The heat of atomization of methane and ethane are 360kJ/mol and 620kJ/mol, respectively the longest wavelength light capable of breaking the C−C bond is (Avodrago constant =6.02×1023mol−1).

# The heat of atomization of methane and ethane are $360\mathrm{kJ}/\mathrm{mol}$ and $620\mathrm{kJ}/\mathrm{mol}$, respectively the longest wavelength light capable of breaking the $\mathrm{C}-\mathrm{C}$ bond is (Avodrago constant $=6.02×{10}^{23}{\mathrm{mol}}^{-1}$).

1. A

$1.49×{10}^{3}\mathrm{nm}$

2. B

$2.48×{10}^{3}\mathrm{nm}$

3. C

$2.48×{10}^{4}\mathrm{nm}$

4. D

$1.49×{10}^{4}\mathrm{nm}$

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### Solution:

Given data as

$\begin{array}{l}{\mathrm{CH}}_{4}\left(\mathrm{g}\right)\to \mathrm{C}\left(\mathrm{g}\right)+4\mathrm{H}\left(\mathrm{g}\right) \mathrm{\Delta }{H}_{1}=4{\epsilon }_{\mathrm{C}-\mathrm{H}}\\ {\mathrm{C}}_{2}{\mathrm{H}}_{6}\left(\mathrm{g}\right)\to 2\mathrm{C}\left(\mathrm{g}\right)+6\mathrm{H}\left(\mathrm{g}\right) \mathrm{\Delta }{H}_{2}={\epsilon }_{\mathrm{C}-\mathrm{C}}+6{\epsilon }_{\mathrm{C}-\mathrm{H}}\end{array}$

From  $\mathrm{\Delta }{H}_{1}$, we get   ${\epsilon }_{\mathrm{C}-\mathrm{H}}=\left(360/4\right)\mathrm{kJ}{\mathrm{mol}}^{-1}=90\mathrm{kJ}{\mathrm{mol}}^{-1}$

From  $\mathrm{\Delta }{H}_{2}$, we get   ${\epsilon }_{\mathrm{C}-\mathrm{H}}=\mathrm{\Delta }{H}_{2}-6{\epsilon }_{\mathrm{C}-\mathrm{H}}=\left(620-6×90\right){\mathrm{kJ}}^{-1}=80\mathrm{kJ}{\mathrm{mol}}^{-1}$

For a single $\mathrm{C}-\mathrm{H}$  bond, we have

$E=\frac{{\epsilon }_{\mathrm{C}}-\mathrm{H}}{{N}_{\mathrm{A}}}=\frac{\left(80\mathrm{kJ}{\mathrm{mol}}^{-1}\right)}{\left(6.02×{10}^{23}{\mathrm{mol}}^{-1}\right)}=1.33×{10}^{-22}\mathrm{kJ}=1.33×{10}^{-14}\mathrm{J}$

The longest wavelength required to break this bond is

$\begin{array}{l}\lambda =\frac{hc}{E}=\frac{\left(6.626×{10}^{-34}\mathrm{Js}\right)\left(3×{10}^{8}{\mathrm{ms}}^{-1}\right)}{\left(1.33×{10}^{-19}\mathrm{J}\right)}=1.495×{10}^{-6}\mathrm{m}\\ =1.495×{10}^{3}\mathrm{nm}\end{array}$  