The heat of atomisation of methane and ethane | vedclass

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(B) For methane: $CH_{4(g)} \rightarrow C_{(g)} + 4H_{(g)}$; $\Delta_{r}H = x \ kJ \ mol^{-1}$.Since there are $4$ $C-H$ bonds,the energy of one $C-H$

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$\frac{N_{A}hc}{250(4y-6x)}$

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(B) For methane: $CH_{4(g)} ightarrow C_{(g)} + 4H_{(g)}$; $\Delta_{r}H = x \ kJ \ mol^{-1}$.Since there are $4$ $C-H$ bonds,the energy of one $C-H$ bond is $\varepsilon_{C-H} = \frac{1000x}{4} \ J \ mol^{-1}$.For ethane: $C_2H_{6(g)} ightarrow 2C_{(g)} + 6H_{(g)}$; $\Delta_{r}H = y \ kJ \ mol^{-1}$.Ethane contains $1$ $C-C$ bond and $6$ $C-H$ bonds,so $1000y = \varepsilon_{C-C} + 6 	imes \varepsilon_{C-H}$.Substituting $\varepsilon_{C-H} = \frac{1000x}{4}$,we get $1000y = \varepsilon_{C-C} + 6 	imes (\frac{1000x}{4}) = \varepsilon_{C-C} + 1500x$.Thus,$\varepsilon_{C-C} = 1000y - 1500x = 1000(y - 1.5x) = 1000(\frac{2y-3x}{2}) = 500(2y-3x) = 250(4y-6x) \ J \ mol^{-1}$.The energy required to break one $C-C$ bond is $E = \frac{\varepsilon_{C-C}}{N_{A}} = \frac{250(4y-6x)}{N_{A}}$.Since $E = \frac{hc}{\lambda}$,the longest wavelength is $\lambda = \frac{hc}{E} = \frac{hc \cdot N_{A}}{250(4y-6x)}$.

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 :

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