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 \((\mathrm{R})_{\mathrm{o}}\)

Short Answer

Expert verified
The rate constant for the reaction at the given temperature can be determined directly from the slope of the plot of \(1 /(\mathrm{R})\) versus \(t\). The initial reaction rate \((\mathrm{R})_{\mathrm{o}}\) can be obtained by taking the reciprocal of the intercept of the graph.

Step by step solution

01

Determine the rate constant

For a second-order reaction, the slope of a plot of \(1 /(\mathrm{R})\) versus \(t\) is equal to the rate constant (k). Therefore, the student can simply use the numerical value of the slope as the rate constant, k.
02

Determine \((\mathrm{R})_{\mathrm{o}}\)

The intercept on the vertical axis of the graph (at \(t = 0\)) represents \(1 /(\mathrm{R})_{\mathrm{o}}\), where \((\mathrm{R})_{\mathrm{o}}\) is the initial reaction rate. Hence, to find \((\mathrm{R})_{\mathrm{o}}\), one needs to simply take the reciprocal of the intercept.

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 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

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.

Sign-up for free