/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 110 The following mechanism has been... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 110

The following mechanism has been proposed for the gasphase reaction of chloroform (CHCl_ ) and chlorine:$$\begin{array}{l}{\text { Step } 1 : \mathrm{Cl}_{2}(g) \frac{k_{1}}{k_{-1}} 2 \mathrm{Cl}(g) \text { (fast) }} \\\ {\text { Step } 2 : \mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g) \text { (slow) }}\end{array}$$ $$ { Step } \quad3 : \quad \mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{CCl}_{4} \quad(\text { fast })$$ (a) What is the overall reaction? (b) What are the intermedi- ates 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

Expert verified
a) Overall reaction: \(Cl_2(g) + CHCl_3(g) 鉄� HCl(g) + CCl_4\) b) Intermediates: Cl(g) and CCl鈧�(g) c) Molecularity: unimolecular (Step 1), bimolecular (Step 2), bimolecular (Step 3) d) Rate-determining step: Step 2: Cl(g) + CHCl鈧�(g) 鉄� HCl(g) + CCl鈧�(g) e) Rate law: \(Rate = k\sqrt{[Cl_2]}[CHCl_3]\)

Step by step solution

01

(Step 1: Adding the elementary reactions to find the overall reaction)

We can find the overall reaction by adding up all the elementary reactions and cancelling out the species that appear on both sides of the reaction: Step 1: Cl鈧�(g) 鈬� 2 Cl(g) Step 2: Cl(g) + CHCl鈧�(g) 鉄� HCl(g) + CCl鈧�(g) Step 3: Cl(g) + CCl3(g) 鉄� CCl鈧� Now we add up the reactions, cancelling intermediates (in this case, Cl and CCl鈧�) which appear on both sides: Cl鈧�(g) + CHCl鈧�(g) 鉄� HCl(g) + CCl鈧� Hence, the overall reaction is: \(Cl_2(g) + CHCl_3(g) 鉄� HCl(g) + CCl_4\)
02

(Step 2: Identifying the intermediates in the mechanism)

The intermediates are the chemical species that are produced in one elementary reaction and consumed in another. In this mechanism, the intermediates are Cl(g) and CCl鈧�(g), as they appear on both sides of the reactions.
03

(Step 3: Determining the molecularity of each of the elementary reactions)

Molecularity describes the number of molecules participating in an elementary reaction: Step 1: Cl鈧�(g) 鈬� 2 Cl(g) (unimolecular as one molecule, Cl鈧�, is involved) Step 2: Cl(g) + CHCl鈧�(g) 鉄� HCl(g) + CCl鈧�(g) (bimolecular because two molecules, Cl and CHCl鈧�, are involved) Step 3: Cl(g) + CCl3(g) 鉄� CCl鈧� (bimolecular because two molecules, Cl and CCl鈧�, are involved)
04

(Step 4: Identifying the rate-determining step)

The rate-determining step is the slowest step in the mechanism. In this mechanism, Step 2 is mentioned as slow, so this is the rate-determining step: Step 2: Cl(g) + CHCl鈧�(g) 鉄� HCl(g) + CCl鈧�(g)
05

(Step 5: Predicting the rate law for the mechanism)

The rate-determining step is the slowest step, so the rate law is determined by this step. Using the balanced equation for Step 2, the rate law for this mechanism would be: \(Rate = k_2[Cl][CHCl_3]\) However, Cl is an intermediate, so we must find its concentration using the equilibrium established in Step 1 (fast equilibrium step): \(K_{eq} = \frac{k_1}{k_{-1}} = \frac{[Cl]^2}{[Cl_2]}\) Solving for [Cl] in terms of [Cl鈧俔: \([Cl] = \sqrt{K_{eq}[Cl_2]}\) Now we can substitute the expression for [Cl] in the rate law: \(Rate = k_2\sqrt{K_{eq}[Cl_{2}]}[CHCl_3]\) Now the rate law is expressed in terms of reactants and constants only: \(Rate = k\sqrt{[Cl_2]}[CHCl_3]\) where \(k=k_2\sqrt{K_{eq}}\) In summary: a) Overall reaction: \(Cl_2(g) + CHCl_3(g) 鉄� HCl(g) + CCl_4\) b) Intermediates: Cl(g) and CCl鈧�(g) c) Molecularity: unimolecular (Step 1), bimolecular (Step 2), bimolecular (Step 3) d) Rate-determining step: Step 2: Cl(g) + CHCl鈧�(g) 鉄� HCl(g) + CCl鈧�(g) e) Rate law: \(Rate = k\sqrt{[Cl_2]}[CHCl_3]\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elementary Reactions
An elementary reaction is a basic single-step chemical reaction where reactants convert directly into products, without undergoing a complex sequence of events. Each of these reactions has specific reactants and products, occurring under a particular set of conditions. They are the foundational building blocks of complex reaction mechanisms that we often see in chemistry.
For instance, the gas-phase reaction between chloroform (CHCl_3) and chlorine (Cl_2) consists of a series of elementary steps. Understanding each of these steps provides insight into how reactants transform into products over the course of the overall reaction.
Rate-Determining Step
The rate-determining step (RDS) acts as the 'traffic jam' in a chemical reaction mechanism. It's the slowest step in the sequence of events that occur during a chemical reaction. Identifying the rate-determining step is crucial for understanding the kinetics of a reaction, as it directly influences the overall rate at which the reaction proceeds. In our example with chloroform and chlorine, Step 2 is identified as the rate-determining step because it's marked as 'slow', meaning it dictates the time scale over which the reaction occurs. Thus, when calculating the reaction rate or discerning kinetic properties, we focus on the reactants and conditions that affect this rate-limiting step.
Reaction Intermediates
Reaction intermediates are species that are formed in one step of a reaction mechanism and consumed in subsequent steps. They are neither the initial reactants nor the final products of the overall chemical reaction. Instead, they are temporary, often unstable, and serve as a bridge between the reactants and products. In the chloroform and chlorine reaction, both Cl(g) and CCl鈧�(g) are intermediates. They are generated in the initial and middle steps but do not appear in the final equation of the overall reaction, highlighting their transitory nature during the reaction process.
Reaction Molecularity
Reaction molecularity tells us the number of reactant particles involved in an elementary reaction. It is a direct reflection of how reactants collide and interact with each other to form products. There are different types of molecularities such as unimolecular (involving one reactant molecule), bimolecular (involving two reactant molecules), and termolecular (involving three reactant molecules). In our textbook exercise, Step 1 is a unimolecular reaction where a chlorine molecule splits into two chlorine atoms, while Steps 2 and 3 are bimolecular because they involve the collisions between two different species鈥擟l and either CHCl鈧� or CCl鈧�.
Rate Law
The rate law is an expression that relates the rate of a reaction to the concentration of the reactants and is grounded in the rate-determining step of the reaction mechanism. It cannot be deduced only from the stoichiometry of the overall reaction but requires understanding of the reaction mechanism. In our chloroform and chlorine reaction, the rate law derived from Step 2, which is the rate-determining step, takes into account the concentration of the reactants involved in that specific step only. It's also influenced by the concentrations of any intermediates, like Cl(g), which are determined from other parts of the mechanism, such as the equilibrium condition of the fast step (Step 1).
Since intermediates are often very short-lived and hard to measure directly, the rate law can be modified using conditions from other steps where the intermediate is at equilibrium to express it solely in terms of the initial reactants and constant parameters.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\begin{array}{l}{\text { (a) What is meant by the term elementary reaction? }} \\ {\text { (b) What is the difference between a unimolecular }} \\\ {\text { and a bimolecular elementary reaction? (c) What is a }}\end{array}\) \(\begin{array}{l}{\text {reaction mechanism?}(\mathbf{d}) \text { What is meant by the term rate- }} \\ {\text { determining step? }}\end{array}\)

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2} .\) The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{s}^{-1} .\) Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from 1.00 \(\mathrm{L}\) of 0.600 \(\mathrm{MN}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of 20.0 \(\mathrm{h}\) if the gas is collected in a \(10.0-\mathrm{L}\) container. (Assume that the products do not dissolve in chloroform.)

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in 0.1 \(\mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$\mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)$$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 M,\) the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\) , (a) What is the rate constant, \(k ?\) units of \(s^{-1}\) . (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to 0.100\(?\)

Consider a hypothetical reaction between \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) that is first order in \(\mathrm{A},\) zero order in \(\mathrm{B},\) and second order in C. (a) Write the rate law for the reaction. (b) How does the rate change when [A] is doubled and the other reactant concentrations are held constant? (c) How does the rate change when [B] is tripled and the other reactant concentrations are held constant? (d) How does the rate change when \([C]\) is tripled and the other reactant concentrations are held constant? (e) By what factor does the rate change when the concentrations of all three reactants are tripled? (f) By what factor does the rate change when the concentrations of all three reactants are cut in half?

You have studied the gas-phase oxidation of HBr by \(\mathrm{O}_{2}\) : $$4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g)$$ You find the reaction to be first order with respect to HBr and first order with respect to \(\mathrm{O}_{2}\) . You propose the following mechanism: $$ \begin{array}{c}{\operatorname{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{HOOBr}(g)} \\ {\operatorname{HOOBr}(g)+\operatorname{HBr}(g) \longrightarrow 2 \mathrm{HOBr}(g)} \\\ {\operatorname{HOBr}(g)+\operatorname{HBr}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g)}\end{array}$$ (a) Confirm that the elementary reactions add to give the overall reaction. (b) Based on the experimentally determined rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.