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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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Most popular questions from this chapter

For a first order reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C},\) if the half-life of \(\mathrm{A}\) at \(25^{\circ} \mathrm{C}\) is $3.05 \times 10^{4} \mathrm{~s},\( what is the rate constant \)k$ at this temperature? What percentage of A will not have reacted after one day?

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}$

The addition of NO accelerates the decomposition of $\mathrm{N}_{2} \mathrm{O}$, possibly by the following mechanism: $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) & \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) & \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Is NO serving as a catalyst or an intermediate in this reaction? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism?

The following mechanism has been proposed for the reaction of \(\mathrm{NO}\) with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and $\mathrm{H}_{2} \mathrm{O}$ : $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{NO}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g) \\ \mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction. (b) Write a rate law for each elementary reaction in the mechanism.(c) Identify anyintermediatesin the mechanism. (d) The observed rate law is rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\). If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?

A certain enzyme catalyzes a biochemical reaction. In water, without the enzyme, the reaction proceeds with a rate constant of $6.50 \times 10^{-4} \mathrm{~min}^{-1}\( at \)37^{\circ} \mathrm{C} .$ In the presence of the enzyme in water, the reaction proceeds with a rate constant of $1.67 \times 10^{4} \mathrm{~min}^{-1}\( at \)37^{\circ} \mathrm{C}$. Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.
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