The enthalpy change for the following process at \( 25 \mathrm{C} \) and under constant pressure...
The enthalpy change for the following process at \( 25 \mathrm{C} \) and under constant pressure at 1 atm are as follows:
\[
\begin{array}{ll}
\mathrm{CH}_{4}(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{H}(\mathrm{g}) & \Delta_{\mathrm{r}} \mathrm{H}=396 \mathrm{kcal} / \mathrm{mole} \\
\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) \longrightarrow 2 \mathrm{C}(\mathrm{g})+6 \mathrm{H}(\mathrm{g}) & \Delta_{\mathrm{r}} \mathrm{H}=676 \mathrm{kcal} / \mathrm{mole}
\end{array}
\]
Calculate \( \mathrm{C}-\mathrm{C} \) bond energy in \( \mathrm{C}_{2} \mathrm{H}_{6} \) \& heat of formation of \( \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) \)
Given : \( \Delta_{\text {sub }} \mathrm{C}(\mathrm{s})=171.8 \mathrm{kcal} / \mathrm{mole} \)
B.E. \( (\mathrm{H}-\mathrm{H})=104.1 \mathrm{kcal} / \mathrm{mole} \)
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