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A compound decomposes by a first-order reaction. The concentration of compound decreases from \(0.1180 M\) to \(0.0950 M\) in \(5.2\) min. What fraction of the compound remains after \(7.1 \mathrm{~min} ?\)

Short Answer

Expert verified
72% of the compound remains after 7.1 min.

Step by step solution

01

Understand First-Order Kinetics

A first-order reaction means the rate of reaction is directly proportional to the concentration of one reactant. The integrated rate law for a first-order reaction is \( [A] = [A]_0 e^{-kt} \), where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant.
02

Calculate the Rate Constant

We need to find the rate constant \(k\) using the concentrations provided. The formula is rearranged to \( k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]}\right) \). Substitute \([A]_0 = 0.1180 M\), \([A] = 0.0950 M\), and \(t = 5.2 \) min: \[ k = \frac{1}{5.2} \ln\left(\frac{0.1180}{0.0950}\right) \]. Calculate \(k\) to find its value.
03

Substitute and Simplify

Calculate \( k \) using the natural logarithm: \[ k = \frac{1}{5.2} \times \ln\left(\frac{0.1180}{0.0950}\right) = \frac{1}{5.2} \times 0.2144 = 0.0412 \text{ min}^{-1} \] (approximately).
04

Determine Remaining Concentration at 7.1 Min

Now that we have \(k\), determine the concentration after 7.1 min using \([A] = [A]_0 e^{-kt}\): \[ [A] = 0.1180 \times e^{-0.0412 \times 7.1} \]. Solve this equation.
05

Calculate Fraction Remaining

Compute the remaining concentration using the equation: \[ [A] = 0.1180 \times e^{-0.29252} \approx 0.0850 M \].The fraction of compound remaining is \(\frac{0.0850}{0.1180}\).
06

Simplify to Fraction

Calculate the fraction: \( \text{Fraction remaining} = \frac{0.0850}{0.1180} \approx 0.720 \). So about 72% of the compound remains.

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Key Concepts

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

Rate Constant
In the realm of chemical kinetics, the rate constant is a fundamental component that helps us understand how quickly a chemical reaction proceeds. For a first-order reaction, which depends solely on the concentration of one reactant, the rate constant is denoted by \( k \). It plays a pivotal role in the reaction's rate law, serving as a proportionality factor.
To determine \( k \) for a first-order reaction, we can use the equation:
  • \( k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]}\right) \)
This equation helps us compute the rate constant using observed concentrations of the substance over time (\( t \)), where \([A]_0\) is the initial concentration, and \([A]\) is the concentration at time \( t \).
In many practical scenarios, like the decomposition reaction in the original exercise, calculating \( k \) allows chemists to predict how much of a substance will remain after a certain period, or how long it will take for the substance to decrease to a certain concentration. The rate constant, thus, provides insights into the efficiency and speed of reactions.
Integrated Rate Law
The integrated rate law for a first-order reaction offers a powerful tool to determine the concentration of a reactant at any given time. It provides a clear picture of how the concentration of the reactant diminishes over time in a reaction.
The integrated rate law for a first-order process is expressed as:
  • \( [A] = [A]_0 e^{-kt} \)
In this formula, \([A]_0\) is the initial concentration of the reactant, and \([A]\) represents the concentration at any time \( t \). The symbol \( e \) is the base of the natural logarithm, and \( k \) is the rate constant.
By using this equation, one can determine the concentration of a substance at any specified time point \( t \), as was performed in the exercise solution to find out how much remained after 7.1 minutes. This equation allows us to visualize the exponential decay of concentration typical for first-order kinetics, making it invaluable for both academic and industrial chemistry applications.
Decomposition Reaction
Decomposition reactions are fascinating chemical processes where a single compound breaks down into two or more simpler products. These reactions are a cornerstone in understanding chemical kinetics, especially in contexts such as thermal decomposition or photodecomposition.
In the original exercise, we analyzed a decomposition reaction following first-order kinetics. This means the reaction rate is directly related to the concentration of the compound that is decomposing. When examining such reactions, it's crucial to understand how concentration changes over time, which can often be accurately modeled with first-order integrated rate laws.
These reactions can be highly valuable in various industries and scientific research fields. Knowing how much of a compound will remain after a certain time can guide decisions in environmental management, pharmaceuticals, and materials science. The study of decomposition reactions not only helps us comprehend chemical behavior but also aids in harnessing these chemical processes for technological and industrial advancements.

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Most popular questions from this chapter

The reaction $$ 2 \mathrm{NOCl}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ has rate-constant values for the reaction of NOCl of \(9.3 \times\) \(10^{-6} / \mathrm{s}\) at \(350 \mathrm{~K}\) and \(6.9 \times 10^{-4} / \mathrm{s}\) at \(400 \mathrm{~K}\). Calculate activation energy for the reaction. What is the rate constant at \(425 \mathrm{~K} ?\)

There is often one step in a reaction mechanism that is rate determining. What characteristic of such a step makes it rate determining? Explain.

For the reaction of nitric oxide, NO, with chlorine, \(\mathrm{Cl}_{2}\), $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g) $$ the observed rate law is $$ \text { Rate }=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right] $$ What is the reaction order with respect to nitric oxide? with respect to \(\mathrm{Cl}_{2}\) ? What is the overall order?

Ozone decomposes to oxygen gas. $$ 2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) $$ A proposed mechanism for this decomposition is $$ \mathrm{O}_{3} \underset{k_{-1}}{\stackrel{k_{1}}{ }} \mathrm{O}_{2}+\mathrm{O} $$ (fast, equilibrium) $$ \mathrm{O}_{3}+\mathrm{O} \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{O}_{2} $$ What is the rate law derived from this mechanism?

In the presence of excess thiocyanate ion, \(\mathrm{SCN}^{-}\), the following reaction is first order in chromium(III) ion, \(\mathrm{Cr}^{3+}\); the rate constant is \(2.0 \times 10^{-6} / \mathrm{s}\). $$ \mathrm{Cr}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \longrightarrow \mathrm{Cr}(\mathrm{SCN})^{2+}(a q) $$ What is the half-life in hours? How many hours would be required for the initial concentration of \(\mathrm{Cr}^{3+}\) to decrease to each of the following values: \(25.0 \%\) left, \(12.5 \%\) left, \(6.25 \%\) left, \(3.125 \%\) left?

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