The average bond enthalpies of the \(\mathrm{C}-\mathrm{C}\) and \(\mathrm{C}-\mathrm{H}\) bonds are \(348 \mathrm{~kJ} / \mathrm{mol}\) and $413 \mathrm{~kJ} / \mathrm{mol}$, respectively. (a) What is the maximum wavelength that a photon can possess and still have sufficient energy to break the \(\mathrm{C}-\mathrm{H}\) and \(\mathrm{C}-\mathrm{C}\) bonds, respectively? (b) Given the fact that \(\mathrm{O}_{2}, \mathrm{~N}_{2},\) and \(\mathrm{O}\) in the upper atmosphere absorb most of the light with wavelengths shorter than $240 \mathrm{nm}$, would you expect the photodissociation of \(\mathrm{C}-\mathrm{C}\) and \(\mathrm{C}-\mathrm{H}\) bonds to be significant in the lower atmosphere?

Step by step solution

01

Planck's Equation

Planck's equation is given by: \[E = h \nu\] Where E is the energy of the photon, h is Planck's constant (6.626 x 10^(-34) J.s), and ν is the frequency of the photon. The frequency of a photon is related to its wavelength (λ) by the speed of light (c): \[\nu = \frac{c}{\lambda}\] Where c = 3 x 10^8 m/s. Substituting this equation into Planck's equation, we get: \[E = \frac{hc}{\lambda}\]

02

Calculate wavelength for C-C bond

Given the bond enthalpy of the C-C bond is 348 kJ/mol, we first need to convert this to energy per photon: \[\frac{348\,\text{kJ}}{\text{mol}} \times \frac{1000\,\text{J}}{1\,\text{kJ}} \times \frac{1 \text{mol}}{6.022 \times 10^{23}\,\text{atoms}} = 5.78 \times 10^{-19}\,\text{J}\] Now we can use Planck's equation to find the maximum wavelength: \[\lambda_{\mathrm{C-C}} = \frac{hc}{E_{\mathrm{C-C}}} = \frac{(6.626 \times 10^{-34}\,\text{J.s})(3 \times 10^8\,\text{m/s})}{5.78 \times 10^{-19}\,\text{J}} = 343\,\text{nm}\]

03

Calculate wavelength for C-H bond

Similarly, do the same for the C-H bond with a bond enthalpy of 413 kJ/mol: \[\frac{413\,\text{kJ}}{\text{mol}} \times \frac{1000\,\text{J}}{1\,\text{kJ}} \times \frac{1 \text{mol}}{6.022 \times 10^{23}\,\text{atoms}} = 6.86 \times 10^{-19}\,\text{J}\] And find the maximum wavelength: \[\lambda_{\mathrm{C-H}} = \frac{hc}{E_{\mathrm{C-H}}} = \frac{(6.626 \times 10^{-34}\,\text{J.s})(3 \times 10^8\,\text{m/s})}{6.86 \times 10^{-19}\,\text{J}} = 289\,\text{nm}\]

04

Compare the wavelengths to atmospheric absorption

We are given that wavelengths shorter than 240 nm are absorbed in the upper atmosphere. The calculated wavelengths for the C-C and C-H bonds were 343 nm and 289 nm, respectively. Both of these values are greater than 240 nm.

05

Determine the significance of photodissociation

Since the maximum wavelengths required to break C-C and C-H bonds are greater than 240 nm, the photons with these wavelengths can pass through the upper atmosphere without being absorbed. This means that photodissociation of C-C and C-H bonds can be significant in the lower atmosphere.

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