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Chemistry Enthalpy Question

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine the standard enthalpy change for the reaction below.

    4NH3 + 3O2 --> 2N2 + 6H2O

    Given:
    Standard heat of formation of NH3: -45.9 kJ/mol

    Standard heat of formation of H2O: -285.83 kJ/mol


    2. Relevant equations



    3. The attempt at a solution
    I had this question on a quiz and got points taken off, but I can't figure out why. Here's my solution:

    { (6 x -285.83 kJ/mol) + (2 x 0) } - { (3 x 0) + (4 x -45.9 kJ/mol)} = -1531 kJ/mol
     
  2. jcsd
  3. Mar 13, 2009 #2

    alxm

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    Because the standard heat of formation of gaseous water is -241.826 kJ/mol.
    I'm not sure where you got the -285.83 from, but I believe it's the liquid value.
     
  4. Mar 13, 2009 #3

    Borek

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    Staff: Mentor

    Reaction (as entered) doesn't state whether it should be gaseous or liquid water. My thermo is somewhat rusty, but isn't standard state of water liquid?
     
  5. Mar 13, 2009 #4

    alxm

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    It is. But it wouldn't be wrong then! :)

    I assumed the gas-phase states since, well, it's a gas-phase reaction.
    If the problem was stated without specifying H2O (l) or H2O (g) then either would be correct, depending on whether you mean the 'standard enthalpy' in the strict sense, or the 'reaction enthalpy at standard pressure and temperature'.
     
  6. Mar 13, 2009 #5

    Borek

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    There is always a slight possibility that there was a mistake in the test :smile:
     
  7. Mar 15, 2009 #6
    Just to clarify, the reaction was wriiten without states in the problem, and there were no states specified in the heat of formation given either.

    Was I supposed to divide the value I got by four?
     
  8. Mar 15, 2009 #7

    Borek

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    That would be enthalpy change for a mole of ammonia, not for the reaction as written. I am not sure if there exeist some precise convention, I have seen it discussed (without any decisive conclusions) by chemistry teachers.
     
  9. Mar 15, 2009 #8
    Yeah, I always thought you didn't divide by anything when it was for an actual reaction. It wouldn't make sense here because it doesn't specify per mole of ammonia vs. per mole of oxygen.

    Anyways, thanks for your help.
     
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