/*! 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 are the units of the rate c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 68

What are the units of the rate constant for a thirdorder reaction?

Short Answer

Expert verified
The units of the rate constant for a thirdorder reaction are \(M^{-2}T^{-1}\).

Step by step solution

01

Understanding reaction rates

Reaction rate can also be expressed in terms of the decrease in the concentration of one of the reactants or the increase in the concentration of one of the products during a unit of time.
02

Expressing reaction rates in terms of reactant concentration and rate constant

The rate of reaction can be expressed as the change in molar concentration of the reactant in a unit time and mathematically, Rate = \(-\frac{d[\text{{Reactant}}]}{dt}\) where \([\text{{Reactant}}]\) is the molar concentration of the reactant. For a third order reaction, Rate = \(k[\text{{Reactant}}]^3\) where k is the rate constant.
03

Determining units of the third order rate constant

From the equation in Step 2, we infer that the dimensions of the rate constant 'k' can be determined by using the dimensional analysis. The dimension of 'Rate' is \([M T^{-1}]\), 'M' being the molar concentration and 'T' being the time. So, the unit of 'k' for third order reactions should be \(M^{-2}T^{-1}\)

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!

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

For a first-order reaction, how long will it take for the concentration of reactant to fall to one-eighth its original value? Express your answer in terms of the half-life \(\left(t_{\frac{1}{2}}\right)\) and in terms of the rate constant \(k\)

Write the Arrhenius equation and define all terms.

Distinguish between homogeneous catalysis and heterogeneous catalysis. Describe some important industrial processes that utilize heterogeneous catalysis.

The bromination of acetone is acid-catalyzed: \(\mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{Br}_{2} \frac{\mathrm{H}^{+}}{\text {catalyst }} \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{Br}+\mathrm{H}^{+}+\mathrm{Br}\) The rate of disappearance of bromine was measured for several different concentrations of acetone, bromine, and \(\mathrm{H}^{+}\) ions at a certain temperature: $$ \begin{array}{lcllc} & & & & {\text { Rate of }} \\ & & & & \text { Disappearance } \\ & {\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]} & {\left[\mathrm{Br}_{2}\right]} & {\left[\mathrm{H}^{+}\right]} & \text {of } \mathrm{Br}_{2}(\mathrm{M} / \mathrm{s}) \\ \hline \text { (a) } & 0.30 & 0.050 & 0.050 & 5.7 \times 10^{-5} \\ \text {(b) } & 0.30 & 0.10 & 0.050 & 5.7 \times 10^{-5} \\ \text {(c) } & 0.30 & 0.050 & 0.10 & 1.2 \times 10^{-4} \\ \text {(d) } & 0.40 & 0.050 & 0.20 & 3.1 \times 10^{-4} \\ \text {(e) } & 0.40 & 0.050 & 0.050 & 7.6 \times 10^{-5} \end{array} $$ (a) What is the rate law for the reaction? (b) Determine the rate constant.

How does a catalyst increase the rate of a reaction?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

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.