/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 The central idea of the collisio... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 19

The central idea of the collision model is that molecules must collide in order to react. Give two reasons why not all collisions of reactant molecules result in product formation.

Short Answer

Expert verified
One reason why not all collisions between reactant molecules result in product formation is insufficient energy. The colliding molecules may not have enough energy to overcome the activation energy barrier, which is the minimum energy required for a reaction to occur. Another reason is the incorrect orientation of the colliding molecules. A productive collision requires the reactant molecules to collide in a specific orientation that allows their atoms to rearrange and form new bonds, leading to the formation of products. If the orientation is not appropriate, the reaction will not proceed.

Step by step solution

01

Reason 1: Insufficient energy

One reason why not all collisions between reactant molecules result in the formation of products is that the colliding molecules may not have enough energy to overcome the activation energy barrier. Activation energy is the minimum energy required for a reaction to occur; if the colliding molecules' combined kinetic energy is less than the activation energy, the reaction will not proceed and the molecules will simply bounce off each other.
02

Reason 2: Incorrect orientation

Another reason why not all collisions between reactant molecules result in the formation of products is that the molecules may not collide in the proper orientation. A productive collision requires the reactant molecules to collide in such a way that their atoms can rearrange and form new bonds (leading to the formation of products). If the orientation of the colliding molecules is not appropriate for this rearrangement to occur, the reaction will not proceed and the molecules will separate without reacting.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

A certain reaction has the following general form: $$\mathrm{aA} \longrightarrow \mathrm{bB}$$ At a particular temperature and \([\mathrm{A}]_{0}=2.80 \times 10^{-3} \mathrm{M},\)concentration versus time data were collected for this reaction, and a plot of \(1 /[\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(+3.60 \times 10^{-2} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to \(7.00 \times 10^{-4} \mathrm{M} ?\)

The rate law for a reaction can be determined only from experiment and not from the balanced equation. Two experimental procedures were outlined in Chapter \(11 .\) What are these two procedures? Explain how each method is used to determine rate laws.

Consider two reaction vessels, one containing A and the other containing \(\mathrm{B},\) with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$\begin{aligned} &k_{A}=4.50 \times 10^{-4} \mathrm{s}^{-1}\\\ &k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{s}^{-1} \end{aligned}$$how much time must pass to reach a condition such that \([\mathrm{A}]=\) \(4.00[\mathrm{B}] ?\)

How does temperature affect \(k,\) the rate constant? Explain.

The reaction $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{OH}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{Br}^{-}$$ in a certain solvent is first order with respect to \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and zero order with respect to OH \(^{-} .\) In several experiments, the rate constant \(k\) was determined at different temperatures. A plot of \(\ln (k)\) versus \(1 / T\) was constructed resulting in a straight line with a slope value of \(-1.10 \times 10^{4} \mathrm{K}\) and \(y\) -intercept of 33.5. Assume \(k\) has units of \(\mathrm{s}^{-1}\).a. Determine the activation energy for this reaction. b. Determine the value of the frequency factor \(A\) c. Calculate the value of \(k\) at \(25^{\circ} \mathrm{C}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Physical 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.