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. $$ \begin{array}{ll}{\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{N}(g)} & {\Delta H=941 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}^{+}(g)+\mathrm{e}^{-}} & {\Delta H=1501 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{N}(g) \longrightarrow \mathrm{N}^{+}(g)+\mathrm{e}^{-}} & {\Delta H=1402 \mathrm{kJ} / \mathrm{mol}}\end{array} $$ 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 \mathrm{kJ} / \mathrm{mol})$

Step by step solution

01

Observe the given enthalpy changes and reactions

We have the following reactions with their respective enthalpy changes: 1. \(N_2(g) \longrightarrow 2N(g)\) with \(\Delta H = 941 \ kJ/mol\) 2. \(N_2(g) \longrightarrow N_2^+(g) + e^-\) with \(\Delta H = 1501 \ kJ/mol\) 3. \(N(g) \longrightarrow N^+(g) + e^-\) with \(\Delta H = 1402 \ kJ/mol\)

02

Calculate the energy of the 25 nm light

We know from the Planck's equation that the energy of the light is given by: \(E = \dfrac{hc}{\lambda}\) where h is the Planck's constant (6.626 x 10^{-34} Js), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength. Let's calculate the energy of the 25 nm light (convert nm to meters): \(E = \dfrac{(6.626 \times 10^{-34} \ Js)(3 \times 10^8 \ m/s)}{25 \times 10^{-9} m} = 7.95 \times 10^{-19} \ J\) Then, we convert this energy to kJ/mol: \(E = 7.95 \times 10^{-19} \ J \times \dfrac{10^3 \ kJ}{1 \ J} \times \dfrac{1 \ mol}{6.022 \times 10^{23}} = 1319 \ kJ/mol\)

03

Determine the possible reactions

The energy of 25 nm light is 1319 kJ/mol. We compare this energy to the ΔH of the given reactions: 1. For Reaction 1: \(1319 \ kJ/mol > 941 \ kJ/mol\), so the dissociation of N2 into atomic nitrogen is possible. 2. For Reaction 2: \(1319 \ kJ/mol < 1501 \ kJ/mol\), so the ionization of N2 is not possible. 3. For Reaction 3: \(1319 \ kJ/mol < 1402 \ kJ/mol\), so the ionization of N is not possible. The species that could form in the flask during irradiation are N2(g) and N(g), but not N2+(g) or N+(g). b. Find the range of wavelengths

04

Lower limit for atomic nitrogen production

First, we need to find the lower limit of the wavelength range to produce atomic nitrogen, which corresponds to the energy required for the first reaction (N2 dissociation): \(E_{1} = 941 \ kJ/mol\) Now, we calculate the corresponding wavelength for this energy, using the Planck's equation, and converting J to kJ and atoms to mol: \(\lambda_{1} = \dfrac{hc \times 6.022 \times 10^{23}}{941 \times 10^3}\) \(\lambda_{1} = 127 \ nm\)

05

Upper limit for no ion production

Next, we need to find the upper limit of the wavelength range that will not produce any ions. This corresponds to the energy for the third reaction (ionization of N): \(E_{3} = 1402 \ kJ/mol\) Calculating the corresponding wavelength for this energy: \(\lambda_{3} = \dfrac{hc \times 6.022 \times 10^{23}}{1402 \times 10^3}\) \(\lambda_{3} = 86 \ nm\) So, the range of wavelengths that will produce atomic nitrogen in the flask without producing any ions is from \(86 \ nm\) to \(127 \ nm\). c. Explain ionization energy difference

06

Factors affecting ionization energy

The ionization energy depends on factors such as the effective nuclear charge on the electron being removed and the distance between the nucleus and the electron. In general, higher the effective nuclear charge and smaller the distance between the nucleus and electron, higher is the ionization energy.

07

Ionization energy comparison

In the case of N2, the electrons are held in bonds between the two nitrogen atoms. When ionizing N2 (removing an electron), we are essentially breaking that bond. The electrons in N2 are closer to the nuclei of both nitrogen atoms, and the effective nuclear charge is also larger, as it is shared by two nitrogen atoms. Therefore, it requires more energy to remove an electron from N2 as compared to atomic nitrogen. For atomic nitrogen, it has a smaller effective nuclear charge, and the outermost electron is not involved in a bond between two nitrogen atoms. Hence, it requires less energy to remove an electron from atomic nitrogen compared to N2. That's why the first ionization energy of N2 (1501 kJ/mol) is greater than that of atomic nitrogen (1402 kJ/mol).

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