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

The decomposition of \(\mathrm{A}_{2} \mathrm{~B}_{2}\) to \(\mathrm{A}_{2}\) and \(\mathrm{B}_{2}\) at \(38^{\circ} \mathrm{C}\) was monitored as a function of time. A plot of \(1 /\left[\mathrm{A}_{2} \mathrm{~B}_{2}\right]\) vs. time is linear, with slope \(0.137 / M \cdot \mathrm{min}\) (a) Write the rate expression for the reaction. (b) What is the rate constant for the decomposition at \(38^{\circ} \mathrm{C} ?\) (c) What is the half-life of the decomposition when \(\left[\mathrm{A}_{2} \mathrm{~B}_{2}\right]\) is \(0.631 \mathrm{M} ?\) (d) What is the rate of the decomposition when \(\left[\mathrm{A}_{2} \mathrm{~B}_{2}\right]\) is \(0.219 \mathrm{M}\) ? (e) If the initial concentration of \(\mathrm{A}_{2} \mathrm{~B}_{2}\) is \(0.822 \mathrm{M}\) with no products present, then what is the concentration of \(\mathrm{A}_{2}\) after \(8.6\) minutes?

Short Answer

Expert verified
(a) The rate expression for the reaction is Rate = k[A2B2]. (b) The rate constant for the decomposition at 38°C is 0.137 M⁻¹·min⁻¹. (c) The half-life of the decomposition when [A2B2] = 0.631 M is approximately 5.06 minutes. (d) The rate of the decomposition when [A2B2] = 0.219 M is approximately 0.030 M·min⁻¹. (e) The concentration of A2 after 8.6 minutes when the initial concentration of A2B2 is 0.822 M is approximately 0.402 M.

Step by step solution

01

(a) Determine the Rate Expression

As a plot of \(1/\left[\mathrm{A}_{2}\mathrm{~B}_{2}\right]\) vs. time is linear, it indicates that the reaction is first order with respect to A₂B₂. Let's write the rate of the reaction: Rate = k[A2B2]
02

(b) Find the Rate Constant

To find the rate constant, we'll use the slope given in the problem. Since the plot of \(1/\left[\mathrm{A}_{2}\mathrm{~B}_{2}\right]\) vs. time is linear, the slope of this plot (\(m\)) is equivalent to the rate constant for the decomposition (k): k = 0.137 M⁻¹·min⁻¹
03

(c) Calculate the Half-life

For a first-order reaction, the half-life (t₁/₂) is given by the formula: \(t_{1/2} = \frac{0.693}{k}\) We can plug in the rate constant found earlier: \(t_{1/2} = \frac{0.693}{0.137\thinspace M^{-1} \cdot min^{-1}} \approx 5.06\thinspace min\)
04

(d) Find the Rate of Decomposition

To find the rate of decomposition when [A₂B₂] is 0.219 M, we can use the rate expression determined earlier: Rate = k[A₂B₂] = (0.137 M⁻¹·min⁻¹)(0.219 M) ≈ 0.030\thinspace M·min⁻¹
05

(e) Calculate the Concentration of A₂

To find the concentration of A₂ after 8.6 minutes, we'll use the integrated rate law for a first-order reaction: \(\left[\mathrm{A}_{2}\right] = \left[\mathrm{A}_{2}\mathrm{~B}_{2}\right]_{0} \cdot \left(1 - e^{-kt}\right)\) Let's plug in the values for the initial concentration of A₂B₂ (0.822 M), the rate constant (0.137 M⁻¹·min⁻¹), and time (8.6 min): \(\left[\mathrm{A}_{2}\right] = 0.822\mathrm{M} \cdot \left(1 - e^{-0.137\cdot8.6\mathrm{M^{-1}\cdot min^{-1}\cdot min}}\right) \approx 0.402\thinspace M\) Therefore, the concentration of A₂ after 8.6 minutes is approximately 0.402 M.

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 Rate
The rate of a chemical reaction is a measure of how fast the reactants are converted into products. Understanding reaction rate is crucial because it helps us control and optimize chemical processes.
  • It is often expressed as the change in concentration of a substance over time, and the units are typically moles per liter per second (mol/L/s).
  • The reaction rate can depend on several factors, including concentration, temperature, physical state, and catalysts.
In the given exercise, we deal with the decomposition of \(\mathrm{A}_{2}\mathrm{~B}_{2}\) into \(\mathrm{A}_{2}\) and \(\mathrm{B}_{2}\). The rate of this reaction is determined by monitoring how the concentration of \(\mathrm{A}_{2}\mathrm{~B}_{2}\) changes over time. The linear nature of the plot of \(1/\left[\mathrm{A}_{2}\mathrm{~B}_{2}\right]\) versus time indicates specific characteristics of the reaction order, which brings us to the next concept.
First-Order Reaction
A first-order reaction is one where the rate is directly proportional to the concentration of one reactant. Therefore, it only requires one reactant concentration to determine the reaction rate. This simplicity makes first-order reactions particularly significant in chemical kinetics.
  • The mathematical expression for a first-order reaction typically follows: \(\text{Rate} = k[\text{A}]\), where \(k\) is the rate constant, and \([\text{A}]\) is the concentration of the reactant.
  • For a first-order reaction, a plot of the reciprocal of the concentration (e.g., \(1/[\text{Concentration}]\)) versus time is linear, as seen in this exercise with \(1/[\mathrm{A}_{2}\mathrm{~B}_{2}]\).
Understanding first-order reactions is invaluable, especially when calculating parameters such as reaction half-life or predicting changes in reactant concentration over time.In the exercise, the fact that the plot of \(1/[\mathrm{A}_{2}\mathrm{~B}_{2}]\) versus time is linear stands as evidence of the reaction being first-order with respect to \(\mathrm{A}_{2}\mathrm{~B}_{2}\). This implicates a straightforward relationship between the concentration and the rate, making calculations easier.
Rate Constant
The rate constant, \(k\), is a fundamental part of the rate law, determining the speed of a reaction at a given concentration of reactants. Each reaction has its own rate constant, and it is influenced by conditions like temperature.
  • Units of the rate constant vary depending on the order of the reaction. For a first-order reaction, the units are typically \(\text{M}^{-1}\cdot\text{min}^{-1}\).
  • It can be determined experimentally, often from the slope of a linear plot as in the exercise. Here, the slope (0.137 \(\text{M}^{-1}\cdot\text{min}^{-1}\)) directly provides the rate constant \(k\).
A rate constant gives insight into how quickly a reaction will occur. In the exercise, knowing \(k\) allows us to predict how the concentration of \(\mathrm{A}_{2}\mathrm{~B}_{2}\) will decrease over time. Furthermore, it helps in calculating other kinetic parameters like the reaction half-life, providing a complete picture of the reaction kinetics. Understanding the rate constant is essential, since it links the calculated rate to actual reactant concentrations and time.

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

Ammonia is produced by the reaction between nitrogen and hydro- gen gases. (a) Write a balanced equation using smallest whole-number coefficients for the reaction. (b) Write an expression for the rate of reaction in terms of \(\Delta\left[\mathrm{NH}_{3}\right]\). (c) The concentration of ammonia increases from \(0.257 \mathrm{M}\) to \(0.815 \mathrm{M}\) in \(15.0 \mathrm{~min} .\) Calculate the average rate of reaction over this time interval.

For the reaction $$5 \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 3 \mathrm{Br}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}$$ it was found that at a particular instant bromine was being formed at the rate of \(0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). At that instant, at what rate is (a) water being formed? (b) bromide ion being oxidized? (c) \(\mathrm{H}^{+}\) being consumed?

The uncoiling of deoxyribonucleic acid (DNA) is a first-order reaction. Its activation energy is \(420 \mathrm{~kJ}\). At \(37^{\circ} \mathrm{C}\), the rate constant is \(4.90 \times 10^{-4} \mathrm{~min}^{-1}\). (a) What is the half-life of the uncoiling at \(37^{\circ} \mathrm{C}\) (normal body temperature)? (b) What is the half-life of the uncoiling if the organism has a temperature of \(40^{\circ} \mathrm{C}\left(\approx 104^{\circ} \mathrm{F}\right)\) ? (c) By what factor does the rate of uncoiling increase (per \({ }^{\circ} \mathrm{C}\) ) over this temperature interval?

Cold-blooded animals decrease their body temperature in cold weather to match that of their environment. The activation energy of a certain reaction in a cold-blooded animal is \(65 \mathrm{~kJ} / \mathrm{mol} .\) By what percentage is the rate of the reaction decreased if the body temperature of the animal drops from \(35^{\circ} \mathrm{C}\) to \(22^{\circ} \mathrm{C} ?\)

The decomposition of \(\mathrm{Y}\) is a zero-order reaction. Its half-life at \(25^{\circ} \mathrm{C}\) and \(0.188 M\) is 315 minutes. (a) What is the rate constant for the decomposition of Y? (b) How long will it take to decompose a \(0.219 \mathrm{M}\) solution of \(\mathrm{Y}\) ? (c) What is the rate of the decomposition of \(0.188 \mathrm{M}\) at \(25^{\circ} \mathrm{C}\) ? (d) Does the rate change when the concentration of \(\mathrm{Y}\) is increased to \(0.289 \mathrm{M}\) ? If so, what is the new rate?

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