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Hess' Law

  1. Dec 10, 2004 #1
    Having some issues with this.

    "What is the standard enthalpy of combustion of C2H6 in kj/mol?
    H2(g) + 1/2O2(g) --> H2O(l) H0= –286 kJ
    C2H4(g) + H2(g) --> C2H6(g) H0= –137 kJ
    C2H4(g) + 3O2(g) --> 2CO2(g) + 2H2O(l) H0= –1412 kJ"

    Now, the answer is presumedly -31561 kj. But I'm getting no where a number that close.

    Since C2H6 is combusted, I want the end-result to be:
    (2)C2H6 + (7)O2 --> (6)H2O + (4)CO2 (right?)
    So I would turn around the 2nd one to:
    C2H6(g) --> C2H4(g) + H2(g)
    and then:
    2 * [C2H6(g) --> C2H4(g) + H2(g) ]

    2 * [H2(g) + (1/2)O2 --> H2O(l)]

    2 * [C2H4(g) + 3O2(g) --> 2CO2(g) + 2H2O(l)]
    Which gets me 3122 kJ, or 1561kj/mol.
    Did I do something wrong? Miss a step? Where could the other 30,000 kj come from?
     
    Last edited: Dec 10, 2004
  2. jcsd
  3. Dec 10, 2004 #2

    Gokul43201

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    Your answer (-1561 kJ/mol) seems to be right. There must be a typo in the provided answer.

    Why are there question marks (?) at the end of each reaction in your question ?
     
  4. Dec 11, 2004 #3

    chem_tr

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    I am not sure, but please look carefully the given data; they say kj, but want kj over moles. So you may have to multiply the findings with Avogadro's number to obtain how much enthalpy is needed for one mole.

    Secondly, the numbers -31561 and -1561 are very close, except the 3 in thi first one; as Gokul said, a typo is probably present here.
     
  5. Dec 11, 2004 #4

    Gokul43201

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    I did this independently and got the same answer (H = -1561 kJ/mol), but I have one minor concern. Typically, the heat of combustion is specified for any hydrocarbon with the products being CO2 and H2O(g). You have H2O(l) in your equation, so I looked up the molar enthalphy of vaporization of water , but 3 moles of water certainly does not contribute 30,000 kJ, so that's not it.

    Tooren, you've done everything correctly - don't worry.
     
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