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The standard enthalpy of formation of propene, CH3 -CH = CH2 is -13.2 kJ mol-1.

Enthalpy of sublimation (atomisation) of graphite is 716 kJmol-1.

Bond enthalpy H-H C-H C-C
ΔH0kJ mol-1 436.4 414 350

Calculate bond enthalpy of C = C.

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Best answer

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

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