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The data in the table give the temperature dependence of the rate constant for the reaction \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow\) \(2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) .\) Plot these data in the appropriate way to derive the activation energy for the reaction. $$\begin{aligned}&\\\&\begin{array}{ll}\hline T(\mathrm{K}) & k\left(\mathrm{s}^{-1}\right) \\ \hline 338 & 4.87 \times 10^{-3} \\\328 & 1.50 \times 10^{-3} \\\318 & 4.98 \times 10^{-4} \\\308 & 1.35 \times 10^{-4} \\\298 & 3.46 \times 10^{-5} \\\273 & 7.87 \times 10^{-7} \\\\\hline\end{array}\end{aligned}$$

Short Answer

Expert verified
Plot \( \ln(k) \) vs. \( \frac{1}{T} \), the slope is \(-\frac{E_a}{R}\), and use it to find \( E_a \).

Step by step solution

01

Understanding the Arrhenius Equation

The Arrhenius equation relates the rate constant \( k \) of a reaction to the temperature \( T \) and the activation energy \( E_a \). It is given by \[ k = A e^{-\frac{E_a}{RT}} \] where \( A \) is the pre-exponential factor, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. For a linear form, we take the natural log: \[ \ln(k) = \ln(A) - \frac{E_a}{R} \frac{1}{T} \]. This implies that a plot of \( \ln(k) \) versus \( \frac{1}{T} \) will yield a straight line with a slope of \( -\frac{E_a}{R} \).
02

Convert Temperature Data

Convert the given temperatures from the table into \( \frac{1}{T} \) (the inverse of the temperature in Kelvin). This will be the x-axis values for our plot.
03

Convert Rate Constants to Natural Logarithm

Take the natural logarithm of the rate constants \( k \) given in the table. These will be the y-axis values for our plot.
04

Plot the Data

Plot \( \ln(k) \) on the y-axis against \( \frac{1}{T} \) on the x-axis. Each pair from the table represents a point on this graph. You should expect the data to line up in a linear fashion if the Arrhenius equation correctly models this reaction.
05

Determine the Slope

Using the plot, determine the slope of the line. The slope is equal to \(-\frac{E_a}{R}\). The gas constant \( R \) is \( 8.314 \, \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \).
06

Calculate the Activation Energy

Once you have the slope from the plot, calculate the activation energy \( E_a \) using the relationship: \[ E_a = -\text{slope} \times R \]. Substitute the value of R to find \( E_a \).

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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 denoted as \( k \), is a fundamental parameter in reaction kinetics that quantifies the rate at which a reaction proceeds. It is an inherent property of the chemical reaction and varies with temperature. This variation is crucial because it exemplifies how reactions speed up or slow down under different thermal conditions.

The value of the rate constant is directly linked to the conditions under which the reaction occurs, especially temperature. As the temperature increases, the molecules have more energy and collide more frequently and with greater energy, which generally leads to an increase in the rate constant. This affects the overall speed of the reaction, emphasizing the importance of \( k \) in chemical kinetics.

In practice, knowing the rate constant helps predict the time needed for a reaction to reach a certain completion level, which is essential for both laboratory experiments and industrial processes. Often, experimental data like the one from our exercise is used to calculate \( k \) at various temperatures and understand the reaction's behavior.
Arrhenius Equation
The Arrhenius equation is a fundamental equation in chemical kinetics that describes how the rate constant \( k \) changes with temperature. This equation highlights the exponential dependence of the rate constant on the inverse of the temperature. It is formulated as:

\[ k = A e^{-\frac{E_a}{RT}} \]

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.

One of the most informative ways of utilizing the Arrhenius equation is its linear form, allowing for an easy graphical determination of the activation energy \( E_a \). By rearranging the equation, we obtain:
\[ \ln(k) = \ln(A) - \frac{E_a}{R} \times \frac{1}{T} \]
This linear equation implies that a plot of \( \ln(k) \) versus \( \frac{1}{T} \) will yield a straight line with the slope \( -\frac{E_a}{R} \), from which \( E_a \) can be calculated.
Reaction Kinetics
Reaction kinetics is the branch of chemistry concerned with understanding the rates of chemical reactions and the factors affecting them. It involves studying how different variables such as concentration, temperature, and catalysts influence the speed of a reaction.

The core of reaction kinetics is to establish a mathematical relationship between the concentration of reactants and the time it takes for a reaction to progress. This relationship allows for predicting how fast a reaction will proceed under various conditions. In the context of our exercise, the focus is on temperature's effect on the reaction rate.

By analyzing kinetic data, scientists can deduce information about reaction mechanisms. For instance, if a reaction follows a particular rate law, it can suggest which molecules collide and rearrange to form products. Reaction kinetics also informs about the efficiency and possible optimization of chemical reactions, crucial for industrial applications.
Temperature Dependence
Temperature dependence in chemical kinetics refers to the crucial role temperature plays in influencing the rate of a chemical reaction. With an increase in temperature, molecules move faster, resulting in more frequent and energetic collisions. This elevated collision rate often leads to a higher reaction rate.

The Arrhenius equation embodies this temperature dependence by showing that even small changes in temperature can lead to significant changes in the rate constant \( k \). This sensitivity is due to the exponential nature of the equation, where the rate constant increases with increasing temperature, assuming that the pre-exponential factor and activation energy remain constant.
Understanding temperature dependence is essential for controlling reactions, especially in industrial settings where temperature control is used to optimize production rates and yields. For example, in our exercise, plotting \( \ln(k) \) against \( \frac{1}{T} \) allows us to observe how the rate constant varies with temperature and subsequently determine the activation energy needed to initiate the reaction.

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Most popular questions from this chapter

The rate equation for the hydrolysis of sucrose to fructose and glucose $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})$$ is " \(-\Delta[\text { sucrose }] / \Delta t=k\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] .\) After \(2.57 \mathrm{h}\) at \(27^{\circ} \mathrm{C}\) the sucrose concentration decreased from \(0.0146 \mathrm{M}\) to \(0.0132 \mathrm{M} .\) Find the rate constant, \(k.\)

A reaction has the experimental rate equation "Rate = \(k[\mathrm{A}]^{2} .\) How will the rate change if the concentration of \(\mathrm{A}\) is tripled? If the concentration of A is halved?

The decomposition of \(\mathrm{CO}_{2}\) is first order with respect to the concentration of \(\mathrm{CO}_{2}.\) $$2 \mathrm{CO}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ Data on this reaction are provided in the table below. $$\begin{array}{lc}\hline\left[\mathrm{CO}_{2}\right](\mathrm{mol} / \mathrm{L}) & \text { Time }(\mathrm{s}) \\\\\hline 0.38 & 0 \\\0.27 & 12 \\\\\hline\end{array}$$ (a) Write the rate equation for this reaction. (b) Use the data to determine the value of \(k\) (c) What is the half-life of \(\mathrm{CO}_{2}\) under these conditions?

The following statements relate to the reaction with the following rate law: Rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right].\) $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})$$ Determine which of the following statements are true. If a statement is false, indicate why it is incorrect. (a) The reaction must occur in a single step. (b) This is a second-order reaction overall. (c) Raising the temperature will cause the value of \(k\) to decrease. (d) Raising the temperature lowers the activation energy for this reaction. (e) If the concentrations of both reactants are doubled, the rate will double. (f) Adding a catalyst in the reaction will cause the initial rate to increase.

A three-step mechanism for the reaction of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and \(\mathrm{H}_{2} \mathrm{O}\) is proposed: Step 1 Slow $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}+\mathrm{Br}^{-}$$ Step 2 Fast $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}_{2}^{+}$$ Step 3 Fast $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{HBr}$$ (a) Write an equation for the overall reaction. (b) Which step is rate-determining? (c) What rate law is expected for this reaction?

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