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For the reaction \(2 \mathrm{C}_{2} \mathrm{F}_{4} \longrightarrow \mathrm{C}_{4} \mathrm{F}_{8},\) a graph of \(1 /\left[\mathrm{C}_{2} \mathrm{F}_{4}\right]\) versus time gives a straight line with a slope of \(+0.04 \mathrm{L} / \mathrm{mol} \cdot \mathrm{s} .\) What is the rate law for this reaction?

Short Answer

Expert verified
Rate law: \( \text{rate} = 0.04 [\text{C}_2\text{F}_4]^2 \).

Step by step solution

01

Identify Reaction Order

A straight-line graph of \(1/[\text{C}_2\text{F}_4]\) versus time indicates a second-order reaction. For a second-order reaction, the integrated rate law has the form \(1/[A]_t = kt + 1/[A]_0\), which matches the given graph.
02

Determine the Rate Constant

From the slope of the graph, we know that it represents the rate constant \(k\). Therefore, the rate constant \(k\) is \(0.04 \text{ L/mol} \cdot \text{s}\).
03

Write the Rate Law

For a second-order reaction \(2 \text{C}_2\text{F}_4 \rightarrow \text{C}_4\text{F}_8\), the rate law is written as: \( \text{rate} = k [\text{C}_2\text{F}_4]^2 \). Using the determined \(k = 0.04 \text{ L/mol} \cdot \text{s}\), the rate law is \( \text{rate} = 0.04 [\text{C}_2\text{F}_4]^2 \).

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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, often symbolized as \( k \), is a crucial component in the study of chemical kinetics. It is a proportionality factor in the rate law that helps predict the speed of a reaction at any given concentration of reactants. In our exercise, the reaction \(2 \text{C}_2\text{F}_4 \rightarrow \text{C}_4\text{F}_8\) illustrates this concept well. Here, the slope of the line from the graph of \(1/[\text{C}_2\text{F}_4]\) versus time gives us the rate constant \( k \).To calculate \( k \) for a second-order reaction, you simply identify the slope of the line. In this case, we found that \( k \) equals \(0.04 \text{ L/mol} \cdot \text{s}\). This value tells us how rapidly the reaction proceeds under the given conditions. The units of \( \text{L/mol} \cdot \text{s} \) are essential in confirming that the rate constant is appropriate for a second-order reaction.
Integrated Rate Law
The integrated rate law offers a mathematical relationship between the concentration of reactants and time. For a second-order reaction like \(2 \text{C}_2\text{F}_4 \rightarrow \text{C}_4\text{F}_8\), the integrated rate law is expressed as:\[\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}\]where:
  • \([A]_t\) is the concentration of the reactant at time \( t \).
  • \([A]_0\) is the initial concentration of the reactant.
  • \( k \) is the rate constant.
This form of the rate law is particularly useful because it allows us to determine the concentration of a reactant at any given time, assuming we know the initial concentration and the rate constant. In the context of our exercise, plotting \(1/[\text{C}_2\text{F}_4]\) versus time resulted in a straight line, confirming the reaction's second-order nature and enabling us to derive \( k \) from the graph's slope.
Reaction Order
Understanding the reaction order is vital in grasping how different factors influence the reaction rate. The order of a reaction refers to the power to which the concentration of a reactant is raised in the rate law expression. In the case of our reaction \(2 \text{C}_2\text{F}_4 \rightarrow \text{C}_4\text{F}_8\), the reaction order is determined to be second-order.For second-order reactions, the rate of reaction is directly proportional to the square of the concentration of one reactant, as shown in the rate law:\[\text{rate} = k [\text{C}_2\text{F}_4]^2\]This rate law tells us that if you double the concentration of \(\text{C}_2\text{F}_4\), the rate of reaction increases by a factor of four. Recognizing a second-order reaction from the graph of \(1/[\text{C}_2\text{F}_4]\) versus time is key, as the linear relationship directly indicates the second-order nature of the reaction.

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Most popular questions from this chapter

At temperatures below \(500 \mathrm{K},\) the reaction between carbon monoxide and nitrogen dioxide $$ \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) $$ has the following rate equation: Rate \(=k\left[\mathrm{NO}_{2}\right]^{2} .\) Which of the three mechanisms suggested here best agrees with the experimentally observed rate equation? Mechanism 1 \(\quad\) single, elementary step $$\mathrm{NO}_{2}+\mathrm{CO} \longrightarrow \mathrm{CO}_{2}+\mathrm{NO}$$ Mechanism \(2 \quad\) Two steps $$\begin{aligned}&\text { Slow } \quad \mathrm{NO}_{2}+\mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\mathrm{NO}\\\&\text { Fast } \quad \mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}\end{aligned}$$ Mechanism 3 \(\quad\) Two steps $$\begin{aligned}&\text { Slow } \quad \mathrm{NO}_{2} \longrightarrow \mathrm{NO}+\mathrm{O}\\\&\text { Fast } \quad \mathrm{CO}+\mathrm{O} \longrightarrow \mathrm{CO}_{2}\end{aligned}$$

The radioactive isotope \(^{64} \mathrm{Cu}\) is used in the form of \(\mathrm{cop}\) per(II) acetate to study Wilson's disease. The isotope has a half-life of \(12.70 \mathrm{h}\). What fraction of radioactive copper (II) acetate remains after \(64 \mathrm{h} ?\)

Data in the table were collected at \(540 \mathrm{K}\) for the following reaction: $$\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g})$$ (a) Derive the rate equation. (b) Determine the reaction order with respect to each reactant. (c) Calculate the rate constant, giving the correct units for \(k.\) $$\begin{array}{lll}\hline \text { Initial Concentration }(\mathrm{mol} / \mathrm{L}) & {\text { Initial Rate }} \\\\\hline[\mathrm{C} 0] & {\left[\mathrm{NO}_{2}\right]} & (\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}) \\\\\hline 5.1 \times 10^{-4} & 0.35 \times 10^{-4} & 3.4 \times 10^{-8} \\\5.1 \times 10^{-4} & 0.70 \times 10^{-4} & 6.8 \times 10^{-8} \\\5.1 \times 10^{-4} & 0.18 \times 10^{-4} & 1.7 \times 10^{-8} \\\1.0 \times 10^{-3} & 0.35 \times 10^{-4} & 6.8 \times 10^{-8} \\\1.5 \times 10^{-3} & 0.35 \times 10^{-4} & 10.2 \times 10^{-8} \\\\\hline\end{array}$$

A Two molecules of the unsaturated hydrocarbon 1,3-butadiene \(\left(\mathrm{C}_{4} \mathrm{H}_{6}\right)\) form the "dimer" \(\mathrm{C}_{8} \mathrm{H}_{12}\) at higher temperatures. $$2 \mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{g}) \longrightarrow \mathrm{C}_{8} \mathrm{H}_{12}(\mathrm{g})$$ Use the following data to determine the order of the reaction and the rate constant, \(k\). (Note that the total pressure is the pressure of the unreacted \(\mathrm{C}_{4} \mathrm{H}_{6}\) at any time and the pressure of the \(\mathrm{C}_{8} \mathrm{H}_{12} .\)) $$\begin{array}{cl}\hline \text { Time (min) } & \text { Total Pressure (mm Hg) } \\\\\hline 0 & 436 \\\3.5 & 428 \\\11.5 & 413 \\\18.3 & 401 \\\25.0 & 391 \\\32.0 & 382 \\\41.2 & 371 \\\\\hline\end{array}$$

We know that the decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}.\) $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ with a half-life of 245 min at \(600 \mathrm{K}\). If you begin with a partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) of \(25 \mathrm{mm}\) Hg in a 1.0 -L. flask, what is the partial pressure of each reactant and product after 245 min? What is the partial pressure of each reactant after \(12 \mathrm{h} ?\)

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