Question

The rate law for the decomposition of ozone to molecular oxygen: $$ 2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) $$ is $$ \text { rate }=k \frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]} $$ The mechanism proposed for this process is: $$ \begin{array}{l} \mathrm{O}_{3} \stackrel{k_{1}}{\rightleftarrows} \mathrm{O}+\mathrm{O}_{2} \\\ \mathrm{O}+\mathrm{O}_{3} \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{O}_{2} \end{array} $$ Derive the rate law from these elementary steps. Clearly state the assumptions you use in the derivation. Explain why the rate decreases with increasing \(\mathrm{O}_{2}\) concentration.

Short answer

The rate law is derived as \( \text{rate} = \frac{k_2 k_1}{k_{-1}} \cdot \frac{[\mathrm{O}_{3}]^2}{[\mathrm{O}_{2}]} \). Increased \([\mathrm{O}_{2}]\) lowers the rate.

Step-by-step solution

Identify the Slow Step

The rate-determining step is the slow step in a reaction mechanism. From the given mechanism: \(\mathrm{O} + \mathrm{O}_{3} \rightarrow 2 \mathrm{O}_{2}\) is the slow step, as indicated in typical problems like this.

Write the Rate Law for the Slow Step

For the slow step, the rate is expressed as: \[ \text{rate} = k_2 [\mathrm{O}][\mathrm{O}_{3}] \] This is based on the reaction \( \mathrm{O} + \mathrm{O}_{3} \rightarrow 2 \mathrm{O}_{2} \).

Assume Equilibrium in the Fast Preceding Step

The fast step \( \mathrm{'suggesting how it is reversible'}: \mathrm{O}_{3} \rightleftarrows \mathrm{O} + \mathrm{O}_{2} \) is assumed to be in equilibrium. From equilibrium: \[ k_1 [\mathrm{O}_{3}] = k_{-1} [\mathrm{O}][\mathrm{O}_{2}] \] Simplify this to express [O]: \[ [\mathrm{O}] = \frac{k_1}{k_{-1}} \cdot \frac{[\mathrm{O}_{3}]}{[\mathrm{O}_{2}]} \]

Substitute [O] into the Rate Law

Substitute the expression of \([\mathrm{O}]\) from Step 3 into the rate expression from Step 2: \[ \text{rate} = k_2 \left( \frac{k_1}{k_{-1}} \cdot \frac{[\mathrm{O}_{3}]}{[\mathrm{O}_{2}]} \right) [\mathrm{O}_{3}] \] Simplifying this gives: \[ \text{rate} = \left( \frac{k_2 k_1}{k_{-1}} \right) \frac{[\mathrm{O}_{ my physics can be }]}{[\mathrm{O}_{2}]} \]

Analyze the Final Rate Law

The derived rate law \( \text{rate} = \frac{k_2 k_1}{k_{-1}} \cdot \frac{[\mathrm{O}_{3}]^2}{[\mathrm{O}_{2}]} \) matches the given rate law, showing that the derivation is consistent with the provided information. The reaction rate decreases with increasing \([\mathrm{O}_{2}]\) because the term \((\=[\mathrm{O}_{2}])\) in the denominator reduces the rate.

Key concepts

Reaction Mechanism

A reaction mechanism is a step-by-step pathway by which a chemical reaction occurs. It breaks down the overall reaction into a series of elementary reactions or steps.
Each step involves a small change, resulting in the formation of intermediates, which may not be seen in the final equation.

• Elementary Step: A single reaction step, involving potentially reactive species, like molecules or atoms. Each step has a molecularity, or number of species involved.
• Intermediates: Species formed in one step and consumed in another. They are not a part of the overall reaction.
• Rate-Determining Step: The slowest step in the mechanism. Think of it like the bottleneck in a process; it controls the overall rate of the reaction.

This framework helps explain how and why a reaction proceeds at a particular rate. For example, in the ozone decomposition, the slow step involving \( \mathrm{O} + \mathrm{O}_3 \rightarrow 2 \mathrm{O}_2 \) directly affects the reaction rate.

Steady-State Approximation

The steady-state approximation is used to simplify complex reaction mechanisms by assuming that the concentration of intermediates remains relatively constant over time.
This assumption allows us to focus on the main reactants and products, simplifying the mathematical analysis.

• Fast Preceding Step: Often the steps before the slowest step are fast and reversible. They rapidly establish an equilibrium that allows for the steady-state assumption.
• Applying Steady-State: In the case of ozone decomposition, \( \mathrm{O} \) is considered steady, since it is formed and used quickly. This helps derive expressions for intermediate concentrations.
• Equilibrium Expression: From the fast step, we derive \( [\mathrm{O}] = \frac{k_1}{k_{-1}} \cdot \frac{[\mathrm{O}_3]}{[\mathrm{O}_2]} \) to substitute back into the rate law.

Using steady-state is particularly useful when intermediates are present briefly and rapidly rearrange.

Elementary Reactions

Elementary reactions are basic, single-step reactions where reactants convert directly to products. They tell us the exact stoichiometry of the reacting species in the step.
These reactions can be unimolecular, bimolecular, or even termolecular, depending on the number of molecules involved.

• Characteristics: Each elementary step has a clear meaning and specific rate law. The rate of an elementary reaction is directly proportional to the concentration of the reactants involved in that step.
• Ozone Decomposition Example: The reactions \( \mathrm{O}_3 \rightleftarrows \mathrm{O} + \mathrm{O}_2 \) and \( \mathrm{O} + \mathrm{O}_3 \rightarrow 2 \mathrm{O}_2 \) show how single steps add up to the overall process.
• Mathematical Link: For an elementary reaction like \( A \rightarrow B \), the rate law can be written directly from the reaction equation as \( \text{rate} = k[A] \).

Understanding elementary reactions helps build a clear picture of the kinetics and dynamics underlying complex chemical systems.