Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 17

The initial rate of a reaction doubles as the concentration of one of the reactants is quadrupled. What is the order of this reactant? If a reactant has a \(-1\) order, what happens to the initial rate when the concentration of that reactant increases by a factor of two?

Short Answer

Expert verified
The order of reactant A is 1, as the initial rate doubles when the concentration of the reactant quadruples. If a reactant has a -1 order, when the concentration of that reactant increases by a factor of two, the initial rate will be halved.

Step by step solution

01

Understand the rate law expression

The rate law expression relates the rate of reaction to the concentration of the reactants. It is given by: Rate \(= k[A]^m\) Where 'Rate' is the initial rate of the reaction, 'k' is the rate constant, \([A]\) is the concentration of reactant A, and 'm' is the order of reactant A.
02

Set up the given information and the rate law expressions

We are given that the initial rate doubles when the concentration of reactant A is quadrupled. Therefore, we can set up the following expressions: \(Rate_1 = k[A_1]^m\) \(Rate_2 = k[A_2]^m = 2Rate_1\) Where \([A_2]\) is four times the concentration of \([A_1]\): \([A_2] = 4[A_1]\)
03

Find the order of the reactant

Now, we can write the rate law expressions for the two cases and divide the second equation by the first to eliminate 'k': \(\dfrac{Rate_2}{Rate_1} = \dfrac{k[A_2]^m}{k[A_1]^m}\) \(\dfrac{2Rate_1}{Rate_1} = \dfrac{k(4[A_1])^m}{k[A_1]^m}\) \(2 = (4)^m\) To find 'm', we can take the logarithm of both sides: \(\log_4{2} = m\) \(m = 1\) The order of reactant A is 1.
04

Analyze the impact of a -1 order and increasing concentration by a factor of 2

Now, we need to determine what happens to the initial rate when the concentration of reactant A increases by a factor of 2 and the order is -1. Let's use the rate law expression: Rate \(= k[A]^m\) Here, \(m = -1\) and \([A'] = 2[A]\), where \([A']\) is the new concentration of reactant A. We can write the expression for the new rate: New Rate \(= k[A']^{-1}\) \(New Rate = k(2[A])^{-1}\) \(New Rate = k\dfrac{1}{2[A]}\) Since the rate is inversely proportional to the concentration, when the concentration of reactant A is doubled, the rate of the reaction will be halved.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 antivenins, 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 $$ \text {Rate} =k[\text { acetylcholine receptor-toxin complex }] $$ If the activation energy of this reaction at \(37.0^{\circ} \mathrm{C}\) is 26.2 \(\mathrm{kJ} /\) mol and \(A=0.850 \mathrm{s}^{-1},\) what is the rate of reaction if you have \(\mathrm{a} 0.200-\mathrm{M}\) solution of receptor-toxin complex at \(37.0^{\circ} \mathrm{C} ?\)

The rate law for the reaction $$ \begin{array}{c}{\mathrm{Cl}_{2}(g)+\mathrm{CHCl}_{3}(g) \longrightarrow \mathrm{HCl}(g)+\mathrm{CCl}_{4}(g)} \\ {\text { Rate }=k\left[\mathrm{Cl}_{2}\right]^{1 / 2}\left[\mathrm{CHCl}_{3}\right]}\end{array} $$ What are the units for \(k,\) assuming time in seconds and concentration in mol/L?

Two isomers (A and B) of a given compound dimerize as follows: $$ \begin{array}{l}{2 \mathrm{A} \stackrel{k_{1}}{\longrightarrow} A_{2}} \\ {2 \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{B}_{2}}\end{array} $$ Both processes are known to be second order in reactant, and \(k_{1}\) is known to be 0.250 \(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) at $25^{\circ} \mathrm{C}\( . In a particular experiment \)\mathrm{A}\( and \)\mathrm{B}$ were placed in separate containers at \(25^{\circ} \mathrm{C},\) where \([\mathrm{A}]_{0}=1.00 \times 10^{-2} M\) and $[\mathrm{B}]_{0}=2.50 \times 10^{-2} M\( It was found that after each reaction had progressed for \)3.00 \mathrm{min},[\mathrm{A}]=3.00[\mathrm{B}]$ . In this case the rate laws are defined as $$ \begin{array}{l}{\text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k_{1}[\mathrm{A}]^{2}} \\ {\text { Rate }=-\frac{\Delta[\mathrm{B}]}{\Delta t}=k_{2}[\mathrm{B}]^{2}}\end{array} $$ a. Calculate the concentration of \(\mathrm{A}_{2}\) after 3.00 \(\mathrm{min}\) . b. Calculate the value of \(k_{2}\) . c. Calculate the half-life for the experiment involving A.

Define what is meant by unimolecular and bimolecular steps. Why are termolecular steps infrequently seen in chemical reactions?

The rate law of a reaction can only be determined from experiment. Two experimental procedures for determining rate laws were outlined in Chapter 12. What are the two procedures and how are they used to determine the rate laws?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

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.

Sign-up for free