The following mechanism has been proposed for the gasphase reaction of chloroform 1CHCl32 and chlorine: Step 1: Cl 2 (g) k 1 ⇌ k -1 2 Cl(g) (fast) Step 2: Cl(g) + CHCl 3 (g) k 2 → HCl(g) + CCl 3 (g) (slow) Step 3: Cl(g0 + CCl3(g) k 3 → CCl 4 (fast) (e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

Welcome back everyone to another video, consider this three step mechanism for a reaction part C. What is the predicted rate law? We given our mechanism, it consists of three steps and we're going to break it down. First of all, we have to recall that the slow step of the mechanism is the ray determining step or R DS. So we're going to use a shorthand notation and state that this is the way determining step and we begin writing the rate law as rate equals. Now, we noticed that the rate constant is K three. So we're going to include K three and multiply it by the concentration of the species we have in our reaction, raised the power of their stom meric efficiency because this is a mechanism. So we have chlorine and trichloromethane CC three. OK. Now, this would be pretty simple. But what we have to notice is that laine is actually an intermediate because in the first step, it is produced and then it is consumed in the second step. And the third step, right, we're producing two moles of chlorine and then consuming two moles of chlorine, which is essentially tells us that it is an intermediate and we cannot have an intermediate in our rate expression. However, CCL three is not an intermediate because we don't have it anywhere else other than in the second step of the mechanism. So our goal is to just express a concentration of chlorine in terms of something which is not an intermediate. And we can use the first step for that purpose because we have an equilibrium, we can say that the rate of the forward reaction which is K one multiplied by the concentration of the atomic fluorine must be equal to the rate of the reverse reaction that is K two multiplied by the concentration of chlorine to the power of two. And from here, we can easily show that the concentration of chlorine is simply K one divided by K two multiplied by the concentration of diatomic chlorine, right. We also have to take square root of that. And mathematically, we can also separate our rate constants as a square root B, square root of K one divided by K two multiplied by the concentration of chlorine, the power of one half, which is the same thing as square root. And now we can plug it in back. So the rate law becomes, first of all, we have K three, then we have square root of K one divided by K two. Then we have the concentration of chlorine, the power of one half. And now we finish it with the concentration of trichloromethane CHCL three. Finally, what we usually do is just combine our weight constants into a single weight constant. So the expression looks better and we can say that the final answer would be weight equals K. So we combine all of them into a single K multiplied by the concentration of chlorine is the power of one half multiplied by the concentration of trichloromethane. That'll be our final answer. We can label it just to make sure we understand where it is and conclude the video. Thank you for watching.

Ch.14 - Chemical Kinetics

All textbooks

Brown 15th Edition

Ch.14 - Chemical Kinetics

Problem 110e

Chapter 14, Problem 110e

The following mechanism has been proposed for the gasphase reaction of chloroform 1CHCl32 and chlorine:
Step 1: Cl₂(g) k₁⇌ k_-1 2 Cl(g) (fast)
Step 2: Cl(g) + CHCl₃(g) k₂→ HCl(g) + CCl₃(g) (slow)
Step 3: Cl(g0 + CCl3(g) k₃→ CCl₄ (fast)
(e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

Verified step by step guidance

Identify the rate-determining step (RDS) in the mechanism. The slowest step, Step 2, is the rate-determining step, which involves Cl(g) and CHCl3(g).

Write the rate law based on the RDS. Since Step 2 is the slow step, the rate law is: rate = k2[Cl][CHCl3].

Recognize that Cl(g) is an intermediate and its concentration needs to be expressed in terms of reactants. Use the pre-equilibrium assumption for Step 1, where Cl2(g) ⇌ 2 Cl(g) is fast and reversible.

Apply the equilibrium expression for Step 1: K_eq = [Cl]^2 / [Cl2]. Solve for [Cl] in terms of [Cl2]: [Cl] = sqrt(K_eq * [Cl2]).

Substitute the expression for [Cl] from Step 4 into the rate law from Step 2: rate = k2[sqrt(K_eq * [Cl2])][CHCl3]. Simplify to find the overall rate law, which will show that the reaction order is not an integer.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Reaction Mechanism

A reaction mechanism is a step-by-step description of the pathway through which reactants are converted into products. It outlines individual elementary steps, each with its own rate, and helps in understanding how the overall reaction occurs. In this case, the mechanism involves multiple steps, including fast and slow reactions, which influence the overall rate law.

Rate Law

The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. It is determined by the slowest step in the reaction mechanism, known as the rate-determining step. The rate law can include fractional orders, indicating that the reaction does not follow simple stoichiometry, as suggested in the question.

Elementary Steps

Elementary steps are individual reactions that occur in a reaction mechanism. Each step has its own rate constant and can be classified as either fast or slow. The slow step typically dictates the overall rate law, while fast steps can establish equilibrium conditions, as seen in the first step of the provided mechanism involving the dissociation of Cl2.

Recommended video:

Guided course

03:06

Reaction Mechanism Overview

Related Practice

Textbook Question

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.

Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)

Step 2: O(g) + O₃(g) → 2 O₂ (slow)

(a) Write the balanced equation for the overall reaction.

Textbook Question

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.

Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)

Step 2: O(g) + O₃(g) → 2 O₂ (slow)

(b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.)

Textbook Question

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.

Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)

Step 2: O(g) + O₃(g) → 2 O₂ (slow)

(d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

Textbook Question

In a hydrocarbon solution, the gold compound (CH₃)₃AuPH₃ decomposes into ethane (C₂H₆) and a different gold compound, (CH₃)AuPH₃. The following mechanism has been proposed for the decomposition of (CH₃)₃AuPH₃:

Step 1: (CH₃)₃AuPH₃ k₁⇌k_-1 (CH₃)₃Au + PH₃ (fast)

Step 2: (CH₃)₃Au k₂→ C₂H₆ + (CH₃)Au (slow)

Step 3: (CH₃)Au + PH₃ k₃→ (CH₃)AuPH₃ (fast)

(a) What is the overall reaction?

(b) What are the intermediates in the mechanism?

Textbook Question

Step 1: (CH₃)₃AuPH₃ k₁⇌k_-1 (CH₃)₃Au + PH₃ (fast)

Step 2: (CH₃)₃Au k₂→ C₂H₆ + (CH₃)Au (slow)

Step 3: (CH₃)Au + PH₃ k₃→ (CH₃)AuPH₃ (fast)