Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Using a graphing calculator or standard graphing software, such as that on the Web site for this book, make an appropriate Arrhenius plot of the data shown here for the decomposition of iodoethane into ethene and hydrogen iodide, \(\mathrm{C}_{2} \mathrm{H} \mathrm{H}_{5} \mathrm{I}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{HI}(\mathrm{g})\), and determine the activation energy for the reaction. (b) What is the value of the rate constant at \(400^{\circ} \mathrm{C}\) ? $$ \begin{array}{lcccc} T(\mathbf{K}) & 660 & 680 & 720 & 760 \\ k\left(\mathbf{s}^{-1}\right) & 7.2 \times 10^{-4} & 2.2 \times 10^{-3} & 1.7 \times 10^{-2} & 0.11 \end{array} $$

Short Answer

Expert verified
Calculate the activation energy by constructing an Arrhenius plot, finding the slope, then calculating with \(E_{a} = -slope \times R\). Use this \(E_{a}\) and the Arrhenius equation to find \(k\) at 400C.

Step by step solution

01

- Understanding Arrhenius Equation

The Arrhenius equation provides a relationship between the rate constant k and the temperature T: \( k = Ae^{-\frac{E_{a}}{RT}} \) where \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. For an Arrhenius plot, we take the natural logarithm of both sides: \( \ln(k) = \ln(A) - \frac{E_{a}}{R}\left(\frac{1}{T}\right) \). This yields a linear equation of the form \(y = mx + b\) where \(y = \ln(k)\), \(m = -\frac{E_{a}}{R}\), \(x = \frac{1}{T}\), and \(b = \ln(A)\).
02

- Plotting the Data

Plot an Arrhenius plot with the given data points. The x-axis should represent \(\frac{1}{T}\) (in Kelvin) and the y-axis should represent \(\ln(k)\). Use the temperatures and rate constants given in the table to find the points to plot.
03

- Calculating the Slope

Using the graph or a linear regression tool, determine the slope of the line. This slope is equal to the negative activation energy divided by the gas constant \(R\) (\(slope = -\frac{E_{a}}{R}\)).
04

- Calculating Activation Energy

Once the slope is found, calculate the activation energy \(E_{a}\) by rearranging the formula for the slope: \(E_{a} = -slope \times R\). Use the value \(R = 8.314 \text{ J/(mol K)}\).
05

- Determine the Rate Constant at 400C

To find the rate constant \(k\) at \(400^\circ C\), convert the temperature to Kelvin \((400 + 273.15 = 673.15K)\). Then use the Arrhenius equation with the previously calculated \(E_{a}\) and \(A\) (determined from the \(y\)-intercept of the plot): \( k = Ae^{-\frac{E_{a}}{RT}} \). Solving for \(k\) will give the rate constant at \(400^\circ C\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Activation Energy
Activation energy, symbolized as \(E_a\), is a crucial concept in chemical kinetics. It represents the minimum amount of energy that reacting particles must possess for a chemical reaction to occur. You can visualize activation energy as a barrier that reactants need to overcome to transform into products. A higher \(E_a\) means that fewer molecules will have sufficient energy to react at a given temperature, leading to a slower reaction rate.

The Arrhenius equation shows how activation energy affects the rate constant, \(k\), of a reaction. As \(E_a\) increases, more energy is required to initiate the reaction, decreasing the rate constant. Conversely, a lower \(E_a\) suggests that the reaction can proceed more readily, resulting in a higher rate constant. Understanding \(E_a\) allows chemists to predict how changes in conditions might affect reaction rates.
Rate Constant
The rate constant, \(k\), is a numerical value that relates the reaction rate to the concentrations of reactants for a given reaction at a specific temperature. The magnitude of \(k\) determines the speed of a chemical reaction under certain conditions. A larger rate constant indicates a faster reaction, meaning that product formation occurs more quickly for a given concentration of reactants.

Rate constants are typically determined experimentally and vary with temperature. This variation is described by the Arrhenius equation, where \(k = Ae^{-\frac{E_{a}}{RT}}\). In this expression, \(A\) is the pre-exponential factor reflecting the frequency of collisions with proper orientation, \(E_{a}\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. Understanding the rate constant gives scientists a quantitative grasp on the pace at which reactions happen.
Arrhenius Equation
The Arrhenius equation is a fundamental formula in chemical kinetics that relates the rate constant \(k\) of a chemical reaction to the temperature \(T\) and the activation energy \(E_a\). Formulated as \(k = Ae^{-\frac{E_{a}}{RT}}\), this equation implies that for a given chemical reaction, the rate constant increases with temperature and decreases with an increase in activation energy.

To analyze reaction rates using this equation, scientists often manipulate it into a linear form: \(\ln(k) = \ln(A) - \frac{E_{a}}{R}\left(\frac{1}{T}\right)\), which aligns with the equation of a straight line. When plotted on an Arrhenius plot, with \(\ln(k)\) on the y-axis and \(\frac{1}{T}\) on the x-axis, the slope of the line (\(m\)) represents \(-\frac{E_{a}}{R}\) and the intercept (\(b\)) corresponds to \(\ln(A)\). By analyzing this plot, chemists can determine the activation energy and pre-exponential factor for the reaction.
Linear Regression
Linear regression is a statistical method that can be used to determine the relationship between variables and predict the value of one variable based on the value of another. In the context of chemical kinetics, linear regression is employed to analyze the Arrhenius plot, yielding the activation energy and pre-exponential factor from experimental data.

By plotting \(\ln(k)\) against \(\frac{1}{T}\), a linear regression can be performed to fit a straight line through the points. The slope of this line is related to the negative activation energy and the y-intercept is associated with the natural logarithm of the pre-exponential factor. Utilizing linear regression in this way allows for precise calculation of these important kinetic parameters, thereby enhancing the understanding of the reaction's temperature dependence.
Temperature Dependence of Reaction Rates
The temperature dependence of reaction rates is a phenomenon where the rate at which a chemical reaction proceeds changes with temperature. Generally, as the temperature increases, the reaction rate increases. This is because higher temperatures provide reactants with more kinetic energy, allowing more particles to surpass the activation energy barrier needed to react.

The Arrhenius equation precisely models this temperature dependence, demonstrating the exponential increase in the rate constant with temperature, factoring in the constants for activation energy and the gas constant. Temperature is a critical parameter in the calculation of reaction rates, as even small temperature changes can have a significant impact due to the exponential relationship. Grasping the influence of temperature on reaction rates is key in optimizing industrial processes, controlling reaction mechanisms, and predicting the behavior of reactions under various conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Derive an expression for the half-life of the reactant A that decays by a third-order reaction with rate constant \(k\).

Determine which of the following statements about catalysts are true. If the statement is false, explain why. (a) A heterogeneous catalyst works by binding one or more of the molecules undergoing reaction to the surface of the catalyst. (b) Enzymes are naturally occurring proteins that serve as catalysts in biological systems. (c) The equilibrium constant for a reaction is greater in the presence of a catalyst. (d) A catalyst changes the pathway of a reaction in such a way that the reaction becomes more exothermic.

The first-order rate constant for the photodissociation of \(\mathrm{A}\) is \(6.85 \times 10^{-2} \mathrm{~min}^{-1}\). Calculate the time needed for the concentration of A to decrease to (a) \(\frac{1}{8}[\mathrm{~A}]_{0} ;\) (b) \(10 . \%\) of its initial concentration; (c) one-third of its initial concentration.

All radioactive decay processes follow first-order kinetics. The half-life of the radioactive isotope tritium \(\left({ }^{3} \mathrm{H}\right.\), or \(\left.\mathrm{T}\right)\) is \(12.3\) years. How much of a \(25.0\)-mg sample of tritium would remain after \(10.9\) years?

In the reaction \(\mathrm{CH}_{3} \mathrm{Br}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{aq})+\) \(\mathrm{Br}^{-}(\mathrm{aq})\), when the \(\mathrm{OH}^{-}\)concentration alone was doubled, the rate doubled; when the \(\mathrm{CH}_{3} \mathrm{Br}\) concentration alone was increased by a factor of \(1.2\), the rate increased by a factor of \(1.2\). Write the rate law for the reaction.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free