Question

The following mechanism has been proposed for the gas-phase reaction of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) and chlorine: Step 1: \(\mathrm{Cl}_{2}(g) \underset{k_{1}}{\stackrel{k_{1}}{\rightleftharpoons}} 2 \mathrm{Cl}(g) \quad\) (fast) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g) \quad\) (slow) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CCl}_{4} \quad\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

Short answer

The overall reaction is: \(\mathrm{CHCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{HCl}(g) + \mathrm{CCl}_{4}(g)\). The intermediates are \(\mathrm{Cl}(g)\) and \(\mathrm{CCl}_{3}(g)\). The molecularity of each elementary reaction is: Step 1: unimolecular, Step 2: bimolecular, Step 3: bimolecular. The rate-determining step is Step 2. The rate law predicted by this mechanism is: \(Rate = k [\mathrm{CHCl}_{3}] [\mathrm{Cl}_{2}]^{\frac{1}{2}}\), where \(k=k_{3}\sqrt{K_{1}}\).

Step-by-step solution

Identify the overall reaction

To identify the overall reaction, add all the reactions and cancel any species that appear on both sides of the equations. Step 1: \(\mathrm{Cl}_{2}(g) \underset{k_{1}}{\stackrel{k_{1}}{\rightleftharpoons}} 2 \mathrm{Cl}(g)\) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g) \) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CCl}_{4}(g)\) The overall reaction is: \(\mathrm{CHCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{HCl}(g) + \mathrm{CCl}_{4}(g)\)

Identify intermediates

Intermediates are species that are formed in one step and consumed in another step. In this mechanism, we can see that \(\mathrm{Cl}(g)\) and \(\mathrm{CCl}_{3}(g)\) are intermediates.

Determine the molecularity of each reaction

Molecularity is the number of reactant molecules involved in each elementary reaction. Step 1: \(\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}(g)\) - Molecularity: 1 (unimolecular) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \longrightarrow \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g)\) - Molecularity: 2 (bimolecular) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \longrightarrow \mathrm{CCl}_{4}(g)\) - Molecularity: 2 (bimolecular)

Identify the rate-determining step

The rate-determining step is the slowest step in the mechanism, which in this case is Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g)\)

Determine the rate law

The rate law for the rate-determining step is given by: $$Rate = k_{3}[\mathrm{Cl}][\mathrm{CHCl}_{3}]$$ However, since \(\mathrm{Cl}\) is an intermediate, we need to express its concentration in terms of the other reactants. Use the equilibrium expression for Step 1: $$K_{1} = \frac{k_{1}}{k_{-1}} = \frac{[\mathrm{Cl}]^{2}}{[\mathrm{Cl}_{2}]}$$ Solving for the concentration of \(\mathrm{Cl}\) and substituting into the rate law: $$[\mathrm{Cl}] = \sqrt{K_{1} [\mathrm{Cl}_{2}]}$$ $$Rate = k_{3}[\mathrm{CHCl}_{3}] \sqrt{K_{1} [\mathrm{Cl}_{2}]}$$ The rate law predicted by this mechanism is: $$Rate = k [\mathrm{CHCl}_{3}] [\mathrm{Cl}_{2}]^{\frac{1}{2}}$$ where \(k=k_{3}\sqrt{K_{1}}\).

Key concepts

Chemical Kinetics

Chemical kinetics focuses on how reactants transform into products over time, providing an understanding of the speed or rate of a chemical reaction. It involves the study of reaction rates and the factors affecting them, such as temperature, concentration of reactants, and the presence of catalysts.

By applying the principles of kinetics, chemists can design experiments to measure reaction rates and use mathematical equations to describe these rates. For example, the reaction between chloroform and chlorine gas involves multiple steps, each having its own rate. To determine the overall reaction rate, chemists need to know the rates at which each step occurs, their molecularity, and which step controls the overall pace of the reaction.

Understanding kinetics helps in several practical applications, such as optimizing industrial chemical processes, developing pharmaceuticals, and reducing environmental pollutants. To optimize a reaction, chemists may adjust the conditions so that reactants are converted to products as efficiently as possible.

Rate-Determining Step

The rate-determining step is the slowest step in a multi-step chemical reaction mechanism, and it essentially sets the pace for the overall reaction. Since the entire process can only proceed as fast as its slowest component, understanding the rate-determining step is crucial for predicting and controlling reaction rates.

In the given example of chloroform reacting with chlorine, the slow step (Step 2) is the reaction between Cl(g) and CHCl3(g). This step will dictate how fast or slow the overall reaction will proceed. Knowing this allows us to concentrate our efforts on this step if we were to find ways to speed up the reaction, for instance by adding a catalyst or increasing the temperature.

Identifying the rate-determining step is also central to deriving the rate law for a reaction, which mathematically relates the rate of a reaction to the concentrations of the reactants. Such an understanding is vital for predicting how the reaction rate will change with varying reactant concentrations.

Reaction Intermediates

Reaction intermediates are species that are produced in one step of a reaction mechanism and used up in a subsequent step. They are the 'middle-men' of a reaction sequence, not present in the reactants or the final products but essential for the reaction path. In our chloroform-chlorine reaction mechanism, both Cl(g) and CCl3(g) act as intermediates.

Intermediates are often very reactive and have a short lifespan. They can give insights into the progression of a reaction and the types of bonds that are formed and broken along the reaction path. Identifying intermediates is also crucial when deducing the mechanism of a reaction as it allows us to piece together plausible steps in a mechanism.

Because intermediates are transient and may have low concentrations that are difficult to measure directly, chemists often infer their presence and behavior from the study of the overall reaction and the rates of its elementary steps.