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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. $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{HCl}(g) $$ The reaction is first order. In an experiment, the initial concentration of ethyl chloride was \(0.00100 M\). After heating at \(500^{\circ} \mathrm{C}\) for \(155 \mathrm{~s}\), this was reduced to \(0.00067 M\). What was the concentration of ethyl chloride after a total of \(256 \mathrm{~s}\) ?

Short Answer

Expert verified
The concentration after 256 s is approximately 0.0005151 M.

Step by step solution

01

Understand the Reaction's Order and Formula

The reaction is first order, meaning the rate depends linearly on the concentration of the reactant. The integrated rate law for a first-order reaction is given by \[ ext{ln} rac{[A]_0}{[A]} = kt \]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 \(k\)

We know \([A]_0 = 0.00100 \, M\) and \([A] = 0.00067 \, M\) after \(t = 155 \, s\).Using the first-order formula:\[ \text{ln}\left(\frac{0.00100}{0.00067}\right) = k \times 155 \]\[ 0.4014 = 155k \]Solving for \(k\):\[ k = \frac{0.4014}{155} \approx 0.00259 \, s^{-1} \]
03

Determine Concentration at \(t = 256 \, s\)

Now use the rate constant to find the concentration after \(t = 256 \, s\):\[ \text{ln}\left(\frac{0.00100}{[A]}\right) = 0.00259 \times 256 \]\[ \text{ln}\left(\frac{0.00100}{[A]}\right) = 0.6630 \]Solve for \([A]\):\[ \frac{0.00100}{[A]} = e^{0.6630} \]\[ \frac{0.00100}{[A]} \approx 1.9401 \]\[ [A] \approx \frac{0.00100}{1.9401} \approx 0.0005151 \, M \]
04

Conclusion

The concentration of ethyl chloride after \(256 \, s\) is approximately \(0.0005151 \, M\).

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Key Concepts

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

Integrated Rate Law
The integrated rate law for first-order reactions is an essential tool in determining the concentration of a reactant at any given time. For a first-order reaction, the rate at which the reaction proceeds depends directly on the concentration of a single reactant. This is represented by the equation:
  • \( \text{ln} \frac{[A]_0}{[A]} = kt \)
where:
  • \([A]_0\) is the initial concentration of the reactant.
  • \([A]\) is the concentration of the reactant at time \(t\).
  • \(k\) is the rate constant, which is specific to each reaction.
  • \(t\) is the elapsed time.
This equation allows chemists to calculate either the remaining concentration of a reactant or the time taken for a certain amount of reactant to be consumed. It provides a logarithmic relationship between time and concentration, reflecting how the concentration decreases over time.
Understanding the integrated rate law is fundamental to grasping how reactions evolve, especially in scenarios where exact timing and concentration are crucial, such as in industrial processes.
Rate Constant
The rate constant, \(k\), is a crucial factor in quantifying the speed of a reaction. For first-order reactions, this constant helps us understand how quickly a reaction progresses at a specific temperature. It remains unchanged as long as the temperature is constant. The units of \(k\) in a first-order reaction are typically \(s^{-1}\), emphasizing that it is a time-based measure. Determining \(k\) involves experimental data:
  • Input the initial and remaining concentrations of the reactant.
  • Use time taken for these changes into the integrated rate law formula.
For example, consider in our exercise where \([A]_0 = 0.00100 \, M\) and \([A] = 0.00067 \, M\) after \(t = 155 \, s\). By substituting these into the integrated rate law, we find \(k\) as \(0.00259 \, s^{-1}\).
Acknowledging the rate constant's role allows chemists to compare and predict behavior across different reaction conditions, making it a pivotal component in chemical kinetics.
Decomposition Reaction
A decomposition reaction involves the breakdown of a single compound into two or more simpler substances. These processes are often endothermic, requiring heat to proceed, as observed with ethyl chloride decomposing into ethylene and hydrogen chloride:
  • \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g) + \mathrm{HCl}(g) \)
This type of reaction is significant in industrial applications and chemical syntheses. The study of decomposition reactions, especially those exhibiting a first-order behavior, helps us predict how quickly a substance will break down under certain conditions. By knowing the rate constant and employing the integrated rate law, chemists can design processes to control the timing and yield of desired products in decomposition. For instance, in our example, by heating ethyl chloride at \(500^\circ C\), we initiate its decomposition, which can be tracked over time to ensure efficient transition to the products.

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Most popular questions from this chapter

Compare physical adsorption and chemisorption (chemical adsorption).

Consider the reaction \(3 \mathrm{~A} \longrightarrow 2 \mathrm{~B}+\mathrm{C}\). a. One rate expression for the reaction is Rate of formation of \(\mathrm{C}=+\frac{\Delta[\mathrm{C}]}{\Delta t}\) Write two other rate expressions for this reaction in this form. b. Using your two rate expressions, if you calculated the average rate of the reaction over the same time interval, would the rates be equal? c. If your answer to part b was no, write two rate expressions that would give an equal rate when calculated over the same time interval.

The thermal decomposition of nitryl chloride, \(\mathrm{NO}_{2} \mathrm{Cl}\) $$ 2 \mathrm{NO}_{2} \mathrm{Cl}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is thought to occur by the mechanism shown in the following equations: $$ \begin{aligned} \mathrm{NO}_{2} \mathrm{Cl} & \stackrel{k_{1}}{\longrightarrow} \mathrm{NO}_{2}+\mathrm{Cl} & & \text { (slow step) } \\ \mathrm{NO}_{2} \mathrm{Cl}+\mathrm{Cl} & \stackrel{k_{2}}{\longrightarrow} \mathrm{NO}_{2}+\mathrm{Cl}_{2} & & \text { (fast step) } \end{aligned} $$ What rate law is predicted by this mechanism?

What is the molecularity of each of the following elementary reactions? a. \(\mathrm{O}+\mathrm{O}_{2}+\mathrm{N}_{2} \longrightarrow \mathrm{O}_{3}+\mathrm{N}_{2}^{*}\) b. \(\mathrm{NO}_{2} \mathrm{Cl}+\mathrm{Cl} \longrightarrow \mathrm{NO}_{2}+\mathrm{Cl}_{2}\) c. \(\mathrm{Cl}+\mathrm{H}_{2} \longrightarrow \mathrm{HCl}+\mathrm{H}\) d. \(\mathrm{CS}_{2} \longrightarrow \mathrm{CS}+\mathrm{S}\)

The reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\) is found to bero-order. If it takes \(3.3 \times 10^{2}\) seconds for an initial concentration of \(\mathrm{A}\) to go from \(0.50 M\) to \(0.25 M\), what is the rate constant for the reaction?

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