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Write Brief Answer · Q38

Q.Calculate the enthalpy of hydrogenation of ethylene from the following data. Bond energies of C-H, C-C, C=C and H-H are 414, 347, 618 and 435 kJ mol−1^{-1}.

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Step 1. C2H4+H2→C2H6C_2H_4+H_2\rightarrow C_2H_6: bonds BROKEN (reactant side) are ethylene's 1 C=C and 4 C-H, plus H2_2's 1 H-H: 618+4(414)+435=618+1656+435=2709618+4(414)+435=618+1656+435=2709 kJ.

Step 2. Bonds FORMED (product side) are ethane's 1 C-C and 6 C-H: 347+6(414)=347+2484=2831347+6(414)=347+2484=2831 kJ. …

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