/*! 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 68 What is the molecularity of each... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 68

What is the molecularity of each of the following elementary reactions? Write the rate law for each. \(\begin{array}{l}{\text { (a) } 2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)} \\ {\mathrm{CH}_{2}} \\ {\text { (b) } \mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}(g)} \\ {\text { (c) } \mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)}\end{array}\)

Short Answer

Expert verified
(a) The molecularity of the reaction \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\) is 2. The rate law is: Rate = k[NO]^2. (b) The molecularity of the reaction \(\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}(g)\) is 1. The rate law is: Rate = k[H鈧侰-CH鈧俔. (c) The molecularity of the reaction \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)\) is 1. The rate law is: Rate = k[SO鈧僝.

Step by step solution

01

(a) Identify Molecularity and Rate Law for Reaction 1:

In the first elementary reaction, we have: \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\) The molecularity of this reaction is 2, as there are two molecules of NO reacting. The rate law for this reaction is given by: Rate = k[NO]^2 Where Rate is the reaction rate and k is the rate constant.
02

(b) Identify Molecularity and Rate Law for Reaction 2:

In the second elementary reaction, we have: \(\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}(g)\) The molecularity of this reaction is 1, as there is only one molecule of H鈧侰-CH鈧� reacting. The rate law for this reaction is given by: Rate = k[H鈧侰-CH鈧俔 Where Rate is the reaction rate and k is the rate constant.
03

(c) Identify Molecularity and Rate Law for Reaction 3:

In the third elementary reaction, we have: \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)\) The molecularity of this reaction is 1, as there is only one molecule of SO鈧� reacting. The rate law for this reaction is given by: Rate = k[SO鈧僝 Where Rate is the reaction rate and k is the rate constant.

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.

Molecularity
Molecularity in chemical kinetics refers to the number of molecules participating as reactants in an elementary reaction. Understanding molecularity helps us analyze simple reactions, which consist of discrete steps, to form a comprehensive picture of complex reactions.
There are three primary types of molecularity:
  • Unimolecular: A single molecule undergoes a transformation, as seen in the reaction \(SO_{3}(g) \rightarrow SO_{2}(g) + O(g)\). This is a unimolecular reaction because only one molecule is involved.
  • Bimolecular: Two molecules come together to react, such as in \(2 NO(g) \rightarrow N_{2}O_{2}(g)\). Here, two molecules of NO are involved, making it a bimolecular reaction.
  • Termolecular: This involves three molecules reacting, although such reactions are rare because the simultaneous collision of three molecules is less probable.
Understanding these classifications helps in predicting the behavior of reactions and facilitates the development of accurate rate laws.
Rate Law
The rate law of a chemical reaction is an equation that relates the rate of reaction to the concentration of reactants. For elementary reactions, the rate law is directly related to the molecularity.
The general form of a rate law is given by: \[\text{Rate} = k[A]^m[B]^n\]where:
  • \(k\) is the rate constant, a unique value for each reaction at a given temperature.
  • \([A]\) and \([B]\) are the concentrations of the reactants.
  • \(m\) and \(n\) are the orders of reaction for each reactant, determined by the stoichiometry of the elementary step.
For example, in the reaction \(2 NO(g) \rightarrow N_{2}O_{2}(g)\), the rate law is \(\text{Rate} = k[NO]^2\), directly reflecting its bimolecular nature. Similarly, \(\text{Rate} = k[SO_{3}]\) represents a unimolecular reaction, showing that only one molecule of \(SO_{3}\) contributes to the rate.
Elementary Reactions
Elementary reactions are fundamental steps in a reaction mechanism. These reactions occur in a single event or step at the molecular level. Understanding these reactions is crucial for developing accurate reaction mechanisms.
Characteristics of elementary reactions include:
  • Stoichiometry directly reflects the molecularity: The coefficients in the balanced equation indicate how many molecules of each reactant are involved. For example, \(2 NO(g) \rightarrow N_{2}O_{2}(g)\) shows the involvement of two molecules of NO.
  • Elementary reactions are not generalized: Unlike overall reactions, elementary reactions are defined by specific collisions and transformations.
  • Rate laws can be directly determined: Because elementary reactions occur as single-step processes, their rate laws are directly linked to their molecularity and stoichiometric coefficients.
Understanding elementary reactions helps chemists build full kinetic models by piecing together how complex reactions might proceed from simpler steps.

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 activation energy of an uncatalyzed reaction is 95 \(\mathrm{kJ} / \mathrm{mol} .\) The addition of a catalyst lowers the activation energy to 55 \(\mathrm{kJ} / \mathrm{mol}\) . Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C},\) (b) \(125^{\circ} \mathrm{C} ?\)

The gas-phase decomposition of ozone is thought to occur by the following two- step mechanism. \(\begin{array}{ll}{\text { Step } 1 :} & {\mathrm{O}_{3}(g) \Longrightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g) \text { (fast) }} \\ {\text { Step } 2 :} & {\mathrm{O}(g)+\mathrm{O}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(g) \quad(\text { slow })}\end{array}\) (a) Write the balanced equation for the overall reaction. (b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.) (c) Is O a catalyst or an intermediate? (d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a \(1 \times 10^{5}\) -fold increase in the reaction rate?

Which of the following linear plots do you expect for a reaction \(A \longrightarrow\) products if the kinetics are (a) zero order, (b) first order, or (c) second order? [Section 14.4\(]\)

As described in Exercise 14.41 , the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at 660 \(\mathrm{K}\) is \(4.5 \times 10^{-2} \mathrm{s}^{-1}\) .half-life for this reaction? (b) If you start with 0.050\(M \mathrm{I}_{2}\) at this temperature, how much will remain after 5.12 s assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

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.