Question:
For the reaction $\mathrm{C}_{2} \mathrm{H}_{6} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{H}_{2}$ the reaction enthalpy $\Delta_{\mathrm{r}} \mathrm{H}=$ $\mathrm{kJmol}^{-1}$ (Round off to the Nearest Integer). [Given : Bond enthalpies in $\mathrm{kJ} \mathrm{mol}^{-1}: \mathrm{C}-\mathrm{C}$ : $347, \mathrm{C}=\mathrm{C}: 611 ; \mathrm{C}-\mathrm{H}: 414, \mathrm{H}-\mathrm{H}: 436]$
Solution:
(128)
$\Delta_{\mathrm{r}} \mathrm{H}=\left[\epsilon_{\mathrm{C}-\mathrm{C}}+2 \epsilon_{\mathrm{C}-\mathrm{H}}\right]-\left[\epsilon_{\mathrm{C}=\mathrm{C}}+\epsilon_{\mathrm{H}-\mathrm{H}}\right]$
$=[347+2 \times 414]-[611+436]$
= 128