Problem 110 - Chemistry The Central Science In Si Units

Question

The following mechanism has been proposed for the gasphase reaction of chloroform (CHCl \(_{3}\) ) and chlorine: Step 1: \(\mathrm{Cl}_{2}(g) \stackrel{\mathrm{k}_{1}}{\mathrm{k}_{-1}} 2 \mathrm{Cl}(g)\) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g)\) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{CCl}_{4}\) (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?

Step-by-Step Solution

Answer

(a) Overall: \(\mathrm{CHCl}_{3}(g) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(g) + \mathrm{HCl}(g)\). (b) Intermediates: \(\mathrm{Cl}\) and \(\mathrm{CCl}_{3}\). (c) Molecularity: Step 1, 2, and 3 are bimolecular. (d) Rate-determining: Step 2. (e) Rate law: \(\text{Rate} = k [\mathrm{Cl}_{2}]^{1/2}[\mathrm{CHCl}_{3}]\).

Step 1: Identify the Overall Reaction

To find the overall reaction, add all the elementary steps given in the mechanism. Cancel out any species that appear on both sides of the reaction equations. Here, \(2 \mathrm{Cl}(g)\), \( \mathrm{CCl}_{3}(g)\), and \( \mathrm{HCl}(g)\) cancel out from reactant and product sides respectively. The net reaction results in: \[\mathrm{CHCl}_{3}(g) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(g) + \mathrm{HCl}(g).\]

Step 2: Identify the Intermediates

Intermediates are species that are formed and consumed during the course of the mechanism but do not appear in the overall reaction. From the steps, \(\mathrm{Cl}(g)\) and \(\mathrm{CCl}_{3}(g)\) qualify as intermediates since they are formed and used up within the reaction mechanism.

Step 3: Determine Molecularity of Each Step

Molecularity refers to the number of reactant molecules involved in an elementary reaction. Step 1 is bimolecular as it involves \(\mathrm{Cl}_{2}\) dissociating into two \(\mathrm{Cl}\) atoms. Step 2 is also bimolecular as it involves \(\mathrm{Cl}(g)\) reacting with \(\mathrm{CHCl}_{3}(g)\). Step 3 is bimolecular as \(\mathrm{Cl}(g)\) reacts with \(\mathrm{CCl}_{3}(g)\).

Step 4: Identify the Rate-Determining Step

The rate-determining step is the slowest step in the reaction mechanism and limits the rate of the overall reaction. Here, Step 2 involving \(\mathrm{Cl}(g)\) and \(\mathrm{CHCl}_{3}(g)\) is the slow step, thus it is the rate-determining step.

Step 5: Predict the Rate Law from Mechanism

The rate law for a reaction is derived from the rate-determining step. Since Step 2 is the rate-determining step, its rate law can be expressed as \(\text{Rate} = k_2 [\mathrm{Cl}][\mathrm{CHCl}_{3}]\). To express rate entirely in terms of initial reactants, substitute \([\mathrm{Cl}]=\frac{k_1}{k_{-1}} [\mathrm{Cl}_{2}]\) from Step 1, which gives: \[\text{Rate} = k [\mathrm{Cl}_{2}]^{1/2}[\mathrm{CHCl}_{3}].\]

Key Concepts

Overall Reaction

In chemical reactions, especially those involving multiple steps, identifying the overall reaction is crucial. The overall reaction encompasses all substances consumed and produced but leaves out any intermediates. This means we look at the starting compounds and the end products. In the given sequence of reactions between chloroform (CHCl₃) and chlorine (Cl₂), the overall reaction is determined by adding up all the individual steps. We start with CHCl₃ and Cl₂ and after all the intermediate steps, we end with CCl₄ and HCl. Thus, the overall reaction is:\[\mathrm{CHCl}_{3}(g) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(g) + \mathrm{HCl}(g).\]This reaction shows the conversion of chloroform and chlorine into carbon tetrachloride and hydrochloric acid.

Intermediates

Intermediates play an essential role in multi-step reaction mechanisms. They are species that are generated in one step and consumed in another. Thus, intermediates do not appear in the overall chemical equation. They "bridge the gap" between reactants and products without being present in the final equation. In the proposed mechanism, the intermediates are Cl(g) and CCl₃(g). These intermediates are formed during the breakdown of Cl₂ and are used up in subsequent steps. They facilitate the reaction but do not impact the overall stoichiometry of the starting and ending compounds.

Molecularity

Molecularity is a concept that describes the number of molecules participating in an elementary reaction step. It is different from order of a reaction, as molecularity refers only to elementary reactions and is always an integer. In the provided mechanism, each step has different molecularity:

• Step 1 has a molecularity of two (bimolecular) because it involves the dissociation of Cl₂ into two Cl atoms.
• Step 2 also has a molecularity of two, involving Cl reacting with CHCl₃.
• Step 3 is again bimolecular as it involves Cl reacting with CCl₃ to form the final product CCl₄.

Understanding molecularity helps in predicting how reaction rates might change under different concentrations of reactants in elementary steps.

Rate-Determining Step

The rate-determining step (RDS) is the slowest step in a reaction mechanism. It acts as a bottleneck, controlling the overall speed of the reaction. It's like the narrowest part of a funnel slowing down the whole process. Identifying this step is crucial because the kinetics of the entire reaction depend on it. In the chloroform-chlorine reaction mechanism, Step 2 is identified as the rate-determining step. This step involves the reaction between Cl(g) and CHCl₃(g), highlighting that the formation of CCl₃ and HCl is the slowest, and thus limits the speed at which products form. Knowing the RDS allows chemists to devise strategies to increase the overall reaction rate, such as using catalysts.

Rate Law

The rate law describes how the rate of a chemical reaction depends on the concentration of reactants. For complex reactions, the rate law is derived from the rate-determining step. In our scenario, since Step 2 is the slowest step, its equation gives the rate law:\[ \text{Rate} = k_2 [\mathrm{Cl}][\mathrm{CHCl}_{3}] \]However, since Cl is an intermediate, we express its concentration in terms of observable reactants. By considering the equilibrium in Step 1, we can substitute \[ [\mathrm{Cl}] = \frac{k_1}{k_{-1}} [\mathrm{Cl}_{2}]^{1/2} \]into the rate law to adjust for intermediates concentration:\[ \text{Rate} = k [\mathrm{Cl}_{2}]^{1/2}[\mathrm{CHCl}_{3}] \]This equation predicts how changes in the concentrations of Cl₂ and CHCl₃ affect the reaction rate, thus guiding the optimization of reaction conditions.