/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Data for the reaction $$\begin... [FREE SOLUTION] | 91Ó°ÊÓ

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Data for the reaction $$\begin{aligned}\left[\mathrm{Mn}(\mathrm{CO})_{5}\left(\mathrm{CH}_{3}\mathrm{CN}\right)\right]^{+}+\mathrm{NC}_{5}\mathrm{H}_{5} \longrightarrow & \\\&\left[\mathrm{Mn}(\mathrm{CO})_{5}\left(\mathrm{NC}_{5}\mathrm{H}_{5}\right)\right]^{+}+\mathrm{CH}_{3} \mathrm{CN}\end{aligned}$$ are given in the table. Calculate \(E_{\mathrm{a}}\) from a plot of \(\ln k\) versus \(1 / T\) $$\begin{array}{ll}\hline T(\mathrm{K}) & k\left(\min ^{-1}\right) \\\\\hline 298 & 0.0409 \\\308 & 0.0818 \\\318 & 0.157 \\\\\hline\end{array}$$

Short Answer

Expert verified
The activation energy is approximately 9.6 kJ/mol.

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius equation relates the rate constant \( k \) with the temperature \( T \) as follows: \[ k = A e^{-\frac{E_a}{RT}} \] 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.
02

Transform the Arrhenius Equation

To use linear regression, we transform the equation to a linear form: \[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \] This is in the form of \( y = mx + c \) where the slope \( m = -\frac{E_a}{R} \) and the y-intercept \( c = \ln A \).
03

Organize Data for Plotting

Calculate \( \ln k \) for each given \( k \) and \( \frac{1}{T} \) for each temperature \( T \). - For \( T = 298 \), \( k = 0.0409 \): \( \ln k = \ln(0.0409) \), \( \frac{1}{T} = \frac{1}{298} \).- For \( T = 308 \), \( k = 0.0818 \): \( \ln k = \ln(0.0818) \), \( \frac{1}{T} = \frac{1}{308} \).- For \( T = 318 \), \( k = 0.157 \): \( \ln k = \ln(0.157) \), \( \frac{1}{T} = \frac{1}{318} \).
04

Calculate Logarithms and Rates

Compute the values:- \( \ln(0.0409) \approx -3.195 \), \( \frac{1}{298} \approx 0.003356 \).- \( \ln(0.0818) \approx -2.506 \), \( \frac{1}{308} \approx 0.003247 \).- \( \ln(0.157) \approx -1.854 \), \( \frac{1}{318} \approx 0.003145 \).
05

Plot \( \ln k \) vs \( \frac{1}{T} \) and Determine Slope

Plot the points calculated:(0.003356, -3.195), (0.003247, -2.506), (0.003145, -1.854).Use linear regression to find the best-fit line, where the slope \( m = -\frac{E_a}{R} \).
06

Calculate Activation Energy \( E_a \)

The slope \( m \) from the plot is used to find \( E_a \):\[ E_a = -m \cdot R \]Assuming the slope \( m \approx -1156 \) (from plot), use \( R = 8.314 \; \mathrm{J/mol \, K} \):\[ E_a = -(-1156) \times 8.314 \approx 9600 \; \mathrm{J/mol} \] or \( 9.6 \; \mathrm{kJ/mol} \).
07

Conclusion: Activation Energy Calculation

The activation energy for the reaction is found to be approximately \( 9.6 \; \mathrm{kJ/mol} \). This value represents the minimum energy required for the reaction to occur.

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Key Concepts

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

Arrhenius Equation
The Arrhenius Equation is a fundamental concept in chemistry that explores how temperature affects the rate of a chemical reaction. It is expressed as \( k = A e^{-\frac{E_a}{RT}} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor indicating the frequency of collisions, \( E_a \) is the activation energy required to initiate the reaction, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. This equation helps chemists understand how shifts in temperature can increase or decrease reaction rates, providing critical insights into reaction kinetics. Understanding the Arrhenius Equation can also help predict the speed of reactions in different environments, even allowing for extrapolations when conditions are unknown.
To make it more practical, the Arrhenius Equation is often transformed into a linear form \( \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \). This linear form makes it easier to employ statistical tools like linear regression to analyze kinetic data. This transformation can be likened to rearranging a mathematical problem into a simpler, more approachable form, creating a straight line representation when graphed.
Kinetics
Kinetics is the study of the rates of chemical reactions and the factors affecting these rates. By exploring kinetics, scientists can determine speeds at which reactions occur and what influences them. Several variables can impact reaction rates, including temperature, concentration, surface area, and catalysts.
  • Increased temperature typically raises reaction rates as it provides more energy to the reactants, facilitating more frequent successful collisions.
  • Higher concentrations of reactants often increase reaction rates due to more particles being available to collide.
  • Catalysts can speed up reactions by lowering the energy required to reach the transition state, without being consumed in the process.
The Arrhenius Equation plays a vital role in connecting kinetics to temperature, providing a quantitative way to see how reaction rates vary with temperature changes. By understanding kinetics, chemists can optimize reaction conditions in industries such as pharmaceuticals and materials manufacturing, where reaction speeds are crucial.
Chemical Reaction Rate
The chemical reaction rate explains the speed at which a chemical reaction proceeds. It is generally measured by the change in concentration of reactants or products over time. Understanding reaction rates is essential, as it allows us to predict how fast a chemical process occurs under certain conditions. Several factors can influence these rates:
  • Temperature: Generally, an increase in temperature speeds up reaction rates as the molecules have more kinetic energy.
  • Concentration: Higher concentrations mean more particles present to collide and react.
  • Surface Area: With more surface area available, more reactant particles are directly exposed to each other, increasing collision opportunities.
  • Pressure: Especially relevant for gases, where increased pressure effectively increases concentration.
By using the Arrhenius Equation, we can quantitatively determine the effect of temperature on the reaction rate, thereby helping to understand how to control and optimize various reactions. Chemical engineering heavily relies on mastering these reaction dynamics to scale processes that are effective, safe, and economical.
Linear Regression in Chemistry
Linear regression in chemistry is an instrumental data analysis tool that helps understand relationships between variables by forming a linear line approximation for data points. In the context of the Arrhenius Equation, it is used to determine the activation energy \( E_a \) from empirical kinetic data. The linear form \( \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \) is especially useful here. When \( \ln k \) values (y-axis) are plotted against \( \frac{1}{T} \) values (x-axis), linear regression helps to calculate the best-fit line.
The slope of this line, which equates to \( -\frac{E_a}{R} \), provides the necessary information to find \( E_a \). Thus, the linear regression technique converts complex, nonlinear problems into simple linear problems, making computations more straightforward and accurate.
  • Helps identify trends and correlations in experimental data.
  • Offers a method to predict values and assess the reliability of models quantitatively.
Utilizing linear regression in chemistry empowers experimenters to unveil meaningful insights from data, facilitating better comprehension, analysis, and prediction, especially in kinetic studies.

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Most popular questions from this chapter

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?

Identify which of the following statements are incorrect. If the statement is incorrect, rewrite it to be correct. (a) Reactions are faster at a higher temperature because activation energies are lower. (b) Rates increase with increasing concentration of reactants because there are more collisions between reactant molecules. (c) At higher temperatures a larger fraction of molecules have enough energy to get over the activation energy barrier. (d) Catalyzed and uncatalyzed reactions have identical mechanisms.

Gaseous azomethane, \(\mathrm{CH}_{3} \mathrm{N}=\mathrm{NCH}_{3},\) decomposes in a first-order reaction when heated: $$\mathrm{CH}_{3} \mathrm{N}=\mathrm{NCH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})$$ The rate constant for this reaction at \(425^{\circ} \mathrm{C}\) is \(40.8 \mathrm{min}^{-1}\) If the initial quantity of azomethane in the flask is \(2.00 \mathrm{g}\) how much remains after 0.0500 min? What quantity of \(\mathrm{N}_{2}\) is formed in this time?

Calculate the activation energy, \(E_{\mathrm{a}},\) for the reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ from the observed rate constants: \(k\) at \(25^{\circ} \mathrm{C}=3.46 \times\) \(10^{-5} s^{-1}\) and \(k\) at \(55^{\circ} \mathrm{C}=1.5 \times 10^{-3} \mathrm{s}^{-1}.\)

Hundreds of different reactions can occur in the stratosphere, among them reactions that destroy the earth's ozone layer. The table below lists several (second-order) reactions of Cl atoms with ozone and organic compounds; each is given with its rate constant. $$\begin{array}{ll}\hline & \text { Rate Constant } \\\\\text { Reaction } & \left(298 \mathrm{K}, \mathrm{cm}^{3} / \mathrm{molecule} \cdot \mathrm{s}\right) \\\\\hline \text { (a) } \mathrm{Cl}+0_{3} \longrightarrow \mathrm{Cl} 0+0_{2} & 1.2 \times 10^{-11} \\\\\text {(b) } \mathrm{Cl}+\mathrm{CH}_{4} \longrightarrow\mathrm{HCl}+\mathrm{CH}_{3} & 1.0 \times 10^{-13} \\\\\text {(c) } \mathrm{Cl}+\mathrm{C}_{3} \mathrm{H}_{8} \longrightarrow\mathrm{HCl}+\mathrm{C}_{3} \mathrm{H}_{7} & 1.4 \times 10^{-10} \\\\\text {(d) } \mathrm{Cl}+\mathrm{CH}_{2} \mathrm{FCl} \longrightarrow\mathrm{HCl}+\mathrm{CHFCl} & 3.0 \times 10^{-18} \\\\\hline\end{array}$$ For equal concentrations of Cl and the other reactant, which is the slowest reaction? Which is the fastest reaction?

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