Question

The bond dissociation energy of a typical C-F bond in a

chlorofluorocarbon is approximately $\mathbf{440}\mathbf{kJ}{\mathbf{mol}}^{-\mathbf{1}}$. Calculate the maximum wavelength of light that can photodissociate a molecule of ${\mathbf{CCl}}_2{\mathbf{F}}_2$, breaking such a C-F bond.

Short answer

The wavelength of light that are capable of dissociating a molecule of C-F is

.$\mathbf{4}\mathbf{.}\mathbf{514}\mathbf{\times }{\mathbf{10}}^{-28}\mathbf{m}$

Step-by-step solution

Bond Dissociation energy of C-F bond

The energy that operates to break a bond resulting in formation of various ‘fragments’is called bond dissociation energy. Since bond breaking process requires energy,it is called endothermic reaction.

Given information

The given data has values as follows:

• $\mathbf{h}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{\times }{\mathbf{10}}^{-34}\mathbf{Js}$ (Planck’s constant)
• role="math" localid="1663694385569" $\mathbf{c}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{9979}\mathbf{\times }{\mathbf{10}}^8{\mathbf{ms}}^{-1}$(Speed of light)
• $\lambda$is the wavelength.
• $\mathbf{E}=\mathbf{208}\mathbf{kJ}{\mathbf{mol}}^{-\mathbf{1}}$(Bond dissociation energy of$\mathbf{Cl}-\mathbf{F}$bond)

Wavelength of CUF bond

From table 3.3, the bond dissociation energy of typical Cl-F bond, is approximately

^{${\text{252kJmol}}^{\text{-1}}$}

The energy of a radiation is related to its wavelength is given by the equation

$E=\frac{hc}{\lambda }$

This gives the wavelength $\lambda$, as:

$\begin{array}{c}\text{λ=}\frac{\text{hc}}{\text{E}}\\ \text{λ=}\frac{{\text{6.626×10}}^{\text{34}}{\text{×2.99879×10}}^{\text{8}}}{\text{440}}\\ {\text{λ=4.514×10}}^{\text{-28}}\text{m}\end{array}$

Therefore, wavelength of light that are capable of dissociating a molecule of C-F is

.$4.514\times {10}^{-28}\text{m}$