9.90_CP - Chemistry: Molecular Nature Of Matter And Change

Question

carbon- carbon bonds form the “backbone” of nearly every organic and biological molecule. The average bond energy of the C-C bond is 347kJ/mol. Calculate the frequency and wavelength of the least energetic photon that can break an average C-C bond. In what region of the electromagnetic spectrum is this radiation?

Step-by-Step Solution

Answer

$\mathbf{8}\mathbf{.}\mathbf{696}\mathbf{\times }{\mathbf{10}}^{14}\mathbf{Hz}$ frequency and 344.98nm wavelength of the least energetic photon can break an average C-C bond. 

This radiation is in the UV region of the electromagnetic spectrum.

energy of the single bond

Firstly, we have to calculate the least energy that is required to break the single bond in C-C. average bond energy divided by the Avogadro number is the least energy required to break a single bond. 

     $\begin{array}{l}{\mathbf{E}}_{\mathbf{single}\mathbf{bond}}\mathbf{=}\frac{{\mathbf{E}}_{\mathrm{average}}}{{\mathbf{N}}_A}\\ \mathbf{=}\frac{\mathbf{347}\mathbf{\times }{\mathbf{10}}^3\mathbf{J}\mathbf{/}\mathbf{mol}}{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{23}{\mathbf{mol}}^{-1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{763}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}\end{array}$             

                                                   

 

So,$\mathbf{5}\mathbf{.}\mathbf{763}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}$  is the least energy that is required to break the single bond in C-C.

Frequency calculation

Frequency is the number of waves that pass a certain point in a specified amount of time.

 

Now, the frequency of the least energetic photon can be calculated by using the Einstein- plank equation.

 

     $\begin{array}{l}E=\mathrm{h\upsilon }\\ \mathbf{\upsilon }\mathbf{=}\frac{E}{h}\mathbf{=}\frac{\mathbf{5}\mathbf{.}\mathbf{762}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}}{\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{\times }{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{s}}\\ \mathbf{=}\mathbf{8}\mathbf{.}\mathbf{696}\mathbf{\times }{\mathbf{10}}^{14}\mathbf{Hz}\end{array}$                

 

calculation of wavelength

Wavelength can be calculated by using the formula

 $\begin{array}{l}\mathbf{\upsilon }\mathbf{=}\frac{c}{\lambda }\\ \mathbf{\lambda }\mathbf{=}\frac{c}{\upsilon }\end{array}$

              

Where c is the speed of light. Value of c is $\mathbf{3}\mathbf{\times }{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}$. 

 $\begin{array}{l}\mathbf{\lambda }\mathbf{=}\frac{c}{\upsilon }\mathbf{=}\frac{\mathbf{3}\mathbf{\times }{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{8}\mathbf{.}\mathbf{696}\mathbf{\times }{\mathbf{10}}^{14}\mathbf{Hz}}\\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{34498}\mathbf{\times }{\mathbf{10}}^{-6}\mathbf{m}\\ \mathbf{=}\mathbf{344}\mathbf{.}\mathbf{98}\mathbf{nm}\end{array}$

 

So, 344.98nm wavelength is required to break an average C-C bond.

calculation of wavenumber

$\begin{array}{l}\mathbf{wavenumber}\mathbf{(}\mathbf{c}{\mathbf{m}}^{-1}\mathbf{)}\mathbf{=}\frac{{\mathbf{10}}^7}{\mathrm{wavelength}\left(\mathrm{nm}\right)}\\ \mathbf{=}\frac{{\mathbf{10}}^7}{\mathbf{344}\mathbf{.}\mathbf{98}\mathbf{nm}}\mathbf{=}\mathbf{28987}\mathbf{.}\mathbf{18}{\mathbf{cm}}^{-1}\end{array}$

Since, the range of the UV region of electromagnetic radiation is 200-400nm or $\mathbf{25000}\mathbf{-}\mathbf{50000}{\mathbf{cm}}^{-1}$, from the calculation value of wavelength and wavenumber it is clear that the radiation is in the UV region.