   # The rate law for the reaction Cl 2 ( g ) +CHCl 3 ( g ) → HCl ( g ) +CCl 4 ( g ) is Rate = k [ Cl 2 ] 1 / 2 [ CHCl 3 ] What are the units for k , assuming time in seconds and concentration in mol/L? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 11, Problem 28E
Textbook Problem
1 views

## The rate law for the reaction Cl 2 ( g ) +CHCl 3 ( g ) → HCl ( g ) +CCl 4 ( g ) is Rate   =   k [ Cl 2 ] 1 / 2 [ CHCl 3 ] What are the units for k, assuming time in seconds and concentration in mol/L?

Interpretation Introduction

Interpretation: A reaction between Cl2 and CHCl3 and its rate equation is given. The unit for the rate constant k is to be stated.

Concept introduction: The rate of a reaction is directly proportional to the concentration of the reactant. For a reaction, XProducts ,

Rate[X]nRate=k[X]n

The constant k in the above expression is the rate constant.

To determine: The unit for the rate constant k for the given reaction.

### Explanation of Solution

Given

The stated reaction is,

Cl2(g)+CHCl3(g)HCl(g)+CCl4(g)

The rate equation for the given reaction is,

Rate=k[Cl2]1/2[CHCl3]

Where,

• k is the rate constant.

Concentration is expressed in moles per liter (mol/L) and time in seconds (s) .

The rate of a reaction is calculated by the formula,

Rateofreaction=ChangeintheconcentrationChangeintime

Substitute the given units of concentration and time in the above expression.

Rateofreaction=mol/Ls=molL1s1

The rate equation for the given reaction is,

Rate=k[Cl2]1/2[CHCl3]

Substitute the units of rate, unit of concentration of the reactants in the above expression

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 