/*! 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 28 The following reaction has a fir... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 28

The following reaction has a first-order rate law:$$ \begin{aligned} &\mathrm{Co}(\mathrm{CN})_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)^{2-}(a q)+\mathrm{I}(a q) \longrightarrow \mathrm{Co}(\mathrm{CN})_{5} \mathrm{I}^{3-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \\ &\text { Rate }=k\left[\mathrm{Co}(\mathrm{CN})_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)^{2-}\right] \end{aligned} $$Suggest a possible reaction mechanism, and show that your mechanism agrees with the observed rate law.

Short Answer

Expert verified
A two-step mechanism with a fast equilibrium and a slow, rate-determining step explains the rate law observations.

Step by step solution

01

Understand the Reaction and Rate Law

The given reaction involves the conversion of a substrate, \(\mathrm{Co}(\mathrm{CN})_{5}(\mathrm{H}_{2}\mathrm{O})^{2-}\), into a product where water is replaced by iodine, forming \(\mathrm{Co}(\mathrm{CN})_{5}\mathrm{I}^{3-}\). The rate law provided is first-order concerning \(\mathrm{Co}(\mathrm{CN})_{5}(\mathrm{H}_{2}\mathrm{O})^{2-}\), indicating the reaction rate depends linearly on its concentration.
02

Suggest Reaction Mechanism

For first-order reactions, a single-step mechanism (unimolecular) or multi-step with a rate-determining step that involves the decomposition of the starting material is plausible. Here, the water molecule in \(\mathrm{Co}(\mathrm{CN})_{5}(\mathrm{H}_{2}\mathrm{O})^{2-}\) could first dissociate to form an intermediate, such as \(\mathrm{Co}(\mathrm{CN})_{5}^{3-}\), which then reacts with \(\mathrm{I}^-\) to form the product \(\mathrm{Co}(\mathrm{CN})_{5}\mathrm{I}^{3-}\).
03

Define the Mechanism Steps

1. \(\mathrm{Co}(\mathrm{CN})_{5}(\mathrm{H}_{2}\mathrm{O})^{2-} \rightleftharpoons \mathrm{Co}(\mathrm{CN})_{5}^{3-} + \mathrm{H}_{2}\mathrm{O}\) (fast equilibrium)2. \(\mathrm{Co}(\mathrm{CN})_{5}^{3-} + \mathrm{I}^- \rightarrow \mathrm{Co}(\mathrm{CN})_{5}\mathrm{I}^{3-}\) (slow, rate-determining step)
04

Show the Mechanism Consistency with the Given Rate Law

The proposed two-step mechanism suggests that the intermediate \(\mathrm{Co}(\mathrm{CN})_{5}^{3-}\) reacts with \(\mathrm{I}^-\). The second step being slow and rate-determining means the rate of the overall reaction is dependent on the formation of \(\mathrm{Co}(\mathrm{CN})_{5}^{3-}\). This step is consistent with the observed rate law \(\text{Rate} = k[\mathrm{Co}(\mathrm{CN})_{5}(\mathrm{H}_{2}\mathrm{O})^{2-}]\) as it reflects the concentration of the initial reactant.

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.

First-Order Rate Law
In chemical kinetics, a first-order rate law indicates that the rate of reaction is directly proportional to the concentration of one reactant. For a reaction characterized by such a rate law, even small changes in the concentration of the reactant can significantly impact the reaction rate.
The mathematical expression of a first-order rate law is \[\text{Rate} = k[ ext{reactant}]\]Here, \( k \) is the rate constant, a unique value for each reaction at a given temperature.
This indicates that doubling the reactant's concentration doubles the rate of reaction.
In the given exercise, the rate law is: \[\text{Rate} = k\left[\mathrm{Co}(\mathrm{CN})_{5}(\mathrm{H}_{2} \mathrm{O})^{2-}\right]\]Since the reaction is first-order with respect to \(\mathrm{Co}(\mathrm{CN})_{5}(\mathrm{H}_{2} \mathrm{O})^{2-}\), this substance predominantly influences the reaction rate.
Intermediate
An intermediate is a species that forms in one step of a reaction mechanism and is consumed in another, never appearing as a final product. It exists for a short time and does not accumulate in the reaction mixture.
In the suggested reaction mechanism for the exercise, the intermediate is \(\mathrm{Co}(\mathrm{CN})_{5}^{3-}\). During the reaction, the initial substance \(\mathrm{Co}(\mathrm{CN})_{5}(\mathrm{H}_{2} \mathrm{O})^{2-}\) quickly reaches an equilibrium with \(\mathrm{Co}(\mathrm{CN})_{5}^{3-}\) and \(\mathrm{H}_{2}\mathrm{O}\).
The intermediate then reacts further with iodine to form \(\mathrm{Co}(\mathrm{CN})_{5}\mathrm{I}^{3-}\), the final product.
  • Intermediates play a crucial role in connecting the initial reactants to the final products.
  • Understanding them allows us to decipher complex reaction pathways and predict how changes in conditions might influence the reaction.
Rate-Determining Step
The rate-determining step (RDS) in a reaction mechanism is the slowest step, limiting the rate of the entire process. Think of it as a bottleneck; even if subsequent steps are faster, the overall reaction pace cannot exceed this rate.
In the proposed mechanism for the exercise, the conversion of \(\mathrm{Co}(\mathrm{CN})_{5}^{3-}\) and \(\mathrm{I}^-\) to \(\mathrm{Co}(\mathrm{CN})_{5}\mathrm{I}^{3-}\) is identified as the RDS. This selection stems from the slow nature of this step.
Since this step dictates the reaction pace, the rate law corresponds to the concentrations involved in the RDS. In the problem, the concentration of \(\mathrm{Co}(\mathrm{CN})_{5}(\mathrm{H}_{2} \mathrm{O})^{2-}\) forms the basis for the rate law, which aligns with the connection between the first step formation of the intermediate and the RDS.
  • Identifying the RDS is vital for understanding and manipulating reaction kinetics, as it provides insights into which factors most significantly control the reaction rate.
  • Alterations in conditions affecting the RDS can lead to marked changes in reaction efficiency.

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

The rate of the reaction \(\mathrm{A}+\mathrm{B}_{2} \rightarrow \mathrm{AB}+\mathrm{B}\) is directly proportional to the concentration of \(\mathrm{B}_{2}\), independent of the concentration of \(A\), and directly proportional to the concentration of a substance \(\mathrm{C}\). (a) What is the rate law? (b) Write a mechanism that agrees with the experimental facts. (c) What is the role of \(C\) in this reaction? Why doesn't \(C\) appear in the chemical equation for the overall reaction?

Polytetrafluoroethylene (Teflon) decomposes when heated above \(500{ }^{\circ} \mathrm{C}\). Rate constants for the decomposition are \(2.60 \times 10^{-4} \mathrm{~s}^{-1}\) at \(530^{\circ} \mathrm{C}\) and \(9.45 \times 10^{-3} \mathrm{~s}^{-1}\) at \(620^{\circ} \mathrm{C}\). (a) What is the activation energy in \(\mathrm{kJ} / \mathrm{mol} ?\) (b) What is the half-life of this substance at \(580^{\circ} \mathrm{C} ?\)

Consider the following mechanism for the decomposition of nitramide \(\left(\mathrm{NH}_{2} \mathrm{NO}_{2}\right)\) in aqueous solution: $$ \mathrm{NH}_{2} \mathrm{NO}_{2}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{NHNO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ \(\mathrm{NHNO}_{2}^{-}(a q) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{OH}(a q)\) (a) Write the chemical equation for the overall reaction. (b) Identify the catalyst and the reaction intermediate. (c) How will the rate of the overall reaction be affected if \(\mathrm{HCl}\) is added to the solution?

The half-life of a typical peptide bond (the \(\mathrm{C}-\mathrm{N}\) bond in a protein backbone) in neutral aqueous solution is about 500 years. When a protease enzyme acts on a peptide bond, the bond's halflife is about \(0.010 \mathrm{~s}\). Assuming that these half-lives correspond to first-order reactions, by what factor does the enzyme increase the rate of the peptide bond breaking reaction?

The values of \(E_{\mathrm{a}}=183 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta E=9 \mathrm{~kJ} / \mathrm{mol}\) have been measured for the reaction $$ 2 \mathrm{HI}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ Sketch a potential energy profile for this reaction that shows the potential energy of reactants, products, and the transition state. Include labels that define \(E_{a}\) and \(\Delta E\).
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

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.