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Chapter 11: Problem 29

A solution of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}\) yields by decomposition at \(45^{\circ} \mathrm{C}, 4.8 \mathrm{ml}\) of \(\mathrm{O}_{2}\), 20 min after the start of the experiment and \(9.6 \mathrm{ml}\) of \(\mathrm{O}_{2}\) after a very long time. The decomposition obeys first-order kinetics. What volume of \(\mathrm{O}_{2}\) would have evolved, 40 min after the start? (a) \(7.2 \mathrm{ml}\) (b) \(2.4 \mathrm{ml}\) (c) \(9.6 \mathrm{ml}\) (d) \(6.0 \mathrm{ml}\)

Short Answer

Expert verified
6.0 ml

Step by step solution

01

Understanding First-Order Kinetics

First-order kinetics implies that the rate of reaction is directly proportional to the concentration of a single reactant. The integrated rate law for a first-order reaction is given by \(\ln\left[\frac{\mathrm{A}}{\mathrm{A}_0}\right]= -kt\), where \(\mathrm{A}\) is the concentration of reactant at time \(t\), \(\mathrm{A}_0\) is the initial concentration, and \(k\) is the rate constant.
02

Determine the Rate Constant

Using the data given that 4.8 ml of \(\mathrm{O}_2\) are produced in 20 minutes, we can set up the equation using the integrated rate law for first-order reactions \(\ln\left[\frac{\mathrm{O}_2 \, at \, 20 \, min}{\mathrm{O}_2 \, at \, \infty}\right]= -k \times 20\). Solving for \(k\), we can find the rate constant.
03

Calculate the Volume of \(\mathrm{O}_2\) after 40 Minutes

Now that we have the rate constant \(k\), we use the first-order kinetic equation to find the volume of \(\mathrm{O}_2\) produced at 40 minutes: \(\ln\left[\frac{\mathrm{O}_2 \, at \, 40 \, min}{\mathrm{O}_2 \, at \, \infty}\right]= -k \times 40\). With the initial and final volumes known, we can solve this equation for the volume of \(\mathrm{O}_2\) at 40 minutes.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the branch of chemistry that encompasses the study of the speed or rate at which chemical reactions occur. It involves analyzing how the reaction rate changes under varying conditions and how different variables can influence it. For instance, by altering the temperature, pressure, or reactant concentration, the rate of a reaction can be significantly affected.

Understanding the factors that affect reaction rates is crucial in a wide array of fields ranging from industrial synthesis to pharmaceuticals, where controlling the speed of reactions is essential to the quality and safety of the products. In educational settings, problems related to chemical kinetics allow students to apply mathematical concepts to predict and interpret the behavior of chemical reactions over time.
Integrated Rate Law
The integrated rate law is a mathematical expression that describes the concentration of a reactant as a function of time. It is derived from the rate law equation, which shows the relationship between the rate of a chemical reaction and the concentration of its reactants. For first-order reactions, the integrated rate law takes on a specific form that makes it possible to directly relate the concentration of the reactant at any given time to the initial concentration and the rate constant. The equation for first-order reactions is: \[\begin{equation}\ln\left[\frac{\mathrm{A}}{\mathrm{A}_0}\right]= -kt\end{equation}\]Here, \( \ln \) is the natural logarithm, \( \mathrm{A} \) is the concentration of the reactant at time \( t \), \( \mathrm{A}_0 \) is the initial concentration, and \( k \) is the rate constant. This law is particularly useful for solving problems where we track the progress of the reaction over time, such as determining the concentration of a reactant after a certain period or calculating the half-life of a substance.
Reaction Rate Constant
The reaction rate constant, denoted as \( k \), is a crucial parameter in the field of chemical kinetics. It represents the proportionality constant in the rate law equation and is specific to each chemical reaction at a given temperature. For a first-order reaction, the rate constant can be determined experimentally by measuring the concentration of a reactant at various times and applying the integrated rate law.

The value of \( k \) gives insights into the speed of the reaction: a larger \( k \) indicates a faster reaction, and vice versa. It is important to note that the rate constant is independent of the concentration but is temperature dependent. The units of \( k \) vary with the order of the reaction; for first-order reactions, it is usually expressed in reciprocal seconds \( (s^{-1}) \). Understanding the reaction rate constant is essential for predicting how fast a reaction will proceed under specific conditions and is often the key to optimizing chemical processes.

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Most popular questions from this chapter

The time taken in \(75 \%\) completion of a zero-order reaction is \(10 \mathrm{~h}\). In what time, he reaction will be \(90 \%\) completed? a) \(12.0 \mathrm{~h}\) (b) \(16.6 \mathrm{~h}\) c) \(10.0 \mathrm{~h}\) (d) \(20.0 \mathrm{~h}\)

The suggested mechanism for the reaction: \(\mathrm{CHCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CCl}_{4}(\mathrm{~g})\) \(+\mathrm{HCl}(\mathrm{g})\), is $$ \mathrm{Cl}_{2} \underset{K_{2}}{\stackrel{K_{1}}{\leftrightarrows}} 2 \mathrm{C} \mathrm{C} \text { (fast) } $$ \(\mathrm{CHCl}_{3}+\mathrm{Ci} \stackrel{K_{3}}{\longrightarrow} \mathrm{HCl}+\dot{\mathrm{C}} \mathrm{Cl}_{3}(\mathrm{slow})\) $$ \dot{\mathrm{C}} \mathrm{Cl}_{3}+\mathrm{C} \mathrm{\textrm{l }} \stackrel{K_{4}}{\longrightarrow} \mathrm{CCl}_{4} \text { (fast) } $$ The experimental rate law consistent with the mechanism is (a) rate \(=K_{3}\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]\) (b) rate \(=K_{4}\left[\mathrm{CCl}_{3}\right][\mathrm{Cl}]\) (c) rate \(=K_{\text {eq }}\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]\) (d) rate \(=K_{3} K_{\text {eq }}^{1 / 2}\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]^{1 / 2}\)

For a first-order reaction: \(\mathrm{A} \rightarrow\) Product, the initial concentration of \(\mathrm{A}\) is \(0.1 \mathrm{M}\) and after time \(40 \mathrm{~min}\), it becomes \(0.025 \mathrm{M}\). What is the rate of reaction at reactant concentration \(0.01 \mathrm{M} ?\) (a) \(3.465 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\) (b) \(3.465 \times 10^{-5} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\) (c) \(6.93 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\) (d) \(1.7325 \times 10^{-4} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\)

For a reaction of order \(n\), the integrated form of the rate equation is: \((n-1) \cdot K \cdot t\) \(=\left(C_{0}\right)^{1-n}-(C)^{1-n}\), where \(C_{0}\) and \(C\) are the values of the reactant concentration at the start and after time ' \(t\) '. What is the relationship between \(t_{3 / 4}\) and \(t_{1 / 2}\), where \(t_{3 / 4}\) is the time required for \(C\) to become \(C_{0} / 4\). (a) \(t_{3 / 4}=t_{1 / 2} \cdot\left[2^{n-1}+1\right]\) (b) \(t_{3 / 4}=t_{1 / 2}\left[2^{n-1}-1\right]\) (c) \(t_{3 / 4}=t_{1 / 2}\left[2^{n+1}-1\right]\) (d) \(t_{3 / 4}=t_{1 / 2} \cdot\left[2^{n+1}+1\right]\)

For a reaction \(2 \mathrm{NH}_{3} \rightarrow \mathrm{N}_{2}+3 \mathrm{H}_{2}\), it is observed that \(-\frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{d} t}=K_{1}\left[\mathrm{NH}_{3}\right] ;+\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{d} t}=K_{2}\left[\mathrm{NH}_{3}\right]\) and \(+\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{d} t}=K_{3}\left[\mathrm{NH}_{3}\right]\) The correct relation between \(K_{1}, K_{2}\) and \(K_{3}\) is (a) \(K_{1}=K_{2}=K_{3}\) (b) \(2 K_{1}=3 K_{2}=6 K_{3}\) (c) \(3 K_{1}=6 K_{2}=2 K_{3}\) (d) \(6 K_{1}=3 K_{2}=2 K_{3}\)
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