Problem 97 - Chemistry The Molecular Science

Question

Two mechanisms are proposed for the reaction $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g}) $$ Mechanism $$ \text { 1: } \mathrm{NO}+\mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{3} $$ fast $$ \mathrm{NO}_{3}+\mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2} $$ slow $$ \text { Mechanism } 2: \mathrm{NO}+\mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} $$ fast Show that each mechanism is consistent with the observed rate law: Rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\).

Step-by-Step Solution

Answer

Both mechanisms are consistent with the rate law: \( \text{Rate} = k[\mathrm{NO}]^{2}[\mathrm{O}_{2}] \).

Step 1: Identify the Rate-Determining Step for Mechanism 1

In Mechanism 1, the rate-determining step is the second step: \( \mathrm{NO}_{3} + \mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2} \). Since the rate of the overall reaction is controlled by the slowest step, we will focus on this step to derive the rate law.

Step 2: Express Rate for the Slow Step in Mechanism 1

Since the slow step is the rate-determining step, we write the rate of the reaction as the rate of this slow step: \( \text{Rate} = k_{slow} [\mathrm{NO}_{3}][\mathrm{NO}] \).

Step 3: Relate Fast Equilibrium to Rate Law for Mechanism 1

In the fast equilibrium step: \( \mathrm{NO} + \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{3} \), we apply the equilibrium assumption: \( K = \frac{[\mathrm{NO}_{3}]}{[\mathrm{NO}][\mathrm{O}_{2}]} \). Solving for \([\mathrm{NO}_{3}]\), we get \([\mathrm{NO}_{3}] = K[\mathrm{NO}][\mathrm{O}_{2}]\).

Step 4: Substitute Equilibrium Expression into the Rate Law in Mechanism 1

Substitute \([\mathrm{NO}_{3}] = K[\mathrm{NO}][\mathrm{O}_{2}]\) into the rate expression from Step 2: \( \text{Rate} = k_{slow} (K[\mathrm{NO}][\mathrm{O}_{2}])[\mathrm{NO}] \). Simplify this to \( \text{Rate} = k[\mathrm{NO}]^{2}[\mathrm{O}_{2}] \), where \( k = k_{slow} K \). This matches the observed rate law.

Step 5: Identify the Rate-Determining Step for Mechanism 2

In Mechanism 2, the rate-determining step is not explicitly given, but we need to show consistency, so we propose a rate-determining step: \( \mathrm{N}_{2}\mathrm{O}_{2} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2} \).

Step 6: Express Rate for the Proposed Slow Step in Mechanism 2

Express the rate law as \( \text{Rate} = k_{slow}[\mathrm{N}_{2}\mathrm{O}_{2}][\mathrm{O}_{2}] \).

Step 7: Relate Fast Equilibrium to Rate Law for Mechanism 2

In the fast equilibrium step: \( \mathrm{NO} + \mathrm{NO} \rightleftharpoons \mathrm{N}_{2}\mathrm{O}_{2} \), apply the equilibrium assumption: \( K = \frac{[\mathrm{N}_{2}\mathrm{O}_{2}]}{[\mathrm{NO}]^{2}} \). Solving for \([\mathrm{N}_{2}\mathrm{O}_{2}]\), we get \([\mathrm{N}_{2}\mathrm{O}_{2}] = K[\mathrm{NO}]^{2} \).

Step 8: Substitute Equilibrium Expression into the Rate Law in Mechanism 2

Substitute \([\mathrm{N}_{2}\mathrm{O}_{2}] = K[\mathrm{NO}]^{2}\) into the rate expression from Step 6: \( \text{Rate} = k_{slow}(K[\mathrm{NO}]^{2})[\mathrm{O}_{2}] \). Simplify to \( \text{Rate} = k[\mathrm{NO}]^{2}[\mathrm{O}_{2}] \), where \( k = k_{slow} K \). This again matches the observed rate law.

Key Concepts

Rate-Determining Step

In chemical reaction mechanisms, the "rate-determining step" is crucial. Imagine it as the slowest step that limits the speed of a whole series of reactions. It's like the bottleneck in a busy road. For any reaction mechanism, identifying this step helps in understanding and predicting the overall reaction rate.
In Mechanism 1 of the exercise, the slow step is \( \text{NO}_3 + \text{NO} \rightarrow 2 \text{NO}_2 \). This step dictates the speed of the reaction. Why? Because no matter how fast other steps are, the reaction can only go as fast as this slowest step.
Understanding this helps us derive the rate law. When we talk about the rate law, we refer to how the rate depends on the concentration of reactants. The reaction rate depends primarily on concentrations involved in this key step, as it takes the longest time to proceed. Hence, for Mechanism 1, we base our calculations on this step to match it to the observed rate law: \( \text{Rate} = k[\text{NO}]^2[\text{O}_2] \).

Equilibrium Assumption

The "Equilibrium Assumption" is a helpful concept in reaction mechanisms. It allows us to simplify complicated interactions into manageable equations. When we assume a fast equilibrium, we establish a balance between the forward and backward reactions before the rate-determining step occurs.
In Mechanism 1, the first step is \( \text{NO} + \text{O}_2 \rightleftharpoons \text{NO}_3 \). We assume it reaches equilibrium quickly. This implies that the concentration of \( \text{NO}_3 \) can be expressed in terms of \( \text{NO} \) and \( \text{O}_2 \).
We write:

• Equilibrium constant \( K = \frac{[\text{NO}_3]}{[\text{NO}][\text{O}_2]} \)

Using this, \( [\text{NO}_3] = K [\text{NO}][\text{O}_2] \). This equation helps replace \( [\text{NO}_3] \) in our rate-determining step's rate equation. Thus, it assists in showing consistency with the rate law, tying the fast initial reactions to the overall rate.

Reaction Rate Law

The "Reaction Rate Law" is a mathematical expression that connects the concentration of reactants and the rate at which a reaction proceeds. Simply put, it tells us how changes in concentration affect the speed of a chemical reaction.
For our reaction \( 2 \text{NO} + \text{O}_2 \rightarrow 2 \text{NO}_2 \), the rate law is \( \text{Rate} = k[\text{NO}]^2[\text{O}_2] \). This expression reveals that the reaction rate increases with the square of the concentration of \( \text{NO} \) and directly with the concentration of \( \text{O}_2 \).
In both mechanisms 1 and 2, after applying the equilibrium assumption and recognizing the rate-determining step, the derived rate law aligns well with this observed rate law. These mechanisms illustrate how the reaction progresses at the atomic level and confirm our theoretical understanding using the rate law.