Question

$\Delta H _{f} C _{2} H _{4}=12.5\, kcal$
Heat of atomisation of $C =171\, kcal$
Bond energy of $H _{2}=104.3 \,kcal$
Bond energy $C - H =99.3\, kcal$
What is $C=C$ bond energy?

• A. 140.9 kcal
• B. 49 kcal
• C. 40 kcal
• D. 76 kcal

Answer 1

Answer: (A)

Given:
$2 C _{(\text {graphite })}+2 H _{2}( g ) \longrightarrow C _{2} H _{4}( g ) ; \Delta H _{ f } $
$\Delta H _{f}=$ Bond dissociation enthalpy of reactant $-$ Bond dissociation enthalpy of products
$=12.5=(171 \times 2)+2 \times 104.3-\left(4 \times 99.3+ BE _{ C = C }\right) $
$\therefore BE _{( C = C )}=140.9\, kcal$