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Chapter 13: Problem 61

Ethyl chloride, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\), used to produce tetraethyllead gasoline additive, decomposes, when heated, to give ethylene and hydrogen chloride. The reaction is first order. In an experiment, the initial concentration of ethyl chloride was \(0.00100 \mathrm{M}\). After heating at \(500^{\circ} \mathrm{C}\) for \(155 \mathrm{~s}\), this was reduced to \(0.00067 \mathrm{M}\). What was the concentration of ethyl chloride after a total of \(256 \mathrm{~s}\) ?

Short Answer

Expert verified
The concentration of ethyl chloride after 256 seconds is approximately 0.000502 M.

Step by step solution

01

Understand the First Order Reaction Formula

For a first-order reaction, the relationship between the initial concentration \([A]_0\) and the concentration at time \(t\) \([A]\) is given by the formula: \[ [A] = [A]_0 e^{-kt} \] where \(k\) is the rate constant, and \(t\) is the time.
02

Determine the Rate Constant (k)

From the given information, the initial concentration \([A]_0\) is 0.00100 M, and the concentration at 155 s \([A]\) is 0.00067 M. Use the formula:\[ 0.00067 = 0.00100 e^{-k \times 155} \]Solving for \(k\):1. \(\frac{0.00067}{0.00100} = e^{-155k} \)2. \( ln(0.67) = -155k \)3. \( k = -\frac{ln(0.67)}{155} \)Evaluate \(k\): \[ k \approx 0.00269 \, s^{-1} \].
03

Calculate the Concentration at 256 Seconds

Now that we have \(k\), use it to find the concentration of ethyl chloride at \(256\) seconds.Using the formula:\[ [A] = 0.00100 \times e^{-0.00269 \times 256} \]Calculate \([A]\):1. Evaluate exponent: \(-0.00269 \times 256 = -0.68864\)2. \( [A] = 0.00100 \times e^{-0.68864} \)3. \( [A] \approx 0.00100 \times 0.5020 \)4. \( [A] \approx 0.000502 \, M \)

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Key Concepts

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

Rate Constant
The rate constant \(k\) is a crucial component in the study of chemical kinetics. It is a specific value that indicates how fast a reaction proceeds. For first-order reactions, the rate constant has the unit of time inverse, such as seconds inverse (\(s^{-1}\)).
When you know the rate constant, you can predict how the concentration of reactants changes over time. The rate constant is involved in the exponential equation for first-order reactions: \[ [A] = [A]_0 e^{-kt} \] Here, \( [A]_0 \) is the initial concentration, and \( [A] \) is the concentration at time \( t \).
To find the rate constant, you rearrange this formula and plug in known values. For example: \( k = -\frac{ln[A]/[A]_0}{t} \).
It is calculated from experimental data where you measure how concentration changes over a specific period. In our example, we determined \( k \) as approximately \(0.00269 \, s^{-1}\) based on concentration changes over 155 seconds.
Ethyl Chloride Decomposition
Ethyl chloride (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{Cl}\)) is a chemical compound used in the production of leaded gasoline. It is known to decompose into ethylene (\(\mathrm{C}_2\mathrm{H}_4\)) and hydrogen chloride (\(\mathrm{HCl}\)) when heated. This decomposition is a classic example of a first-order reaction.
In a first-order reaction, the rate at which the reaction occurs is directly proportional to the concentration of only one reactant. This means that as the concentration of ethyl chloride decreases, the reaction rate changes accordingly. The characteristic feature of first-order reactions is that they follow an exponential decay model, captured in the equation: \[ [A] = [A]_0 e^{-kt} \].
Understanding ethyl chloride's decomposition requires observing how its concentration lessens over time, providing insights into how the reaction proceeds and allowing us to calculate the rate constant.
Concentration Calculation
Calculating concentration over time is essential for tracking how a reaction unfolds. For first-order reactions like ethyl chloride decomposition, you use an exponential decay formula. This formula allows you to determine what the concentration will be at any point in time once you have the rate constant.
Using the equation \[ [A] = [A]_0 e^{-kt} \], you can substitute values to find the concentration at a given time. For instance, after 256 seconds in the given scenario, you start with the initial concentration (\(0.00100\) M), plug in the rate constant (\(0.00269 \, s^{-1}\)), and compute based on the time elapsed.
Breaking it down involves evaluating the exponent first, \(-0.00269 \, \times \, 256\), which results in \-0.68864\. Then calculate \(e^{-0.68864}\) to get about 0.5020. Finally, you multiply by the initial concentration to get approximately \(0.000502 \, M\), which is the concentration after 256 seconds.

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Most popular questions from this chapter

At high temperature, the reaction $$\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)$$ is thought to occur in a single step. What should be the rate law in that case?

Relate the rate of decomposition of \(\mathrm{NO}_{2}\) to the rate of formation of \(\mathrm{O}_{2}\) for the following reaction: $$2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$

A reaction of the form aA \(\longrightarrow\) Products is second order with a half-life of \(307 \mathrm{~s}\). What is the rate constant of the reaction if the initial concentration of \(\mathrm{A}\) is \(1.87 \times 10^{-02} \mathrm{~mol} / \mathrm{L} ?\)

Compare physical adsorption and chemisorption (chemical adsorption).

Chlorine dioxide oxidizes iodide ion in aqueous solution to iodine; chlorine dioxide is reduced to chlorite ion. $$2 \mathrm{ClO}_{2}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{ClO}_{2}^{-}(a q)+\mathrm{I}_{2}(a q)$$ The order of the reaction with respect to \(\mathrm{ClO}_{2}\) was determined by starting with a large excess of \(\mathrm{I}^{-}\), so that its concentration was essentially constant. Then $$\text { Rate }=k\left[\mathrm{ClO}_{2}\right]^{m}\left[\mathrm{I}^{-}\right]^{n}=k^{\prime}\left[\mathrm{ClO}_{2}\right]{m}$$ where \(k^{\prime}=k\left[\mathrm{I}^{-}\right]^{n}\). Determine the order with respect to \(\mathrm{ClO}_{2}\) and the rate constant \(k^{\prime}\) by plotting the following data assuming firstand then second-order kinetics. [Data from H. Fukutomi and G. Gordon, J. Am. Chem. Soc., 89,1362 (1967).] \(\begin{array}{cl}\text { Time (s) } & {\left[\text { ClO }_{2}\right] \text { (moll } \text { L) }} \\ 0.00 & 4.77 \times 10^{-4} \\ 1.00 & 4.31 \times 10^{-4} \\ 2.00 & 3.91 \times 10^{-4} \\ 3.00 & 3.53 \times 10^{-4} \\ 5.00 & 2.89 \times 10^{-4} \\ 10.00 & 1.76 \times 10^{-4} \\ 30.00 & 2.4 \times 10^{-5} \\ 50.00 & 3.2 \times 10^{-6}\end{array}\)
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