Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 56: Problem 8

Recall that for a second-order reaction: $$ \frac{1}{(\mathrm{R})}=\frac{1}{(\mathrm{R})_{\mathrm{o}}}+k t $$ a) When \(t=t_{1 / 2}\), what is the value of \((\mathrm{R})\) in terms of \((\mathrm{R})_{0}\) ? b) Show that \(t_{1 / 2}=\frac{1}{k(\mathrm{R})_{\mathrm{o}}}\) for a second- order reaction.

Short Answer

Expert verified
a) When \(t=t_{1 / 2}\), [R] is equal to \([R]_0 / 2\). b) We have shown that for a second-order reaction, \(t_{1 / 2} = \frac{1}{k[R]_{0}}\).

Step by step solution

01

Understand the Second-Order Rate Law

The second-order rate law given is: \[\frac{1}{[R]}=\frac{1}{[R]_{0}}+kt\] Here, [R] is the concentration of R at any time \(t\), [R]_0 is the initial concentration of R and \(k\) is the rate constant. The symbol 't' represents the time elapsed.
02

Calculate [R] when \(t=t_{1 / 2}\)

\(t_{1 / 2}\) is defined as the half-life time of the reaction, i.e. the time it takes for the concentration of a reactant to decrease to half of its original value. Setting \(t\) to \(t_{1 / 2}\) and [R] to \([R]_0 / 2\) in the second-order rate law, we have: \[\frac{2}{[R]_{0}} =\frac{1}{[R]_{0}}+ kt_{1 / 2}\] Solving this will give us the value of [R] in terms of [R]_0 when \(t=t_{1 / 2}\).
03

Derive the Expression for \(t_{1 / 2}\)

From step 2, we can further calculate \(t_{1 / 2}\). Simplifying the equation gives: \[kt_{1 / 2} = \frac{1}{[R]_{0}}\] Solving for \(t_{1 / 2}\), we have: \[t_{1 / 2} = \frac{1}{k[R]_{0}}\] Thus, we have shown that \(t_{1 / 2}\) for a second order reaction is given by \(t_{1 / 2} = \frac{1}{k[R]_{0}}\). This is a key kinetic property of second order reactions.

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

A student obtains data for a second-order reaction at a given temperature, and then makes a graph of \(1 /(\mathrm{R})\) (along the vertical axis) versus \(t\) (along the horizontal axis). He notes that the resulting plot appears to correspond to a straight-line relationship. The student then determines the slope and intercept of the best-fit straight line. Describe how the student could use the slope and/or intercept of the best-fit straight line to determine: a) the rate constant for the reaction at the given temperature. b) the value of \((\mathrm{R})_{\mathrm{o}}\)

A student obtains data for a first-order reaction at a given temperature, and then makes a graph of \(\ln (\mathrm{R})\) (along the vertical axis) versus \(t\) (along the horizontal axis). She notes that the resulting plot appears to correspond to a straight-line relationship. The student then determines the slope and intercept of the best-fit straight line. Describe how the student could use the slope and/or intercept of the best-fit straight line to determine: a) the rate constant for the reaction at the given temperature. b) the value of \((\mathrm{R})_{0}\)

Recall that for a first-order reaction: $$ \ln (\mathrm{R})=\ln (\mathrm{R})_{0}-k t $$ a) When \(t=t_{1 / 2}\), what is the value of \((\mathrm{R})\) in terms of \((\mathrm{R})_{0}\) ? b) Show that \(t_{1 / 2}=\frac{\ln 2}{k}=\frac{0.693}{k}\) for a first-order reaction.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

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