Calculate the enthalpy change of the following reaction
$H_{2} C = C H_{2} (g) + H_{2} (g) \to H_{3} C - C H_{3} (g)$
The bond energy of $C - H, C - C, C = C, H - H$ are 414, 347, 615 and 453 KJ $m o l^{- 1}$ respectively.

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Δ H = Σ bond energy of reactant - Σ bond energy of product to write bond energy use proper symbols E C = C or Δ C = C H = [ 1 ( C = C ) + 4 ( C - H ) + 1 ( H - H ) ] - [ 1 ( C - C ) + 6 ( C - H ) ] = -1 ( C - C ) - 2 ( C - H ) + 1 ( C = C ) + 1 ( H - H ) = - 347 - 2 ( 414 ) + 1 ( 615 ) + 1 ( 435 ) = - 125 kJ.

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