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Heat of formation?

  1. May 6, 2008 #1
    1. The problem statement, all variables and given/known data

    This is a question from a sample exam that I can't figure out. The answer is 1297 kJ/mol, but I don't know how to solve it. Any help is appreciated.


    The following are average bond energies (kJ/mol):
    C−H 413 O−H 463 C=C 614
    C−C 348 O−O 146 C=O 799
    C−O 358 O2 495

    What is the amount of heat released in the complete combustion of ethylene, H2C=CH2?

    2. Relevant equations

    H(reaction) = H(products) - H(reactants)

    3. The attempt at a solution

    I've written the balanced equation...

    C2H4 + 3O2 -> 2H2O + 2CO2

    Now, I think I need to break down the reaction and add all of the energies from broken bonds (reactants) together, and then subtract that sum from the energy released from bonds made (products).

    Hf of reactants :
    C2H4 -> one C=C bond, four C-H bonds = (614 + 4*413) = 2266 kJ/mol * 1mol = 2266 kJ
    O2 -> this is an element in its normal state, so the enthalpy is zero

    Hf of products :
    H2O -> two O-H bonds = (2*463) = 926 kJ/mol * 2mol = 1852 kJ
    CO2 -> two C=O bonds = (2*799) = 1598 kJ/mol * 2mol = 3196 kJ

    I continued like this, but the answer I got was wrong. Am I completely off base? I have a feeling I'm misunderstanding the question and approaching it in the wrong way.
  2. jcsd
  3. May 7, 2008 #2
    This is the raw equation I got [4(413)+614+3(495)]-[4(463)+4(799)]=-1297 kJ/mol

    I'm pretty sure its going to be negative because its Change in H formation=D(reactant bonds)-D(product bonds).

    "Hf of products :
    H2O -> two O-H bonds = (2*463) = 926 kJ/mol * 2mol = 1852 kJ
    CO2 -> two C=O bonds = (2*799) = 1598 kJ/mol * 2mol = 3196 kJ"

    In your original equation there are two O-H bonds "per H2O molecule." But you have 2 H2O molecules, so you'll have 4 O-H bonds. Also, drawing Lewis diagrams help.

    C2H4 + 3O2 -> 2H2O + 2CO2
    So [(4(C−H)+C=C+3(O=O))-(4(O−H)+4(C=O))]
  4. May 7, 2008 #3
    I multiplied by 2 mol...so I think all of my numbers are the same as yours, I just never used the value for O2 (a huge duh on my part). I guess I thought that the "zero enthalpy" applied because O2 (g) is oxygen's natural state at room temperature.

    This was a HUGE help, this problem has been eating up my time all night, and I'm so glad to finally understand where I was going wrong! Thanks!!
  5. May 9, 2008 #4
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