The enthalpy of atomisation of $\mathrm{CH}_{4}$ and $\mathrm{C}_{2} \mathrm{H}_{6}$ are 360 and $620…

The enthalpy of atomisation of $\mathrm{CH}_{4}$ and $\mathrm{C}_{2} \mathrm{H}_{6}$ are 360 and $620 \mathrm{kcal} \mathrm{mol}^{-1}$ respectively. The $\mathrm{C}-\mathrm{C}$ bond energy is expected to be
  1. $210 \mathrm{kcal} \mathrm{mol}^{-1}$
  2. $80 \mathrm{kcal} \mathrm{mol}^{-1}$
  3. $130 \mathrm{kcal} \mathrm{mol}^{-1}$
  4. $180 \mathrm{kcal} \mathrm{mol}^{-1}$

Solution

Atomisation of methane
$\mathrm{CH}_{4}(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{H}(\mathrm{g})$
$\Delta \mathrm{H}=360 \mathrm{kcal}$
$\therefore \mathrm{C}-\mathrm{H}$ bond energy $=\frac{360}{4}$
$=90 \mathrm{kcal} \mathrm{mol}^{-1}$
$\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) \longrightarrow 2 \mathrm{C}(\mathrm{g})+6 \mathrm{H}(\mathrm{g})$
$\Delta \mathrm{H}=620 \mathrm{kcal}$
or $\mathrm{H}_{\mathrm{C}-\mathrm{C}}+6 \mathrm{H}_{\mathrm{C}-\mathrm{H}}=620$
$\therefore \quad \mathrm{H}_{\mathrm{C}-\mathrm{C}}=620-6 \mathrm{H}_{\mathrm{C}-\mathrm{H}}$
$=620-6 \times 90=80 \mathrm{kcal} \mathrm{mol}^{-1}$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

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