Question

Energy from radiation can cause chemical bonds to break. To break the nitrogen-nitrogen bond in \(\mathrm{N}_{2}\) gas, \(941 \mathrm{~kJ} / \mathrm{mol}\) is required. (a) Calculate the wavelength of the radiation that could break the bond. (b) In what spectral range does this radiation occur?

Short answer

Answer: The wavelength of the radiation that could break the nitrogen-nitrogen bond is \(1.276 \times 10^{-7} \mathrm{~m}\) or 127.6 nm, and it occurs in the ultraviolet (UV) spectral range.

Step-by-step solution

(Step 1: Write down the known values)

We are given: - Energy required to break the nitrogen-nitrogen bond: \(941 \mathrm{~kJ/mol}\) - Planck's constant: \(h = 6.626 \times 10^{-34} \mathrm{~Js}\) - Speed of light: \(c = 3.0 \times 10^8 \mathrm{~m/s}\) Note that we need to convert the energy from kJ/mol to Joules.

(Step 2: Convert energy to Joules)

Energy in Joules (J): \(E = 941 \mathrm{~kJ/mol} \times \frac{1000 \mathrm{~J}}{1 \mathrm{~kJ}} \times \frac{1 \mathrm{~mol}}{6.022 \times 10^{23} \mathrm{~molecules}} = 1.560 \times 10^{-18} \mathrm{~J}\) Now we have the energy in Joules, we can calculate the wavelength of the radiation that could break the bond.

(Step 3: Calculate the frequency of the radiation)

Using the energy equation, \(E=hf\) We can solve for the frequency (f): \(f = \frac{E}{h} = \frac{1.560 \times 10^{-18} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{~Js}} = 2.352 \times 10^{15} \mathrm{~Hz}\)

(Step 4: Calculate the wavelength of the radiation)

Now, we can use the speed of light equation to find the wavelength: \(c = \lambda f\) Solving for the wavelength (λ): \(\lambda = \frac{c}{f} = \frac{3.0 \times 10^8 \mathrm{~m/s}}{2.352 \times 10^{15} \mathrm{~Hz}} = 1.276 \times 10^{-7} \mathrm{~m}\) So the wavelength of the radiation that could break the nitrogen-nitrogen bond is \(1.276 \times 10^{-7} \mathrm{~m}\). (a) The wavelength of the radiation that could break the bond is \(1.276 \times 10^{-7} \mathrm{~m}\).

(Step 5: Determine the spectral range of the radiation)

The given wavelength is in meters. To find the spectral range, we can convert it to nanometers: \(\lambda = 1.276 \times 10^{-7} \mathrm{~m} \times \frac{10^9 \mathrm{~nm}}{1 \mathrm{~m}} = 127.6 \mathrm{~nm}\) The wavelength of the radiation is \(127.6 \mathrm{~nm}\). Based on the electromagnetic spectrum, this wavelength falls within the ultraviolet (UV) range (10-400 nm). (b) The radiation occurs in the ultraviolet (UV) spectral range.