The heat of atomization of methane and ethane | vedclass

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(B) In $CH_4,$ the heat of atomization corresponds to $4 	imes BE_{(C-H)} = 360 \ kJ/mol.$ Therefore,$BE_{(C-H)} = 90 \ kJ/mol.$ In $C_2H_6,$ the hea

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$1.49 	imes 10^3 \ nm$

Vedclass · Answer

(B) In $CH_4,$ the heat of atomization corresponds to $4 	imes BE_{(C-H)} = 360 \ kJ/mol.$ Therefore,$BE_{(C-H)} = 90 \ kJ/mol.$ In $C_2H_6,$ the heat of atomization corresponds to $BE_{(C-C)} + 6 	imes BE_{(C-H)} = 620 \ kJ/mol.$ Substituting $BE_{(C-H)} = 90 \ kJ/mol,$ we get $BE_{(C-C)} + 6(90) = 620 \ kJ/mol.$ Therefore,$BE_{(C-C)} = 620 - 540 = 80 \ kJ/mol.$ The energy required per molecule is $E = \frac{80 	imes 10^3 \ J/mol}{6.02 	imes 10^{23} \ molecules/mol} \approx 1.329 	imes 10^{-19} \ J/molecule.$ Using the relation $E = \frac{hc}{\lambda},$ the wavelength is $\lambda = \frac{hc}{E}.$ $\lambda = \frac{6.62 	imes 10^{-34} \ J \cdot s 	imes 3 	imes 10^8 \ m/s}{1.329 	imes 10^{-19} \ J} \approx 1.494 	imes 10^{-6} \ m.$ Converting to nanometers,$\lambda \approx 1.494 	imes 10^3 \ nm.$ Thus,the correct option is $B$.

The heat of atomization of methane and ethane are $360 \ kJ/mol$ and $620 \ kJ/mol,$ respectively. The longest wavelength of light capable of breaking the $C-C$ bond is (Avogadro number $= 6.02 \times 10^{23},$ $h = 6.62 \times 10^{-34} \ J \cdot s$)

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