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Homework Help: Is this right? chem heat of combustion

  1. Jun 18, 2005 #1
    Hi guys can you please check if this is right....thanks in advance

    The heat released by the reaction 2C + 2H2 = CH4 is 74.8kJ/mol. Using this information and a table of standard enthalpy of formation values, determine the heat of combustion for the burning of methane to form CO2 and H2O(g).
    C02 = -393.3 kJ/mol
    H20 = -286.2 kJ/mol

    2C + 2H2 = CH4 H1 = 74.8
    C + O2 = CO2 H 2= -393.3

    H2 +1/2 O2 = H2O H3 = -286..2

    CH4 +3O2 = CO2 + 2H2O -H1 + H2 +2H3 = -1040.5 Kj/mol
  2. jcsd
  3. Jun 18, 2005 #2
    can anyone help me out?
  4. Jun 18, 2005 #3
    I may be incorrect but I got -754.3 kJ mol-1.

    The formation of CH4 is 74.8 kJ mol-1.
    The formation of CO2 and H2O is -679.5 kJ mol-1.
    The formation of O is not counted due to it not changing.

    This means that -74.8 kJ mol-1 + (-679.5 kJ mol-1) = -754.3 kJ mol-1.

    The Bob (2004 ©)
  5. Jun 19, 2005 #4


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    Bob, you are wrong.
    joejo, the H1 is negative -74.8 kJ/mol.
    It should be -890.9 kJ.
    Viet Dao,
  6. Jun 19, 2005 #5
    It was. See???

    The Bob (2004 ©)
  7. Jun 19, 2005 #6
    Now I have had another look, I am going to agree with joejo. Why? because he is right. :biggrin: It is -1040.5 kJ mol-1.


    The Bob (2004 ©)

    Just as a small hate of mine, can you please use the correct capitals?? Cheers. :biggrin:
    Last edited: Jun 19, 2005
  8. Jun 19, 2005 #7


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    . That means [itex]\Delta H_1 = -74.8 \mbox{ kJ/mol}[/itex].
    So [itex]-\Delta H_1 +\Delta H_2 + 2 \Delta H_3 = -890.9 \mbox{ kJ}[/itex].
    Am I missing something?
    Viet Dao,
    Last edited: Jun 19, 2005
  9. Jun 19, 2005 #8
    Although this did not affect what I thought or wrote, it normally gets people's defenses up and does not allow them to see errors. May I ask that you address the issue in a lighter-hearted way???

    Nothing is missed. You are correct. I did the missing. However, both of us have done something wrong. joejo has (perhaps) lost some experience here because we both told him what had to be done. Although I was wrong, as was joejo, our methods were correct. What joejo really needed was some aimed help, not the answer. Of this we are both guilty and here is a point where we must learn to help (not tell the answer and method), not to argue and to have a bit of fun along the way.

    I will take responsibilty if joejo is confused (not insult intended) due to my incorrect posts.

    The Bob (2004 ©)
  10. Jun 20, 2005 #9


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    I just want my posts to be neat, short, but also contain enough information. So others can finish it in a glance. :biggrin: . Anyway, I'll take your advice. Thanks :smile:
    Viet Dao,
  11. Jun 20, 2005 #10
    Exothermic reactions should always be negative enthalpy.
    Last edited: Jun 20, 2005
  12. Jun 20, 2005 #11
    Absolutely correct. So what is a reaction with a positive enthaply?

    The Bob (2004 ©)
  13. Jun 20, 2005 #12
  14. Jun 21, 2005 #13
    Another correct answer. :smile:

    The Bob (2004 ©)
  15. Jun 22, 2005 #14
    i think its still wrong....

    shouldn't it be.... -802.3kJ
  16. Jun 22, 2005 #15

    The Bob (2004 ©)
  17. Jun 22, 2005 #16
    By definition the burning of methane to make CO2 and H2O is...

    CH4 + 2 O2 --> CO2 + 2 H2O

    The heat for this reaction is 2(-286.3) + (-393.3) -(-74.8) - 0 = -891.1 kJ.

  18. Jun 23, 2005 #17
    can anyone help me out?
  19. Jun 23, 2005 #18
    Your answer keeps on changing.

    Which is it?

    The Bob (2004 ©)
  20. Jun 23, 2005 #19
    the bob....read above....i'll post it below just in case u never saw it.

    By definition the burning of methane to make CO2 and H2O is...

    CH4 + 2 O2 --> CO2 + 2 H2O

    The heat for this reaction is 2(-286.3) + (-393.3) -(-74.8) - 0 = -891.1 kJ.

  21. Jun 23, 2005 #20
    I did not miss it. I saw it. I was commenting on this:

    These ARE two different answers, both posted by you.

    The next problem is also by you. Your original post states:
    yet you said in your last post:
    You have got a difference for the formation of water. You said it was 286.2 and now you say it is 286.3. Please make up your mind.

    If the value is 286.2 then me and VietDao29 are right. If it is 286.3 then you are right.

    This is why I asked. Ok?????

    The Bob (2004 ©)

    P.S. May I ask you not to be sarcastic in future. It makes me, if no one else, very on edge and will explain all of the boldness in this post. I was annoyed due to your sarcasm.
  22. Jun 23, 2005 #21
    i wasn't being sarcastic...sorry for the misunderstanding...so was my second answer right...
  23. Jun 23, 2005 #22
    In which case I apologise. I did misunderstand.

    Your answer is correct if the Enthalpy change of Formation for water is -286.3 kJ mol-1. If it is -286.2 kJ mol-1 then VietDao29 and me are correct.

    The Bob (2004 ©)

    P.S. Apologise again.
    Last edited: Jun 23, 2005
  24. Jun 23, 2005 #23
    thank you...just to verify...my second answer (-891.1 kJ.) right?? sorry im getting confused and I can imagine you are too! thanks again
  25. Jun 23, 2005 #24
    Yes, if the enthaply change of formation of water is -286.3 kJ mol-1.

    The Bob (2004 ©)
  26. Jun 23, 2005 #25
    thanks again!! :)
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