/*! 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 4 Define the terms "unimolecular e... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Define the terms "unimolecular elementary reaction" and "bimolecular elementary reaction," and give an example of each.

Short Answer

Expert verified
Unimolecular: single reactant changes. Example: cyclopropane to propene. Bimolecular: two reactants collide. Example: \( \text{H}_2 + \text{I}_2 \rightarrow 2\text{HI} \).

Step by step solution

01

Define Unimolecular Elementary Reaction

A unimolecular elementary reaction is a type of chemical reaction where a single reactant molecule undergoes a transformation to form products. This transformation happens in a single step without any intermediates. These reactions are often described by a first-order rate law, where the rate of reaction depends only on the concentration of the single reactant.
02

Provide Example of a Unimolecular Reaction

An example of a unimolecular elementary reaction is the isomerization of cyclopropane to propene. In this reaction, a cyclopropane molecule rearranges itself to form propene in a single step.
03

Define Bimolecular Elementary Reaction

A bimolecular elementary reaction involves the collision and interaction of two reactant molecules to form products. This reaction happens in a single step and is characterized by a second-order rate law, where the rate of reaction depends on the concentration of both reactants.
04

Provide Example of a Bimolecular Reaction

An example of a bimolecular elementary reaction is the reaction between hydrogen and iodine to form hydrogen iodide: \( \text{H}_2 + \text{I}_2 \rightarrow 2\text{HI} \). This reaction occurs when one molecule of hydrogen and one molecule of iodine collide to produce hydrogen iodide.

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 91Ó°ÊÓ!

Key Concepts

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

Unimolecular Reaction
A unimolecular reaction is an elementary reaction involving a single molecule that transforms into one or more different molecules. This type of reaction occurs in a single, uninterrupted step, without the involvement of any intermediates or additional reactants. The kinetic behavior of unimolecular reactions is often described by a first-order rate law, where the reaction rate is directly proportional to the concentration of the single reactant.
For example, in the isomerization of cyclopropane to propene, one cyclopropane molecule rearranges its atomic structure to form propene in a straightforward, solitary step. This is a classic demonstration of how unimolecular reactions operate, relying solely on the structural change of one reactant molecule.
Bimolecular Reaction
In contrast, a bimolecular reaction involves the interaction of two distinct reactant molecules. This meeting, often described as a collision, leads to the formation of products. These reactions follow a second-order rate law, where the rate depends on the concentrations of both participating reactants.
A quintessential example of a bimolecular reaction is the formation of hydrogen iodide (HI) from hydrogen (\( \text{H}_2 \)) and iodine (\( \text{I}_2 \)). Here, the molecules must collide with each other with sufficient energy to form two molecules of hydrogen iodide. Such reactions are fundamental in understanding reaction mechanisms involving two reactants.
Rate Laws
Rate laws are mathematical expressions that describe the speed of chemical reactions. They are crucial for studying reaction kinetics as they link the rate of a reaction to the concentration of its reactants.
  • First-order reactions, like unimolecular reactions, have rates that are directly proportional to the concentration of one reactant. The rate law can be expressed as \(Rate = k[A]\), where \(k\) is the rate constant and \([A]\) is the concentration of the reactant.
  • Second-order reactions, such as bimolecular reactions, depend on the concentrations of two reactants. Their rate law might look like \(Rate = k[A][B]\), involving two concentrations, \([A]\) and \([B]\).
These equations help predict how fast a reaction will proceed under given conditions, and can be derived from experimental data or theoretical models.
Chemical Kinetics
Chemical kinetics is the branch of chemistry that focuses on understanding the speed or rate of chemical reactions. It explores how different conditions, such as temperature, concentration, and catalysts, affect these rates. This field is essential for designing chemical processes and optimizing reactions for industrial applications.
In kinetics, elementary reactions like unimolecular and bimolecular processes are considered the simplest forms because they occur in single steps. Studying these reactions gives insights into more complex reaction mechanisms, allowing chemists to construct models that predict reaction behavior on a macroscopic scale.
  • The study of kinetics also involves exploring reaction mechanisms, pathways, and the energy changes that occur during reactions.
  • Understanding kinetics is vital for manipulating reaction conditions to achieve desired outcomes quickly and efficiently.
Kinetics provides a detailed picture of how molecules interact and react, making it an integral part of both theoretical and applied chemistry.

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

Express the rate of the reaction $$ 2 \mathrm{~N}_{2} \mathrm{H}_{4}(\ell)+\mathrm{N}_{2} \mathrm{O}_{4}(\ell) \longrightarrow 3 \mathrm{~N}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ in terms of (a) \(\Delta\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]\). (b) \(\Delta\left[\mathrm{N}_{2}\right]\)

A reaction has the experimental rate law, Rate = \(k[\mathrm{~A}]^{2}[\mathrm{~B}]\). If the concentration of \(\mathrm{A}\) is doubled and the concentration of \(\mathrm{B}\) is halved, what happens to the reaction rate?

The rate constant for decomposition of azomethane at $$ \begin{aligned} 425^{\circ} \mathrm{C} \text { is } 0.68 \mathrm{~s}^{-1} & \\ \mathrm{CH}_{3} \mathrm{~N}=\mathrm{NCH}_{3}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) \end{aligned} $$ (a) Based on the units of the rate constant, determine if this reaction is zeroth-, first-, or second-order. (b) If \(2.0 \mathrm{~g}\) azomethane is placed in a \(2.0-\mathrm{L}\) flask and heated to \(425^{\circ} \mathrm{C},\) calculate the mass of azomethane that remains after \(5.0 \mathrm{~s}\). (c) Calculate how long it takes for the mass of azomethane to drop from \(2.0 \mathrm{~g}\) to \(0.24 \mathrm{~g}\). (d) Calculate the mass of nitrogen that would be found in the flask after \(0.50 \mathrm{~s}\) of reaction.

Which of these is appropriate for determining the rate law for a chemical reaction? (a) Theoretical calculations based on balanced equations (b) Measuring the rate of the reaction as a function of the concentrations of the reacting species (c) Measuring the rate of the reaction as a function of temperature

Which of these statements is (are) true? If a statement is false, reword it so that it becomes true. (a) The concentration of a homogeneous catalyst may appear in the rate law. (b) A catalyst is always consumed in the reaction. (c) A catalyst must always be in the same phase as the reactants. (d) A catalyst can change the course of a reaction and allow different products to be produced.

See all solutions

Recommended explanations on Chemistry Textbooks

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.