/*! 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 9 The rate constant \((k)\) depend... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 9

The rate constant \((k)\) depends on which of the following (there may be more than one answer)? a. the concentration of the reactants b. the nature of the reactants c. the temperature d. the order of the reaction Explain.

Short Answer

Expert verified
The rate constant \((k)\) depends on the nature of the reactants (option b) and the temperature (option c). The nature of the reactants influences the value of \(k\) through factors like reactant particle size, chemical bonds, and their physical state. The rate constant also depends on temperature, as an increase in temperature leads to an increase in the rate of the reaction, which can be modeled using the Arrhenius equation: \(k=Ae^{-\frac{E_a}{RT}}\).

Step by step solution

01

Option a: concentration of reactants

The rate constant \((k)\) does not depend on the concentration of reactants. In fact, the rate constant is used alongside the concentration of the reactants to calculate the rate of reaction. If the concentration changes, the rate of reaction will change accordingly, but the rate constant remains the same at a constant temperature.
02

Option b: nature of reactants

The rate constant \((k)\) depends on the nature of the reactants. Different reactions have different rates of reactions, which depend on the physical and chemical properties of the reactants being used. Factors like the size of the reactant particles, their chemical bonds, and whether they are in a solid, liquid, or gaseous state all influence the value of \(k\).
03

Option c: temperature

The rate constant \((k)\) depends on temperature. When the temperature increases, the kinetic energy of particles increases, and more collisions between reactants occur with sufficient energy to overcome the activation energy barrier. As a result, the rate of the reaction increases and so does the rate constant. According to the Arrhenius equation, the relation between the rate constant and temperature can be modeled as \(k=Ae^{-\frac{E_a}{RT}}\), where \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.
04

Option d: order of the reaction

The rate constant \((k)\) does not directly depend on the order of the reaction. The order of the reaction is the sum of the exponents in the reactants to which they are raised in the rate law. The order of a reaction and the rate constant are two separate parameters that describe the reaction. However, the units of the rate constant depend on the order of the reaction. For example, the units of a first-order reaction rate constant are s鈦宦� (inverse seconds), while the units of a second-order reaction rate constant are M鈦宦箂鈦宦� (inverse molarity per second). In conclusion, the rate constant \((k)\) depends on options b: the nature of the reactants, and c: the 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.

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

Consider a reaction of the type \(\mathrm{aA} \longrightarrow\) products, in which the rate law is found to be rate \(=k[\mathrm{~A}]^{3}\) (termolecular reactions are improbable but possible). If the first half-life of the reaction is found to be \(40 . \mathrm{s}\), what is the time for the second half-life? Hint: Using your calculus knowledge, derive the integrated rate law from the differential rate law for a termolecular reaction: $$ \text { Rate }=\frac{-d[\mathrm{~A}]}{d t}=k[\mathrm{~A}]^{3} $$

The type of rate law for a reaction, either the differential rate law or the integrated rate law, is usually determined by which data is easiest to collect. Explain.

Provide a conceptual rationale for the differences in the half-lives of zero-, first-, and second-order reactions.

Consider the reaction $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ If, in a certain experiment, over a specific time period, \(0.0048 \mathrm{~mol}\) \(\mathrm{PH}_{3}\) is consumed in a 2.0-L container each second of reaction, what are the rates of production of \(\mathrm{P}_{4}\) and \(\mathrm{H}_{2}\) in this experiment?

Write the rate laws for the following elementary reactions. a. \(\mathrm{CH}_{3} \mathrm{NC}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CN}(g)\) b. \(\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g)\) c. \(\mathrm{O}_{3}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\) d. \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

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.