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The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride: Reaction 1: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g})+\mathrm{HCl}(\mathrm{g})\) This reaction is very slow in the absence of light. However, \(\mathrm{Cl}_{2}(g)\) can absorb light to form \(\mathrm{Cl}\) atoms: $$ \text { Reaction 2: } \mathrm{Cl}_{2}(g)+h v \longrightarrow 2 \mathrm{Cl}(g) $$ Once the \(\mathrm{Cl}\) atoms are generated, they can catalyze the reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{Cl}_{2}\), according to the following proposed mechanism: Reaction 3: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})\) Reaction 4: \(\mathrm{CH}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{Cl}(g)\) The enthalpy changes and activation energies for these two reactions are tabulated as follows: $$ \begin{array}{lll} \hline \text { Reaction } & \Delta H_{\mathrm{ran}}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & E_{a}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline 3 & +4 & 17 \\ 4 & -109 & 4 \end{array} $$ (a) By using the bond enthalpy for \(\mathrm{Cl}_{2}\) (Table \(8.4\) ), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. In which portion of the electromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g)\), should be placed on your diagram in part (b). Use this result to estimate the value of \(E_{a}\) for the reaction \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow\) \(\mathrm{CH}_{3}(g)+\mathrm{HCl}(g)+\mathrm{Cl}(g) .\) (d) The species \(\mathrm{Cl}(g)\) and \(\mathrm{CH}_{3}(\mathrm{~g})\) in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of \(\mathrm{CH}_{3}\), and verify that is a radical. (e) The sequence of reactions 3 and 4 comprise a radical chain mechanism. Why do you think this is called a "chain reaction"? Propose a reaction that will terminate the chain reaction.

Short Answer

Expert verified
(a) The longest wavelength of light to cause Reaction 2 is found using the relation \(E = \dfrac{hc}{\lambda}\). Convert bond enthalpy of Cl₂(242 kJ/mol) to energy per photon and solve for \(\lambda_{max}\). The portion of the electromagnetic spectrum in which the light is found depends on the obtained value. (b) Create an energy profile for Reactions 3 and 4 using given enthalpy changes and activation energies. (c) Estimate the position of CH₄(g) + Cl₂(g) in the energy profile by comparing the sum of bond enthalpies for reactants and products. Further, estimate the activation energy for the overall reaction. (d) The Lewis structure of CH₃ radical has one carbon atom in the center surrounded by three hydrogen atoms with single bonds. The carbon atom has one unpaired electron, making it a radical: H | H - C - H * (e) "Chain reaction" refers to the self-sustaining sequence of reactions involving radicals. A chain termination reaction could involve the reaction between two radicals, creating a molecule without unpaired electrons, e.g. CH₃(g) + Cl(g) → CH₃Cl(g).

Step by step solution

01

Part (a): Determine the longest wavelength of light to cause Reaction 2

Begin by using the relationship between energy (E), wavelength (λ), and the speed of light (c): \(E = h \nu = \dfrac{hc}{\lambda}\), where h is Planck's constant. The bond enthalpy for Cl2 is given in the problem as 242 kJ/mol. Convert this to energy per photon, and then solve for the longest wavelength. \(Energy\ per\ photon = \dfrac{242\ kJ/mol}{6.022 \times 10^{23}\ photons/mol}\) Now, convert this to energy per photon in Joules (J): \(Energy\ per\ photon\ (J) = \dfrac{242,000\ J/mol}{6.022 \times 10^{23}\ photons/mol\) Finally, use the equation mentioned above to solve for the longest wavelength: \(\lambda_{max} = \dfrac{hc}{Energy\ per\ photon\ (J)}\) The obtained value will determine the portion of the electromagnetic spectrum in which the light is found.
02

Part (b): Sketch an energy profile for Reactions 3 and 4

Based on the provided enthalpy changes and activation energies, create an energy profile for Reactions 3 and 4. Begin with the reactants CH4 and Cl and follow the reaction path through the intermediates CH3 and Cl and finally to the products CH3Cl and HCl. For Reaction 3, note that the enthalpy change is +4 kJ/mol and the activation energy is 17 kJ/mol. For Reaction 4, the enthalpy change is -109 kJ/mol and the activation energy is 4 kJ/mol.
03

Part (c): Estimate the position of reactants and activation energy for overall reaction

To determine the location of CH4(g) + Cl2(g) on the energy profile in part (b), use bond enthalpies: Bond enthalpy of CH4 (broken) = 412 kJ/mol Bond enthalpy of Cl2 (broken) = 242 kJ/mol Sum of the bond enthalpy of broken bonds: 412 + 242 = 654 kJ/mol Now calculate the sum of bond enthalpies for the products: Bond enthalpy of CH3 (formed) = 381 kJ/mol Bond enthalpy of HCl (formed) = 431 kJ/mol Bond enthalpy of a Cl atom (not formed or broken) = 0 kJ/mol Sum of the bond enthalpy of formed bonds: 381 + 431 = 812 kJ/mol Now, find the difference between these sums to place CH4(g) + Cl2(g) on the diagram in part (b): ΔH = 812 - 654 = 158 kJ/mol From this value, estimate the activation energy of the reaction, taking into account the individual activation energies of the reactions 3 and 4 mentioned in the table.
04

Part (d): Draw a Lewis structure of CH3 radical

In a Lewis structure, we represent atoms and their valence electrons. In CH3, there is one carbon atom and three hydrogen atoms. Carbon has 4 valence electrons and hydrogen has one. Place the carbon atom in the center and surround it with the three hydrogen atoms. Now connect carbon to each hydrogen atom with a single bond (represented by a line). This uses 3 valence electrons of carbon. The carbon atom still possesses one unpaired electron, which makes it a radical. CH3 Lewis structure: H | H - C - H * * represents the unpaired electron on the carbon atom.
05

Part (e): Understand the term "chain reaction" and propose a chain termination reaction

A chain reaction is called so because once the radicals are formed (like in reactions 3 and 4), they continually react with other molecules, generating new radicals that propagate the reaction. The entire process is self-sustained until a chain termination step occurs. A chain termination reaction could involve the reaction between two radicals, which results in a molecule without any unpaired electrons. For example: CH3(g) + Cl(g) → CH3Cl(g)

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Key Concepts

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

Atmospheric Chemistry
Atmospheric chemistry plays a crucial role in environmental science, as it involves the study of chemical processes within the Earth's atmosphere. A prime example of such processes is the photochemical reaction between methane and chlorine molecules, which is greatly accelerated by light. In this reaction, ultraviolet radiation can split chlorine molecules into chlorine radicals, a process known as photodissociation. Understanding this concept helps clarify why some reactions that are otherwise slow, suddenly become significant in the presence of sunlight, such as the formation of smog.

This scenario also touches on the interaction between electromagnetic radiation and chemical compounds. Different molecules absorb light at specific wavelengths, which corresponds to certain portions of the electromagnetic spectrum. This is fundamental to predicting the reactivity of molecules in the atmosphere when exposed to sunlight, providing valuable insights into topics like ozone depletion and the formation of photochemical pollutants.
Reaction Kinetics
The study of reaction kinetics is crucial to understanding the speed of chemical reactions and the factors that influence that speed. In the reaction between methane and chlorine, light acts as a catalyst, increasing the rate of the reaction. Reaction kinetics involves investigating such rate changes by analyzing activation energy, which is the minimum energy required to initiate a chemical reaction.

By examining the enthalpy changes and activation energies provided for the reactions, you can predict how the reaction progresses over time. The relatively low activation energy for the radical-generating step in comparison to the overall reaction shows why such steps are fast and why radicals are highly reactive. These concepts of reaction kinetics allow scientists and students to understand the complexities of atmospheric reactions, including those that contribute to environmental phenomena.
Chemical Bond Energy
The energy associated with the breaking and forming of chemical bonds is a central concept in chemistry, known as bond energy or bond enthalpy. In atmospheric reactions, such as the photochemical reactions between methane and chlorine, the bond energies dictate which wavelengths of light are capable of inducing photodissociation. Calculating the longest wavelength that can break the bond in the chlorine molecule involves using the bond energy and applying principles of quantum mechanics like the relationship between a photon’s energy and its wavelength.

Energy diagrams that visualize the path of a reaction highlight the input and release of energy during bond breaking and forming. These diagrams are helpful in depicting the overall energy change of the process and provide a visual aid for the locations of reactants, intermediates, and products in terms of their relative energy levels.
Radical Chain Reactions
Radical chain reactions are a type of chemical reaction mechanism that involves free radicals - highly reactive atoms or molecules with unpaired electrons. These reactions unfold in a series of steps beginning with initiation (the formation of radicals), propagation (where radicals generate more radicals), and termination (where radicals are removed by combining with one another).

In the given exercise, the formation of methyl chloride from methane and chlorine is described as a radical chain reaction. The chlorine radical, produced by the absorption of light, initiates the reaction by abstracting a hydrogen atom from methane, creating another radical which then reacts with chlorine molecule to form methyl chloride and regenerate the chlorine radical. This perpetuates the reaction sequence, forming what is known as a chain reaction. Understanding the nature of radicals and how they drive chain reactions sheds light on numerous chemical processes, from industrial syntheses to biochemical pathways.

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Most popular questions from this chapter

Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction in order to achieve a \(1 \times 10^{5}\) -fold increase in the reaction rate?

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\). When \([\mathrm{NO}]=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M}\), the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of \(1.8\) ?

(a) What is meant by the term elementary reaction? (b) What is the difference between a unimolecular and a bimolecular elementary reaction? (c) What is a reaction mechanism?

You have studied the gas-phase oxidation of \(\mathrm{HBr}\) by \(\mathrm{O}_{2}\) : $$ 4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g) $$ You find the reaction to be first order with respect to \(\mathrm{HBr}\) and first order with respect to \(\mathrm{O}_{2}\). You propose the following mechanism: $$ \begin{aligned} \mathrm{HBr}(g)+\mathrm{O}_{2}(g) & \rightarrow \mathrm{HOOBr}(g) \\ \mathrm{HOOBr}(g)+\mathrm{HBr}(g) & \longrightarrow 2 \mathrm{HOBr}(g) \\ \mathrm{HOBr}(g)+\mathrm{HBr}(g) \longrightarrow & \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g) \end{aligned} $$ (a) Indicate how the elementary reactions add to give the overall reaction. (Hint: You will need to multiply the coefficients of one of the equations by 2.) (b) Based on the rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\), is first order. Near room temperature, the rate constant equals \(7.0 \times 10^{-4} \mathrm{~s}^{-1} .\) Calculate the half-life at this temperature. (b) At \(415^{\circ} \mathrm{C}\), \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}\) decomposes in the gas phase, \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g) .\) If the reac- tion is first order with a half-life of \(56.3 \mathrm{~min}\) at this temperature, calculate the rate constant in \(\mathrm{s}^{-1}\).

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