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

For the reaction \(A \longrightarrow\) products, the following data were obtained: \(t=0 \mathrm{s},[\mathrm{A}]=0.715 \mathrm{M} ; 22 \mathrm{s}, 0.605 \mathrm{M}\) 74 s, 0.345 M; 132 s, 0.055 M. (a) What is the order of this reaction? (b) What is the half-life of the reaction?

Short Answer

Expert verified
The order of reaction is first order. The half-life can be calculated using the average value of k obtained from step 1 in formula \( t1/2 = 0.693 / k \).

Step by step solution

01

Calculate the Rate Constants for Each Time Interval

To find the order of the reaction, initially calculate the rate constant (k) for each time interval using the formula for a first order reaction, \( k = (1/t) * ln([A]_{initial} / [A]_{final}) \) where [A] is the molar concentration. For e.g, for \( t = 0 \) to \( t = 22 \) sec, \( k = (1/22) * ln(0.715/0.605) \) Repeat the above step for each time interval and compare the values of k.
02

Analyze the Rate Constants

If the rate constants are approximately the same for each time interval, this indicates that the reaction is indeed first order. If they are not, then might need to test for zero or second order reaction, but for simplicity of this explanation, assume it is a first-order reaction if k keeps approximately constant.
03

Calculate the Half-Life

For a first order reaction, the half-life (t1/2) is given by \( t1/2 = 0.693 / k \). Use the average k obtained from step 1 to evaluate t1/2. You might find that it matches closely with the interval of time it takes for the molar concentration of A to reduce approximately to half of its initial value, further confirming first order kinetics.

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.

Reaction Order
Understanding the reaction order helps us determine how the rate of a chemical reaction depends on the concentration of its reactants. In simple terms, it tells us how the speed of the reaction changes when we change the concentration of the reactants.

* **Zero-order reactions**: For these reactions, the rate is independent of the concentration. Essentially, whenever you change the concentration, it doesn't affect the reaction rate. This means that the reaction occurs at a constant rate until the reactants are completely consumed.
* **First-order reactions**: Here, the rate is directly proportional to the concentration of one reactant. If you double the concentration, the rate also doubles. This indicates a direct relationship between concentration and rate.
* **Second-order reactions**: These depend on the concentration of one reactant squared or on two different reactants each raised to the power of one. So, if the concentration doubles, the rate becomes four times faster in some cases.

Determining the order is crucial because it helps in understanding how quickly a reaction can proceed under given conditions. In the exercise, by checking the consistency of the rate constant, it's suggested that the reaction is first-order.
Rate Constant
The rate constant, denoted as \(k\), is a crucial factor in chemical reactions. It acts as a proportionality constant in the rate equation that relates reactant concentrations to the reaction rate.

For a first-order reaction, the rate equation can be expressed as:
\[ ext{Rate} = k[A] \]

Here, *\([A]\)* represents the concentration of the reactant. The rate constant \(k\) helps quantify how fast the reaction proceeds under certain conditions. Importantly, the value of \(k\) remains constant for a given reaction at a specific temperature and is independent of the concentration.

In the problem given, the calculation of \(k\) for different time intervals allows us to determine the reaction order. If \(k\) stays relatively consistent, it gives us evidence supporting the reaction being first-order. Understanding the rate constant aids in better control and prediction of reaction outcomes.
Half-Life Calculation
Half-life in the context of chemical kinetics refers to the time it takes for half of the reactant concentration to be consumed in a reaction. It is a valuable metric because it gives a time scale for the reaction's progress.

For first-order reactions, the half-life can be calculated using the formula:
\[ t_{1/2} = \frac{0.693}{k} \]

where \( t_{1/2} \) is the half-life and \( k \) is the rate constant. This particular formula indicates that for first-order reactions, the half-life is constant regardless of the initial concentration.

In the given exercise, after determining the average rate constant \(k\), the half-life can be calculated. This value provides insight, indicating that the time required for the concentration to reduce to half its initial value matches observations, confirming the reaction's first-order nature.
First Order Reactions
First-order reactions occupy a special place in chemical kinetics due to their simplicity. In these reactions, the rate is determined by the concentration of a single reactant raised to the power of one. This makes the mathematical treatment straightforward and the predictions reliable.

Characteristics of first-order reactions include:
  • The rate depends directly on the concentration of one reactant.
  • The half-life is independent of the initial concentration, meaning it remains constant throughout the reaction.
  • They often involve processes like radioactive decay and certain types of chemical reactions like isomerization.


In the provided exercise, the process of calculating \(k\) and the resultant consistent values suggest that the reaction is first-order. Such reactions are easier to handle due to these predictable characteristics, making them a fundamental concept in studying chemical kinetics.

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

The following statements about catalysis are not stated as carefully as they might be. What slight modifications would you make in them? (a) A catalyst is a substance that speeds up a chemical reaction but does not take part in the reaction. (b) The function of a catalyst is to lower the activation energy for a chemical reaction.

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 \(k.\)

Ammonia decomposes on the surface of a hot tungsten wire. Following are the half-lives that were obtained at \(1100^{\circ} \mathrm{C}\) for different initial concentrations of \(\mathrm{NH}_{3}:\left[\mathrm{NH}_{3}\right]_{0}=0.0031 \mathrm{M}, t_{1 / 2}=7.6 \mathrm{min} ; 0.0015 \mathrm{M}\) \(3.7 \mathrm{min} ; 0.00068 \mathrm{M}, 1.7 \mathrm{min.}\) For this decomposition reaction, what is (a) the order of the reaction; (b) the rate constant, \(k ?\)

The mechanism proposed for the reaction of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{I}_{2}(\mathrm{g})\) to form \(\mathrm{HI}(\mathrm{g})\) consists of a fast reversible first step involving \(\mathrm{I}_{2}(\mathrm{g})\) and \(\mathrm{I}(\mathrm{g}),\) followed by a slow step. Propose a two-step mechanism for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g}),\) which is known to be first order in \(\mathrm{H}_{2}\) and first order in \(\mathrm{I}_{2}.\)

In the reaction \(A \longrightarrow\) products, 4.40 min after the reac- tion is started, \([\mathrm{A}]=0.588 \mathrm{M}\). The rate of reaction at this point is rate \(=-\Delta[\mathrm{A}] / \Delta t=2.2 \times 10^{-2} \mathrm{M} \mathrm{min}^{-1}.\) Assume that this rate remains constant for a short period of time. (a) What is \([\mathrm{A}] 5.00\) min after the reaction is started? (b) At what time after the reaction is started will \([\mathrm{A}]=0.565 \mathrm{M} ?\)

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.

Sign-up for free