Question

An important reaction in the formation of photochemical smog is the photodissociation of \(\mathrm{NO}_{2} :\) $$\mathrm{NO}_{2}+h \nu \longrightarrow \mathrm{NO}(g)+\mathrm{O}(g)$$ The maximum wavelength of light that can cause this reaction is 420 \(\mathrm{nm}\) . (a) In what part of the electromagnetic spectrum is light with this wavelength found? (b) What is the maximum strength of a bond, in kJ/mol, that can be broken by absorption of a photon of \(420-\mathrm{nm}\) light? (c) Write out the photodissociation reaction showing Lewis-dot structures.

Short answer

(a) Light with a wavelength of 420 nm is found in the visible part of the electromagnetic spectrum. (b) The maximum bond strength that can be broken by a photon with a wavelength of 420 nm is 286 kJ/mol. (c) The photodissociation reaction using Lewis-dot structures is: \[ \text{N}=\text{O} - \text{O}^{\bullet} + h\nu \longrightarrow \text{N}=\text{O} + \text{O}^{\bullet} \]

Step-by-step solution

a) Identifying the part of the electromagnetic spectrum

To understand where the 420 nm wavelength of light is found in the electromagnetic spectrum, we must compare it with the common ranges of wavelengths for different categories of the spectrum. Here are the approximate ranges for visible light: - Ultraviolet light (UV): 100 nm to 400 nm, - Visible light: 400 nm to 700 nm, - Infrared light (IR): 700 nm to 1,000,000 nm. Since 420 nm is between 400 and 700 nm, light with this wavelength is found in the visible part of the electromagnetic spectrum.

b) Calculating the maximum bond strength in kJ/mol

To find the energy associated with a 420 nm photon, we can use the equation: \[E = \frac{hc}{\lambda}\] Where: - E is the energy of the photon - h is the Planck's constant (\(6.626\times10^{-34}\; \text{Js}\)) - c is the speed of light in vacuum (\(2.998\times10^8\; \text{m/s}\)) - \(\lambda\) is the wavelength (420 nm) First, let's convert the wavelength from nanometers (nm) to meters (m), since we have the speed of light in m/s: \[\lambda = 420 \; \text{nm} \times \frac{1 \; \text{m}}{10^9 \; \text{nm}} = 4.20 \times 10^{-7} \; \text{m}\] Now we can find the energy of the photon: \[E = \frac{(6.626\times10^{-34}\; \text{Js})(2.998\times10^8\; \text{m/s})}{4.20 \times 10^{-7} \; \text{m}}\] \[E = 4.74\times10^{-19} \; \text{J}\] Next, let's convert this energy from Joules (J) to kJ/mol: \[E = 4.74\times10^{-19} \; \text{J} \times \frac{1 \; \text{kJ}}{10^3 \; \text{J}} \times \frac{6.022 \times 10^{23} \; \text{photons}}{1 \; \text{mol}}\] \[E = 286 \; \text{kJ/mol}\] Therefore, the maximum bond strength that can be broken by a photon with a wavelength of 420 nm is 286 kJ/mol.

c) Writing the photodissociation reaction with Lewis-dot structures

To display the photodissociation reaction using Lewis-dot structures, we simply need to draw the Lewis-dot structures for the molecules involved in the reaction: \[ \text{NO}_2 + h\nu \longrightarrow \text{NO} + \text{O} \] The Lewis-dot structures are as follows: - NO₂: N is connected to an O by a double bond and to another O by a single bond. N has one more lone electron. - NO: N is connected to O by a double bond. N has one more lone electron. - O: O atom has six valence electrons as lone pairs. With these structures, the photodissociation reaction can be represented as: \[ \text{N}=\text{O} - \text{O}^{\bullet} + h\nu \longrightarrow \text{N}=\text{O} + \text{O}^{\bullet} \]

Key concepts

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, which vary in wavelength and frequency. This broad range includes, from longest wavelength to shortest: radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays, and gamma rays. In the context of photodissociation, we find that light with a wavelength of 420 nm falls within the visible spectrum, specifically in the violet-blue region. This segment of the spectrum is crucial for many chemical reactions, as the energy provided by visible light is often enough to break certain chemical bonds, leading to significant effects such as the formation of photochemical smog mentioned in the exercise.

Photon Energy Calculation

The energy of a photon can be calculated using the equation
\(E = \frac{hc}{\lambda}\)
where \(E\) represents the energy of the photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of light. By plugging in the given values, as shown in the step-by-step solution, we calculate the maximum energy a single photon can impart when its wavelength is 420 nm. Understanding how to calculate photon energy is integral for predicting the extent to which different types of electromagnetic radiation can drive photodissociation and similar processes.

Lewis-dot Structures

Lewis-dot structures are a representation of molecules that illustrate the valence electrons as dots around the elements. They provide a visual way to predict the number and types of bonds that an atom can form, based on the electrons available for bonding. When exploring the photodissociation of NO₂, Lewis-dot structures help us visualize how the molecule breaks apart into NO and O. This graphical representation makes it clear that a single photon can disrupt the electronic structure sufficiently to sever the nitrogen-oxygen bond, illustrating the potency of photon-induced reactions at the molecular level.