Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 45

Explain why (a) A reaction rate cannot be calculated from the collision frequency alone. (b) The rate of a chemical reaction may increase dramatically with temperature, whereas the collision frequency increases much more slowly. (c) The addition of a catalyst to a reaction mixture can have such a pronounced effect on the rate of a reaction, even if the temperature is held constant.

Short Answer

Expert verified
Reaction rates cannot be calculated from collision frequency alone because not all collisions result in reactions. Only those with sufficient energy and correct orientation will be effective. Reaction rates increase dramatically with temperature not just because of increased collision frequency, but also thanks to a stark upsurge in the number of molecules with necessary activation energy. A catalyst can increase reaction rates without needing temperature fluctuation as it provides an alternate reaction pathway with a lower activation energy.

Step by step solution

01

Justifying the inadequacy of collision frequency alone to determine reaction rate

A chemical reaction occurs when reactant molecules collide with each other. However, not all collisions result in reactions. This is because, for a reaction to occur, the colliding molecules must have a certain minimum amount of energy called activation energy and must also be correctly oriented. Therefore, only those collisions which meet this criterion are said to be effective or fruitful. Hence, collision frequency alone is not sufficient to determine the rate of a reaction.
02

Understanding temperature's effect on reaction rate versus collision frequency

While it's true that increasing the temperature increases the collision frequency, it also escalates the proportion of molecules having energy equal to or greater than the activation energy. This dramatic increase in the number of molecules with requisite energy, as a result of temperature hike, outweighs the more modest increase in collision frequency alone, resulting in a significant increase in reaction rate.
03

Effect of a catalyst on reaction rate, holding temperature constant

A catalyst works by providing an alternate reaction pathway with a lower activation energy than the uncatalyzed reaction. Thus, more reactant molecules will have the necessary energy to react in the presence of a catalyst, even without a boost in temperature. As a result, the addition of a catalyst can significantly increase the rate of reaction while maintaining constant temperature.

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

Ammonia decomposes on the surface of a hot tungsten wire. Following are the half-lives that were obtained at \(1100^{\circ} \mathrm{C}\) for different initial concentrations of \(\mathrm{NH}_{3}:\left[\mathrm{NH}_{3}\right]_{0}=0.0031 \mathrm{M}, t_{1 / 2}=7.6 \mathrm{min} ; 0.0015 \mathrm{M}\) \(3.7 \mathrm{min} ; 0.00068 \mathrm{M}, 1.7 \mathrm{min.}\) For this decomposition reaction, what is (a) the order of the reaction; (b) the rate constant, \(k ?\)

For the first-order reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ \(t_{1 / 2}=22.5 \mathrm{h}\) at \(20^{\circ} \mathrm{C}\) and \(1.5 \mathrm{h}\) at \(40^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) If the Arrhenius constant \(A=2.05 \times 10^{13} \mathrm{s}^{-1}\) determine the value of \(k\) at \(30^{\circ} \mathrm{C}\).

For the reversible reaction \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D},\) the enthalpy change of the forward reaction is \(+21 \mathrm{kJ} / \mathrm{mol}\) The activation energy of the forward reaction is \(84 \mathrm{kJ} / \mathrm{mol}.\) (a) What is the activation energy of the reverse reaction? (b) In the manner of Figure 14-10, sketch the reaction profile of this reaction.

For the reaction \(A \longrightarrow\) products, the following data were obtained: \(t=0 \mathrm{s},[\mathrm{A}]=0.715 \mathrm{M} ; 22 \mathrm{s}, 0.605 \mathrm{M}\) 74 s, 0.345 M; 132 s, 0.055 M. (a) What is the order of this reaction? (b) What is the half-life of the reaction?

The first-order reaction \(A \longrightarrow\) products has a halflife, \(t_{1 / 2},\) of 46.2 min at \(25^{\circ} \mathrm{C}\) and \(2.6 \mathrm{min}\) at \(102^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) At what temperature would the half-life be 10.0 min?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

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