/*! 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 53 For the reaction of iodine atoms... [FREE SOLUTION] | 91Ó°ÊÓ

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For the reaction of iodine atoms with hydrogen molecules in the gas phase, these rate constants were obtained experimentally. \(2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})\) 2 \begin{tabular}{lc} \hline\(T(\mathrm{~K})\) & \(10^{-5} \mathrm{k}\left(\mathrm{L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}\right)\) \\ \hline 417.9 & 1.12 \\ 480.7 & 2.60 \\ 520.1 & 3.96 \\ 633.2 & 9.38 \\ 666.8 & 11.50 \\ 710.3 & 16.10 \\ 737.9 & 18.54 \\ \hline \end{tabular} (a) Calculate the activation energy and frequency factor for this reaction. (b) Estimate the rate constant of the reaction at \(400.0 \mathrm{~K}\).

Short Answer

Expert verified
(a) Calculate \( E_a \) and \( A \) using linear regression on \( \ln k \) vs. \( 1/T \). (b) Estimate \( k \) at 400 K with the Arrhenius equation.

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius equation relates the rate constant \( k \) of a reaction to the temperature \( T \) and is given by \( k = A e^{-E_a/(RT)} \), where \( A \) is the frequency factor, \( E_a \) is the activation energy, and \( R \) is the universal gas constant (8.314 J/mol·K). By taking the natural logarithm of both sides, we obtain \( \ln k = \ln A - \frac{E_a}{RT}\). This provides a linear form that can be used to extract \( E_a \) and \( A \).
02

Transform Data into Linear Form

From the Arrhenius equation \( \ln k = \ln A - \frac{E_a}{RT} \), plot \( \ln k \) against \( 1/T \). First, calculate \( \ln k \) and \( 1/T \) for each temperature given.\( \ln k \) values (from \( 10^{-5} \text{k} \)): - 417.9 K: \( \ln(1.12 \times 10^{-5}) = -11.3984 \)- 480.7 K: \( \ln(2.60 \times 10^{-5}) = -10.5565 \)- etc.\( 1/T \) values:- 417.9 K: \( 1/417.9 = 0.002393 \)- 480.7 K: \( 1/480.7 = 0.002080 \)- etc.
03

Plot and Determine the Slope

Plot each calculated \( \ln k \) against \( 1/T \) to form a straight line. The slope of this line is \( -E_a/R \). Use linear regression to find the best fit line, and extract the slope from this line to calculate \( E_a \).
04

Calculate Activation Energy \( E_a \)

Using the slope from the linear regression, calculate the activation energy:\[ \text{slope} = -\frac{E_a}{R} \]Rearrange to find \( E_a \):\[ E_a = -\text{slope} \times R \]
05

Calculate Frequency Factor \( A \)

Once \( E_a \) is known, the intercept of the line (\( \ln A \)) provides the logarithm of the frequency factor \( A \). Calculate \( A \) by rearranging:\[ \ln k = \ln A - \frac{E_a}{RT} \]
06

Estimate Rate Constant at 400 K

Now that both \( A \) and \( E_a \) have been determined, use the Arrhenius equation to estimate the rate constant at 400 K:\[ k = A e^{-E_a/(R \times 400)} \] Substitute \( A \), \( E_a \), and \( R \) to compute \( k \) at 400 K.

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Key Concepts

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

activation energy
Activation energy is a crucial concept in understanding how chemical reactions occur and how fast they proceed. In essence, it refers to the minimum energy that reactant molecules must possess to transform into products. Imagine it as the energy barrier that the molecules need to overcome for the reaction to take place.
In the context of the Arrhenius equation, the activation energy, denoted by \(E_a\), plays a significant role. When the natural logarithm of the rate constant \(k\) is plotted against the reciprocal of the temperature \(1/T\), the resulting line's slope can yield the value of \(E_a\).
This is expressed through the equation \[\ln k = \ln A - \frac{E_a}{RT}\] where \(R\) is the universal gas constant. By determining the slope of the line from such a plot, which is \(-E_a/R\), we can calculate the activation energy as \(E_a = -\text{slope} \times R\).
Understanding activation energy helps in manipulating reaction conditions such as temperature, as lower activation energy typically means the reaction can proceed more rapidly under milder conditions.
frequency factor
The frequency factor, represented as \(A\) in the Arrhenius equation, is a measure of the number of times that reactant molecules collide with the correct orientation per unit time. It is sometimes referred to as the pre-exponential factor.
This factor gives insight into the reaction's likelihood in terms of favorable collisions leading to a successful reaction. In practical terms, it indicates how often molecules hit each other in ways that might cause a reaction, assuming they have ample energy to overcome the activation energy barrier.
When we graph \(\ln k\) versus \(1/T\) and derive a straight line, the y-intercept of this line is \(\ln A\). Once we calculate \(\ln A\), finding the frequency factor involves taking the exponential of the intercept, resulting in \(A = e^{\ln A}\).
Grasping this concept allows chemists to predict how reaction rates might change under different conditions, particularly in scenarios where temperatures remain constant, but molecular arrangements can influence reaction outcomes.
rate constant estimation
Estimating the rate constant \(k\) at different temperatures is an integral part of understanding chemical kinetics and is a direct application of the Arrhenius equation. Using the derived values of the activation energy \(E_a\) and the frequency factor \(A\), it's possible to estimate how fast a reaction proceeds at temperatures not initially tested.
Given the temperature, the rate constant is calculated with the formula \[k = A e^{-E_a/(RT)}\] where \(R\) is the universal gas constant and \(T\) is the temperature in Kelvin.
For instance, if you need to estimate the rate constant at 400 K, you would simply substitute the known values of \(A\), \(E_a\), and \(R\) into the equation to find \(k\) at that temperature.
  • This approach provides a powerful forecasting tool to determine how fast a reaction occurs under various thermal conditions.
  • Such estimations are not only academically interesting but also have real-world implications in industrial processes where controlled reaction rates can impact safety and efficiency.
With these calculations, one can plan better regarding what temperatures might need adjustments when aiming for optimal reaction speeds.

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Most popular questions from this chapter

In a time-resolved picosecond spectroscopy experiment, Sheps, Crowther, Carrier, and Crim (Journal of Physical Chemistry \(A,\) Vol. 110,\(2006 ;\) pp. \(3087-3092\) ) generated chlorine atoms in the presence of pentane. The pentane was dissolved in dichloromethane, \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\). The chlorine atoms are free radicals and are very reactive. After a nanosecond the chlorine atoms have reacted with pentane molecules, removing a hydrogen atom to form \(\mathrm{HCl}\) and leaving behind a pentane radical with a single unpaired electron. The equation is \(\mathrm{Cl} \cdot(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{12}(\mathrm{dcm}) \longrightarrow \mathrm{HCl}(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{11} \cdot(\mathrm{dcm})\) where \((\mathrm{dcm})\) indicates that a substance is dissolved in dichloromethane. Measurements of the concentration of chlorine atoms were made as a function of time at three different concentrations of pentane in the dichloromethane. These results are shown in the table. of Chlorine Atoms (M) \begin{tabular}{clcc} & \multicolumn{3}{c} { Concentration of Chlorine Atoms (m) } \\ Time (ps) & \multicolumn{4}{c} { for Different \(\mathrm{C}_{5} \mathrm{H}_{12}\) Concentrations } \\ \cline { 3 - 5 } & \(0.15-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.30-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.60-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) \\ \hline 100.0 & \(4.42 \times 10^{-5}\) & \(3.11 \times 10^{-5}\) & \(1.77 \times 10^{-5}\) \\ 140.0 & \(3.823 \times 10^{-5}\) & \(2.39 \times 10^{-5}\) & \(1.15 \times 10^{-5}\) \\\ 180.0 & \(3.38 \times 10^{-5}\) & \(1.94 \times 10^{-5}\) & \(8.48 \times 10^{-6}\) \\\ 220.0 & \(3.03 \times 10^{-5}\) & \(1.49 \times 10^{-5}\) & \(6.04 \times 10^{-6}\) \\\ 260.0 & \(2.68 \times 10^{-5}\) & \(1.19 \times 10^{-5}\) & \(4.12 \times 10^{-6}\) \\\ 300.0 & \(2.42 \times 10^{-5}\) & \(9.45 \times 10^{-6}\) & \(3.14 \times 10^{-6}\) \\\ 340.0 & \(2.08 \times 10^{-5}\) & \(7.75 \times 10^{-6}\) & \(2.38 \times 10^{-6}\) \\\ 380.0 & \(1.91 \times 10^{-5}\) & \(6.35 \times 10^{-6}\) & \(1.75 \times 10^{-6}\) \\\ 420.0 & \(1.71 \times 10^{-5}\) & \(4.58 \times 10^{-6}\) & \(1.61 \times 10^{-6}\) \\\ 460.0 & \(1.53 \times 10^{-5}\) & \(3.77 \times 10^{-6}\) & \(9.98 \times 10^{-7}\) \\\ \hline \end{tabular} (a) Determine the order of the reaction with respect to chlorine. (b) Determine whether the reaction rate depends on the concentration of pentane in dichloromethane. If so, determine the order of the reaction with respect to pentane. (c) Explain why the concentration of pentane in dichloromethane does not affect the data analysis that you performed in part (a). (d) Write the rate law for the reaction and calculate the rate of reaction for a concentration of chlorine atoms equal to \(1.0 \mu \mathrm{M}\) and a pentane concentration of \(0.23 \mathrm{M}\). (e) Sheps, Crowther, Carrier, and Crim found that the rate of formation of \(\mathrm{HCl}\) matched the rate of disappearance of Cl. From this they concluded that there were no intermediates and side reactions were not important. Explain the ic

When substrates are present at relatively high concentration and are catalyzed by enzymes, the effect on reaction rate of changing substrate concentration can be described by zeroth-order kinetics. Calculate by what factor the rate of an enzyme-catalyzed reaction changes when the substrate concentration is changed from \(1.5 \times 10^{-2} \mathrm{M}\) to \(4.5 \times 10^{-2} \mathrm{M}\)

Indicate whether each of these statements is true or false. Change the wording of each false statement to make it true. (a) It is possible to change the rate constant for a reaction by changing the temperature. (b) The reaction rate remains constant as a first-order reaction proceeds at a constant temperature. (c) The rate constant for a reaction is independent of reactant concentrations. (d) As a second-order reaction proceeds at a constant temperature, the rate constant changes.

Which of these is appropriate for determining the rate law for a chemical reaction? (a) Theoretical calculations based on balanced equations (b) Measuring the rate of the reaction as a function of the concentrations of the reacting species (c) Measuring the rate of the reaction as a function of temperature

Which of these mechanisms (more than one may be chosen) is compatible with the rate law? Rate \(=k\left[\mathrm{Cl}_{2}\right]^{3 / 2}[\mathrm{CO}]\) $$ \text { (a) } \frac{1}{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{Cl} $$ \(\mathrm{Cl}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{Cl}_{3}\) fast \(\mathrm{Cl}_{3}+\mathrm{CO} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl} \quad\) slow $$ \mathrm{Cl} \rightleftharpoons \frac{1}{2} \mathrm{Cl}_{2} $$ (b) \(\mathrm{Cl}_{2}+\mathrm{CO} \longrightarrow \mathrm{CCl}_{2}+\mathrm{O}\) slow \(\mathrm{O}+\mathrm{Cl}_{2} \longrightarrow \mathrm{Cl}_{2} \mathrm{O}\) fast $$ \mathrm{Cl}_{2} \mathrm{O}+\mathrm{CCl}_{2} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl}_{2} $$ fast s (c) \(\frac{1}{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{Cl}\) $$ \mathrm{Cl}+\mathrm{CO} \rightleftharpoons \mathrm{COCl} $$ fast $$ \begin{array}{l} \mathrm{COCl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl} \\\ \mathrm{Cl} \rightleftharpoons \frac{1}{2} \mathrm{Cl}_{2} \end{array} $$ slow (d) \(\mathrm{Cl}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{COCl}+\mathrm{Cl}\) fast \(\mathrm{COCl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl}\) slow \(\mathrm{Cl}+\mathrm{Cl} \longrightarrow \mathrm{Cl}_{2}\) fast

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