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Chapter 12: Problem 98

A certain reaction has the form \(\mathrm{aA} \longrightarrow\) Products At a particular temperature, concentration versus time data were collected. A plot of \(1 /[\mathrm{A}]\) versus time (in seconds) gave a straight line with a slope of \(6.90 \times 10^{-2}\). What is the differential rate law for this reaction? What is the integrated rate law for this reaction? What is the value of the rate constant for this reaction? If \([\mathrm{A}]_{0}\) for this reaction is \(0.100 M\), what is the first half-life (in seconds)? If the original concentration (at \(t=0\) ) is \(0.100 M\), what is the second half-life (in seconds)?

Short Answer

Expert verified
The reaction is a second-order reaction with a differential rate law of \(\dfrac{d[\mathrm{A}]}{dt} = -k[\mathrm{A}]^2\). The integrated rate law is \(\dfrac{1}{[\mathrm{A}]} - \dfrac{1}{[\mathrm{A}]_0} = kt\). The rate constant, k, is \(6.90 \times 10^{-2}\,\text{s}^{-1}\). The first and second half-lives are 14.49 seconds and 28.99 seconds, respectively.

Step by step solution

01

Determine the order of the reaction

Since the plot of \(1/[\mathrm{A}]\) versus time gives a straight line, we can conclude that the reaction is a second-order reaction. Second-order reactions have the form: \(\dfrac{d[\mathrm{A}]}{dt} = -k[\mathrm{A}]^2\), where k is the rate constant.
02

Write the differential rate law

Now that we know the order of the reaction, we can write the differential rate law: \(\dfrac{d[\mathrm{A}]}{dt} = -k[\mathrm{A}]^2\)
03

Obtain the integrated rate law

We can find the integrated rate law by integrating the differential rate law: \(\int \dfrac{d[\mathrm{A}]}{[\mathrm{A}]^2} = -k \int dt\) This yields: \(\dfrac{1}{[\mathrm{A}]} - \dfrac{1}{[\mathrm{A}]_0} = kt\)
04

Calculate the value of the rate constant

We are given the slope of the plot as \(6.90 \times 10^{-2}\). Since the straight line represents the integrated rate law, the slope is equal to the rate constant k: \(k = 6.90 \times 10^{-2}\,\text{s}^{-1}\)
05

Find the first half-life

The first half-life can be found using the initial concentration and the rate constant. For a second-order reaction, the half-life is: \(t_{1/2} = \dfrac{1}{k[\mathrm{A}]_0}\) We're given the initial concentration as \([\mathrm{A}]_0 = 0.100\,\text{M}\), so: \(t_{1/2} = \dfrac{1}{(6.90 \times 10^{-2}\,\text{s}^{-1})(0.100\,\text{M})} = 14.49\,\text{s}\)
06

Find the second half-life

To find the second half-life, we need to find the new initial concentration for the next half-life period. After the first half-life, the remaining concentration would be half of the initial concentration: \([\mathrm{A}]_{1/2} = \dfrac{[\mathrm{A}]_0}{2} = \dfrac{0.100\,\text{M}}{2} = 0.050\,\text{M}\) Now, use the new initial concentration to find the second half-life: \(t_{1/2} = \dfrac{1}{k[\mathrm{A}]_{1/2}} = \dfrac{1}{(6.90 \times 10^{-2}\,\text{s}^{-1})(0.050\,\text{M})} = 28.99\,\text{s}\) So, the second half-life is 28.99 seconds.

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Most popular questions from this chapter

Sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) decomposes to sulfur dioxide \(\left(\mathrm{SO}_{2}\right)\) and chlorine \(\left(\mathrm{Cl}_{2}\right)\) by reaction in the gas phase. The following pressure data were obtained when a sample containing \(5.00 \times 10^{-2}\) mol sulfuryl chloride was heated to \(600 . \mathrm{K}\) in a \(5.00 \times 10^{-1}-\mathrm{L}\) container. Defining the rate as \(-\frac{\Delta\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]}{\Delta t}\), a. determine the value of the rate constant for the decomposition of sulfuryl chloride at \(600 . \mathrm{K}\). b. what is the half-life of the reaction? c. what fraction of the sulfuryl chloride remains after \(20.0 \mathrm{~h}\) ?

Consider the hypothetical reaction \(\mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(g) \longrightarrow\) \(2 \mathrm{AB}(g)\), where the rate law is: $$ -\frac{\Delta\left[\mathrm{A}_{2}\right]}{\Delta t}=k\left[\mathrm{~A}_{2}\right]\left[\mathrm{B}_{2}\right] $$ The value of the rate constant at \(302^{\circ} \mathrm{C}\) is \(2.45 \times 10^{-4} \mathrm{~L} / \mathrm{mol}\). \(\mathrm{s}\), and at \(508^{\circ} \mathrm{C}\) the rate constant is \(0.891 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\). What is the activation energy for this reaction? What is the value of the rate constant for this reaction at \(375^{\circ} \mathrm{C}\) ?

The mechanism for the gas-phase reaction of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon dioxide is thought to be $$ \begin{aligned} &\mathrm{NO}_{2}+\mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\mathrm{NO} \\\ &\mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2} \end{aligned} $$ Write the rate law expected for this mechanism. What is the overall balanced equation for the reaction?

The activation energy for the reaction $$ \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) $$ is \(125 \mathrm{~kJ} / \mathrm{mol}\), and \(\Delta E\) for the reaction is \(-216 \mathrm{~kJ} / \mathrm{mol}\). What is the activation energy for the reverse reaction \(\left[\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{CO}(g)\right] ?\)

Consider the hypothetical reaction $$ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} $$ where the rate law is $$ \text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{~A}][\mathrm{B}]^{2} $$ An experiment is carried out where \([\mathrm{A}]_{0}=1.0 \times 10^{-2} \mathrm{M}\), \([\mathrm{B}]_{0}=3.0 M\), and \([\mathrm{C}]_{0}=2.0 M .\) The reaction is started, and after \(8.0\) seconds, the concentration of \(\mathrm{A}\) is \(3.8 \times 10^{-3} \mathrm{M}\). a. Calculate the value of \(k\) for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after \(13.0\) seconds. d. Calculate the concentration of \(\mathrm{C}\) after \(13.0\) seconds.
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