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Chapter 7: Problem 52

The first order reaction, \(2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\), has a rate constant equal to \(0.76 \mathrm{~s}^{-1}\) at \(1000 \mathrm{~K}\). How long will it take for the concentration of \(\mathrm{N}_{2} \mathrm{O}\) to decrease to \(42 \%\) of its initial concentration? a. \(3.1 \mathrm{~s}\) b. \(0.18 \mathrm{~s}\) c. \(1.1 \mathrm{~s}\) d. \(2.4 \mathrm{~s}\)

Short Answer

Expert verified
It takes approximately 1.1 seconds.

Step by step solution

01

Identify Reaction Type and Use the Correct Formula

This reaction is a first-order reaction. For first-order reactions, the rate law is described by the formula: \[ ext{ln}( [A]_0 / [A] ) = kt \] where \([A]_0\) is the initial concentration,\([A]\) is the concentration at time \(t\),\(k\) is the rate constant andt is the time.
02

Set Up the Problem

In this problem, we are given:- The rate constant \(k = 0.76 \, \text{s}^{-1}\)- The concentration of \(\mathrm{N}_{2} \mathrm{O}\) decreases to \(42\%\) of its initial concentration, so \([A] = 0.42 \, [A]_0\).
03

Substitute Values into the Formula and Solve for t

Substitute the known values into the first-order reaction rate law formula:\[ \ln\left( \frac{[A]_0}{0.42 \, [A]_0} \right) = 0.76 \, t \] Simplifying gives:\[ \ln\left( \frac{1}{0.42} \right) = 0.76 \, t \]Calculate \(\ln(2.38095)\) which approximately equals \(0.8673\).
04

Solve for t

Solving for \(t\), we get:\[ 0.8673 = 0.76 \, t \]Divide both sides by \(0.76\):\[ t = \frac{0.8673}{0.76} \approx 1.1417 \, \text{s} \]Therefore, it takes about \(1.1417\) seconds, which rounds to \(1.1\) seconds.

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Key Concepts

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

Rate Constant
In first-order reaction kinetics, the rate constant, often denoted as \( k \), is a crucial parameter. It provides information on the speed at which the reaction proceeds. The rate constant is specific to the reaction and depends on factors like temperature and the nature of the reactants.
A higher rate constant indicates a faster reaction. Conversely, a lower rate constant suggests a slower process. For the reaction \(2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\), the rate constant is given as \(0.76 \mathrm{~s}^{-1}\) at \(1000 \mathrm{~K}\).
This means that, at this temperature, the reaction will proceed with moderate speed. The rate constant remains constant for a given reaction at a specific temperature.
Concentration Change
Understanding concentration change is vital in monitoring a reaction over time. In the context of a first-order reaction, the change in concentration follows an exponential decay pattern.
This is captured by the formula \( \ln( [A]_0 / [A] ) = kt \), where \([A]_0\) is the initial concentration and \([A]\) is the concentration at time \( t \).
When we say the concentration decreases to \(42\%\) of its initial value, it means \([A] = 0.42 \, [A]_0\). This reflects the fraction of reactant remaining after a certain period.
The exponential nature of this relationship is fundamental to many real-world scenarios, such as chemical breakdowns, drug decay in the body, and more.
Reaction Rate Law
The reaction rate law describes the relationship between the concentrations of reactants and the rate of the reaction. For first-order reactions, it is given by the integrated rate equation \( \ln( [A]_0 / [A] ) = kt \).
This equation can predict how long it takes for a reactant concentration to decline to a certain level. Importantly, it considers only the concentration of the reactant itself, in this case, \( \mathrm{N}_{2} \mathrm{O} \).
The rate law provides a direct way to calculate time \( t \) required for the concentration to hit a specified target, given the rate constant \( k \).
This concept is fundamental in fields like pharmacokinetics, environmental science, and any area dealing with reactive systems. It highlights that the reaction rate is directly proportional to the amount of reactant present.

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Most popular questions from this chapter

The equation tris(1,10-phenanthroline) iron(II) in acid solution takes place according to the equation: \(\mathrm{Fe}(\text { phen })_{3}^{2+}+3 \mathrm{H}_{3} \mathrm{O}^{+}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow\) $$ \left.\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{2+}+3 \text { (Phen }\right) \mathrm{H}^{+} $$ If the activation energy (Ea) is \(126 \mathrm{~kJ} / \mathrm{mol}\) and the rate constant at \(30^{\circ} \mathrm{C}\) is \(9.8 \times 10^{-3} \mathrm{~min}^{-1}\), what is the frequency factor (A)? a. \(9.5 \times 10^{18} \mathrm{~min}^{-1}\) b. \(2.5 \times 10^{19} \mathrm{~min}^{-1}\) c. \(55 \times 10^{19} \mathrm{~min}^{-1}\) d. \(5.0 \times 10^{19} \mathrm{~min}^{-1}\)

The following set of data was obtained by the method of initial rates for the reaction: \(2 \mathrm{HgCl}_{2}(\mathrm{aq})+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(\mathrm{aq}) \rightarrow\) \(2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(\mathrm{~s})\) What is the rate law for the reaction? \begin{tabular}{lll} \hline\(\left[\mathrm{HgCl}_{2}\right], \mathrm{M}\) & {\(\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right], \mathrm{M}\)} & Rate, \(\mathrm{M} / \mathrm{s}\) \\ \hline \(0.10\) & \(0.10\) & \(1.3 \times 10^{-7}\) \\ \(0.10\) & \(0.20\) & \(5.2 \times 10^{-7}\) \\ \(0.20\) & \(0.20\) & \(1.0 \times 10^{-6}\) \\ \hline \end{tabular} a. Rate \(=\mathrm{k}\left[\mathrm{HgCl}_{2}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2}\) b. Rate \(=\mathrm{k}\left[\mathrm{HgCl}_{2}\right]^{2}\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]\) c. Rate \(=\mathrm{k}\left[\mathrm{HgCl}_{2}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2-}\) d. Rate \(=\mathrm{k}\left[\mathrm{HgCl}_{2}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{-1}\)

(A): Hydrolysis of methyl acetate in aqueous solution is a pseudo first order reaction. (R): In this reaction concentration of \(\mathrm{H}_{2} \mathrm{O}\) remains nearly constant during the course

The following data pertains to the reaction between A and \(B\)\begin{tabular}{llll} \hline S. & {\([\mathrm{A}]\)} & {\([\mathrm{B}]\)} & Rate \\ No. & \(\mathrm{mol} \mathrm{L}^{-1}\) & \(\mathrm{~mol} \mathrm{~L}^{-1}\) & \(\mathrm{Mol} \mathrm{L}^{-1} \mathrm{t}^{-1}\) \\ \hline 1 & \(1 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(2 \times 10^{-4}\) \\ 2 & \(2 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(4 \times 10^{-4}\) \\ 3 & \(2 \times 10^{-2}\) & \(4 \times 10^{-2}\) & \(8 \times 10^{-4}\) \\ \hline \end{tabular} Which of the following inferences are drawn from the above data? (1) Rate constant of the reaction is \(10^{-4}\) (2) Rate law of the reaction is \(\mathrm{k}[\mathrm{A}][\mathrm{B}]\) (3) Rate of reaction increases four times by doubling the concentration of each reactant. Select the correct answer the codes given below: a. 1 and 3 b. 2 and 3 c. 1 and 2 d. 1,2 and 3

The bromination of acetone that occurs in acid solution is represented by this equation. \(\mathrm{CH}_{3} \mathrm{COCH}_{3}(\mathrm{aq})+\mathrm{Br}_{2}\) (aq) \(\rightarrow\) \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{Br}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\) These kinetic data were obtained from given reaction concentrations. Initial concentrations, (M) \(\begin{array}{lll}{\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]} & {\left[\mathrm{Br}_{2}\right]} & {\left[\mathrm{H}^{+}\right]} \\ 0.30 & 0.05 & 0.05 \\ 0.30 & 0.10 & 0.05 \\\ 0.30 & 0.10 & 0.10 \\ 0.40 & 0.05 & 0.20\end{array}\) Initial rate, disappearance of \(\mathrm{Br}_{2}, \mathrm{Ms}^{-1}\) \(5.7 \times 10^{-5}\) \(5.7 \times 10^{-5}\) \(1.2 \times 10^{-4}\) \(3.1 \times 10^{-4}\) Based on these data, the rate equation is: a. Rate \(=\mathrm{k}\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{Br}_{2}\right]\left[\mathrm{H}^{+}\right]^{2}\) b. Rate \(=\mathrm{k}\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{Br}_{2}\right]\left[\mathrm{H}^{+}\right]\) c. Rate \(=\mathrm{k}\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{H}^{+}\right]\) d. Rate \(=\mathrm{k}\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{Br}_{2}\right]\)
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