Given :
Bond enthalpy |
H-H |
C-H |
C-C |
ΔH0kJ mol-1 |
436.4 |
414 |
350 |
Bond enthalpy of C = C = ΔH0C = C = ?
For the formation of propene,
(CH3 – CH = CH2),
13.2 = [3 × 716 + 3 × 436.4] – [6 × 414 + 350 + \(ΔH^0_{C=C}\)]
= [2148 + 1309.2] – [2484 + 350 +\(ΔH^0_{C=C}\)]
= [3457.2] – [2834 + \(ΔH^0_{C=C}\)]
= 3457.2 – 2834 – \(ΔH^0_{C=C}\)
\(ΔH^0_{C=C}\) = 3457.2 – 2834 – 13.2
= 610 kJmol-1
∴ Bond enthalpy of C = C = \(ΔH^0_{C=C}\)
= 610 kJ mol-1