Question

A flask containing gaseous \(\mathrm{N}_{2}\) is irradiated with \(25-\mathrm{nm}\) light. a. Using the following information, indicate what species can form in the flask during irradiation. $$ \begin{aligned} \mathrm{N}_{2}(g) & \longrightarrow 2 \mathrm{~N}(g) & \Delta H &=941 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{N}_{2}(g) & \longrightarrow \mathrm{N}_{2}^{+}(g)+\mathrm{e}^{-} & \Delta H &=1501 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{N}(g) & \longrightarrow \mathrm{N}^{+}(g)+\mathrm{e}^{-} & \Delta H &=1402 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ b. What range of wavelengths will produce atomic nitrogen in the flask but will not produce any ions? c. Explain why the first ionization energy of \(\mathrm{N}_{2}(1501 \mathrm{~kJ} / \mathrm{mol})\) is greater than the first ionization energy of atomic nitrogen \((1402 \mathrm{~kJ} / \mathrm{mol})\).

Short answer

The species that can form in the flask during irradiation are none, as the energy of the 25 nm light (13.3 kJ/mol) is too low to cause any of the given reactions. The range of wavelengths that will produce atomic nitrogen but not produce any ions is 662 nm - 998 nm. The first ionization energy of N₂ is greater than that of atomic nitrogen because in the molecular form (N₂), the electrons are more tightly bound to the molecules due to strong triple bonds between the nitrogen atoms, making it more difficult to remove an electron from N₂ as compared to a single N atom.

Step-by-step solution

a. Determination of formed species

First, we need to calculate the energy of the irradiated light. We can do that using the Planck's equation: $$E = \dfrac{hc}{\lambda}$$ where \(E\) is the energy, \(h\) is Planck's constant (\(6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\)), \(c\) is the speed of light (\(2.998 \times 10^8 \mathrm{m/s}\)) and \(\lambda\) is the wavelength of the light: \(25 \mathrm{nm} = 25 \times 10^{-9} \mathrm{m}\). Calculating the energy, $$E = \dfrac{(6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s})(2.998 \times 10^8 \mathrm{m/s})}{25 \times 10^{-9} \mathrm{m}} = 7.975 \times 10^{-18}\mathrm{J}$$ To convert Joules to kJ/mol, $$E = \dfrac{7.975 \times 10^{-18} \mathrm{J}}{1 \times 10^{-3} \mathrm{kJ/J}} \times \dfrac{1 \mathrm{mol}}{6.022 \times 10^{23}} = 13.3 \mathrm{kJ/mol}$$ Now, compare the energy of the light with the given enthalpy changes: 1. Breaking N₂ into atomic N (ΔH = 941 kJ/mol) will not happen as the energy of light is way too low (13.3 kJ/mol). 2. Ionizing N₂ (ΔH = 1501 kJ/mol) will not happen for the same reason. 3. Ionizing atomic N (ΔH = 1402 kJ/mol) will not happen either. Thus, no species will form in the flask during irradiation. The energy of the light is too low.

b. Wavelength range for producing atomic nitrogen without ions

To determine the range of wavelengths that will produce atomic nitrogen but not produce any ions, consider the following energy requirements: 1. Breaking N₂ into atomic N requires a minimum energy of ΔH = 941 kJ/mol. 2. Ionization of N₂ requires a minimum energy of ΔH = 1501 kJ/mol. The wavelength range should be calculated such that: 941 kJ/mol ≤ E < 1501 kJ/mol Calculate the wavelengths corresponding to 941 kJ/mol and 1501 kJ/mol using the Planck's equation: Minimum wavelength (corresponds to 1501 kJ/mol): $$\lambda_{min} = \dfrac{hc}{E} = \dfrac{(6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s})(2.998 \times 10^8 \mathrm{m/s})}{1501 \times 10^3 \mathrm{J/mol} \times \dfrac{1 \mathrm{mol}}{6.022 \times 10^{23}}} = 6.62 \times 10^{-10} \mathrm{m} = 662 \mathrm{nm}$$ Maximum wavelength (corresponds to 941 kJ/mol): $$\lambda_{max} = \dfrac{hc}{E} = \dfrac{(6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s})(2.998 \times 10^8 \mathrm{m/s})}{941 \times 10^3 \mathrm{J/mol} \times \dfrac{1 \mathrm{mol}}{6.022 \times 10^{23}}} = 9.98 \times 10^{-10} \mathrm{m} = 998 \mathrm{nm}$$ The range of wavelengths that will produce atomic nitrogen but not produce any ions is 662 nm - 998 nm.

c. Explanation for different ionization energies

The ionization energy of a molecule or atom is the minimum energy required to remove an electron from its valence shell, forming a positive ion. The first ionization energy of N₂ is greater than the first ionization energy of atomic nitrogen because in the molecular form (N₂), the electrons are more tightly bound to the molecules due to strong triple bonds between the nitrogen atoms. This makes it more difficult to remove an electron from N₂ as compared to a single N atom. Moreover, in N₂, the extra stability provided by the triple bond results in a lower energy state, making its electrons less accessible. In the atomic state, nitrogen has unpaired electrons that can be more easily removed because they are less stabilized than the electrons within the triple bond of N₂. Therefore, the first ionization energy of N₂ is higher than that of atomic nitrogen.

Key concepts

Understanding Planck's Equation

Planck's equation is a fundamental concept in quantum mechanics, relating the energy of a photon to its frequency. In its simplest form, the equation is expressed as:
\[\begin{equation}E = \frac{hc}{\/lambda}\end{equation}\]
Where:

• \(E\) is the energy of the photon in joules (J),
• \(h\) is Planck's constant (40.6 times 10^{-34}\t J \cdot s),
• \(c\) is the speed of light in a vacuum (42.998 \times 10^{8}\t m/s),
• \(\lambda\) is the wavelength of the light in meters (m).

Determining the energy of light is essential in chemistry because it allows us to understand the interactions of light with atoms and molecules, such as the ionization of gases. For example, in our textbook exercise, we use Planck's equation to calculate whether a 25-nm light has sufficient energy to cause ionization in gaseous nitrogen.

Wavelength and Energy Relationship

The relationship between the wavelength of light and its energy is inversely proportional, meaning that as the wavelength decreases, the energy increases, and vice versa. This is depicted in Planck's equation, where energy is inversely related to wavelength. In the context of our exercise, understanding this relationship helps decipher what processes can occur when gaseous nitrogen is irradiated with light of a specific wavelength.
Higher energy photons (with shorter wavelengths) can break molecular bonds or ionize atoms, leading to the formation of various species. For instance, to produce atomic nitrogen from 42 but avoid ionization, one needs to find the sweet spot in the wavelength spectrum – not too high to prevent ionization and not too low to have enough energy to break the molecular bonds. This explains the importance of selecting the right wavelength range for inducing specific chemical reactions.

Molecular vs Atomic Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or molecule. There's a distinct difference between the ionization energies of atoms and molecules due to their chemical structure. Molecular ionization energy is often higher than atomic ionization energy. This is because in a molecule, such as nitrogen gas (42_), the electrons are shared between atoms, forming chemical bonds. These bonds can impart greater stability to the molecule, making it more difficult to remove an electron.
Specifically, in our exercise, the ionization energy for molecular nitrogen is higher than for atomic nitrogen primarily due to the triple bond's strength in the nitrogen molecule. Understanding this concept is crucial for explaining why higher energy is needed to ionize 42_ compared to atomic nitrogen. This knowledge also guides us in applications such as spectroscopy, where knowing the difference between molecular and atomic ionization energies can help in identifying substances.