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Chapter 17: Problem 23

Azomethane, \(\mathrm{CH}_{3} \mathrm{~N}_{2} \mathrm{CH}_{5}(g)\), decomposes according to the equation $$ \mathrm{CH}_{3} \mathrm{~N}_{2} \mathrm{CH}_{3}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{3}(g)+\mathrm{N}_{2}(g) $$ Given that the decomposition is a first-order process with \(k=4.0 \times 10^{-4} \mathrm{~s}^{-1}\) at \(300^{\circ} \mathrm{C}\), calculate the fraction of azomethane that remains after \(1.0\) hour.

Short Answer

Expert verified
23.7% of azomethane remains after 1 hour.

Step by step solution

01

Understand the given

The decomposition reaction is a first-order process, which means the rate equation follows the first-order kinetics: \( \ln[A]_t = \ln[A]_0 - kt \), where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant given as \(4.0 \times 10^{-4}\ s^{-1}\). We need to find the fraction of azomethane remaining after \(1.0\) hour.
02

Convert time

Since the rate constant \(k\) is given in \(s^{-1}\), convert 1.0 hour into seconds. \(1.0\) hour is \(3600\) seconds because \(1\) hour \(= 60\) minutes and \(1\) minute \(= 60\) seconds. Thus, \(1.0\) hour \(= 3600\) seconds.
03

Apply first-order kinetics equation

Substitute the known values into the first-order kinetics equation to find \(\ln[A]_t / [A]_0\). Using \(k=4.0 \times 10^{-4}\ s^{-1}\) and \(t=3600\ s\), the equation becomes: \[ \ln \left(\frac{[A]_t}{[A]_0}\right) = -kt \] \[ \ln \left(\frac{[A]_t}{[A]_0}\right) = -(4.0 \times 10^{-4} \ s^{-1}) \times (3600\ s) \]
04

Calculate the expression

Evaluate the expression for \(-kt\): \[ -kt = -(4.0 \times 10^{-4} \times 3600) = -1.44 \]
05

Solve for remaining fraction

Use the result from the previous step:\[ \ln \left(\frac{[A]_t}{[A]_0}\right) = -1.44 \]Exponentiate both sides to solve for \(\frac{[A]_t}{[A]_0}\):\[ \frac{[A]_t}{[A]_0} = e^{-1.44} \]
06

Calculate the remaining fraction

Compute \(e^{-1.44}\) to find the fraction of azomethane remaining after 1 hour:\[ e^{-1.44} \approx 0.237 \] Thus, approximately 23.7% of azomethane remains after 1 hour.

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Key Concepts

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

Understanding Rate Constant in First-Order Reactions
In chemical kinetics, the rate constant, represented as \( k \), is a crucial parameter that determines the speed of a reaction. For a first-order reaction, like the decomposition of azomethane, the rate constant has units of \( \text{s}^{-1} \). This tells us how quickly the reaction occurs per second.
First-order reactions follow a specific rate law: \( k[A]^1 \).
  • The rate is directly proportional to the concentration of only one reactant.
  • In our case with azomethane, as the rate constant is \( 4.0 \times 10^{-4} \ \text{s}^{-1} \), it indicates that for each second, a small fraction of azomethane molecules are decomposing.
Understanding this helps us predict how the concentration of reactants decrease over time during the reaction.
Decomposition Reaction Mechanism
A decomposition reaction involves a single compound breaking down into two or more products. The decomposition of azomethane is a classic example, where:
  • The molecule \( \text{CH}_3 \text{N}_2 \text{CH}_3 \) breaks down into \( \text{CH}_3 \text{CH}_3 \) (ethane) and \( \text{N}_2 \) (nitrogen gas).
  • This type of reaction is essential because it often releases energy and forms stable products.
  • It's also characterized by having only one reactant, which simplifies the understanding and calculation of its kinetics, as we focus on the destruction of the initial compound.
The first-order nature of azomethane's decomposition means that the rate at which this breakdown occurs depends solely on the concentration of azomethane, which gradually decreases over time.
Time Conversion and Impact on Chemical Kinetics
In chemical equations, especially when dealing with rate constants that have time-based units like \( \text{s}^{-1} \), converting units of time is critical. Given that azomethane's rate constant is expressed per second, calculations involving longer periods require conversion for consistency.
For example, converting 1 hour into seconds helps in plugging the right value into equations. Since \( 1 \) hour is \( 60 \) minutes and each minute is \( 60 \) seconds, the total for an hour becomes \( 3600 \) seconds.
  • This conversion ensures that when you apply the rate law equation, you maintain consistent units throughout the calculation, avoiding errors.
  • Such conversions are foundational in kinetic studies, ensuring precision when predicting reaction behaviors over any given timeline.
So, whether predicting the remaining concentration of a material or adjusting for experimental parameters, understanding time conversion is essential.

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Most popular questions from this chapter

Determine the rate law for the reaction described by $$ \mathrm{NO}(g)+\mathrm{H}_{2}(g) \rightarrow \text { products } $$ from the initial-rate data tabulated below: \begin{tabular}{ccc} \hline Initial pressure of \(P_{\mathrm{H}_{2}} /\) Torr & Initial pressure of \(P_{\mathrm{No}} /\) Torr & Initial rate of reaction/Torr \(\cdot \mathrm{s}^{-1}\) \\ \hline 400 & 159 & 34 \\ 400 & 300 & 125 \\ 289 & 400 & 160 \\ 205 & 400 & 110 \\ 147 & 400 & 79 \\ \hline \end{tabular} Calculate the value of the rate constant for this reaction.

How does doubling the concentration of a reactant change the rate of a reaction that is first order in that reactant? How does it change the rate of a reaction that is second order in that reactant?

Strontium-90 is a radioactive isotope that is produced in nuclear explosions. It decays by \(\beta\) -emission with a half-life of \(29.1\) years. Suppose that an infant ingests strontium-90 in mother's milk. Calculate the fraction of the ingested strontium-90 that remains in the body when the infant reaches 74 years of age, assuming no loss of strontium-90 except by radioactive decay.

The rate law for the reaction described by the equation $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightarrow 2 \mathrm{NOBr}(g) $$ is rate of reaction \(=\left(1.3 \times 10^{-9} \mathrm{M}^{-2} \cdot \mathrm{min}^{-1}\right)[\mathrm{NO}]^{2}\left[\mathrm{Br}_{2}\right]\) Calculate the rate of reaction when \([\mathrm{NO}]_{0}=\left[\mathrm{Br}_{2}\right]_{0}=\) \(3.0 \times 10^{-4} \mathrm{M}\). What is the overall order of this reaction?

The following table gives \(\left[\mathrm{NO}_{2}\right]\) as a function of time for the reaction described by \begin{tabular}{l} \(\mathrm{NO}_{2}(g) \rightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\) \\\ {\(\left[\mathrm{NO}_{2}\right] / \mathbf{M} \quad t / \mathrm{s}\)} \\ \hline \(0.0831 \quad 0\) \\ \(0.0666\) & \(4.2\) \\ \(0.0567\) & \(7.9\) \\ \(0.0497\) & \(11.4\) \\ \(0.0441\) & \(15.0\) \\ \hline \end{tabular} Show that this reaction is second order by plotting \(1 /\left[\mathrm{NO}_{2}\right]\) versus time and determine the value of the rate constant.
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