has gone to completion? (b) What will be the total gas pressure when the partial pressure of \(\mathrm{A}(\mathrm{g})\) has fallen to \(8.00 \times 10^{2} \mathrm{mmHg} ?\)

The first-order reaction \(A \longrightarrow\) products has \(t_{1 / 2}=180 \mathrm{s}\) (a) What percent of a sample of A remains unreacted \(900 \mathrm{s}\) after a reaction has been started? (b) What is the rate of reaction when \([\mathrm{A}]=0.50 \mathrm{M} ?\)

One proposed mechanism for the formation of a double helix in DNA is given by $$\left(S_{1}+S_{2}\right)=\left(S_{1}: S_{2}\right)^{*} \quad \text { (fast) }$$ $$\left(S_{1}: S_{2}\right)^{*} \longrightarrow S_{1}: S_{2} \quad \text { (slow) }$$ where \(S_{1}\) and \(S_{2}\) represent strand 1 and \(2,\) and \(\left(S_{1}: S_{2}\right)^{*}\) represents an unstable helix. Write the rate of reaction expression for the formation of the double helix.
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