Chapter 7, Problem 7.108QP

It requires 799 kJ of energy to break one mole of carbon-oxygen double bonds in carbon dioxide. What wavelength of light does this correspond to per bond? Is there any transition in the hydrogen atom that has at least this quantity of energy to one photon?

Interpretation Introduction

Interpretation:

The energy required to rupture one mole of $\text{C=}\text{O}$ in ${\text{CO}}_{\text{2}}$ is $\text{799KJ}$.  The wavelength of light used to break per bond has to be calculated.

Concept introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion.  By using this he derived a formula for energy levels of electron in H-atom.

$\begin{array}{l}\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}\text{n}\text{=}\text{1,2,3,}\mathrm{......}\text{(For}\text{hydrogen}\text{atom)}\\ \\ {\text{R}}_{\text{H}}\text{is}\text{Rydberg}\text{constant}\text{(2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J)}\text{.}\\ \text{Eis}\text{energy}\text{level}\text{.}\\ \text{n}\text{is}\text{principal}\text{quantum}\text{number}\text{.}\end{array}$

Relation between frequency and wavelength is,

$\text{C}\text{=}\text{λν}$

C is the speed of light.

$\text{ν}$ is the frequency.

$\text{λ}$ is wavelength.

$\text{E}\text{=}\text{hν}$

h is Planck’s constant ( $\text{6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s}$ ) which relates energy and frequency.

$\text{ν}$ is the frequency.

E is energy of light particle.

The distance between any two similar points of a wave is called wavelength

Figure 1

$\text{λ}$ is wavelength.

Frequency is defined as number of wavelengths of a wave that can pass through a point in one second.

Answer to Problem 7.108QP

The wavelength of light used to break per bond is $\text{1}\text{.6342}\text{×}{\text{10}}^{\text{-18}}\text{J}$.

Explanation of Solution

To calculate: The wavelength of light used to break per bond.

The energy of each photon is calculated as follows,

$\text{E}\text{=}\frac{\text{799 kJ}}{\text{1 mol}}\text{×}\frac{\text{1000 J}}{\text{1 kJ}}\text{×}\frac{\text{1mol}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\text{=}\text{1}\text{.326}\text{×}{\text{10}}^{\text{-18}}\text{J}$

The wave length of light used to break per bond is

$\begin{array}{l}\text{ν}\text{=}\frac{\text{E}}{\text{h}}\text{=}\frac{\text{1}\text{.326}\text{×}{\text{10}}^{\text{-18}}\text{J}}{\text{6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s}}\text{=}\text{2}\text{.001}\text{×}{\text{10}}^{\text{15}}\text{/s}\\ \\ \text{λ}\text{=}\frac{\text{c}}{\text{ν}}\text{=}\frac{\text{3}\text{.00}\text{×}{\text{10}}^{\text{8}}\text{m/s}}{\text{2}\text{.001}\text{×}{\text{10}}^{\text{15}}\text{/s}}\text{=}\text{1}\text{.50}\text{×}{\text{10}}^{\text{-7}}\text{m}\text{(150 nm)}\end{array}$

The obtained wavelength is in range of near UV region of electromagnetic spectrum which corresponds to lyman series of H-atom.  The minimum energy required in this region for transition corresponds to $\text{n = 2 to n = 1}$.

$\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}$

$\begin{array}{l}\text{ΔE}\text{=}\text{hν}\text{=}{\text{E}}_{\text{1}}\text{-}{\text{E}}_{\text{2}}\\ \text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{1}}^{\text{2}}}\text{-}\frac{{\text{-R}}_{\text{H}}}{{\text{2}}^{\text{2}}}\\ \text{=}\frac{{\text{-R}}_{\text{H}}}{\text{1}}\text{-}\frac{{\text{-R}}_{\text{H}}}{\text{4}}\\ \text{=}\frac{{\text{-3R}}_{\text{H}}}{\text{4}}\text{=}\frac{\text{-(3)(2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{)}}{\text{4}}\text{=}\text{-1}\text{.6342}\text{×}{\text{10}}^{\text{-18}}\text{J}\end{array}$

The obtained energy is more than required energy to break the $\text{C}\text{=}\text{O}$  in carbon dioxide.

Conclusion

By using the formula for energy levels of electron in H-atom, the wavelength of light used to break per bond was calculated.