Problem 32 - Chemistry The Central Science

Question

An important reaction in the formation of photochemical \(\operatorname{smog}\) 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 420 nm wavelength light is found in the visible light region of the electromagnetic spectrum. (b) The maximum bond strength that can be broken by a photon of 420 nm light is about \(2.85 \cdot 10^2 \ \text{kJ/mol}\). (c) The photodissociation reaction with Lewis-dot structures is: : O = N - O : + h \nu → : O = N : + : O :

Step 1: Find the region in the electromagnetic spectrum

To find the region in which the 420 nm wavelength light lies, we can refer to the electromagnetic spectrum and find the corresponding region. It is known that visible light has wavelengths between 380 nm and 750 nm. Since 420 nm lies within this range, the electromagnetic spectrum region corresponds to visible light.

Step 2: Calculate the photon energy

To calculate the photon energy, we can use the formula \[E= h \frac{c}{\lambda}\] where E is the energy, h is Planck's constant \(6.63 \cdot 10^{-34} \ \text{Js}\), c is the speed of light \(3.00 \cdot 10^{8} \ \text{m/s}\), and \(\lambda\) is the wavelength (420 nm). First, let's convert the wavelength from nm to m: \[420 \ \text{nm} = 420 \cdot 10^{-9} \ \text{m}\] Now, we can plug in the values into the formula, \[E = 6.63 \cdot 10^{-34} \ \text{Js} \cdot \frac{3.00 \cdot 10^{8} \ \text{m/s}}{420 \cdot 10^{-9} \ \text{m}} = 4.74 \cdot 10^{-19} \ \text{J}\] The energy of a photon with a wavelength of 420 nm is approximately \(4.74 \cdot 10^{-19} \ \text{J}\).

Step 3: Convert the energy to kJ/mol and find the bond energy

To find the bond strength in kJ/mol, we first need to convert the photon energy from J to kJ: \[4.74 \cdot 10^{-19} \ \text{J} \cdot \frac{1 \ \text{kJ}}{10^3 \ \text{J}} = 4.74 \cdot 10^{-22} \ \text{kJ}\] Now, we can convert the energy per photon to energy per mole using Avogadro's number (\(6.022 \cdot 10^{23} \ \text{mol}^{-1}\)), \[4.74 \cdot 10^{-22} \ \text{kJ} \cdot \frac{6.022 \cdot 10^{23} \ \text{mol}^{-1}}{1 \ \text{molecule}} = 2.85 \cdot 10^2 \ \text{kJ/mol}\] Thus, the maximum bond strength that can be broken by a photon of 420 nm light is about \(2.85 \cdot 10^2 \ \text{kJ/mol}\).

Step 4: Draw the Lewis-dot structures for the photodissociation reaction.

For the photodissociation reaction, \[\text{NO}_2 + h \nu \longrightarrow \text{NO} + \text{O}\] The corresponding Lewis-dot structures are: : O = N - O : + h \nu → : O = N : + : O : We have now identified the region in the electromagnetic spectrum for the 420 nm light (visible light), calculated the maximum bond strength that can be broken by this photon energy (\(2.85 \cdot 10^2 \ \text{kJ/mol}\)), and drawn the Lewis-dot structures for the photodissociation reaction.

Key Concepts

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, ranging from gamma rays, which have the shortest wavelengths, to radio waves, which have the longest wavelengths. The spectrum is divided into several regions based on the wavelength of the electromagnetic waves. Notably, the visible light spectrum falls between ultraviolet (UV) and infrared (IR) regions, with wavelengths from approximately 380 nm to 750 nm.

Visible light is what we can see with our human eyes and it's further divided into colors, each with its specific wavelength range. For example, light with a wavelength of 420 nm is perceived as violet and sits at the lower end of the visible light spectrum. Understanding which part of the spectrum a wavelength belongs to helps in areas such as photodissociation, where the energy of the light is used to break chemical bonds, as is the case with NO2.

Photon Energy Calculation

The energy of a photon can be calculated using the formula:
\[E = h \frac{c}{\lambda}\]
where:\

• \(E\) is the energy of the photon,
• \(h\) is Planck's constant (approximately \(6.63 \times 10^{-34} \text{ Js}\)),
• \(c\) is the speed of light (about \(3.00 \times 10^8 \text{ m/s}\)),
• and \(\lambda\) is the wavelength of the light.

By substituting the appropriate values into the formula, you can determine the amount of energy in joules (J) that a single photon carries. To understand how this relates to chemical processes, you can convert this energy into kilojoules per mole (kJ/mol), which is the common unit used to express bond energies in molecules. This conversion involves Avogadro's number because one mole of a substance contains exactly that many particles (approximately \(6.022 \times 10^{23}\)). Thus, knowing the photon energy is crucial in estimating the maximum strength of a chemical bond that can be broken by light of a certain wavelength.

Lewis-dot Structures

Lewis-dot structures offer a visual representation of the valence electrons in atoms and molecules, which are crucial for understanding chemical bonding and reactions. In these structures, dots are used to represent valence electrons around the chemical symbols for the elements. Lines or pairs of dots between atoms represent bonds, indicating shared electrons.

Example of a Lewis-dot Structure

For the nitrogen dioxide (\(NO_2\)) molecule involved in the photodissociation reaction, a Lewis-dot structure illustrates the location of the unpaired electron, which is significant in understanding its reactivity. During photodissociation, the absorption of light provides energy to break specific chemical bonds, as depicted in the reaction:\[\text{NO}_2 + h u \longrightarrow \text{NO} + \text{O}\]The Lewis-dot structures before and after the reaction show the transfer of valence electrons, which results in the formation of different species – in this case, nitric oxide (NO) and a free oxygen atom (O). This type of representation is helpful in visualizing the changes during the reaction and is a fundamental aspect of understanding chemistry at the molecular level.