Question

In a study of nitrosyl halides, a chemist proposes the following mechanism for the synthesis of nitrosyl bromide: \(\mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(g) \rightleftharpoons \operatorname{NOBr}_{2}(g)\) [fast] \(\operatorname{NOBr}_{2}(g)+\mathrm{NO}(g) \longrightarrow 2 \mathrm{NOBr}(g)\) [slow] If the rate law is rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{Br}_{2}\right]\), is the proposed mechanism valid? If so, show that it satisfies the three criteria for validity.

Short answer

The proposed mechanism is valid as it satisfies the requirement for the rate law and the validity criteria of reaction mechanisms.

Step-by-step solution

Identify the Proposed Mechanism

The proposed mechanism consists of two steps: 1. \(\text{NO(g) + Br}_2\text{(g)} \rightleftharpoons \text{NOBr}_2\text{(g)}\) [fast] 2. \(\text{NOBr}_2\text{(g) + NO(g)} \rightarrow 2\text{NOBr(g)}\) [slow]

Write the Rate Law for the Slow Step

Since the second step is the slow step, it determines the rate of the reaction. The rate law for this step is: \(\text{rate} = k_2 [\text{NOBr}_2][\text{NO}]\)

Express Intermediate Concentration

Express the concentration of the intermediate \(\text{NOBr}_2\) using the equilibrium expression from the fast step: \[K_{eq} = \frac{[\text{NOBr}_2]}{[\text{NO}][\text{Br}_2]} \] \([\text{NOBr}_2] = K_{eq}[\text{NO}][\text{Br}_2]\)

Substitute Intermediate into Rate Law

Substitute \([\text{NOBr}_2]\) from the fast step into the rate law for the slow step: \[ \text{rate} = k_2 (K_{eq}[\text{NO}][\text{Br}_2])[\text{NO}] \= k' [\text{NO}]^2[\text{Br}_2] \] \(\text{where } k' = k_2 K_{eq}\)

Compare with Given Rate Law

The derived rate law \(\text{rate} = k' [\text{NO}]^2[\text{Br}_2]\) matches the given rate law \(\text{rate} = k [\text{NO}]^2[\text{Br}_2]\).

Validate Criteria for a Valid Mechanism

Check the three criteria for a valid mechanism: 1. The sum of the steps gives the overall reaction. 2. The mechanism is consistent with the rate law. 3. The mechanism must involve plausible steps. Since these are satisfied, the proposed mechanism is valid.

Key concepts

rate law

A rate law is an equation that links the reaction rate with the concentrations of reactants. For the given mechanism, the rate law is provided as \(\text{rate} = k[\text{NO}]^2[\text{Br}_2]\). This suggests that the rate depends on the concentration of \text{NO} raised to the power of two and the concentration of \text{Br}_2 raised to the power of one.
The general form of a rate law is written as \(\text{rate} = k[A]^m[B]^n\), where \m\ and \ are the orders of reaction with respect to reactants \A\ and \B\, respectively. In this example, \A = \text{NO}\ and \B = \text{Br}_2\.
The exponents (2 for \text{NO} and 1 for \text{Br}_2) come from the reaction mechanism and are not just the stoichiometric coefficients of the balanced chemical equation. It's key to understand that these orders must be determined experimentally or through mechanistic reasoning.

intermediate concentration

Intermediate species are those that are formed in one step of a mechanism and used up in another step, without appearing in the overall reaction. In our mechanism, \text{NOBr}_2 is an intermediate.
To use intermediates in the rate law, we need to express their concentration in terms of the reactants. For our mechanism, we have the equilibrium expression from the fast step:

\[ K_{eq} = \frac{[\text{NOBr}_2]}{[\text{NO}][\text{Br}_2]} \]

From this, we solve for the intermediate concentration:
\([\text{NOBr}_2] = K_{eq}[\text{NO}][\text{Br}_2]\)
By substituting \([\text{NOBr}_2]\) in the rate law of the slow step, we make the rate law dependent only on the concentrations of the initial reactants \(\text{NO} \ and \ \text{Br}_2\). This transformation is crucial, as intermediates should not appear in the final rate law.

equilibrium expression

An equilibrium expression relates the concentrations of reactants and products at equilibrium for a reversible reaction. This expression is derived from the law of mass action.
In the mechanism, the fast step \(\text{NO}(\text{g})+\text{Br}_2(\text{g}) \rightleftharpoons \text{NOBr}_2(\text{g})\) is at equilibrium. The equilibrium constant \(K_{eq}\) for this step is given by:

\[ K_{eq} = \frac{[\text{NOBr}_2]}{[\text{NO}][\text{Br}_2]} \]

This equation allows us to express the concentration of the intermediate \(\text{NOBr}_2\) in terms of the reactants \(\text{NO}\ and \ \text{Br}_2\). It serves as a bridge between the reaction steps, enabling us to connect the observed rate law with the proposed mechanism.

reaction steps

The reaction mechanism is a series of elementary steps that describe the pathway from reactants to products. Here, the mechanism is broken down into two steps:

• The first step: \(\text{NO(g) + Br}_2\text{(g)} \rightleftharpoons \text{NOBr}_2\text{(g)}\) - fast equilibrating reaction.
• The second step: \(\text{NOBr}_2\text{(g) + NO(g)} \rightarrow 2\text{NOBr(g)}\) - slow rate-determining step.

In the first step, an intermediate \(\text{NOBr}_2\) is formed, which quickly establishes equilibrium with \(\text{NO}\ and \ \text{Br}_2\). The second step, which is slower, uses the intermediate to produce the final product \(\text{NOBr}\).
The overall reaction needs to be the sum of these steps, and the rate law derived from these steps should match the experimentally observed rate law. This confirms the validity of the proposed mechanism.