Problem 65 - Chemistry and Chemical Reactivity

Question

The ozone in the earth's ozone layer decomposes according to the equation $$2 \mathrm{O}_{3}(\mathrm{g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{g})$$ The mechanism of the reaction is thought to proceed through an initial fast equilibrium and a slow step: Step 1: Fast, reversible \(\quad \mathrm{O}_{3}(\mathrm{g}) \rightleftarrows \mathrm{O}_{2}(\mathrm{g})+\mathrm{O}(\mathrm{g})\) Step 2: Slow \(\mathrm{O}_{3}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{O}_{2}(\mathrm{g})\) Show that the mechanism agrees with this experimental rate law: $$\text { Rate }=-(1 / 2) \Delta\left[\mathrm{O}_{3}\right] / \Delta t=k\left[\mathrm{O}_{3}\right]^{2} /\left[\mathrm{O}_{2}\right]$$

Step-by-Step Solution

Answer

The mechanism aligns with the rate law when considering the equilibrium of intermediates.

Step 1: Identify the Rate-Determining Step

The rate-determining step is the slowest step in the reaction mechanism. Here, it is given by Step 2: \( \mathrm{O}_{3}(\mathrm{g}) + \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{O}_{2}(\mathrm{g}) \). Thus, the rate equation will be based on this step.

Step 2: Write the Rate Law for the Slow Step

The rate law for Step 2, the slow step, can be written as: \( \text{Rate} = k [\mathrm{O}_{3}][\mathrm{O}] \), where \( k \) is the rate constant of the slow step.

Step 3: Express the Intermediate Concentration

The intermediate \( \mathrm{O} \) is formed in the fast equilibrium Step 1: \( \mathrm{O}_{3}(\mathrm{g}) \rightleftharpoons \mathrm{O}_{2}(\mathrm{g}) + \mathrm{O}(\mathrm{g}) \). At equilibrium, the rate of forward reaction equals the rate of the reverse reaction, allowing us to set \( k_1[\mathrm{O}_{3}] = k_{-1}[\mathrm{O}_{2}][\mathrm{O}] \). Thus, \( [\mathrm{O}] = \frac{k_1}{k_{-1}} \frac{[\mathrm{O}_{3}]}{[\mathrm{O}_{2}]} \).

Step 4: Substitute Intermediate Concentration in Rate Law

Substitute the expression for \( [\mathrm{O}] \) into the rate law obtained in Step 2: \( \text{Rate} = k [\mathrm{O}_{3}] \frac{k_1}{k_{-1}} \frac{[\mathrm{O}_{3}]}{[\mathrm{O}_{2}]} \). Simplifying this gives \( \text{Rate} = k' \frac{[\mathrm{O}_{3}]^2}{[\mathrm{O}_{2}]} \), where \( k' = k \frac{k_1}{k_{-1}} \).

Step 5: Match to Experimental Rate Law

Compare the derived rate law from the mechanism, \( \text{Rate} = k' \frac{[\mathrm{O}_{3}]^2}{[\mathrm{O}_{2}]} \), with the given experimental rate law \( \text{Rate} = -\frac{1}{2} \Delta[\mathrm{O}_{3}]/\Delta t = k \frac{[\mathrm{O}_{3}]^2}{[\mathrm{O}_{2}]} \). The two match, confirming the mechanism agrees with the experimental rate law when constants are adjusted.

Key Concepts

Understanding Rate Law in Reaction Mechanisms

In the context of chemical reactions, a rate law is a mathematical expression that relates the rate of a reaction to the concentration of the reactants. It is crucial in understanding how fast a reaction occurs. The general form of a rate law is:

• Rate = k ([A]^m[B]^n...)

where:

• k is the rate constant
• [A] and [B] are the concentrations of reactants
• m and n are the reaction orders with respect to each reactant

In the given exercise, the experimental rate law is expressed as:\[\text{Rate} = k \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]}.\]This suggests that the rate of decomposition of ozone (\mathrm{O}_3) depends on the concentration of ozone itself and inversely on the concentration of diatomic oxygen (\mathrm{O}_2). This relationship allows chemists to predict how changes in reactant concentrations affect reaction speed. Understanding this concept helps in refining chemical processes and reactions to achieve desired outcomes efficiently.

Exploring the Rate-Determining Step

The rate-determining step in a reaction mechanism is like the bottleneck in a production line. It is the slowest step and controls the overall rate of the reaction. Identifying this step is essential because it dictates how the entire reaction behaves under different conditions.
In our example, the rate-determining step is identified as the slow reaction:\[\mathrm{O}_3(\mathrm{g}) + \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{O}_2(\mathrm{g}),\]This step is crucial because the reaction rate depends on the formation and consumption of the intermediate (\mathrm{O}).
The rate law for this step is given by:

• Rate = k [\mathrm{O}_3][\mathrm{O}]

From this, we can see the reaction rate is influenced by the concentrations of \mathrm{O}_3 and the intermediate \mathrm{O}.

Equilibrium Dynamics in Reaction Steps

Equilibrium dynamics play a significant role in reaction mechanisms, especially when an intermediate is involved. Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in stable concentrations of reactants and products.
In this mechanism, step 1 is a fast, reversible equilibrium:\[\mathrm{O}_3(\mathrm{g}) \rightleftharpoons \mathrm{O}_2(\mathrm{g}) + \mathrm{O}(\mathrm{g})\]Here, the concentrations of \mathrm{O}_3, \mathrm{O}_2, and \mathrm{O} reach an equilibrium state, allowing us to express the concentration of the intermediate \mathrm{O} in terms of the equilibrium constant (k_1/k_{-1}):

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

Substituting this expression into the rate law for the rate-determining step helps align the theoretical rate law derived from the mechanism with the experimental rate law. This connection highlights the importance of equilibrium considerations in accurately describing complex reaction mechanisms.