Question

A flask containing gaseous \(\mathrm{N}_{2}\) is irradiated with 25 -nm light. a. Using the following information, indicate what species can form in the flask during irradiation. $$\begin{array}{ll} \mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{N}(g) & \Delta E=941 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}^{+}(g)+\mathrm{e}^{-} & \Delta E=1501 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{N}(g) \longrightarrow \mathrm{N}^{+}(g)+\mathrm{e}^{-} & \Delta E=1402 \mathrm{kJ} / \mathrm{mol} \end{array}$$ 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 \(\mathrm{N}_{2}(1501 \mathrm{kJ} /\) mol) is greater than the first ionization energy of atomic nitrogen (1402 kJ/mol).

Short answer

a. During irradiation, all three processes can occur in the flask: \(N_2(g) \rightarrow 2 N(g)\), \(N_2(g) \rightarrow N_2^+(g) + e^-\), and \(N(g) \rightarrow N^+(g) + e^-\). b. The range of wavelengths that will produce atomic nitrogen without ions is greater than or equal to 128 nm. c. The first ionization energy of N2 is greater than the first ionization energy of atomic nitrogen because more energy is required to break the triple covalent bond present in the N2 molecule.

Step-by-step solution

a. Species formed in the flask during irradiation

We are given the energy of the 25 nm light. To find out which processes can occur during the irradiation, we need to first convert the wavelength (25 nm) into energy using Planck's equation: \[E = hf\] Where h is Planck's constant (6.63 x 10^(-34) J.s), f is frequency and E is energy. To convert between frequency (f) and wavelength (λ), the following equation can be used: \[c = f\lambda\] Where c is the speed of light (3 x 10^8 m/s), λ is the wavelength (25 nm), and f is frequency. Step 1: Calculate the energy of the 25 nm light. First, convert λ to meters: λ = 25 nm x 10^(-9) m/nm = 2.5 x 10^(-8) m Next, calculate the frequency, f: f = c / λ = (3 x 10^8 m/s) / (2.5 x 10^(-8) m) = 1.2 x 10^16 Hz Finally, calculate the energy: E = hf = (6.63 x 10^(-34) J.s)(1.2 x 10^16 Hz) = 7.96 x 10^(-18) J Step 2: Compare the energy of the light to the energies required for different species formation processes. Convert the given energies for different species formation processes from kJ/mol to J: ΔE1 = 941 kJ/mol x 10^3 J/1 kJ x 1 mol/6.022 x 10^23 species = 1.56 x 10^(-18) J ΔE2 = 1501 kJ/mol x 10^3 J/1 kJ x 1 mol/6.022 x 10^23 species = 2.49 x 10^(-18) J ΔE3 = 1402 kJ/mol x 10^3 J/1 kJ x 1 mol/6.022 x 10^23 species = 2.33 x 10^(-18) J Now, we need to determine which processes will occur by comparing the energy of the light (7.96 x 10^(-18) J) with the energies required for different species formation processes (ΔE1, ΔE2, and ΔE3). As 7.96 x 10^(-18) J > ΔE1 (1.56 x 10^(-18) J), N2(g) → 2 N(g) can occur. As 7.96 x 10^(-18) J > ΔE2 (2.49 x 10^(-18) J), N2(g) → N2+(g) + e- can occur. As 7.96 x 10^(-18) J > ΔE3 (2.33 x 10^(-18) J), N(g) → N+(g) + e- can occur. Thus, during irradiation, all three processes can occur in the flask.

b. Wavelength range for producing atomic nitrogen without ions

To find the range of wavelengths that will produce atomic nitrogen in the flask without producing any ions, we need to find the minimum and maximum energy required. A wavelength that can only provide energy for the first reaction (N2(g) → 2 N(g)), will be our desired wavelength range. Minimum energy = ΔE1 = 1.56 x 10^(-18) J Step 3: Calculate the wavelength corresponding to the minimum energy. Using Planck's equation E = hf, and knowing the speed of light c = fλ, we have: \[E=\dfrac{hc}{\lambda}\] Now, we can rearrange for λ: \[\lambda=\dfrac{hc}{E}\] Substitute the known values: \[\lambda=\dfrac{(6.63 \times 10^{-34}\ \text{J}\cdot\text{s})(3 \times 10^8\ \text{m/s})}{1.56 \times 10^{-18}\ \text{J}}\] After calculating, we get: \[\lambda \approx 1.28 \times 10^{-7}\ \text{m} = 128\ \text{nm}\] Therefore, the range of wavelengths that will produce atomic nitrogen without ions is greater than or equal to 128 nm.

c. Explaining the difference in first ionization energies

To explain why the first ionization energy of N2 (1501 kJ/mol) is greater than the first ionization energy of atomic nitrogen (1402 kJ/mol), we need to understand what is happening at the molecular level. In N2 molecule, the two nitrogen atoms are bound in a strong triple covalent bond. This triple bond gives N2 molecule its stability and makes it much less reactive. To ionize N2, this bond must be broken, which requires a significant amount of energy. On the other hand, atomic nitrogen (N) has only one nitrogen atom with unpaired electrons, making it more prone to ionization. Therefore, the first ionization energy of N2 is greater than the first ionization energy of atomic nitrogen because more energy is required to break the triple covalent bond present in N2 molecule.

Key concepts

Atomic transitions

Atomic transitions refer to the process where an electron within an atom or molecule changes from one energy level to another. This can occur through various processes, such as absorption or emission of a photon. In the context of the exercise, when nitrogen molecules in the flask are irradiated with 25-nm light, they absorb energy. This energy is enough to trigger transitions like breaking molecular bonds or ionizing the nitrogen atoms.

• When the energy of the light absorbed is greater than the transition energy, electrons can be moved between different atomic orbits, leading to molecular dissociation or ionization.
• For nitrogen, transitions can result in the breaking of the bond in \( ext{N}_2\) to form two nitrogen atoms. This is an example of an electronic transition that occurs due to the absorption of light energy.

Understanding these transitions is crucial as they govern the behavior of atoms under the influence of light, which is a fundamental aspect of spectroscopy.

Ionization energy

Ionization energy is the energy required to remove an electron from an atom or molecule in its gaseous state. This concept is central to understanding why photons of specific energy can ionize gases. In the provided exercise, we see different ionization energies for \( ext{N}_2\) and atomic nitrogen N.

• The energy needed to ionize \( ext{N}_2\) (1501 kJ/mol) is higher than that needed to ionize atomic nitrogen (1402 kJ/mol).
• This difference in ionization energy is due to the additional energy required to break the strong triple bond present in the \( ext{N}_2\) molecule.

Ionization energies inform us about the stability of a molecule or atom, where higher ionization energy means greater stability. The first ionization energy gives insights into how easily atoms or molecules can lose electrons and become ions.

Photon energy

Photon energy is determined by the wavelength of the light, with shorter wavelengths corresponding to higher energies. For instance, the light with wavelength 25 nm used in the exercise carries significant energy due to its short wavelength. This energy can be calculated using Planck's equation \(E = hf\), where \(h\) is Planck's constant and \(f\) is the frequency.

• The energy of a photon determines what atomic transitions it can induce.
• High-energy photons, such as those with a wavelength of 25 nm, can cause molecular dissociation or even ionization, as seen with the nitrogen gas in the flask receiving sufficient energy to break bonds.

Understanding photon energy is critical in determining which molecular and atomic processes, such as excitation, ionization, or chemical reactions, can occur upon light exposure.

Molecular dissociation

Molecular dissociation is the process by which a molecule splits into smaller components, often atoms, under the influence of energy. In the context of light-matter interaction, irradiation can provide the necessary energy to break the bonds holding atoms together in a molecule.

• In the exercise, dissociation of \( ext{N}_2 \) occurs upon irradiation because the energy supplied by the 25-nm light is sufficient to overcome the bond dissociation energy.
• The specific energy required for dissociation is called the bond dissociation energy, which for \( ext{N}_2 \) is 941 kJ/mol.

By understanding molecular dissociation, we gain insights into how energy transformations can alter chemical and physical states. This underscores the relationship between chemical reactivity and external energy applications, such as light.