Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 14

Consider our substitution reaction between \(\mathrm{OH}^{-}\) and \(\mathrm{CH}_{3} \mathrm{Br}\). Imagine it is occurring in a solution where there are 1000 collisions every second between \(\mathrm{OH}^{-}\) ions and \(\mathrm{CH}_{3}\) Br molecules. Suppose only \(10 \%\) of these collisions are sufficiently energetic to lead to products. Also, assume that the orientation factor is \(0.2\). (a) What does an orientation factor of \(0.2\) mean? (b) What is the rate of this reaction? Give your answer in number of \(\mathrm{CH}_{3} \mathrm{OH}\) molecules formed per second.

Short Answer

Expert verified
(a) An orientation factor of 0.2 means that only 20% of the collisions between OH- and CH3Br have a favorable orientation for the reaction to proceed. (b) The rate of this reaction is 20 CH3OH molecules formed per second, as there are 20 successful collisions between OH- and CH3Br per second.

Step by step solution

01

Understanding orientation factor

The orientation factor represents the fraction of collisions that have the appropriate orientation to lead to a successful reaction. When two molecules collide, they need to approach each other in a specific way to break and form new bonds, otherwise the reaction won't occur. When the orientation factor is 0.2, it means that only 20% of the collisions have favorable orientations for the reaction to proceed.
02

Determine the number of successful collisions per second

First, we need to find out how many collisions are both energetic enough and have the correct orientation for the reaction to proceed. Since 10% of the collisions are sufficiently energetic, we have 1000 * 0.10 = 100 energetic collisions per second. Now, we should consider the orientation factor. Since only 20% of these collisions have the correct orientation, we'll have 100 * 0.2 = 20 successful collisions per second.
03

Calculate the rate of the reaction

The rate of a reaction is essentially the number of successful collisions leading to product formation per unit time. Since we have calculated that there are 20 successful collisions between OH- and CH3Br per second, the rate of the reaction is 20 CH3OH molecules formed per second.

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

For a particular reaction, the absorbed energy is \(800 \mathrm{~kJ}\) to break old bonds, and \(\Delta E_{\mathrm{rxn}}\) is equal to \(-800 \mathrm{~kJ}\). How much energy is released into the surroundings as the product bonds are formed?

If there were no orientation requirement for collisions, would reactions be faster or slower than they are? Explain your answer.

The experimental rate law for the reaction \(\mathrm{A}+\mathrm{A} \rightarrow \mathrm{A}_{2}\) is Rate \(=k[\mathrm{~A}][\mathrm{BC}]\) Two mechanisms have been proposed for the reaction: Step 1: \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB}\) (slow) Step \(2: \mathrm{AB}+\mathrm{A} \rightarrow \mathrm{A}_{2}+\mathrm{B}\) (fast) and Step \(1: \mathrm{A}+\mathrm{BC} \rightarrow \mathrm{AB}+\mathrm{C}\) (slow) Step \(2: \mathrm{A}+\mathrm{AB} \rightarrow \mathrm{B}+\mathrm{A}_{2}\) (fast) Step \(3: \mathrm{B}+\mathrm{C} \rightarrow \mathrm{BC}\) (fast) (a) Show that each mechanism results in the correct overall reaction. (b) Which mechanism is consistent with the rate law? (c) Why does \(\mathrm{BC}\) appear in the rate law but not in the overall reaction?

In a kinetic study of the reaction \(2 \mathrm{~A}(g)+\mathrm{B}(g) \rightarrow \mathrm{P}(g)\) the following rate data were obtained. Write the rate law with proper orders. Give the overall order of the reaction. Finally, state what this problem confirms about the relationship between reactant orders and coefficients in the balanced equation. $$\begin{array}{cccc} & & & \begin{array}{l} \text { Rate of } \\ \text { disappearance } \end{array} \\ \text { Experiment } & {[\mathbf{A}]} & {[\mathrm{B}]} & \text { of A }(\mathrm{M} / \mathrm{s}) \\ \hline 1 & 0.0125 \mathrm{M} & 0.0253 \mathrm{M} & 0.0281 \\ 2 & 0.0250 \mathrm{M} & 0.0253 \mathrm{M} & 0.0562 \\ 3 & 0.0125 \mathrm{M} & 0.0506 \mathrm{M} & 0.1124 \end{array}$$

True or false? A catalyst in a reaction decreases the energy gap between reactants and products. If the statement is false, explain why.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

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