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Standard Heat of Formation Question

  1. Nov 11, 2005 #1
    Ok, on this one I have no idea where to start. I could do all the homework ones, but have no clue even where to start on this. To me it seems backwards, but who knows. Here is the question:
    Calculate the standard heat of formation of acetaldehyde, [tex]CH_{3}CHO[/tex] , given the following information:

    [tex]CH_{3}CHO(g)+5/2O_{2}(g)\rightarrow2H_{2}0(l)+2CO_{2}(g)\triangle H=-1194kJ[/tex]

    [tex]\triangle H{_f^\circ}~H_{2}O (g) [/tex] = -286 kJ/mol

    [tex]\triangle H{_f^\circ}~CO_{2} (g)[/tex] = -394 kJ/mol

    I'd show what I have so far, but honestly I have no idea what I am doing with this one.

    Thanks for the help.:biggrin:

    I'll probably have to end up redoing the LaTeX stuff because I can never seem to get it to work. It always says reload page in a moment in the preview.:confused:
    Last edited: Nov 11, 2005
  2. jcsd
  3. Nov 11, 2005 #2
    Oooh, wait a second.:!!)

    Is it possible to just calculate the heat of reaction using the below equation, rather than breaking it into seperate reactions?

    [tex] \triangle H{_f^\circ}=\sum~n~\triangle H{_f^\circ}(products) ~-~\sum~m~\triangle H{_f^\circ}(reactants) [/tex]

    If so, do you just subtract the reactants from the products?

    Or am I even more lost than I was when I first posted this?:confused:

    If it is possible that way, I come up with:

    [(2x-286)+(2x-394)] - [-1194+(5(0)] = -166kJ
    Last edited: Nov 11, 2005
  4. Nov 12, 2005 #3
    You have the equation/answer right, but make sure that the reaction was carried out in standard conditions. That reaction could be carried out in 300K and 200kPa for all we know. It's probably right anways. If it's not the case we have a whole can of worms to open here.

    By the way, how do you use the equation writer in the forums?
  5. Nov 12, 2005 #4


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    you're answer seems fine, assuming that's gaseous water in the equation.
  6. Nov 12, 2005 #5
    Hmm...maybe I did it wrong then. Because in the original equation it says the water is a liquid. I shouldn't be too hard, it is only from Chem 1210. I don't get any points for it so if it's wrong, I'll find out on Tuesday.

    It is most likely standard conditions, because there wasn't anything special listed in the question. Usually when it is something like that she lists it.

    As far as writing equations, if you go to the main chemistry section, or any of the other sections, there are "stickeys" that have tutorials on how to use it. In the title is should have something to do with LaTeX. Although it usually takes me a few edits until I have the equation right.:rofl:

    Thanks for the help, both of you. It is nice when you actually get a problem right, when it seems like a difficult one. I thought I had no idea what I was doing with it.
  7. Nov 15, 2005 #6
    standard condition means 298k and 1atm.are u sure it is 200kpa
  8. Nov 15, 2005 #7
    Never said it was. You're right Shravan, it's 298K and 1atm.
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