/*! 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 25 What are the units for each of t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 25

What are the units for each of the following if the concentrations are expressed in moles per liter and the time in seconds? a. rate of a chemical reaction b. rate constant for a zero-order rate law c. rate constant for a first-order rate law d. rate constant for a second-order rate law e. rate constant for a third-order rate law

Short Answer

Expert verified
a. The units for the rate of a chemical reaction are mol/Ls. b. The units for the rate constant of a zero-order rate law are mol/Ls. c. The units for the rate constant of a first-order rate law are s鈦宦�. d. The units for the rate constant of a second-order rate law are L/mol路s. e. The units for the rate constant of a third-order rate law are L虏/mol虏路s.

Step by step solution

01

a. Units for the rate of a chemical reaction

The rate of a chemical reaction (represented as 'r') is usually expressed as the change in concentration (鈭哰A]) of the reacting species A over the change in time (鈭唗). Thus: \(r = \frac{-\Delta{[A]}}{\Delta{t}}\) The units for the change in concentration are moles per liter (mol/L) and time in seconds (s). Therefore, the unit for the rate of a chemical reaction will be in mol/Ls.
02

b. Units for the rate constant of a zero-order rate law

In a zero-order rate law, the rate of the reaction (r) is directly proportional to the concentration of the reactant [A]^0: \(r = k [A]^0\) Since [A]^0 = 1, the rate constant k has the same units as the rate of reaction. Thus, the units for a zero-order rate constant k are mol/Ls.
03

c. Units for the rate constant of a first-order rate law

In a first-order rate law, the rate of the reaction (r) is directly proportional to the concentration of the reactant [A]: \(r = k [A]\) To find the units of the rate constant k, we rearrange the equation: \(k = \frac{r}{[A]}\) Since the units of the rate are mol/Ls and the units of the concentration are mol/L, the units for the rate constant k in a first-order rate law are s鈦宦�.
04

d. Units for the rate constant of a second-order rate law

In a second-order rate law, the rate of the reaction (r) is directly proportional to the square of the concentration of the reactant [A]^2: \(r = k [A]^2\) To find the units of the rate constant k, we rearrange the equation: \(k = \frac{r}{[A]^2}\) Since the units of the rate are mol/Ls and the units of the concentration squared are (mol/L)^2, the units for the rate constant k in a second-order rate law are L/mol路s.
05

e. Units for the rate constant of a third-order rate law

In a third-order rate law, the rate of the reaction (r) is directly proportional to the cube of the concentration of the reactant [A]^3: \(r = k [A]^3\) To find the units of the rate constant k, we rearrange the equation: \(k = \frac{r}{[A]^3}\) Since the units of the rate are mol/Ls and the units of the concentration cubed are (mol/L)^3, the units for the rate constant k in a third-order rate law are L虏/mol虏路s.

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

Cobra venom helps the snake secure food by binding to acetylcholine receptors on the diaphragm of a bite victim, leading to the loss of function of the diaphragm muscle tissue and eventually death. In order to develop more potent antivenoms, scientists have studied what happens to the toxin once it has bound the acetylcholine receptors. They have found that the toxin is released from the receptor in a process that can be described by the rate law Rate \(=k[\) acetylcholine receptor-toxin complex \(]\) If the activation energy of this reaction at \(37.0^{\circ} \mathrm{C}\) is \(26.2 \mathrm{~kJ} / \mathrm{mol}\) and \(A=0.850 \mathrm{~s}^{-1}\), what is the rate of reaction if you have a \(0.200 M\) solution of receptor-toxin complex at \(37.0^{\circ} \mathrm{C}\) ?

Describe at least two experiments you could perform to determine a rate law.

The central idea of the collision model is that molecules must collide in order to react. Give two reasons why not all collisions of reactant molecules result in product formation.

A certain reaction has an activation energy of \(54.0 \mathrm{~kJ} / \mathrm{mol}\). As the temperature is increased from \(22^{\circ} \mathrm{C}\) to a higher temperature, the rate constant increases by a factor of \(7.00 .\) Calculate the higher temperature.

Assuming that the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section \(12.7\) is correct, would you predict that the product of the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) would be \(\mathrm{CH}_{2} \mathrm{D}-\mathrm{CH}_{2} \mathrm{D}\) or \(\mathrm{CHD}_{2}-\mathrm{CH}_{3} ?\) How could the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) be used to confirm the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section \(12.7\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

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.