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Chapter 13: Problem 113

The decomposition of nitrogen dioxide, $$2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ has a rate constant of \(0.498 \mathrm{M} / \mathrm{s}\) at \(319^{\circ} \mathrm{C}\) and a rate constant of \(1.81 \mathrm{M} / \mathrm{s}\) at \(354^{\circ} \mathrm{C}\). What are the values of the activation energy and the frequency factor for this reaction? What is the rate constant at \(420^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
Activation energy: 112.5 kJ/mol, frequency factor: 3.92 x 10^9 M/s, rate constant at 420掳C: 12.6 M/s.

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius equation is essential for understanding reaction kinetics. It is given by \[ k = A e^{-\frac{E_a}{RT}} \]where \( k \) is the rate constant, \( A \) is the frequency factor, \( E_a \) is the activation energy, \( R \) is the gas constant (8.314 J/mol路K), and \( T \) is the temperature in Kelvin.
02

Convert Temperatures to Kelvin

To use the Arrhenius equation, temperatures must be in Kelvin. Convert each given temperature:- For \( 319\, ^{\circ} \mathrm{C} \): \[ 319 + 273.15 = 592.15\, \text{K} \]- For \( 354\, ^{\circ} \mathrm{C} \):\[ 354 + 273.15 = 627.15\, \text{K} \]- For \( 420\, ^{\circ} \mathrm{C} \):\[ 420 + 273.15 = 693.15\, \text{K} \]
03

Use Two-Point Arrhenius Equation to Find Activation Energy

For two different temperatures, the Arrhenius equation becomes:\[\ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)\]Substitute \( k_1 = 0.498 \mathrm{M/s} \), \( k_2 = 1.81 \mathrm{M/s} \), \( T_1 = 592.15\, \text{K} \), and \( T_2 = 627.15\, \text{K} \):\[\ln \left( \frac{1.81}{0.498} \right) = \frac{E_a}{8.314} \left( \frac{1}{592.15} - \frac{1}{627.15} \right)\]Calculate the left side: \( \ln (3.635) \approx 1.291 \).Rearrange to solve for \( E_a \):\[ E_a = \frac{1.291 \times 8.314}{ \left(\frac{1}{592.15} - \frac{1}{627.15}\right) } \]
04

Calculate Activation Energy

Continuing from the previous step, calculate:\[\left(\frac{1}{592.15} - \frac{1}{627.15}\right) \approx -0.0000953 \text{ K}^{-1}\]\[ E_a = \frac{1.291 \times 8.314}{-0.0000953} \approx 112508 \text{ J/mol} \]Convert to \( \text{kJ/mol} \):\[ E_a \approx 112.5 \text{ kJ/mol} \]
05

Determine the Frequency Factor

Rearrange the Arrhenius equation to solve for \( A \) using one of the rate constants and its corresponding temperature. We'll use the first set (\( k_1 = 0.498 \mathrm{M/s}, T_1 = 592.15\, \text{K} \)):\[ 0.498 = A e^{-\frac{112500}{8.314 \times 592.15}} \]Solving exponents: \[ A = \frac{0.498}{e^{-22.785}} \approx \frac{0.498}{1.27 \times 10^{-10}} \approx 3.92 \times 10^{9} \text{ M/s} \]
06

Calculate the Rate Constant at 420掳C

Using the Arrhenius equation with \( T = 693.15\, \text{K} \), \( E_a = 112500 \text{ J/mol} \), and \( A = 3.92 \times 10^{9} \text{ M/s} \):\[ k = A e^{-\frac{112500}{8.314 \times 693.15}} \]Calculate exponent:\[ k = 3.92 \times 10^{9} e^{-19.565} \approx 3.92 \times 10^{9} \times 3.21 \times 10^{-9} \approx 12.6 \mathrm{M/s} \]
07

Conclusion

The activation energy for the reaction is \( 112.5 \text{ kJ/mol} \), the frequency factor \( A \) is \( 3.92 \times 10^{9} \mathrm{M/s} \), and the rate constant at \( 420^{\circ} \mathrm{C} \) is \( 12.6 \mathrm{M/s} \).

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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 crucial tool for understanding how reaction rates vary with temperature. It provides a mathematical relationship between the rate constant ( k ) of a chemical reaction and the temperature at which the reaction occurs. The equation is expressed as\[ k = A e^{-\frac{E_a}{RT}} \], where:
  • k is the rate constant, which quantifies the speed of a reaction.
  • A is the frequency factor, representing the number of times reactants approach the activation barrier per unit time.
  • E_a is the activation energy, the minimum energy required to initiate the reaction.
  • R is the gas constant, with a value of 8.314 J/mol路K.
  • T is the absolute temperature measured in Kelvin.
This equation helps chemists predict how changes in temperature affect the rate of reaction, making it an invaluable concept in reaction kinetics.
Reaction kinetics
Reaction kinetics is the study of the rates at which chemical processes occur. It focuses on understanding the factors that affect reaction speed and the sequence of steps (mechanism) through which reactants transform into products. Several factors influence the kinetics of a reaction:
  • Concentration of reactants: Higher concentrations typically lead to faster reactions.
  • Temperature: Higher temperatures increase the kinetic energy of molecules, leading to more frequent and successful collisions.
  • Catalysts: Substances that increase the rate of reaction without being consumed in the process.
  • Physical state of reactants: Reactants in the same physical state (e.g., all gases) usually react faster than those in different states.
Kinetics not only helps in determining the speed and mechanism of reactions but also allows for the optimization of conditions to improve the efficiency of industrial chemical processes.
Frequency factor
The frequency factor ( A ) is a component of the Arrhenius equation representing how often reactants collide with the correct orientation to overcome the activation energy. It is a reflection of the likelihood of reactants forming products during collisions. The frequency factor takes into account:
  • The frequency of collisions between reactant molecules.
  • The orientation of these molecules during collisions.
  • Any steric factors that may affect the effective collision rate.
Normally, the frequency factor has units matching those of the rate constant (typically \( \, \text{M/s} \) for second-order reactions), and it often spans a wide range of values depending on the specific reaction and conditions. Understanding the frequency factor is crucial for designing experiments and predicting reaction behaviors under various conditions.
Rate constant
The rate constant ( k ) is a fundamental part of reaction kinetics, reflecting the reaction rate when the concentrations of all reactants are at their standardized values. It is derived from the Arrhenius equation and depends on the activation energy and the temperature. The rate constant is unique for each reaction and is affected by:
  • The temperature: Higher temperatures generally increase the rate constant.
  • The activation energy: Reactions with lower activation energies have higher rate constants.
  • The frequency factor: This provides a base value that is modified by changing temperature and activation energy.
The rate constant allows chemists to compare the speeds of different reactions and to understand how quickly a reaction will proceed under given conditions. The study of the rate constant is essential in both theoretical and applied chemistry, enabling the design and control of chemical processes.

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Most popular questions from this chapter

It is found that a gas undergoes a zero-order decomposition reaction in the presence of a nickel catalyst. If the rate constant for this reaction is \(8.1 \times 10^{-2} \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})\), how long will it take for the concentration of the gas to change from an initial concentration of \(0.10 \mathrm{M}\) to \(1.0 \times 10^{-2} M ?\)

The reaction \(2 \mathrm{~A}(\mathrm{~g}) \longrightarrow \mathrm{A}_{2}(g)\) is being run in each of the following containers. The reaction is found to be second-order with respect to \(\mathrm{A}\). a. Write the rate law for the reaction. b. Which reaction container will have a greater reaction rate? c. In which container will the reaction have a shorter halflife? d. What are the relative rates of the reactions in each container? e. After a set amount of time has elapsed, which container will contain fewer A atoms?

Chlorine dioxide oxidizes iodide ion in aqueous solution to iodine; chlorine dioxide is reduced to chlorite ion. $$2 \mathrm{ClO}_{2}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{ClO}_{2}^{-}(a q)+\mathrm{I}_{2}(a q)$$ The order of the reaction with respect to \(\mathrm{ClO}_{2}\) was determined by starting with a large excess of \(\mathrm{I}^{-}\), so that its concentration was essentially constant. Then $$\text { Rate }=k\left[\mathrm{ClO}_{2}\right]^{m}\left[\mathrm{I}^{-}\right]^{n}=k^{\prime}\left[\mathrm{ClO}_{2}\right]{m}$$ where \(k^{\prime}=k\left[\mathrm{I}^{-}\right]^{n}\). Determine the order with respect to \(\mathrm{ClO}_{2}\) and the rate constant \(k^{\prime}\) by plotting the following data assuming firstand then second-order kinetics. [Data from H. Fukutomi and G. Gordon, J. Am. Chem. Soc., 89,1362 (1967).] \(\begin{array}{cl}\text { Time (s) } & {\left[\text { ClO }_{2}\right] \text { (moll } \text { L) }} \\ 0.00 & 4.77 \times 10^{-4} \\ 1.00 & 4.31 \times 10^{-4} \\ 2.00 & 3.91 \times 10^{-4} \\ 3.00 & 3.53 \times 10^{-4} \\ 5.00 & 2.89 \times 10^{-4} \\ 10.00 & 1.76 \times 10^{-4} \\ 30.00 & 2.4 \times 10^{-5} \\ 50.00 & 3.2 \times 10^{-6}\end{array}\)

Dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), decomposes when heated in carbon tetrachloride solvent. $$\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+{ }_{2}^{1} \mathrm{O}_{2}(g)$$ If the rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(6.2 \times\) \(10^{-4} / \mathrm{min}\), what is the half-life? (The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5} .\) ) How long would it take for the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to decrease to \(25 \%\) of its initial value? to \(12.5 \%\) of its initial value?

Ethylene oxide, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\), decomposes when heated to give methane and carbon monoxide. The following kinetic data were observed for the reaction at \(688 \mathrm{~K}:\) \(\begin{array}{lll} & \text { Initial } & \\ & \text { Concentration } & \\\ & \text { of Ethylene Oxide } & \text { Initial Rate } \\ \text { Exp. 1 } & 0.00272 M & 5.57 \times 10^{-7} \mathrm{M} / \mathrm{s} \\ \text { Exp. } 2 & 0.00544 M & 1.11 \times 10^{-6} \mathrm{M} / \mathrm{s}\end{array}\) Find the rate law and the value of the rate constant for this reaction.
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