Question

Use bond energies to estimate the maximum wavelength of light that will cause the reaction

${\mathbf{\text{O}}}_{\mathbf{\text{3}}}\stackrel{\mathbf{\text{hv}}}{\mathbf{\to }}{\mathbf{\text{O}}}_{\mathbf{\text{2}}}\mathbf{\text{+ O}}$

Short answer

Using bond energies,the maximum wavelength of light that will cause the reaction ${\text{O}}_{\text{3}}\stackrel{\text{hv}}{\to }{\text{O}}_{\text{2}}\text{+ O}$is $\lambda =330nm$.

Step-by-step solution

Concept Introduction

The wavelength is defined as the distance between any two consecutive identical spots on the waveform. The amplitude of a wave is its height. Because many waves pass a given place in a given period of time, a wave with a short wavelength has a high frequency.

Wavelength equation

UV radiation plays a crucial role in the formation and destruction of ozone.

UV amounts are greatest in the tropical regions;thus, it is not surprising that most of the destruction and production ofozone occurs in the tropical stratosphere.

The energy in light is proportional to its frequency and inversely proportional to the wavelength of the light.

$\text{E =}\frac{\text{h·c}}{\lambda }$

Calculation for wavelength

The ozone bonds are slightly weaker than the oxygen bonds. The energyrequired to break the bond in ozone is$\text{364kJ/mol}$.

$\text{λ=}\frac{\text{h·c}}{\text{E}}$

For one mole, we need to divide the energy required to break the bond inmolecule by the number of molecules in a mole –

role="math" localid="1663822641138" $$\begin{gathered} \lambda =\frac{{\displaystyle {\text{6.63×10}}^{\text{37}}{\text{kJs·3.0·10}}^{17}\text{nm/s}}}{364kJ/mol}\times \left(6.02\times {10}^{23}mol\right) \\ \lambda =330nm \end{gathered}$$

Therefore, the lowest energy light that can break the bond in ozone has a wavelength of$330nm$ .