Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 119

The rate law for a reaction involving \(\mathrm{A}(g)\) as the only reactant is: Rate \(=k[\mathrm{~A}]^{2}\) What happens to the rate when: (a) The volume of the reaction container is halved? (b) The concentration of \(\mathrm{A}\) is tripled?

Short Answer

Expert verified
(a) When the volume of the reaction container is halved, the rate of the reaction increases by 4 times. (b) When the concentration of A is tripled, the rate of the reaction increases by 9 times.

Step by step solution

01

1. Understand the Rate Law equation

The Rate Law equation for a reaction involving A(g) as the only reactant is given by: Rate \(= k[\text{A}]^2\) where: - Rate is the rate of the reaction - k is the rate constant - [A] is the concentration of the reactant A(g) We can see that the rate of this reaction depends on the square of the concentration of A.
02

2. Calculate the effect of halving the volume on the rate (a)

When the volume of the reaction container is halved, the concentration of A doubles, due to the relationship: Concentration \(=\frac{Moles}{Volume}\) We can represent the new concentration as \([A']\) and the new rate as \(Rate'\): \([A'] = 2[A]\) Now we can substitute the new concentration into the Rate Law equation: \(Rate' = k[A']^2 = k(2[A])^2 = 4k[A]^2\) Comparing with the original rate equation, we get: \(Rate' = 4 \cdot Rate\) So when the volume of the reaction container is halved, the rate of the reaction increases by 4 times.
03

3. Calculate the effect of tripling the concentration of A on the rate (b)

When the concentration of A is tripled, we can represent the new concentration as \([A']\) and the new rate as \(Rate'\): \([A'] = 3[A]\) Now we can substitute the new concentration into the Rate Law equation: \(Rate' = k[A']^2 = k(3[A])^2 = 9k[A]^2\) Comparing with the original rate equation, we get: \(Rate' = 9 \cdot Rate\) So when the concentration of A is tripled, the rate of the reaction increases by 9 times.

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

Consider the transition state for a chemical reaction. (a) What is it (define it). (b) Can there be only imminent bond breaking in a transition state? Explain.

True or false? (a) The orders \(x\) and \(y\) in a rate law are written directly from the balancing coefficients from the balanced equation for a reaction. (b) The overall order of a reaction is the sum of the individual orders of the reactants. (c) If you start a reaction with just reactants (no product present), then the rate of the reaction will remain constant with time. (d) If you start a reaction with just reactants (no product present), then the rate of the reaction will decrease with time.

True or false? An energy-downhill reaction can always be expected to be faster than an energyuphill reaction. Explain your answer.

The mechanism for the endothermic reaction \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{X}\) is Step \(1: \mathrm{A}+\mathrm{A} \rightarrow \mathrm{C}+\mathrm{D}(\mathrm{slow})\) Step \(2: \mathrm{B}+\mathrm{D} \rightarrow \mathrm{X}+\mathrm{A}\) (fast) (a) Draw the reaction-energy profile for this reaction and label reactants, products, reaction intermediates, transition states, activation energies, and \(\Delta E_{\mathrm{rxn}} .\) (Hint: First draw a profile for step \(1 .\) Make it a very endothermic reaction, and remember that a slow reaction has a large value for \(E_{\mathrm{a}}\) Then draw a profile for step 2, using the line representing the step 1 products as the reactants line for step \(2 .\) Remember that a fast reaction has a small value for \(E_{\mathrm{a}}\). (b) What is the rate law for this reaction? (c) If you wanted to quadruple the rate of this reaction, by what factor would you have to increase the concentration of \(\mathrm{A}\) ?

What is a catalyst and why is so little catalyst needed to get the desired effect?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

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