Question

The overall reaction for the halogenation of a hydrocarbon (RH) using Br as the halogen is \(\mathrm{RH}(g)+\mathrm{Br}_{2}(g) \rightarrow\) \(\operatorname{RBr}(g)+\operatorname{HBr}(g) .\) The following mechanism has been proposed for this process: \\[ \begin{array}{l} \operatorname{Br}_{2}(g) \stackrel{k_{1}}{\longrightarrow} 2 \operatorname{Br} \cdot(g) \\ \operatorname{Br} \cdot(g)+\operatorname{RH}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{R} \cdot(g)+\operatorname{HBr}(g) \\\ \mathrm{R} \cdot(g)+\operatorname{Br}_{2}(g) \stackrel{k_{3}}{\longrightarrow} \operatorname{RBr}(g)+\operatorname{Br} \cdot(g) \\ \operatorname{Br} \cdot(g)+\mathrm{R} \cdot(g) \stackrel{k_{4}}{\longrightarrow} \operatorname{RBr}(g) \end{array} \\] Determine the rate law predicted by this mechanism.

Short answer

The predicted rate law for the halogenation of a hydrocarbon (RH) using Br as the halogen, with the given reaction mechanism, is \(Rate = 2k_1[Br_2]\).

Step-by-step solution

Analyze the reaction mechanism

We have been given a reaction mechanism with four elementary steps: 1. \(Br_2(g) \stackrel{k_{1}}{\longrightarrow} 2 Br \cdot(g)\) 2. \(Br \cdot(g) + RH(g) \stackrel{k_{2}}{\longrightarrow} R \cdot(g) + HBr(g)\) 3. \(R \cdot(g) + Br_2(g) \stackrel{k_{3}}{\longrightarrow} RBr(g) + Br \cdot(g)\) 4. \(Br \cdot(g) + R \cdot(g) \stackrel{k_{4}}{\longrightarrow} RBr(g)\)

Identify the rate-determining step

The rate-determining step is the slowest step in the reaction mechanism. In this exercise, there is no information provided on which step is the slowest. Therefore, we will assume that the rate-determining step is second step, indicated by the rate constant \(k_2\).

Write the rate law for the rate-determining step

The rate law for the rate-determining step can be expressed as: \(Rate = k_2[Br \cdot][RH]\)

Replace the intermediate concentrations

In the rate law expression, [Br·] is an intermediate concentration. We need to replace this intermediate with the concentrations of the reactants from the other elementary steps. To do this, we find the steady-state approximation for the intermediate. From Step 1: \(\frac{d[Br \cdot]}{dt} = 2k_1[Br_2]-k_2[Br \cdot][RH] = 0\) Solving for [Br·]: \([Br \cdot] = \frac{2k_1}{k_2} \cdot \frac{[Br_2]}{[RH]}\)

Substitute the intermediate into the rate law

Now, we substitute this expression for the intermediate concentration [Br·] back into the rate law expression from Step 3: \(Rate = k_2 \cdot \frac{2k_1}{k_2} \cdot \frac{[Br_2]}{[RH]} \cdot [RH]\) Simplifying, we get: \(Rate = 2k_1[Br_2]\) The predicted rate law for this mechanism is: \(Rate = 2k_1[Br_2]\)

Key concepts

Rate Law

The rate law expresses the relationship between the rate of a chemical reaction and the concentrations of the reactants. It is usually determined experimentally and takes the form of an equation: \(\text{Rate} = k[A]^m[B]^n\), where \(k\) is the rate constant, \(A\) and \(B\) are the concentrations of reactants, and \(m\) and \(n\) are the orders of the reaction with respect to each reactant.

The orders of a reaction indicate how sensitive the rate is to a change in the concentration of the reactant; they do not necessarily correspond to the stoichiometry of the reaction and must be determined either experimentally or through a given mechanism. In our example, the rate law is simplified to just the concentration of \(Br_2\) because the halogenation of a hydrocarbon is believed to be first order with respect to \(Br_2\) and zero order with respect to \(RH\), based on the proposed mechanism and the assumption that the halogen carves through the reaction.

Reaction Mechanism

A reaction mechanism is a series of elementary steps that describe the sequence of events on a molecular level that contribute to the overall reaction. Every elementary step is a simple reaction that follows its own rate law. A complete mechanism should sum up to the overall stoichiometry of the reaction while providing insight into the formation and consumption of intermediates.

In the halogenation example, the mechanism involves a series of steps leading to the formation of RBr and HBr from RH and \(Br_2\). Identifying each elementary step helps chemists understand the order in which bonds are broken and formed, and also which species, like the bromine radicals in this case, are intermediates.

Steady-State Approximation

The steady-state approximation is an assumption used in chemical kinetics to simplify the calculation of reaction rates in complex mechanisms. Under this approximation, the rate of formation and consumption of an intermediate is the same, meaning its concentration remains constant throughout the reaction process after an initial induction period. This is particularly useful for intermediates that are quickly consumed in subsequent steps after they are formed.

Applying the steady-state approximation means we assume that \( \frac{d[Br\cdot]}{dt} = 0 \), enabling us to solve for the concentration of the bromine radical, \([Br\cdot]\), as seen in the given exercise. This is crucial for deriving a rate law that only includes the concentrations of the stable reactants and products that can be readily measured.

Rate-Determining Step

The rate-determining step (RDS) of a reaction mechanism is the slowest step, which therefore determines the overall rate of the reaction. It effectively acts as a bottleneck, limiting the speed at which the product can be formed. In our exercise, we had to assume the second step was the RDS as we were not given any information on the speeds of the different steps.

Once the RDS is identified, it's possible to express the rate law focusing on this specific step. However, one must ensure that any intermediates are eliminated from the rate law expression, as seen in the exercise, by substituting with the concentrations of stable species. The rate law, therefore, reflects the kinetics of the RDS and, by extension, the overall reaction.