Problem 34 - Chemistry: The Central Science in SI Units, Global Edition

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 \(\mathrm{kJ} / \mathrm{mol}\), that can be broken by absorption of a photon of 420 -nm light? \(?\) (c) Write out the photodissociation reaction showing Lewis-dot structures.

Step-by-Step Solution

Answer

(a) The maximum wavelength of light, 420 nm, falls within the visible light range of the electromagnetic spectrum. (b) The maximum strength of a bond that can be broken by the absorption of a photon of 420-nm light is 285 kJ/mol. (c) The photodissociation reaction using Lewis-dot structures is :NO₂ + hν -> :N≡O + ·O.

Step 1: (a) Identify the part of the electromagnetic spectrum where this wavelength is found

First, we need to know the ranges of the electromagnetic spectrum to see where 420 nm falls. The electromagnetic spectrum can be classified into seven categories according to their wavelengths: 1. Radio waves: wavelength (> 1 m) 2. Microwaves: wavelength (1 mm - 1 m) 3. Infrared: wavelength (700 nm - 1 mm) 4. Visible light: wavelength (400 nm - 700 nm) 5. Ultraviolet: wavelength (10 nm - 400 nm) 6. X-rays: wavelength (0.01 nm - 10 nm) 7. Gamma rays: wavelength (< 0.01 nm) Given that our wavelength is 420 nm, it falls within the visible light range of the electromagnetic spectrum.

Step 2: (b) Determine the maximum strength of a bond that can be broken by the absorption of a photon of 420-nm light

To calculate the maximum bond strength, we will use the Planck-Einstein relation and the concept of energy required to break a bond: \(E = h \cdot \nu\) Where: E = energy of the photon (Joules) h = Planck's constant (\(6.626 \times 10^{-34}\) Js) \(\nu\) = frequency (Hz) We can relate frequency (\(\nu\)) and wavelength (\(\lambda\)) with the speed of light (c) using the following equation: \(\nu = \frac{c}{\lambda}\) Now, we can substitute and solve for E: \(E = h \cdot \frac{c}{\lambda}\) To obtain energy per mole of photons, we can multiply E by the Avogadro's number (NA) and convert the energy to kJ/mol. \(E_{mol} = E \cdot N_A\)

Step 3: Calculation of the maximum bond strength that can be broken by the absorption of a photon of 420-nm light

1. Convert the wavelength to meters: \(\lambda = 420 \cdot 10^{-9} m\) 2. Calculate the frequency: \(\nu = \frac{3 \times 10^8 \, \text{m/s}}{420 \cdot 10^{-9}\, \text{m}} = 7.14 \times 10^{14} \, \text{Hz}\) 3. Calculate the energy of one photon: \(E = (6.626 \times 10^{-34} \, \text{Js}) \cdot (7.14 \times 10^{14} \, \text{Hz}) = 4.73 \times 10^{-19} \, \text{J}\) 4. Calculate the energy per mole: \(E_{mol} = (4.73 \times 10^{-19}\, \text{J}) \cdot (6.022 \times 10^{23}\, \text{mol}^{-1}) = 2.85 \times 10^5 \, \text{J/mol}\) 5. Convert the energy to kJ/mol: \(E_{mol} = 2.85 \times 10^5 \, \text{J/mol} \cdot \frac{1\, \text{kJ}}{1000\, \text{J}} = 285\, \text{kJ/mol}\) Thus, the maximum strength of a bond that can be broken by the absorption of a photon of 420-nm light is 285 kJ/mol.

Step 4: (c) Write the photodissociation reaction using Lewis-dot structures

To write the photodissociation reaction using Lewis-dot structures, we need to demonstrate the electron distribution in each molecule of the reaction: NO₂: :N═O: , Oxygen is double bonded to Nitrogen and one unpaired electron on Nitrogen. NO: :N≡O: , Oxygen is triple bonded to Nitrogen. O: ·O , Oxygen has one unpaired electron. Now, we can write the photodissociation reaction using the Lewis-dot structures: :NO₂ + hν -> :N≡O + ·O

Key Concepts

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, each distinguished by different wavelengths. Radiation corresponds to waves of energy across varying spectrums. Understanding where a particular wavelength fits in this spectrum can tell us a lot about the nature of the radiation.

• Radio Waves: These have wavelengths greater than 1 meter and serve primarily for communication.
• Microwaves: Measuring from 1 millimeter to 1 meter, these are often used for heating food.
• Infrared: Lying in the 700 nm to 1 mm range, infrared can be felt as heat.
• Visible Light: This is most pertinent to our everyday observation, with wavelengths from 400 nm to 700 nm, allowing us to see colors.
• Ultraviolet: These rays have wavelengths between 10 nm and 400 nm and are responsible for causing sunburns.
• X-rays: With wavelengths ranging from 0.01 nm to 10 nm, they are crucial for imaging bones.
• Gamma Rays: The shortest waves, with less than 0.01 nm, highly penetrating and emitted by radioactive materials.

The specific wavelength of 420 nm falls into the visible light category, marking it close to violet in color. Understanding this part of the spectrum is essential for analyzing various phenomena like photodissociation.

Photon Energy

The energy of a photon is critical in photochemical reactions. Photon energy can affect molecular bonds, as seen in processes like photodissociation. To determine the energy, we use the key relationship tying together frequency (\(u\)) and wavelength (\(\lambda\)) with the speed of light (\(c\)).

Key Formula

Photon energy (\(E\)) is determined by Planck's equation:
\[E = h \cdot u\]
Here, \(h\) is Planck's constant (\(6.626 \times 10^{-34}\, Js\)). Using the relationship between frequency and wavelength:\[ u = \frac{c}{\lambda}\]where \(c\) is the speed of light (\(3 \times 10^8 m/s\)).

Calculating Energy

Substituting for \(u\) in Planck's equation yields:
\[E = h \cdot \frac{c}{\lambda}\]
Now, to find the energy per mole of 420-nm light, multiply the photon energy by Avogadro's number (\(6.022 \times 10^{23} \ mol^{-1}\)). Converting from Joules to kJ/mol gives you the energy measurement commonly used to express bond strength.
This calculated photon energy tells us about the maximum potential bond strength that can be disrupted in photochemical processes, critical when examining reactions like photodissociation.

Lewis Dot Structures

Lewis dot structures represent the arrangement of electrons in atoms and molecules, emphasizing valence electrons. These are pivotal in understanding reactions, especially photodissociation.

• The principle is to visualize electrons around an atom as dots. Each dot represents a valence electron.
• Pairs of dots between atoms represent covalent bonds.

In the case of \(NO_2\), the nitrogen atom shares electrons with oxygen atoms creating bonds. For nitrogen dioxide represented as :N=O:, nitrogen creates a double bond with one oxygen and holds an unpaired electron.
During photodissociation, absorption of light energy breaks these bonds. In this meaning, the reaction can be illustrated:
:N=O: + \(hv\) → :N≡O: + ·O
This diagram highlights the transition of bonds, with the nitrogen monoxide (\NO) forming a triple bond and the free radical (\(\cdot O\)) depicted with a lone unpaired electron. Understanding these structures is valuable for visualizing electron shifts in chemical bonds influenced by photon energy.