Problem 10 - Chemistry in 30 days

Question

The chemical reaction \(2 \mathrm{O}_{3}(\mathrm{~g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{~g})\) proceeds as follows: \(\mathrm{O}_{3} \rightleftharpoons \mathrm{O}_{2}+\mathrm{O}\) (fast) \(2 \mathrm{O}+\mathrm{O}_{2} \rightarrow 2 \mathrm{O}_{2}\) (slow) The rate law expression is (a) rate \(=k\left[\mathrm{O}_{3}\right]_{2}\) (b) rate \(=k\left[\mathrm{O}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{-1}\) (c) rate \(=k\left[\mathrm{O}_{3}\right]\left[\mathrm{O}_{2}\right]\) (d) none of the above

Step-by-Step Solution

Answer

Option (b) is correct: rate = k[\mathrm{O}_{3}]^2[\mathrm{O}_{2}]^{-1}.

Step 1: Analyze the Reaction Mechanism

The given reaction mechanism involves two steps: the first step, which is fast, involves the dissociation of \([\mathrm{O}_{3}]\) into \([\mathrm{O}_{2} + \mathrm{O}]\), and the second step, which is slow, involves \([2 \mathrm{O} + \mathrm{O}_{2} \rightarrow 2 \mathrm{O}_{2}]\). Since the second step is slow, it is the rate-determining step.

Step 2: Write Expression for Rate-Determining Step

The rate-determining step reaction is \([2 \mathrm{O} + \mathrm{O}_{2} \rightarrow 2 \mathrm{O}_{2}]\). The rate law would typically depend on the reactants in this step, specifically \([\mathrm{O}]\) and \([\mathrm{O}_{2}]\). The intermediate \([\mathrm{O}]\) is not stable and must be expressed in terms of stable species.

Step 3: Equilibrium Expression for Fast Step

Since the first step is fast and at equilibrium, apply the equilibrium concept to \([\mathrm{O}_{3} \rightleftharpoons \mathrm{O}_{2} + \mathrm{O}]\). The equilibrium constant expression can be written as \([K = \frac{[\mathrm{O}_{2}][\mathrm{O}]}{[\mathrm{O}_{3}]}]\). Rearrange to express \([\mathrm{O}]\) in terms of \([\mathrm{O}_{3}]\) and \([\mathrm{O}_{2}]\).

Step 4: Substitute Intermediate Concentration

From Step 3, \([\mathrm{O}] = K \frac{[\mathrm{O}_{3}]}{[\mathrm{O}_{2}]}\). Substitute this relation into the rate law of the slow step \((rate = k'[\mathrm{O}]^2[\mathrm{O}_{2}])\) to express the rate law in terms of \([\mathrm{O}_{3}]\) and \([\mathrm{O}_{2}]\): \(rate = k'\left(K \frac{[\mathrm{O}_{3}]}{[\mathrm{O}_{2}]}\right)^2 [\mathrm{O}_{2}]\).

Step 5: Simplify the Rate Law Expression

Simplifying the expression \(rate = k'K^2 \frac{[\mathrm{O}_{3}]^2}{[\mathrm{O}_{2}]} [\mathrm{O}_{2}]\) gives \(rate = k'' \frac{[\mathrm{O}_{3}]^2}{[\mathrm{O}_{2}]}\). The form of this rate law expression matches option (b), \(rate = k[\mathrm{O}_{3}]^2[\mathrm{O}_{2}]^{-1}\).

Key Concepts

Reaction Mechanism

A reaction mechanism provides a step-by-step sequence of elementary reactions by which a chemical reaction occurs. It is vital because it breaks down complex reactions into simpler steps that can be analyzed individually. In our example, the overall reaction is the decomposition of ozone (\(2 \mathrm{O}_{3} \rightarrow 3 \mathrm{O}_{2}\)). Here, it's essential to recognize that this happens via two distinct steps:

• First, \(\mathrm{O}_{3}\) rapidly dissociates into \(\mathrm{O}_{2}\) and \(\mathrm{O}\).
• Second, two \(\mathrm{O}\) atoms react with an \(\mathrm{O}_{2}\) molecule to form two \(\mathrm{O}_{2}\) molecules.

This mechanism helps us understand how the reaction progresses and indicates that although the overall idea is the decomposition of ozone, the reaction doesn't happen in a single leap.Understanding the reaction mechanism aids in determining the speed of a reaction and the rate law expression.

Equilibrium Constant

The equilibrium constant is a number representing the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. For the fast initial reaction in our mechanism, the equilibrium can be expressed using the concept of equilibrium constant. This step in the mechanism,\(\mathrm{O}_{3} \rightleftharpoons \mathrm{O}_{2} + \mathrm{O}\), quickly reaches equilibrium.
The equilibrium constant (\(K\)) for this step is given as \[K = \frac{[\mathrm{O}_{2}][\mathrm{O}]}{[\mathrm{O}_{3}]}\]This mathematical expression helps us calculate the concentration of an unstable intermediate like \(\mathrm{O}\). This is crucial because the intermediate's concentration impacts the rate-determining step, allowing us to express it in terms of the stable species, \(\mathrm{O}_{3}\) and \(\mathrm{O}_{2}\).

Rate-Determining Step

In multi-step reactions, the rate-determining step is the slowest step. It serves as a bottleneck, controlling the overall rate of the reaction. For the decomposition of ozone, the second step,\(2 \mathrm{O} + \mathrm{O}_{2} \rightarrow 2 \mathrm{O}_{2}\), is labeled slow, thus it is the rate-determining step.

• The rate of this step dictates the overall reaction speed.
• Therefore, the rate law is derived from this step, involving the concentrations of the reactants \(\mathrm{O}\) and \(\mathrm{O}_{2}\).

Understanding which step is rate-determining allows chemists to focus on it when trying to alter the rate of a reaction. For example, if you are designing a chemical process or choosing conditions for an industrial reaction, knowing the rate-determining step could highlight what can be practically adjusted to speed up the reaction.

Intermediate Concentration

Intermediate concentration refers to the concentration of species that appear in the mechanistic steps but not in the overall reaction. Intermediates are often short-lived and exist only during the reaction. In our case, \(\mathrm{O}\) is an intermediate forming in the fast equilibrium step and reacting in the slowest step.
To find its concentration in terms of measurable quantities, we use the equilibrium constant derived expression:\[[\mathrm{O}] = K \frac{[\mathrm{O}_{3}]}{[\mathrm{O}_{2}]}\]This expression is pivotal for substituting the intermediate back into the rate equation of the slow step, helping in complete rate law derivation. It's a technique that allows chemists to express reactions in terms of observable and stable reactants, ensuring the predicted reaction rates match experimental findings.