Bond Enthalpy Question - 2

QuestionEstimate the enthalpy change for the hydrogenation of ethene:C₂H₄(g) + H₂(g) → C₂H₆(g)Mean bond enthalpies / kJ mol^-1: C=C 612, C-C 347, C-H 413, H-H 436Method 1: break everything.Bonds broken: C₂H₄ has 1 C=C and 4 C-H; H₂ has 1 H-H.
612 + 4(413) + 436 = 612 + 1652 + 436 = 2700 kJBonds made: C₂H₆ has 1 C-C and 6 C-H.
347 + 6(413) = 347 + 2478 = 2825 kJΔH = 2700 - 2825 = -125 kJ mol^-1Method 2: only the bonds that change.Four C-H bonds are present on both sides, so they cancel. What really happens is that the C=C becomes a C-C and two new C-H bonds form.Broken: C=C + H-H = 612 + 436 = 1048 kJ
Made: C-C + 2 C-H = 347 + 826 = 1173 kJΔH = 1048 - 1173 = -125 kJ mol^-1Both methods give the same answer. Method 2 is quicker and gives fewer chances for arithmetic errors, but you must be sure which bonds really change.The value found from enthalpy changes of combustion is -137 kJ mol^-1. The difference arises because mean bond enthalpies are averages, not values for these exact molecules.