/*! 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 137 The hypothetical reaction \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The hypothetical reaction \(\mathrm{A}+2 \mathrm{~B} \longrightarrow\) Products has the rate law Rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{3}\). If the reaction is run two separate times, holding the concentration of A constant while doubling the concentration of \(\mathrm{B}\) from one run to the next, how would the rate of the second run compare to the rate of the first run?

Short Answer

Expert verified
The rate of the second run is 8 times the rate of the first run.

Step by step solution

01

Understand the Rate Law Formula

The rate law is given as \( \text{Rate} = k[A]^2[B]^3 \). This formula tells us how the rate depends on the concentration of the reactants \(A\) and \(B\), where \(k\) is the rate constant, \([A]\) is the concentration of \(A\), and \([B]\) is the concentration of \(B\).
02

Analyze the Impact of Doubling \([B]\)

Since we are keeping \([A]\) constant and are doubling \([B]\), we can write the new concentration of \(B\) as \(2[B]\). The original rate is \(k[A]^2[B]^3\). Substituting the new concentration for \(B\), the new rate becomes \(k[A]^2(2[B])^3\).
03

Calculate the New Rate Expression

Substitute \(2[B]\) into the rate law to find the rate of the second run:\[ \text{New Rate} = k[A]^2 (2[B])^3 = k[A]^2 (8[B]^3). \]This simplifies to \[ 8 imes k[A]^2[B]^3. \]
04

Compare the New Rate with the Original Rate

The original rate was given as \(k[A]^2[B]^3\). The new rate, with \([B]\) doubled, is \(8 \times k[A]^2[B]^3\). Thus, the rate of the second run is 8 times the rate of the first run.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Rate Law
The rate law is a fundamental concept in chemical kinetics, expressing the relationship between the rate of a reaction and the concentration of its reactants. In our example reaction, the rate law is given as \( \text{Rate} = k[A]^2[B]^3 \). Here, \( A \) and \( B \) are the reactants, and their concentrations are denoted by \([A]\) and \([B]\), respectively.

The rate constant \(k\) is a specific value for the reaction that remains constant if the temperature does not change. The exponents \(2\) and \(3\) in the rate law are known as the reaction orders with respect to \( A \) and \( B \), respectively. These values indicate how sensitive the reaction rate is to changes in the concentration of each reactant. Specifically, the rate will increase by a factor of \([A]^2\) and \([B]^3\).

This formula helps chemists predict how changing the concentration of the reactants affects the speed of the reaction, which is crucial for designing efficient chemical processes.
Significance of Concentration in Reaction Rates
Concentration plays a pivotal role in determining the speed of chemical reactions. In our reaction example \( \mathrm{A} + 2 \mathrm{B} \rightarrow \text{Products} \), doubling the concentration of \( B \) while keeping \( A \) constant directly affects the rate.

When we double the concentration of \( B \), mathematically, the concentration changes from \([B]\) to \(2[B]\). According to the rate law \( \text{Rate} = k[A]^2[B]^3 \), this modification translates to the expression \( (2[B])^3 \). Consequently, the new rate becomes \( k[A]^2 \times 8[B]^3 \), showing that the rate increases by a factor of 8.

This illustrates the profound influence concentration changes can have on reaction rates, emphasizing the importance of understanding concentration for managing and optimizing chemical reactions.
Overview of Chemical Kinetics
Chemical kinetics involves the study of reaction rates and the factors influencing them. It provides insights into how reactions occur and how fast they proceed, which is crucial in both academic research and industrial applications.

Key aspects of chemical kinetics involve
  • Analyzing the time required for reactants to transform into products.
  • Understanding how variables such as concentration, temperature, and catalysts impact the reaction rate.
  • Developing rate laws that express the reaction rate as a function of reactant concentrations.
In a broader sense, chemical kinetics helps us understand the mechanisms of reactions by breaking down the steps involved in a chemical process.

This field offers a quantitative basis for controlling reactions to achieve desired outcomes more efficiently, shedding light on maximum yield practices, energy requirements, and safety in chemical processes.

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

Benzene diazonium chloride, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NNCl}\), decomposes by a first-order rate law. $$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NNCl} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{N}_{2}(g)$$ If the rate constant at \(20^{\circ} \mathrm{C}\) is \(4.3 \times 10^{-5} / \mathrm{s}\), how long will it take for \(75 \%\) of the compound to decompose?

Sketch a potential-energy diagram for the exothermic, elementary reaction $$\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D}$$ and on it denote the activation energies for the forward and reverse reactions. Also indicate the reactants, products, and activated complex.

Ethyl chloride, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\), used to produce tetraethyllead gasoline additive, decomposes, when heated, to give ethylene and hydrogen chloride. The reaction is first order. In an experiment, the initial concentration of ethyl chloride was \(0.00100 \mathrm{M}\). After heating at \(500^{\circ} \mathrm{C}\) for \(155 \mathrm{~s}\), this was reduced to \(0.00067 \mathrm{M}\). What was the concentration of ethyl chloride after a total of \(256 \mathrm{~s}\) ?

Nitryl bromide, \(\mathrm{NO}_{2} \mathrm{Br}\), decomposes into nitrogen dioxide and bromine. $$2 \mathrm{NO}_{2} \mathrm{Br}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)+\mathrm{Br}_{2}(g)$$ A proposed mechanism is $$\begin{array}{r}\mathrm{NO}_{2} \mathrm{Br} \longrightarrow \mathrm{NO}_{2}+\mathrm{Br} \\\\\mathrm{NO}_{2} \mathrm{Br}+\mathrm{Br} \longrightarrow \mathrm{NO}_{2}+\mathrm{Br}_{2}\end{array}$$ Write the rate law predicted by this mechanism.

A reaction of the form \(a \mathrm{~A} \longrightarrow\) Products is second order with a half-life of \(425 \mathrm{~s}\). What is the rate constant of the reaction if the initial concentration of \(\mathrm{A}\) is \(5.99 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\) ?

See all solutions

Recommended explanations on Chemistry Textbooks

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.