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Bond Energies

  1. Sep 21, 2012 #1
    Calculate the standard enthalpy of ethyne

    C2H2 + 2H2 -------> C2H6

    Assume the following mean bond energies

    C[itex]\equiv[/itex]C = 813 KJ mole-1

    C - C = 364 KJ mole-1

    C - H = 413 KJ mole -1

    H - H = 436 KJ mole -1


    Attempted Solution:

    Bonds broken
    1 C[itex]\equiv[/itex]C = 813 KJ mole-1 ( 813 KJ mole-1)
    2 H - H = 436 KJ mole -1 (2 x 436 KJ mole -1)

    Bonds formed
    4 C - H = 413 KJ mole -1 (4x 413 KJ mole-1)
    1 C - C = 364 KJ mole-1 (364 KJ mole-1)

    ΔHh - Enthalpy of hydrogenation

    ΔHh = ((813 + 2x 436) - ((4x 413) + 364)) KJ mole -1
    ΔHh = 1685 KJ mole-1 - 2016 KJ mole-1
    ΔHh = -331 KJ mole -1

    --
    Is my method and the way I worked it out correct ?
     
  2. jcsd
  3. Sep 21, 2012 #2

    chemisttree

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    Looks good. You might check that by subtracting the two enthalpies of formation for ethyne and ethane. The difference should be about the same as your answer.
     
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