ThermochemistryMedium
Question
The heat of atomisation of methane and ethane are ' x ' $kJ{mol}^{- 1}$ and ' y ' $kJ{mol}^{- 1}$ respectively. The longest wavelength ( $\lambda$ ) of light capable of breaking the $C - C$ bond can be expressed in SI unit as :
Options
A.$\frac{hc}{1000}\left( \frac{y - 6x}{4} \right)^{- 1}$
B.$\frac{N_{A}hc}{250(4y - 6x)}$
C.$\frac{N_{A}hc}{250(y - 6x)}$
D.$N_{A}hc\left( y - \frac{6x}{4} \right)^{- 1}$
Solution
$\ {CH}_{4}(\text{ }g) \rightarrow C(g) + 4H(g);\ \Delta_{r}H = xkJ/$ mole $C_{2}H_{6}(\text{ }g) \rightarrow 2C(g) + 6H(g);\Delta_{r}H = ykJ/mole1000x = 4 \times \varepsilon_{C - H}$
$${1000y = 1 \times \varepsilon_{C - C} + 6 \times \varepsilon_{C - H} }{\varepsilon_{C - C} = \left\lbrack y - \frac{3x}{2} \right\rbrack \times 1000 = \frac{hc}{\lambda}.N_{A} }$$(' $\lambda$ ') wavelength of photon $= \frac{hcN_{A}}{\lbrack 4y - 6x\rbrack \times 250}$
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