/*! 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 27 The rate constant for the second... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The rate constant for the second-order reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ is \(0.80 / M \cdot \mathrm{s}\) at \(10^{\circ} \mathrm{C}\). (a) Starting with a concentration of \(0.086 M,\) calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when \([\mathrm{NOBr}]_{0}=0.072 M\) and \([\mathrm{NOBr}]_{0}=0.054 \mathrm{M}\)

Short Answer

Expert verified
The concentration of NOBr after 22 seconds is 0.0342 M. The half-life when [NOBr]_0 = 0.072 M is 1.7 s and when [NOBr]_0 = 0.054 M, the half-life is 2.3 s.

Step by step solution

01

Calculate the concentration of NOBr after 22 s

We use the second order reaction equation \(\frac{1}{{[NOBr]}} = kt + \frac{1}{{[NOBr]_0}}\). Let \(x = [NOBr]\), then we have to solve the equation for \(x\) in: \(\frac{1}{x} = (0.80/M.s)(22 s) + \frac{1}{0.086 M}\)
02

Solve the equation

Combining the terms on the right-hand side gives \(\frac{1}{x} = 17.6 M^{-1} + 11.6 M^{-1} = 29.2 M^{-1}\), or \(x = [NOBr] = 0.0342 M\)
03

Calculate the half-lives

Using the second order half-life equation \(\(\(t_{1/2} = \frac{1}{{k[A]_0}}\), we can calculate the half-lives for an initial concentration of 0.072 M and 0.054 M, respectively.
04

Half-life for [NOBr]_0 = 0.072 M

\(\(\(t_{1/2} = \frac{1}{{(0.80/M.s)(0.072 M)}} = 1.7 s\)
05

Half-life for [NOBr]_0 = 0.072 M

\(\(\(t_{1/2} = \frac{1}{{(0.80/M.s)(0.054 M)}} = 2.3 s\)

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 91Ó°ÊÓ!

Key Concepts

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

Rate Constant
The rate constant, denoted as \( k \), plays a crucial role in determining how fast a reaction occurs in chemical kinetics. For a second-order reaction, the rate is proportional to the product of the concentrations of the reactants raised to their stoichiometric coefficients. The equation for a second-order reaction is usually expressed as \( ext{Rate} = k[A][B] \), but in the simpler case where only one reactant repeats, it becomes \( ext{Rate} = k[A]^2 \).
In the given exercise, the second-order reaction involves nitric oxide bromide decomposing, and the rate constant is provided as \( 0.80 \, M^{-1} \, s^{-1} \). This rate constant indicates how strongly the concentration of reactants influences the speed of the reaction at a certain temperature, here specified at \( 10^{\circ} \mathrm{C} \). Understanding the meaning of the rate constant is key, as it allows us to predict the reaction behavior under various experimental conditions.
Half-Life
The half-life of a reaction is the time required for the concentration of a reactant to reduce to half its initial value. For second-order reactions, the half-life is not constant and depends on the initial concentration of the reactant. The formula used to calculate the half-life \( t_{1/2} \) for second-order reactions is \( t_{1/2} = \frac{1}{k[A]_0} \).
This exercise highlights how, for different initial concentrations of nitric oxide bromide (\([\mathrm{NOBr}]_0\)), the half-life changes. For \([\mathrm{NOBr}]_0 = 0.072 \, M\), the calculated half-life is approximately \(1.7 \, s\), while for \([\mathrm{NOBr}]_0 = 0.054 \, M\), it is \(2.3 \, s\). This demonstrates the inverse relationship between concentration and half-life in second-order kinetics.
Chemical Kinetics
Chemical kinetics involves the study of reaction rates and the steps that occur in chemical processes. It allows us to understand the factors influencing the speed of a reaction, such as reactant concentration, temperature, and the presence of catalysts.
In the exercise, observing a chemical reaction with nitric oxide bromide offers insights into the kinetics of decomposition. The rate constant already mentioned is central to these kinetic considerations, being sensitive to changes in temperature. Hence, understanding kinetics not only helps in predicting how long a process will take but also assesses the conditions under which it will proceed efficiently with regard to time and resource consumption.
Concentration Calculation
In chemical kinetics, calculating the concentration of reactants or products over time is an essential aspect of understanding a reaction's progress. For second-order reactions, the concentration of a reactant at any time \( t \) can be determined using the integrated rate equation:
\[ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \]
This equation allows us to calculate the remaining concentration of \([NOBr]\) in the given exercise after \(22\) seconds, with an initial concentration of \(0.086 \, M\). By plugging into the equation and solving, the concentration of \([NOBr]\) is shown to decrease to \(0.0342 \, M\). This calculation shows how the concentration diminishes as the reaction progresses over 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

The reaction \(\mathrm{H}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2}+\mathrm{H}\) has been studied for many years. Sketch a potential energy versus reaction progress diagram for this reaction.

A gas mixture containing \(\mathrm{CH}_{3}\) fragments, \(\mathrm{C}_{2} \mathrm{H}_{6}\), molecules, and an inert gas (He) was prepared at \(600 \mathrm{~K}\) with a total pressure of 5.42 atm. The elementary reaction $$ \mathrm{CH}_{3}+\mathrm{C}_{2} \mathrm{H}_{6} \longrightarrow \mathrm{CH}_{4}+\mathrm{C}_{2} \mathrm{H}_{5} $$ has a second-order rate constant of \(3.0 \times 10^{4} / M \cdot\) s. Given that the mole fractions of \(\mathrm{CH}_{3}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) are 0.00093 and \(0.00077,\) respectively, calculate the initial rate of the reaction at this temperature.

Chlorine oxide (ClO), which plays an important role in the depletion of ozone (see Problem 13.101 ), decays rapidly at room temperature according to the equation $$ 2 \mathrm{ClO}(g) \longrightarrow \mathrm{Cl}_{2}(g)+\mathrm{O}_{2}(g) $$ From the following data, determine the reaction order and calculate the rate constant of the reaction. $$ \begin{array}{ll} \hline \text { Time (s) } & {[\mathrm{ClO}](M)} \\ \hline 0.12 \times 10^{-3} & 8.49 \times 10^{-6} \\ 0.96 \times 10^{-3} & 7.10 \times 10^{-6} \\ 2.24 \times 10^{-3} & 5.79 \times 10^{-6} \\ 3.20 \times 10^{-3} & 5.20 \times 10^{-6} \\ 4.00 \times 10^{-3} & 4.77 \times 10^{-6} \\ \hline \end{array} $$

The decomposition of \(\mathrm{N}_{2} \mathrm{O}\) to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is a firstorder reaction. At \(730^{\circ} \mathrm{C}\) the half-life of the reaction is \(3.58 \times 10^{3} \mathrm{~min} .\) If the initial pressure of \(\mathrm{N}_{2} \mathrm{O}\) is 2.10 atm at \(730^{\circ} \mathrm{C},\) calculate the total gas pressure after one half-life. Assume that the volume remains constant.

To carry out metabolism, oxygen is taken up by hemoglobin (Hb) to form oxyhemoglobin (HbO \(_{2}\) ) according to the simplified equation $$ \mathrm{Hb}(a q)+\mathrm{O}_{2}(a q) \stackrel{k}{\longrightarrow} \mathrm{HbO}_{2}(a q) $$ where the second-order rate constant is \(2.1 \times 10^{6} / M \cdot \mathrm{s}\) at \(37^{\circ} \mathrm{C}\). (The reaction is first order in \(\mathrm{Hb}\) and \(\mathrm{O}_{2}\).) For an average adult, the concentrations of \(\mathrm{Hb}\) and \(\mathrm{O}_{2}\) in the blood at the lungs are \(8.0 \times 10^{-6} \mathrm{M}\) and \(1.5 \times 10^{-6} M,\) respectively. (a) Calculate the rate of formation of \(\mathrm{HbO}_{2}\). (b) Calculate the rate of consumption of \(\mathrm{O}_{2}\). (c) The rate of formation of \(\mathrm{HbO}_{2}\) increases to \(1.4 \times 10^{-4} M / \mathrm{s}\) during exercise to meet the demand of increased metabolism rate. Assuming the \(\mathrm{Hb}\) concentration to remain the same, what must be the oxygen concentration to sustain this rate of \(\mathrm{HbO}_{2}\) formation?

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.