/*! 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 97 The half-life of a chemical reac... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 97

The half-life of a chemical reaction at a particular concentration is \(50 \mathrm{~min}\), when the concentration of reactants is doubled, the half-life becomes \(100 \mathrm{~min}\). Find the order. (a) zero (b) first (c) second (d) third

Short Answer

Expert verified
The reaction is second order (c).

Step by step solution

01

Understanding Half-Life and Reaction Order

The half-life of a reaction is the time required for the concentration of a reactant to decrease by half. Reaction order affects how half-life changes with concentration changes. We need to determine the order based on the given changes in half-life at different concentrations.
02

Observing Half-life Change with Concentration

The problem states that when the concentration of the reactant is doubled, the half-life increases from 50 minutes to 100 minutes. For reactions of different orders, the relationship between half-life and concentration can vary.
03

Analyzing Zero Order Reaction

In a zero-order reaction, the half-life decreases as concentration increases, according to the formula: \[ t_{1/2} = \frac{[A]_0}{2k} \] where \([A]_0\) is the initial concentration and \(k\) is the rate constant. Here, doubling concentration should lead to a reduced half-life, not an increased half-life.
04

Analyzing First Order Reaction

In a first-order reaction, the half-life is independent of the concentration, given by the formula: \[ t_{1/2} = \frac{0.693}{k} \]Doubling concentration does not affect the half-life, contradicting our observation.
05

Analyzing Second Order Reaction

For a second-order reaction, the half-life increases with the increase in concentration. The formula for half-life is: \[ t_{1/2} = \frac{1}{k[A]_0} \]Doubling the concentration results in doubling the half-life, which matches the given change from 50 minutes to 100 minutes.
06

Determining the Order

Based on the analysis, only the second-order reaction description fits the observed change in half-life with a change in concentration. Thus, the reaction is of second order.

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!

Key Concepts

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

Half-Life
The half-life of a chemical reaction is the time it takes for the concentration of a reactant to reduce to half its original amount. It is a critical concept in chemistry, especially in the study of reaction kinetics.
Understanding how half-life relates to concentration changes is essential to determine the reaction's order.
  • For some reactions, half-life is constant, regardless of concentration.
  • For others, it may vary depending on how concentration changes.

This relationship helps chemists predict how long it will take for a reaction to reach a certain extent, which is valuable in various industrial and scientific applications.
Zero Order Reaction
In a zero-order reaction, the rate of the reaction is constant, not dependent on the concentration of the reactants. This means that the concentration decreases at a steady rate over time.
The formula for the half-life of a zero-order reaction is:\[t_{1/2} = \frac{[A]_0}{2k}\]Here,
  • \([A]_0\) is the initial concentration
  • \(k\) is the rate constant.
If the initial concentration \([A]_0\) is increased, the half-life becomes shorter because the numerator of the formula grows with concentration, but the entire expression still divides by a constant rate. Conversely, in this problem, the half-life doubled, ruling out a zero-order reaction.
First Order Reaction
In a first-order reaction, the rate depends linearly on the concentration of one reactant. The unique aspect of first-order reactions is that their half-life remains constant regardless of changes in concentration.
The half-life can be calculated with:\[t_{1/2} = \frac{0.693}{k}\]This formula tells us that the half-life
  • Does not depend on the initial concentration.
Thus, doubling the concentration will not alter the half-life, which does not align with the problem's condition where the half-life increases. This inconsistency helps eliminate first-order reactions in this context.
Second Order Reaction
A second-order reaction's rate is proportional to either the square of one reactant's concentration or the product of two reactants' concentrations.
The half-life for a second-order reaction is expressed as:\[t_{1/2} = \frac{1}{k[A]_0}\]From this formula:
  • The half-life increases as the initial concentration decreases.
  • Doubling initial concentration directly doubles the half-life.
When the concentration doubles and the half-life also doubles, it matches what we would expect for a second-order reaction. In this way, examining the half-life's dependency on concentration effectively helps identify second-order reactions, as demonstrated in this exercise.

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

For a first-order reaction, \(t_{0.9}\) is \(138.6\) seconds. Its specific reaction rate constant (in \(\left.\sec ^{-1}\right)\) is (a) \(10^{-2}\) (b) \(10^{-4}\) (c) \(10^{-5}\) (d) \(10^{-6}\)

The rate low for the hydrolysis of thioacetamide, \(\mathrm{CH}_{3} \mathrm{CSNH}_{2}\), CC(=S)NCOCC(N)=O Is rate \(=\mathrm{k}\left[\mathrm{H}^{+}\right]\)[TA], where TA is thioacetamide. In which of the following solutions, will the rate of hydrolysis of thioacetamide (TA) is least at \(25^{\circ} \mathrm{C}\) ? (a) \(0.1 \mathrm{M}\) in \(\mathrm{TA}\) and \(0.20 \mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) (b) \(0.1 \mathrm{M}\) in \(\mathrm{TA}\) and \(0.20 \mathrm{M}\) in \(\mathrm{HNO}_{3}\) (c) \(0.1 \mathrm{M}\) in TA and \(0.20 \mathrm{M}\) in \(\mathrm{HCO}_{2} \mathrm{H}\) (d) \(0.15 \mathrm{M}\) in TA and \(0.15 \mathrm{M}\) in \(\mathrm{HCl}\)

If the initial concentration of reactant in certain reaction is doubled, the half life period of the reaction is also doubled. The order of reaction is (a) zero (b) first (c) second (d) \(1.5\)

A redox reaction is carried out at \(127^{\circ} \mathrm{C}\). If the same reaction is carried out in presence of a catalyst at the same temperature, the rate of reaction is doubled. To what extent is the energy barrier lowered by the catalyst? [Use \(\mathrm{R}=2\) cal \(\mathrm{mol}^{-1} \mathrm{~K}^{-1}\) and \(\left.\log 2=0.301\right]\) (a) \(455 \mathrm{cal}\) (b) \(231 \mathrm{cal}\) (c) \(693 \mathrm{cal}\) (d) \(554 \mathrm{cal}\)

The half-life of a substance in a first-order reaction is 15 minutes. The rate constant is (a) \(2.46 \times 10^{2} \mathrm{~min}^{-1}\) (b) \(4.62 \times 10^{-2} \mathrm{~min}^{-1}\) (c) \(3 \times 10^{-5} \mathrm{~min}^{-1}\) (d) \(3 \times 10^{-4} \mathrm{~min}^{-1}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

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.