Problem 109 - Chemistry The Central Science In Si Units

Question

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: \(\mathrm{O}(g)+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{O}_{2}(g)\) (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,

Step-by-Step Solution

Answer

(a) \(2\mathrm{O}_3(g) \rightarrow 3\mathrm{O}_2(g)\). (b) rate law: \(\text{rate} = k' [\mathrm{O}_3]^2 [\mathrm{O}_2]^{-1}\). (c) \(\mathrm{O}\) is an intermediate. (d) Rate law would change to \(\text{rate} = k [\mathrm{O}_3]^2\).

Step 1: Identify the overall reaction

Since this reaction mechanism has two steps, we need to sum up the steps to get the overall balanced chemical equation. In the first step, \(\mathrm{O}_3\) decomposes into \(\mathrm{O}_2\) and \(\mathrm{O}\). In the second step, \(\mathrm{O}\) reacts with another \(\mathrm{O}_3\) to form \(2\mathrm{O}_2\). Adding up these reactions, we observe that \(\mathrm{O}\) is an intermediate, and thus it gets canceled out, resulting in the overall reaction: \(2\mathrm{O}_3(g) \rightarrow 3\mathrm{O}_2(g)\).

Step 2: Determine the rate-determining step

The rate of a reaction is determined by its slowest step. According to the given mechanism, Step 2 is the slow step. Therefore, the rate law is based on this step: \(\text{rate} = k [\mathrm{O}] [\mathrm{O}_3]\).

Step 3: Relate the intermediate's concentration to reactants

Since \(\mathrm{O}\) is an intermediate, we cannot have it in the rate law. We use the fast equilibrium of Step 1, where \(\mathrm{O}_3 \rightleftharpoons \mathrm{O}_2 + \mathrm{O}\), to express \([\mathrm{O}]\) in terms of \(\mathrm{O}_3\). Equilibrium implies \(k_1[\mathrm{O}_3] = k_{-1}[\mathrm{O}_2][\mathrm{O}]\). Solve for \([\mathrm{O}]\): \([\mathrm{O}] = \frac{k_1}{k_{-1}} \frac{[\mathrm{O}_3]}{[\mathrm{O}_2]}\).

Step 4: Substitute the intermediate's concentration in the rate law

Substitute \([\mathrm{O}]\) from Step 3 back into the rate law equation from Step 2: \(\text{rate} = k' [\mathrm{O}_3]^2 [\mathrm{O}_2]^{-1}\), where \(k'\) is the combination of constants used \(k k_1/k_{-1}\).

Step 5: Classify \(\mathrm{O}\) in the reaction

The chemical species \(\mathrm{O}\) is generated in one elementary step and consumed in another, indicating that it is an intermediate in this mechanism.

Step 6: Consider the single step reaction rate law

If the reaction occurred in a single step, \(2\mathrm{O}_3 \rightarrow 3\mathrm{O}_2\), the rate law would be based on a termolecular reaction (involving three molecules interacting simultaneously). Thus, the rate law would be \(\text{rate} = k [\mathrm{O}_3]^2\), assuming the reaction order matches the stoichiometry of the equation.

Key Concepts

Ozone Decomposition

Ozone decomposition is a crucial chemical reaction within our atmosphere that can be studied through its two-step mechanism. In the first step, ozone (\( \mathrm{O}_{3} \)) breaks down into oxygen molecules (\( \mathrm{O}_{2} \)) and an oxygen atom (\( \mathrm{O} \)). This part of the reaction occurs quickly and establishes an equilibrium.
In the second, slower step, the free oxygen atom reacts with another ozone molecule to produce two oxygen molecules, \( 2 \mathrm{O}_{2} \). By adding the equations from both steps, we obtain the complete balanced equation for the overall reaction:

• \( 2\mathrm{O}_{3} (g) \rightarrow 3\mathrm{O}_{2} (g) \)

Understanding this mechanism is pivotal to grasp how ozone protects us from harmful UV rays and identifies compounds aiding its depletion.

Rate Law

The rate law is an equation that connects the rate of a reaction to the concentration of its reactants. For the decomposition of ozone, the slow step is rate determining because it limits the reaction's velocity.
Given its role, the rate law focuses on this step:

• \( \text{rate} = k [\mathrm{O}][\mathrm{O}_{3}] \)

This expression suggests that the reaction depends on the concentration of both the oxygen atom and ozone. However, because \( \mathrm{O} \) is an intermediate, we need to express its concentration in terms of stable reactants or products. Using the fast equilibrium for the reversible first step, where

• \( k_1[\mathrm{O}_3] = k_{-1}[\mathrm{O}_2][\mathrm{O}] \)

Substitutes the intermediate concentration. This yields:

• \( \text{rate} = k' [\mathrm{O}_{3}]^2 [\mathrm{O}_{2}]^{-1} \)

This equation clarifies how multiple reactants influence the reaction's pace.

Chemical Equilibrium

Chemical equilibrium is an essential concept for analyzing reactions that involve reversible steps, like ozone decomposition. It describes a state where the forward and backward reactions proceed at the same rate, yielding constant concentrations of reactants and products.
In the ozone decomposition mechanism, chemical equilibrium is established in the first step:

• \( \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{O}_{2}(g) + \mathrm{O}(g) \)

This balance allows us to relate the concentration of the intermediate\( \mathrm{O} \)to stable products and reactants. In turn, this relationship is crucial for constructing a proper rate law that doesn't include intermediates.
Understanding equilibrium helps us predict how various conditions, like pressure or temperature changes, might shift the balance, thus impacting the reaction’s rate and extent.

Reaction Intermediates

Reaction intermediates are transient entities appearing during a reaction mechanism but not explicitly shown in the overall balanced equation. They are crucial for understanding complex reactions, like the ozone decomposition.
In the given mechanism, the oxygen atom\( \mathrm{O} \)acts as an intermediate generated in the first step and consumed in the second. Thus, it is not present in the final overall reaction formula:

• \( 2\mathrm{O}_{3} (g) \rightarrow 3\mathrm{O}_{2} (g) \)

Identifying intermediates is vital for deriving the true rate law and understanding each step's role in the process.
Intermediates often provide insights into the reaction's subtleties and are keys to designing better catalytic processes that might enhance or inhibit specific paths based on practical needs.