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ΔH for Gas Phase Reaction

  1. Feb 24, 2007 #1
    Hi everyone, first time poster here. I've lurked on the site in the past, and really appreciate the wealth of resources that are provided here, especially the math section! Anyway, I have a question that I'm sure is very elementary but for some reason I have failed to grasp and therefore has produced bogus results. I'm trying to calculate the ΔH of a gas reaction that is as follows:

    2CH2=CHCH3 + 2NH3 + 3O2 → 2CH2=CHCN + 6H2O

    So what I did was sum up [moles of bonds * bond enthalpy for bonds broken] and subtract [moles of bonds * bond enthalpy for bonds formed] from that. My values looked like this:

    [(2 mols C=C * 598) + (4 mols C-H * 416) + (2 mols C-H * 416) + (2 mols C-C * 356) + (6 mols C-H * 416) + (6 mols N-H * 391) + (3 mols O=O * 498)] - [(4 mols C-H) + (2 mols C=C * 598) + (2 mols C-H * 416) + (2 mols C-C * 356) + (2 mols C-N * 285) + (12 mols H-O * 467)] = 162 kJ

    Just looking at how messy that came out, I knew before I began calculating that my method of approach must have been flawed. So, can anyone point out how to calculate deltaH? My text gives only the most elementary of examples, like CH4 + 2O2 ---> CO2 + 2H2O, so that hasn't helped me with a more advanced problem like this one.

    Thanks in advance to anyone who could give me some advice.:smile:
    Last edited: Feb 24, 2007
  2. jcsd
  3. Feb 24, 2007 #2
    I think that's right. The method of subtracting the sum of one side from the other was the way I was taught (this year) how to determine change in enthaply. There are other ways you can do it to, which involves adding up the enthaplies of formation, ionization, etc. on one side and subtracting it from another, but the average bond energy method is much simplier.

  4. Feb 24, 2007 #3
    Thanks for your response, scott.

    I suppose the method is correct, but I think the problem is how I'm determining the bonds and the moles of said bonds. My course has an online quiz which tells you if you got the question right or wrong when you enter your answer. I entered 162 and was told the answer was incorrect.
  5. Feb 24, 2007 #4
    Take out the redundancies to clear things up. There is no need given the mechanism to calculate C=C broken and C=C formed.
  6. Feb 24, 2007 #5
    Hmm, I see; but if they are redundant, they will just cancel each other out in the calculation anyway, no?
  7. Feb 24, 2007 #6
    I would draw out each molocuel and count up all of the bonds. It seems like it would be pretty easy to just miss one by mistake.

  8. Mar 4, 2010 #7
    In 2CH2=CHCN the CN is a triple bond. You need to use the enthalpy value for that triple bond and you'll get the right answer.
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