Heat of atomisation of CH$_4$(g) and C$_2$H$_6$(g) are $x$ kJ/mol and $y$ kJ/mol respectively. Find the maximum wavelength of photon required to dissociate C–C bond in C$_2$H$_6$.
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Higher bond energy corresponds to shorter wavelength of photon required for bond dissociation.
To find the maximum wavelength of a photon required to dissociate the C–C bond in ethane (C$_2$H$_6$), we need to understand the following concepts and steps:
Heat of atomization is the energy required to dissociate a compound into its constituent atoms. For CH$_4$ (methane), the heat of atomization is given as \(x\) kJ/mol, which represents the energy to break four C–H bonds into C and H atoms.
Similarly, for C$_2$H$_6$ (ethane), the heat of atomization is \(y\) kJ/mol, indicating the energy required to break six C–H bonds and one C–C bond into individual C and H atoms.
We know that the heat of atomization for ethane (\(y\)) includes the energy for six C–H bonds and one C–C bond. The energy for one C-H bond can be calculated as \(x/4\). Therefore, six C–H bonds would require energy \(6(x/4) = \frac{3x}{2}\).
The dissociation energy specifically for the C–C bond can be calculated by subtracting the energy required for the C–H bonds from the total heat of atomization of ethane:
\[
\text{Energy for C–C bond} = y - \frac{3x}{2}
\]
The energy of a photon required to break the C–C bond is given by \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the photon.
Rearranging for the wavelength, we have:
\[
\lambda = \frac{hcN_A}{y - \frac{3x}{2}}
\]
Given the options, the correct formula for the wavelength is \(\frac{hcN_A}{500(2y-3x)}\), which implies that the dissociation energy should be expressed correctly in the form as per the given options.
Therefore, the maximum wavelength of a photon required to dissociate the C–C bond in ethane (C$_2$H$_6$) is given by the expression \(\frac{hcN_A}{500(2y-3x)}\).
