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Chapter 9: Problem 12

The reaction of methyl radicals to form ethane was investigated in a laser flash photolysis experiment at \(300 \mathrm{K}\) \\[ \mathrm{CH}_{3}^{*}+\mathrm{CH}_{3}^{*} \rightarrow \mathrm{C}_{2} \mathrm{H}_{8} \\] The rate constant for this reaction at \(300 \mathrm{K}\) is \(3.7 \times 10^{10} \mathrm{dm}^{3}\) \(\mathrm{mol}^{-1} \mathrm{s}^{-1}\). The concentration of methyl radicals, \(\left[\mathrm{CH}_{3}^{\circ}\right],\) at time \(t=0\) was \(1.70 \times 10^{-8} \mathrm{moldm}^{-3}\). Calculate a value for \(\left[\mathrm{CH}_{3}\right]\) at \(\left.t=1.00 \times 10^{-3} \mathrm{s} \text { . (Section } 9.4\right)\)

Short Answer

Expert verified
The concentration \( \left[\mathrm{CH}_3^*\right] \) at \( t = 1.00 \times 10^{-3} \mathrm{s} \) is \( 1.04 \times 10^{-8} \mathrm{mol\,dm}^{-3} \).

Step by step solution

01

Set Up the Rate Equation

The given reaction is a second-order reaction with respect to methyl radicals \( \mathrm{CH}_3^* \). The rate law for this reaction is expressed as: \[ \text{Rate} = k[\mathrm{CH}_3^*]^2 \] where \( k \) is the rate constant.
02

Use the Integrated Rate Law for a Second-Order Reaction

For a second-order reaction, the integrated rate law is given by: \[ \frac{1}{[\mathrm{CH}_3^*]} = \frac{1}{[\mathrm{CH}_3^*]_0} + kt \] where \( [\mathrm{CH}_3^*] \) is the concentration at time \( t \), \( [\mathrm{CH}_3^*]_0 \) is the initial concentration, and \( k \) is the rate constant.
03

Plug in the Given Values

We have the following values: \( [\mathrm{CH}_3^*]_0 = 1.70 \times 10^{-8} \mathrm{mol\,dm}^{-3} \), \( k = 3.7 \times 10^{10} \mathrm{dm}^3 \mathrm{mol}^{-1} \mathrm{s}^{-1} \), and \( t = 1.00 \times 10^{-3} \mathrm{s} \). Substituting these into the integrated rate law equation: \[ \frac{1}{[\mathrm{CH}_3^*]} = \frac{1}{1.70 \times 10^{-8}} + 3.7 \times 10^{10} \times 1.00 \times 10^{-3} \]
04

Solve for \( [\mathrm{CH}_3^*] \)

Calculate the right side of the equation: \[ \frac{1}{1.70 \times 10^{-8}} = 5.88 \times 10^{7} \] \[ 3.7 \times 10^{10} \times 1.00 \times 10^{-3} = 3.7 \times 10^{7} \] Add these two results: \[ 5.88 \times 10^{7} + 3.7 \times 10^{7} = 9.58 \times 10^{7} \] Now take the reciprocal to find \( [\mathrm{CH}_3^*] \): \[ [\mathrm{CH}_3^*] = \frac{1}{9.58 \times 10^{7}} = 1.04 \times 10^{-8} \mathrm{mol\,dm}^{-3} \]

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Key Concepts

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

Second-Order Reaction
A second-order reaction is a type of chemical reaction that depends on the concentration of one second-order reactant or on the concentrations of two first-order reactants. In the case of the reaction investigated in the exercise, the rate is second-order with respect to the methyl radicals, \(\mathrm{CH}_3^*\). This means the rate of the reaction depends quadratically on the concentration of \(\mathrm{CH}_3^*\).
This relationship is expressed through the rate law:
  • \(\text{Rate} = k[\mathrm{CH}_3^*]^2\)
Here, \(k\) is the rate constant, and the rate of the reaction increases with the square of the concentration of the methyl radicals. Understanding this relationship is crucial, as it helps to predict how changes in concentration can affect the speed of the reaction. This is essential for controlling reaction conditions in practical scenarios.
Rate Constant
The rate constant, denoted as \(k\), is a crucial factor in defining how fast a reaction proceeds under specific conditions. It's a proportionality constant that links the reaction rate to the concentrations of the reactants raised to their respective powers in the rate law. For the second-order reaction involving methyl radicals, the rate constant is given as \(3.7 \times 10^{10} \, \mathrm{dm}^3 \mathrm{mol}^{-1} \mathrm{s}^{-1}\).
  • A larger rate constant indicates a faster reaction at a given concentration.
  • The units for the rate constant differ depending on the order of the reaction: for a second-order reaction, it's typically \(\mathrm{dm}^3 \mathrm{mol}^{-1} \mathrm{s}^{-1}\).
Knowing the rate constant allows chemists to predict how quickly a reaction will progress, which is important for applications like designing reactors or understanding mechanisms at a molecular level.
Integrated Rate Law
The integrated rate law for a second-order reaction provides a mathematical relationship that describes the concentration of reactants over time. This is especially useful when our goal is to determine concentrations at various times during a reaction. For a second-order reaction such as the methyl radical dimerization, the integrated rate law is given by:
  • \[\frac{1}{[\mathrm{CH}_3^*]} = \frac{1}{[\mathrm{CH}_3^*]_0} + kt\]
This equation shows that the reciprocal of the reactant concentration increases linearly over time.
By using the integrated rate law, you can solve for the concentration of the reactant at any time \(t\) if you know the initial concentration \([\mathrm{CH}_3^*]_0\), the rate constant \(k\), and the time \(t\). This is useful in reactions that may not be directly measurable at all stages due to reactant consumption or other limiting factors.

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Most popular questions from this chapter

The addition of bromine to propene is an elementary reaction with a rate constant, \(k\) \\[ \mathrm{CH}_{2}=\mathrm{CHCH}_{3}+\mathrm{Br}_{2} \rightarrow \mathrm{CH}_{2} \mathrm{BrCHBrCH}_{3} \\] Kinetic studies were carried out at \(298 \mathrm{K}\) using excess \(\mathrm{Br}_{2}\). For \(\left[\mathrm{Br}_{2}\right]_{0}=0.20 \mathrm{mol} \mathrm{dm}^{-3},\) the pseudo-first order rate constant, \(k^{\prime}\) for the reaction was found to be \(900 \mathrm{s}^{-1}\). What is the value of \(k\) at \(298 \mathrm{K} ? \text { (Section } 9.4)\)

The rate constant for the decomposition of a compound in solution is \(2.0 \times 10^{-4} \mathrm{s}^{-1}\). If the initial concentration is \(0.02 \mathrm{moldm}^{-3},\) what will the concentration of the compound be after \(10 \text { min? (Section } 9.4)\)

Rate constants at a series of temperatures were obtained for the decomposition of azomethane \\[ \mathrm{CH}_{3} \mathrm{N}_{2} \mathrm{CH}_{3} \rightarrow 2 \mathrm{CH}_{3}^{*}+\mathrm{N}_{2} \\] $$\begin{array}{llllll} \hline T / \mathrm{K} & 523 & 541 & 560 & 576 & 593 \\ k / 10^{-6} \mathrm{s}^{-1} & 1.8 & 15 & 60 & 160 & 950 \end{array}$$ Use the data in the table to find the activation energy, \(E_{a}\), for the reaction. (Section \(9.7)\)

At low substrate concentrations, the initial rate of an enzyme catalysed reaction was found to be directly proportional to the initial substrate concentration, [S]o, and directly proportional to the initial enzyme concentration, \([\mathrm{E}]_{0} .\) (Section 9.9 ) (a) Outline the series of experiments that led to these results. (b) Write a rate equation for the reaction under these conditions and explain the observed kinetics in terms of the mechanism for the reaction. What is the rate-determining step under these conditions? At much higher substrate concentrations, the initial rate was found to be constant and independent of the initial substrate concentrations. The initial enzyme concentration was the same in each experiment. (c) Write a rate equation for the reaction under these conditions and explain the observed kinetics in terms of the mechanism for the reaction. What is the rate-determining step under these conditions?

The reaction between \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{OH}^{-}\) was investigated at \(298 \mathrm{K}\) using the initial rate method \\[ \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{HPO}_{4}^{2-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{aq}) \\] The following results were obtained. (Section 9.5 ) $$\begin{array}{llll} & \begin{array}{l} \text { Initial rate } / 10^{-3} \\ \text {moldm }^{-3} \text {min }^{-1} \end{array} & \begin{array}{l} {[\mathrm{OH}]_{0} / 10^{-3}} \\ \text {moldm }^{-3} \end{array} & \begin{array}{l} {\left[\mathrm{H}_{2} \mathrm{PO}_{4}\right]_{0} / 10^{-3}} \\ \mathrm{mol} \mathrm{dm}^{-3} \end{array} \\ \hline \text { Experiment 1 } & 2.0 & 0.40 & 3.0 \\ \text { Experiment 2 } & 3.7 & 0.55 & 3.0 \\ \text { Experiment 3 } & 7.1 & 0.75 & 3.0 \end{array}$$ (a) Plot a log-log graph to determine the order of reaction with respect to \(\mathrm{OH}^{-}(\mathrm{aq})\) (b) What further experiments would you need to do to find the order with respect to \(\mathrm{H}_{2} \mathrm{PO}_{4}-?\)
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