Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A kinetic study of the reaction \(A \longrightarrow\) products yields the data: \(t=0 \mathrm{s},[\mathrm{A}]=2.00 \mathrm{M} ; 500 \mathrm{s}, 1.00 \mathrm{M}; 1500 \mathrm{s}, 0.50 \mathrm{M} ; 3500 \mathrm{s}, 0.25 \mathrm{M} .\) Without performing detailed calculations, determine the order of this reaction and indicate your method of reasoning.

Short Answer

Expert verified
The given reaction is a first-order reaction.

Step by step solution

01

Understanding Rate Laws and Reaction Order

Rate laws express the rate of a reaction in terms of the concentration of reactants. The rate of a reaction is generally represented as -d[A]/dt = k[A]^n, where 'A' is the reactant, 'k' is the rate constant, 'n' is the reaction order and 't' is time.
02

Observing The Given Data

The kinetic data provided for the reaction shows that when the time triples (from 500s to 1500s, and from 1500s to 3500s), the concentration of 'A' halves each time (from 2.00M to 1.00M, then to 0.50M, and finally to 0.25M).
03

Identifying The Reaction Order

The constant halving of concentration 'A' each time the time interval triples is characteristic of a first-order reaction. This is because, for first order reactions, the rate of the reaction is proportional to the first power of reactant concentration. The data fits this pattern, indicating that the reaction is first order with respect to 'A'.

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!

Key Concepts

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

Rate Laws
The study of how the speed of chemical reactions changes with varying conditions is a core principle of chemical kinetics. At the heart of this analysis lies the rate law, a mathematical equation that conveys the relationship between the concentration of reactants and the rate at which they transform into products.

Rate laws can be represented by the equation \( -d[A]/dt = k[A]^n \), where \( [A] \) designates the molar concentration of the reactant \( A \), \( k \) is the rate constant that provides the speed of the reaction at a particular temperature, and \( n \) stands for the reaction order. This order is an exponent that shows the dependence of the rate on the concentration of \( A \) and is determined empirically. Different reactions may exhibit zero, first, second, or mixed-order behaviors depending on how their rates relate to the concentration of reactants.

If the concentration of the reactant does not affect the rate, the reaction is zero-order (\( n = 0 \)). For a first-order reaction (\( n = 1 \)), the rate doubles if the concentration of the reactant doubles. A second-order reaction (\( n = 2 \)) would see the rate quadruple if the reactant concentration doubled.
Kinetic Study
A kinetic study involves analyzing how a chemical reaction proceeds over time to determine the speed and the mechanism of the reaction. By measuring changes in reactant or product concentrations at various times, chemists can unravel the complexity of the reaction pathway and deduce the reaction order from the rate law.

In practice, this involves conducting experiments and gathering data like concentration changes over time, as seen in the exercise problem. When interpreting this data, patterns emerge that indicate how the reaction rate is affected by the concentration of reactants. It's methodical detective work—plotting data on graphs, calculating rates, and comparing these rates under different conditions all contribute to building an understanding of a reaction's kinetics.

Critical to any kinetic study is maintaining controlled conditions. For instance, temperature is kept constant because it significantly affects the rate constant. Once data is collected, graphs such as concentration versus time for a first-order reaction or 1/concentration versus time for a second-order reaction, can reveal the reaction kinetics clearly and facilitate the determination of the rate law.
First-Order Reaction
For a first-order reaction, the rate is directly proportional to the concentration of one reactant. This direct proportionality implies that as the concentration of the reactant decreases by a certain factor, the time required for that decrease is constant.

Looking at the kinetic data provided in the exercise, the concentration of reactant \( A \) is halved repetitively over regular time intervals (500s to 1500s to 3500s). Such regular halving is indicative of a first-order reaction because it suggests a constant rate of reaction per unit concentration of \( A \). It aligns with the mathematical characteristic of a first-order rate law, where a plot of the natural logarithm of the concentration of \( A \) versus time yields a straight line, a hallmark of first-order kinetics.

Additionally, the integrated rate law for a first-order reaction, \( \ln[A] = -kt+ \ln[A]_0 \), where \( [A]_0 \) is the initial concentration of the reactant, shows that the time required to reach half the initial concentration (also known as the half-life) is constant regardless of the starting concentration. This property is unique to first-order reactions and becomes a powerful tool for identifying this type of reaction order from experimental data.

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

In three different experiments, the following results were obtained for the reaction \(A \longrightarrow\) products: \([\mathrm{A}]_{0}=1.00 \mathrm{M}, t_{1 / 2}=50 \mathrm{min} ;[\mathrm{A}]_{0}=200 \mathrm{M}, t_{1 / 2}=\) \(25 \min ;[\mathrm{A}]_{0}=0.50 \mathrm{M}, t_{1 / 2}=100 \mathrm{min} .\) Write the rate equation for this reaction, and indicate the value of \(k.\)

For the reaction \(\mathrm{A}+2 \mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D},\) the rate law is rate of reaction \(=k[\mathrm{A}][\mathrm{B}]\) (a) Show that the following mechanism is consistent with the stoichiometry of the overall reaction and with the rate law. $$\begin{array}{l} \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{I} \quad(\text { slow }) \\ \mathrm{I}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D} \quad(\text { fast }) \end{array}$$ (b) Show that the following mechanism is consistent with the stoichiometry of the overall reaction, but not with the rate law. $$\begin{array}{c} 2 \mathrm{B} \stackrel{k_{1}}{\mathrm{k}_{1}} \mathrm{B}_{2} \text { (fast) } \\\ \mathrm{A}+\mathrm{B}_{2} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}+\mathrm{D} \text { (slow) } \end{array}$$

The following three-step mechanism has been proposed for the reaction of chlorine and chloroform. $$\begin{aligned} & \text { (1) } \quad \mathrm{Cl}_{2}(\mathrm{g}) \stackrel{k_{1}}{\rightleftharpoons_{k-1}} 2 \mathrm{Cl}(\mathrm{g})\\\ & \text { (2) } \quad \mathrm{Cl}(\mathrm{g})+\mathrm{CHCl}_{3}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{HCl}(\mathrm{g})+\mathrm{CCl}_{3}(\mathrm{g})\\\ &\text { (3) } \quad \mathrm{CCl}_{3}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) \stackrel{k_{3}}{\longrightarrow} \mathrm{CCl}_{4}(\mathrm{g}) \end{aligned}$$ The numerical values of the rate constants for these steps are \(k_{1}=4.8 \times 10^{3} ; \quad k_{-1}=3.6 \times 10^{3} ; \quad k_{2}=1.3 \times 10^{-2} ; k_{3}=2.7 \times 10^{2} .\) Derive the rate law and the magnitude of \(k\) for the overall reaction.

A first-order reaction, \(\mathrm{A} \longrightarrow\) products, has a halflife of \(75 \mathrm{s},\) from which we can draw two conclusions. Which of the following are those two (a) the reaction goes to completion in 150 s; (b) the quantity of \(A\) remaining after 150 s is half of what remains after 75 s; (c) the same quantity of A is consumed for every 75 s of the reaction; (d) one- quarter of the original quantity of A is consumed in the first 37.5 s of the reaction; (e) twice as much A is consumed in 75 s when the initial amount of \(\mathrm{A}\) is doubled; (f) the amount of \(\mathrm{A}\) consumed in 150 s is twice as much as is consumed in 75 s.

Derive a plausible mechanism for the following reaction in aqueous solution, \(\mathrm{Hg}_{2}^{2+}+\mathrm{Tl}^{3+} \longrightarrow 2 \mathrm{Hg}^{2+}+\mathrm{Tl}^{+}\) for which the observed rate law is: rate \(=k\left[\mathrm{Hg}_{2}^{2+1}\right]\) \(\left.[\mathrm{T}]^{3+}\right] /\left[\mathrm{Hg}^{2+}\right].\)

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