Q9.78CP - Chemistry: Molecular Nature Of Matter And Change

Question

By using photons of specific wavelengths, chemists can dissociate gaseous Hi to produce H atoms at certain speeds. When HI dissociates, the H atoms move away rapidly. Whereas the heavier I atoms move away more slowly.

1. What is the longest wavelength (in nm) that can dissociate a molecule of HI?
2. If a photon of 254 nm is used, what is the excess energy (in J) over that needed for dissociation?
3. If the excess energy is carried away by the H atom as kinetic energy, what is the speed (in m/s)?

Step-by-Step Solution

Answer

1. The longest wavelength that can dissociate a molecule oh HI is 405.78nm
2. The excess energy that is needed for dissociation is $\mathbf{2}\mathbf{.}\mathbf{92}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}$ .
3. The speed is 590 m/s if the extra energy is transferred to the H atom as kinetic energy.

Longest wavelength

The bond energy of H-I 

 

 $\mathbf{E}\left(\mathrm{HI}\right)\mathbf{=}\mathbf{295}\mathbf{kJ}\mathbf{/}\mathbf{mol}$

To find out the longest wavelength that can dissociate a molecule of HI, we calculate the energy of one molecule of HI

 $\mathbf{E}\mathbf{(}\mathbf{HI}{\mathbf{)}}_{\mathrm{molecule}}\mathbf{=}\frac{E(H-I)}{{\mathbf{N}}_A}$

 

Where ${\mathbf{N}}_A$ is Avogadro’s number. value of  ${\mathbf{N}}_A$ is  $\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{23}\mathbf{molecules}\mathbf{/}\mathbf{mol}$ 

So,

 $\begin{array}{c}\mathbf{E}\mathbf{(}\mathbf{HI}{\mathbf{)}}_{\mathrm{molecule}}\mathbf{=}\frac{\mathbf{295}\mathbf{kJ}\mathbf{/}\mathbf{mol}\mathbf{\times }{\mathbf{10}}^3\mathbf{J}\mathbf{/}\mathbf{kJ}}{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{23}\mathbf{molecules}\mathbf{/}\mathbf{mol}}\\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{8987}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}\mathbf{/}\mathbf{molecule}\end{array}$

 

Therefore, the maximum wavelength of the HI molecule can be calculated as

 $\mathbf{\lambda }\mathbf{=}\frac{\mathrm{hc}}{\mathbf{E}\mathbf{(}\mathbf{HI}{\mathbf{)}}_{\mathrm{molecule}}}$

 

Where h is Planck’s constant. Value of h is  $\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{\times }{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{sec}$ . And c is the speed of light. Value of c is  $\mathbf{3}\mathbf{\times }{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{sec}$ .

 

 $\begin{array}{c}\mathbf{\lambda }\mathbf{=}\frac{\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{\times }{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{s}\mathbf{\times }\mathbf{3}\mathbf{\times }{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{4}\mathbf{.}\mathbf{8987}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}\mathbf{.}\mathbf{Mol}}\\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{05}\mathbf{\times }{\mathbf{10}}^{-7}\mathbf{m}\\ \mathbf{=}\mathbf{405}\mathbf{.}\mathbf{78}\mathbf{nm}\end{array}$

Thus, the longest wavelength that can dissociate a molecule oh HI is 405.78nm 

 

Excess energy

Given wavelength = 254nm

First of all, find the energy corresponding to this wavelength and then subtract the bond dissociation energy of HI.

So, the energy corresponding to this wavelength

  $\begin{array}{l}\mathbf{E}\mathbf{=}\frac{\mathrm{hc}}{\lambda }\\ \mathbf{=}\frac{\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{\times }{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{s}\mathbf{\times }\mathbf{3}\mathbf{\times }{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{254}\mathbf{\times }{\mathbf{10}}^{-9}\mathbf{m}}\end{array}$

     

Excess energy = energy corresponding to 254nm wavelength- bond dissociation

                                                                                            The energy of a molecule oh HI 

     

 $\begin{array}{l}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{82}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{8987}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}\\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{92}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}\end{array}$

Thus, the excess energy over that needed for dissociation is  $\mathbf{2}\mathbf{.}\mathbf{92}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}$.

Speed

If the excess energy is carried away by the H atom as kinetic energy, the speed is calculated by the formula  $\mathbf{KE}\mathbf{=}\frac{1}{2}{\mathbf{mv}}^2$.  where m is the H atom's mass. The value of m is $\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{-24}\mathbf{g}$. And v is the speed.

Therefore,

 $\mathbf{KE}\mathbf{=}\frac{1}{2}{\mathbf{mv}}^2$

                      

$\begin{array}{l}\mathbf{2}\mathbf{.}\mathbf{92}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{=}\frac{1}{2}\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{-24}\mathbf{\times }{\mathbf{v}}^2\\ {\mathbf{v}}^2\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{92}\mathbf{\times }{\mathbf{10}}^{-19}}{\mathbf{8}\mathbf{.}\mathbf{365}\mathbf{\times }{\mathbf{10}}^{-24}}\\ {\mathbf{v}}^2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{349}\mathbf{\times }{\mathbf{10}}^{-5}\\ \mathbf{v}\mathbf{=}\mathbf{590}\mathbf{m}\mathbf{/}\mathbf{s}\end{array}$ 

The speed is 590 m/s if the extra energy is transferred to the H atom as kinetic energy.