Calculate enthalpy change for following reaction. \(\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{H}_3 \mathrm{C}-\mathrm{CH}_{3(\mathrm{~g})}\) [The bond energy of \(\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}\) and \(\mathrm{H}-\mathrm{H}\) is \(414,347,615\) and 435 kJ respectively]

For the reaction,
\(\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{H}_3 \mathrm{C}-\mathrm{CH}_{3(\mathrm{~g})} \)
\( \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=\sum \Delta \mathrm{H}_{\text {reactant }}^{\circ}-\sum \Delta \mathrm{H}_{\text {product }}^{\circ} \)
\( =\left[4 \Delta \mathrm{H}_{(\mathrm{C}-\mathrm{H})}^{\circ}+\Delta \mathrm{H}_{(\mathrm{C=C})}^{\circ}+\Delta \mathrm{H}_{(\mathrm{H}-\mathrm{H})}^{\circ}\right] \)
\( -\left[6 \Delta \mathrm{H}_{(\mathrm{C}-\mathrm{H})}^{\circ}+\Delta \mathrm{H}_{(\mathrm{C}-\mathrm{C})}^{\circ}\right] \)
\( =[(4 \times 414 \mathrm{~kJ})+615 \mathrm{~kJ}+435 \mathrm{~kJ}]-\)\([(6 \times 414 \mathrm{~kJ}) \) \( +347 \mathrm{~kJ}] \)
\( =(1656 \mathrm{~kJ}+615 \mathrm{~kJ}+435 \mathrm{~kJ})-\)\((2484 \mathrm{~kJ}+347 \mathrm{~kJ}) \)
\( =2706 \mathrm{~kJ}-2831 \mathrm{~kJ}=-125 \mathrm{~kJ}\)