Consider this three-step mechanism for a reaction: Cl 2 (g) k 1 ⇌k 2 2 Cl (g) Fast Cl (g) + CHCl 3 (g) →k 3 HCl (g) + CCl 3 (g) Slow Cl (g) + CCl 3 (g) →k 4 CCl 4 (g) Fast c. What is the predicted rate law?

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.15 - Chemical Kinetics

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Tro 5th Edition

Ch.15 - Chemical Kinetics

Problem 77c

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Chapter 15, Problem 77c

Consider this three-step mechanism for a reaction:
Cl₂ (g) k₁⇌k₂ 2 Cl (g) Fast
Cl (g) + CHCl₃ (g) →k₃ HCl (g) + CCl₃ (g) Slow
Cl (g) + CCl₃ (g) →k₄ CCl₄ (g) Fast
c. What is the predicted rate law?

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Identify the slow step in the mechanism, as it is the rate-determining step. In this case, the slow step is: Cl (g) + CHCl_3 (g) → HCl (g) + CCl_3 (g).

Write the rate law based on the slow step. The rate law is determined by the reactants involved in this step: rate = k_3[Cl][CHCl_3].

Since Cl is an intermediate, express its concentration in terms of the reactants and products of the fast equilibrium step: Cl_2 (g) ⇌ 2 Cl (g).

Use the equilibrium expression for the fast step to express [Cl] in terms of [Cl_2]: K_eq = [Cl]^2 / [Cl_2], where K_eq is the equilibrium constant for the fast step.

Substitute the expression for [Cl] from the equilibrium step into the rate law from the slow step to obtain the overall rate law in terms of the initial reactants.

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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 how a chemical reaction occurs at the molecular level. It outlines the individual elementary steps that lead to the overall reaction, including the formation and consumption of intermediates. Understanding the mechanism is crucial for predicting the rate law, as it reveals which steps are rate-determining and how reactants are transformed into products.

Rate Law

The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. It is typically formulated as rate = k[A]^m[B]^n, where k is the rate constant, and m and n are the orders of the reaction with respect to reactants A and B. The rate law can be derived from the slowest step in the reaction mechanism, which dictates the overall reaction rate.

Elementary Steps

Elementary steps are the individual reactions that occur in a reaction mechanism. Each step represents a single molecular event, such as the collision of molecules or the breaking/forming of bonds. The rate of the overall reaction is influenced by the slowest elementary step, known as the rate-determining step, which controls how quickly the products are formed from the reactants.