/*! 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 39 (a) For a generic second-order r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B},\) what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part (a)? (c) How do the half-lives of first-order and second-order reactions differ?

Short Answer

Expert verified
(a) For a generic second-order reaction, plotting \(\frac{1}{[A]_t}\) against time (\(t\)) will yield a straight line. (b) The slope of the straight line is the rate constant, \(k\). (c) The half-life of a first-order reaction remains constant, independent of the reactant concentration, while the half-life of a second-order reaction depends on the initial concentration of the reactant.

Step by step solution

01

Part (a): Determine the y-axis quantity for the graph

We start by writing the rate equation for a second-order reaction: \(rate = k \times [A]^2\) Where \(k\) is the rate constant, and \([A]\) represents the concentration of reactant A. We are asked to find a quantity that will yield a straight line when graphed against time. To do this, we can convert the rate equation into an integrated rate law. For a second-order reaction, the integrated rate law is: \(\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\) Where \([A]_t\) is the concentration of A at a specific time, \([A]_0\) is the initial concentration of A, and \(t\) is time. Here, if we plot \(\frac{1}{[A]_t}\) against time (\(t\)), we will get a straight line.
02

Part (b): Find the slope of the straight line

To find the slope of the straight line from part (a), we can rewrite the integrated rate law in the form of a linear equation: \(\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}\) This equation is in the form of \(y = mx + c\), where \(y = \frac{1}{[A]_t}\), \(x = t\), \(m = k\), and \(c = \frac{1}{[A]_0}\). The slope of the straight line, or the coefficient of \(t\), is the rate constant, \(k\).
03

Part (c): Comparison of half-lives between first-order and second-order reactions

The half-life, \(t_{1/2}\), is the amount of time it takes for the concentration of the reactant to reduce to half its initial value. The half-life of a reaction depends on the order of the reaction. For a first-order reaction, the half-life is given by: \(t_{1/2}^{(1)} = \frac{0.693}{k}\) Here, we can see that the half-life is constant and does not depend on the initial concentration. For a second-order reaction, the half-life is given by: \(t_{1/2}^{(2)} = \frac{1}{k[A]_0}\) In this case, the half-life depends on the initial concentration of the reactant. As the initial concentration increases, the half-life becomes smaller. In summary, the half-life of a first-order reaction remains constant, independent of the reactant concentration, while the half-life of a second-order reaction depends on the initial concentration of the reactant.

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 the Rate Equation in Second-Order Reactions
In chemical kinetics, the rate equation is a mathematical expression that describes the speed of a reaction in terms of the concentration of reactants. For a second-order reaction like \( \mathrm{A} \longrightarrow \mathrm{B} \), the rate equation is expressed as:\[ \text{rate} = k \times [A]^2 \]Here, \( [A] \) represents the concentration of reactant A, and \( k \) is the rate constant, a measure of the reaction's speed. The superscript '2' in \( [A]^2 \) indicates that the reaction rate is proportional to the square of the concentration of A. This means that even a small change in \([A]\) can noticeably affect the reaction rate.The rate equation serves as a foundational concept for understanding how the concentration of substances in a reaction influences the speed at which the reaction proceeds. This quadratic relationship is specific to second-order reactions, making them distinct from zero and first-order reactions.
Exploring the Integrated Rate Law for Second-Order Reactions
The integrated rate law provides insight into the concentration changes of reactants over time in a chemical reaction. For a second-order reaction, the integrated rate law is a pivotal tool and is given by:\[ \frac{1}{[A]_t} - \frac{1}{[A]_0} = kt \]In this equation, \([A]_t\) is the concentration of reactant A at time \(t\), and \([A]_0\) is the initial concentration. Rearranging this equation, we can write it in a linear form:\[ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \]This format is akin to the equation of a straight line \( y = mx + c \), where:
  • \( y = \frac{1}{[A]_t} \)
  • \( x = t \)
  • \( m = k \) (the slope)
  • \( c = \frac{1}{[A]_0} \) (the y-intercept)
By plotting \( \frac{1}{[A]_t} \) versus time, we obtain a straight line. The slope of this line (the rate constant \(k\)) reflects how quickly the reaction proceeds. This visualization helps us track how concentrations decrease over time, providing a clearer picture of reaction dynamics.
Comparing Half-life in First-Order and Second-Order Reactions
The half-life of a reaction is a crucial concept in kinetics. It represents the time it takes for the concentration of a reactant to drop to half its initial value. The behavior of half-life varies significantly between different reaction orders.For a first-order reaction, the half-life is constant and does not depend on the initial concentration of the reactant. It is calculated as:\[ t_{1/2}^{(1)} = \frac{0.693}{k} \]This constancy makes first-order reactions easy to predict, as the time required for half of the reactant to react remains the same throughout the process.In contrast, for a second-order reaction, the half-life is dependent on the initial concentration, represented by:\[ t_{1/2}^{(2)} = \frac{1}{k[A]_0} \]As the initial concentration \([A]_0\) increases, the half-life decreases. This dependency highlights a key distinction: second-order reactions have varying half-lives based on starting concentrations. Therefore, each successive half-cycle takes less time as the reactant concentration decreases. Understanding these distinctions helps in predicting the behavior and duration of chemical reactions more accurately.

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 colored dye compound decomposes to give a colorless product. The original dye absorbs at \(608 \mathrm{nm}\) and has an extinction coefficient of \(4.7 \times 10^{4} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at that wavelength. You perform the decomposition reaction in a \(1-\mathrm{cm}\) cuvette in a spectrometer and obtain the following data: $$ \begin{array}{rl} \hline \text { Time (min) } & \text { Absorbance at } 608 \mathrm{nm} \\ \hline 0 & 1.254 \\ 30 & 0.941 \\ 60 & 0.752 \\ 90 & 0.672 \\ 120 & 0.545 \end{array} $$ From these data, determine the rate law for the reaction "dye \(\longrightarrow\) product" and determine the rate constant.

Consider the gas-phase reaction between nitric oxide and bromine at \(273^{\circ} \mathrm{C}: 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{NOBr}(g) .\) The following data for the initial rate of appearance of NOBr were obtained: (a) Determine the rate law. (b) Calculate the average value of the rate constant for the appearance of NOBr from the four data sets. (c) How is the rate of appearance of NOBr related to the rate of disappearance of \(\mathrm{Br}_{2}\) ? (d) What is the rate of disappearance of \(\mathrm{Br}_{2}\) when \([\mathrm{NO}]=0.075 \mathrm{M}\) and \(\left[\mathrm{Br}_{2}\right]=0.25 \mathrm{M} ?\)

Molecular iodine, \(\mathrm{I}_{2}(g)\), dissociates into iodine atoms at \(625 \mathrm{~K}\) with a first-order rate constant of \(0.271 \mathrm{~s}^{-1}\). (a) What is the half-life for this reaction? (b) If you start with \(0.050 \mathrm{M} \mathrm{I}_{2}\) at this temperature, how much will remain after 5.12 s assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2} ?\)

You study the effect of temperature on the rate of two reactions and graph the natural logarithm of the rate constant for each reaction as a function of \(1 / T\). How do the two graphs compare (a) if the activation energy of the second reaction is higher than the activation energy of the first reaction but the two reactions have the same frequency factor, and \((b)\) if the frequency factor of the second reaction is higher than the frequency factor of the first reaction but the two reactions have the same activation energy? [Section 14.5\(]\)

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. (b) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathrm{c}) \mathrm{What}\) is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M},\) while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M}\) ?

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.