The standard enthalpy of atomization of ethane according to the equation $\mathrm{C}_2…

The standard enthalpy of atomization of ethane according to the equation $\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) \rightarrow 2 \mathrm{C}(\mathrm{g})+6 \mathrm{H}(\mathrm{g})$ is $622 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If standard mean $\mathrm{C}-\mathrm{H}$ bond dissociation enthalpy is $90 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$, the standard mean dissociation enthalpy of $\mathrm{C}-\mathrm{C}$ bond (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is
  1. 540
  2. 90
  3. 85
  4. 82

Solution

$\begin{aligned} & \Delta_{\mathrm{a}} \mathrm{H}^{\circ}=\Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\circ}+\Delta_{\mathrm{C}-\mathrm{C}} \mathrm{H}^{\circ} \\ & 622=[6 \times 90]+\left[\Delta_{\mathrm{C}-\mathrm{C}} \mathrm{H}^{\circ}\right]=540+\Delta_{\mathrm{C}-\mathrm{C}} \dot{\mathrm{H}}^{\circ} \\ & \Rightarrow \Delta_{\mathrm{C}-\mathrm{C}} \mathrm{H}^{\circ}=622-540=82 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$

Asked in: AP EAMCET 2023 (19 May Shift 1)

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