/*! 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 64 The mechanism for the gas-phase ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 64

The mechanism for the gas-phase reaction of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon dioxide is thought to be $$ \begin{array}{c}{\mathrm{NO}_{2}+\mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\mathrm{NO}} \\ {\mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}}\end{array} $$ Write the rate law expected for this mechanism. What is the overall balanced equation for the reaction?

Short Answer

Expert verified
The rate law for the given reaction mechanism is: Overall Rate = (k鈧/k鈧)(Rate鈧)[CO] Where Rate鈧 = k鈧乕NO鈧俔虏 and k鈧 and k鈧 are the rate constants for the elementary steps. The overall balanced equation for the reaction is: 2NO鈧 + CO 鈫 2NO + CO鈧

Step by step solution

01

Write the rate law for each elementary step

To find the rate law for each elementary step, we need to look at the coefficients of the reactants in the step. For an elementary reaction, the rate law can be directly obtained from the reaction equation. 1. For the first elementary step, NO鈧 + NO鈧 鈫 NO鈧 + NO, we have two molecules of NO鈧 involved. Rate鈧 = k鈧乕NO鈧俔虏 2. For the second elementary step, NO鈧 + CO 鈫 NO鈧 + CO鈧, we have one molecule of NO鈧 and one molecule of CO involved. Rate鈧 = k鈧俒NO鈧僝[CO]
02

Identify the reaction intermediates

Reaction intermediates are species that are produced during one elementary step and consumed during another, so they don't appear in the overall balanced equation. In this case, NO鈧 is the reaction intermediate, as it is produced in the first step and consumed in the second step.
03

Write the overall rate law

Since NO鈧 is a reaction intermediate, we need to express [NO鈧僝 in terms of other concentrations. From the first step, we have: [NO鈧僝 = (Rate鈧)/(k鈧乕NO鈧俔) Now we can substitute this into the Rate鈧 expression: Overall Rate = Rate鈧 = k鈧俒(Rate鈧)/(k鈧乕NO鈧俔)][CO] This simplifies to: Overall Rate = (k鈧/k鈧)(Rate鈧)[CO]
04

Write the overall balanced equation

Now we can find the overall balanced equation for the reaction by adding the two elementary steps: (NO鈧 + NO鈧 鈫 NO鈧 + NO) + (NO鈧 + CO 鈫 NO鈧 + CO鈧) The NO鈧 intermediate is produced in the first step and consumed in the second step. Therefore, it does not appear in the overall balanced equation. The equation simplifies to: NO鈧 + NO鈧 + CO 鈫 NO + NO鈧 + CO鈧 Further simplifying it, we have: 2NO鈧 + CO 鈫 2NO + CO鈧 So, the overall balanced equation is: 2NO鈧 + CO 鈫 2NO + CO鈧

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.

Rate Law
The rate law helps us understand how the concentration of reactants influences the rate of a chemical reaction.
In the reaction mechanism, each elementary step has its own rate law. An elementary reaction is a single reaction step with its own participating molecules.
In such a mechanism, the rate law can be directly written from the stoichiometry of the reactants involved. For example:
  • The first step in the given reaction mechanism: \(\text{NO}_2 + \text{NO}_2 \rightarrow \text{NO}_3 + \text{NO}\) has two \(\text{NO}_2\) molecules. Thus, its rate law is \(\text{Rate}_1 = k_1 [\text{NO}_2]^2\).
  • For the second step: \(\text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2\), the molecules \(\text{NO}_3\) and \(\text{CO}\) participate, leading to the rate law: \(\text{Rate}_2 = k_2 [\text{NO}_3][\text{CO}]\).
It's important to note that reaction intermediates, such as \(\text{NO}_3\) in this mechanism, must be expressed in terms of the concentrations of stable species when determining the overall rate law.
Elementary Reaction
An elementary reaction is a fundamental concept in the study of reaction mechanisms. These are single events where reactants convert to products in a single step with no intermediates. The rate law for such a reaction can be directly extracted from its molecularity, which is the number of molecules colliding in that step.
In the mechanism you analyzed:
  • The step \( \text{NO}_2 + \text{NO}_2 \rightarrow \text{NO}_3 + \text{NO} \) is an example of a bimolecular reaction as it involves two molecules of \(\text{NO}_2\).
  • The step \( \text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2 \) is also a bimolecular reaction, involving one \(\text{NO}_3\) and one \(\text{CO}\) molecule.
The stoichiometry of these elementary reactions is essential in deducing the rate laws and understanding the overall mechanism of a complex reaction.
Reaction Intermediates
Reaction intermediates are species formed in one step of a mechanism and consumed in another step, never appearing in the overall balanced equation of the reaction. In your mechanism, \(\text{NO}_3\) is a key intermediate.
Intermediates like \(\text{NO}_3\) are crucial for constructing a correct overall rate law:
  • \(\text{NO}_3\) is produced during the first step \(\text{NO}_2 + \text{NO}_2 \rightarrow \text{NO}_3 + \text{NO}\) and consumed in the second \(\text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2\).
  • In the overall rate law, you must eliminate intermediates like \(\text{NO}_3\) by substituting from other steps, as seen with \(\text{NO}_3 = \frac{\text{Rate}_1}{k_1[\text{NO}_2]}\).
Understanding reaction intermediates allows chemists to piece together how and why reactions proceed the way they do.
Overall Balanced Equation
The overall balanced equation provides the simplified stoichiometric representation of all reactants and products in a reaction, without intermediates. To derive this, add each elementary step and cancel out intermediates.
For the reaction mechanism you are exploring:
  • The elementary steps are combined: \(\text{NO}_2 + \text{NO}_2 \rightarrow \text{NO}_3 + \text{NO}\) and \(\text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2\).
  • By adding these, the intermediates \(\text{NO}_3\) cancel, as it is formed and then consumed across steps.
  • The resulting overall balanced equation is: \(2\text{NO}_2 + \text{CO} \rightarrow 2\text{NO} + \text{CO}_2\).
This balanced equation reflects the total change from initial reactants to final products, omitting the transient intermediates.

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

Which of the following statement(s) is(are) true? a. The half-life for a zero-order reaction increases as the reaction proceeds. b. A catalyst does not change the value of \(\Delta \mathrm{E}\) . c. The half-life for a reaction, aA \(\longrightarrow\) products, that is first order in A increases with increasing \([\mathrm{A}]_{0} .\) d. The half-life for a second-order reaction increases as the reaction proceeds.

The rate law of a reaction can only be determined from experiment. Two experimental procedures for determining rate laws were outlined in Chapter 12. What are the two procedures and how are they used to determine the rate laws?

The thiosulfate ion \(\left(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\right)\) is oxidized by iodine as follows: $$ 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)+2 \mathrm{I}^{-}(a q) $$ In a certain experiment, \(7.05 \times 10^{-3}\) mol/L of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is consumed in the first 11.0 seconds of the reaction. Calculate the rate of consumption of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) . Calculate the rate of production of iodide ion.

How does temperature affect k, the rate constant? Explain.

A certain reaction has the following general form: $$ \mathrm{aA} \longrightarrow \mathrm{bB} $$ At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} M,\) con- centration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{min}^{-1}\) . a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to \(2.50 \times 10^{-3} M ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

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.