Question

In \(\mathrm{CF}_{3} \mathrm{Cl}\) the \(\mathrm{C}-\mathrm{Cl}\) bond- dissociation energy is \(339 \mathrm{~kJ} / \mathrm{mol}\). In \(\mathrm{CCl}_{4}\) the \(\mathrm{C}-\mathrm{Cl}\) bond-dissociation energy is \(293 \mathrm{~kJ} / \mathrm{mol}\). What is the range of wavelengths of photons that can cause \(\mathrm{C}-\mathrm{Cl}\) bond rupture in one molecule but not in the other?

Short answer

The range of wavelengths of photons that can cause C-Cl bond rupture in CF3Cl but not in CCl4 is between \(3.54 \times 10^{-7} \mathrm{~m}\) (or 354 nm) and \(4.07 \times 10^{-7} \mathrm{~m}\) (or 407 nm).

Step-by-step solution

Convert bond dissociation energies to Joules per molecule

To work with energy per molecule rather than per mole, we will convert the given energies from kJ/mol to J/molecule using the Avogadro's number (6.022 x 10²³ molecules/mol). For CF3Cl: \( 339 \mathrm{~kJ/mol} \times \dfrac{10^3 \mathrm{~J}}{1 \mathrm{~kJ}} \times \dfrac{1 \mathrm{~mol}}{6.022 \times 10^{23} \mathrm{~molecules}} = 5.63 \times 10^{-19} \mathrm{~J/molecule} \) For CCl4: \( 293 \mathrm{~kJ/mol} \times \dfrac{10^3 \mathrm{~J}}{1 \mathrm{~kJ}} \times\dfrac{1 \mathrm{~mol}}{6.022 \times 10^{23} \mathrm{~molecules}} = 4.87 \times 10^{-19} \mathrm{~J/molecule} \)

Use Planck's equation to find the energy range of photons

Planck's equation relates the energy (E) of a photon to its frequency (ν) or wavelength (λ) through the Planck's constant (h, \(6.626 \times 10^{-34} \mathrm{~J \cdot s}\)) and the speed of light in vacuum (c, \(3 \times 10^8 \mathrm{~m/s}\)). \(E = h \nu = \cfrac{hc}{\lambda}\) To find the range of wavelengths corresponding to the energy range, we note that the bond rupture in CF3Cl but not in CCl4: \( 4.87 \times 10^{-19} \mathrm{~J} < E < 5.63 \times 10^{-19} \mathrm{~J} \)

Calculate the range of wavelengths

Rearrange Planck's equation for the wavelength: \( \lambda = \cfrac{hc}{E} \) Calculate the range of wavelengths that will cause bond rupture in CF3Cl but not in CCl4: \(\lambda\) minimum for CF3Cl: \( \lambda_{min} = \cfrac{6.626 \times 10^{-34} \mathrm{~J\cdot s} \times 3 \times 10^8 \mathrm{~m/s}}{5.63 \times 10^{-19} \mathrm{~J}} = 3.54 \times 10^{-7} \mathrm{~m} \) \(\lambda\) maximum for CCl4: \( \lambda_{max} = \cfrac{6.626 \times 10^{-34} \mathrm{~J\cdot s} \times 3 \times 10^8 \mathrm{~m/s}}{4.87\times 10^{-19} \mathrm{~J}} = 4.07 \times 10^{-7} \mathrm{~m} \)

Present the range of wavelengths

The range of wavelengths of photons that can cause C-Cl bond rupture in CF3Cl but not in CCl4 is between \(3.54 \times 10^{-7} \mathrm{~m}\) (or 354 nm) and \(4.07 \times 10^{-7} \mathrm{~m}\) (or 407 nm).