Problem 99 - Chemistry: The Central Science

Question

The following mechanism has been proposed for the gas-phase reaction of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) and chlorine: Step 1: \(\mathrm{Cl}_{2}(g) \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftarrows}} 2 \mathrm{Cl}(g) \quad\) (fast) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g)\) (slow) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CCl}_{4}\) (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? (Hint: The overall reaction order is not an integer.)

Step-by-Step Solution

Answer

The overall reaction is \(Cl_2(g) + CHCl_3(g) \longrightarrow HCl(g) + CCl_4(g)\). The intermediates in the mechanism are \(Cl\) and \(CCl_3\). The molecularity of the elementary reactions is unimolecular for Step 1, and bimolecular for Steps 2 and 3. The rate-determining step is Step 2: \(Cl(g) + CHCl_3(g) \longrightarrow HCl(g) + CCl_3(g)\). The rate law predicted by this mechanism is Rate = \(k [Cl_2]^\frac{1}{2} [CHCl_3]\), where \(k = k_3\sqrt{K_{eq}}\).

Step 1: Identify the overall reaction

To find the overall reaction, we need to sum up all the given steps: Step 1: \(Cl_2 \rightleftarrows 2 Cl\) (fast) Step 2: \(Cl + CHCl_3 \longrightarrow HCl + CCl_3\) (slow) Step 3: \(Cl + CCl_3 \longrightarrow CCl_4\) (fast) Adding the three steps together and cancelling out the reactive intermediates on both sides, we get: \(Cl_2 + CHCl_3 \longrightarrow HCl + CCl_4\) The overall reaction is: \(Cl_2(g) + CHCl_3(g) \longrightarrow HCl(g) + CCl_4(g)\)

Step 2: Identify the intermediates

Intermediates are formed in one step of the mechanism and consumed in the subsequent steps. In this mechanism, we can see that there are two intermediates: 1. \(Cl\) is produced in Step 1 and consumed in Steps 2 and 3. 2. \(CCl_3\) is produced in Step 2 and consumed in Steps 3.

Step 3: Determine the molecularity

The molecularity of an elementary reaction is determined by the number of molecules involved in the reaction. In this mechanism, we have: 1. Step 1: One molecule of \(Cl_2\) is involved - Unimolecular 2. Step 2: One molecule of \(Cl\) and one molecule of \(CHCl_3\) are involved - Bimolecular 3. Step 3: One molecule of \(Cl\) and one molecule of \(CCl_3\) are involved - Bimolecular

Step 4: Find the rate-determining step

The rate-determining step is the slowest step in the mechanism. In this case, Step 2 is the slowest step: \(Cl(g) + CHCl_3(g) \longrightarrow HCl(g) + CCl_3(g)\)

Step 5: Predict the rate law

Since Step 2 is the rate-determining step, the rate law is determined by this step. The rate law for this step will be: Rate = \(k_3 [Cl][CHCl_3]\) However, we need to eliminate the intermediate, which is \(Cl\). Since Step 1 is a fast equilibrium step, we have: \(K_{eq} = \frac{k_1}{k_{-1}} = \frac{[Cl]^2}{[Cl_2]}\) This allows us to solve for the intermediate's concentration: \([Cl] = \sqrt{K_{eq}[Cl_2]}\) Substituting this expression for \([Cl]\) into the rate law, we have: Rate = \(k_3 \sqrt{K_{eq}[Cl_2]}[CHCl_3]\) So the rate law predicted by this mechanism is: Rate = \(k [Cl_2]^\frac{1}{2} [CHCl_3]\) where \(k = k_3\sqrt{K_{eq}}\).

Key Concepts

Elementary Reactions

Elementary reactions are the basic building blocks of a complex chemical reaction, much like individual steps in a recipe. In our chloroform-chlorine reaction example, the mechanism is composed of three such elementary steps. Each step represents a single event at the molecular level where reactants directly form products without any intermediates in between. These steps can either be unimolecular, involving a single molecule breaking down or changing state, bimolecular, involving the collision and reaction of two species, or, though less common, termolecular, involving three reactant species. Identifying these steps is crucial because they tell us the sequence of the actual chemical change, offering a clearer picture of how the reaction proceeds on a microscopic level.

Reaction Intermediates

Imagine you are building a puzzle. Some pieces act as bridges, connecting different parts of the picture. In chemical reactions, these pieces are called reaction intermediates; they are formed in one step and consumed in another, never appearing in the final reaction equation. For the reaction of chloroform and chlorine, intermediates such as the Cl and CCl₃ radicals are vital for the process to move forward. However, because they are not present in the initial or final state of the reaction, you won't see them in the overall equation. Recognizing intermediates helps chemists understand the complexity of a reaction and design catalysts that can alter their stability, thus controlling the reaction rate.

Molecularity

Molecularity refers to the number of reactant particles involved in an elementary step. Think of it as the number of players on a sports team during a play. Our example mechanism has steps with different molecularities: Step 1 is a unimolecular step involving the dissociation of a Cl₂ molecule. Steps 2 and 3 are bimolecular, meaning two reactant particles are colliding. The molecularity provides insight into the reaction's complexity; high molecularity often indicates a lower likelihood of the step occurring since it's statistically less probable that many particles will meet at the same time.

Rate-Determining Step

In a relay race, the team is only as fast as its slowest runner. Similarly, in a reaction mechanism, the rate-determining step is like that slowest runner. It limits the overall pace at which the reaction progresses. In the chloroform and chlorine reaction, the slow step is the second one, where a Cl atom reacts with CHCl₃. This is the step that dictates how fast the products can form. Understanding this concept is pivotal for chemists when they aim to optimize industrial processes or conduct kinetic studies. By identifying and potentially accelerating the rate-determining step, the overall reaction can be made more efficient.

Rate Law

Rate law expresses the mathematical relationship between the concentration of reactants and the speed at which they convert into products. For the students out there, this might look similar to the equation of a trend line that fits your data in science class. The rate law for our discussed reaction is different than what you would guess by just looking at the overall equation. It's derived from the rate-determining step and adjusted to account for intermediates, which should not be part of the final expression. Comprehending rate law is not just about plugging numbers into an equation; it involves understanding the interplay of reaction steps to accurately predict how changes in conditions will affect the speed of the reaction.