Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Heat of Combustion, Enthalpy

  1. Sep 29, 2015 #1
    Dear PF Forum,
    I have read this link in Wiki.
    But I don't understand what it means. Can someone help me?

    A: Heat of combustion of CH4 is 50.09 MJ/kg
    B: Enthalpy of water: -285.88 KJ/mol
    C: Enthalpy of CO2: -393.509 KJ/mol

    Then, I'll try to do a simple calculation.

    A: CH4 heat of combustion is 50.09 MJ/kg.
    What does it mean?
    Does it mean that if we burn 1 kg of CH4 completely
    CH4 + 2O2 -> CO2 + 2H2O
    It will produce 50.09 mega joules?

    B: Enthalpy of water is 285.88 kj/mol
    What does it mean?
    If we combine 2 moles of Hydrogen and 1 moles of Oxygen, it will explode and gives 285.88 kilo joules?

    C: Then I did some calculation...
    The heat of combustion of
    1 moles CH4 + 2 moles O2 -> 1 moles CO2 + 2 moles H2O
    16 gr CH4 + 32 gr O2 -> 44 gr CO2 + 36 gr H2O will gives 50.09 MJ/kg * 16 gr = 801.44 KJ

    The enthalpy of
    1 moles H2O: -285.88 KJ
    1 moles CO2: -393.519 KJ

    Combining those two:
    2 moles H2O: -571.76 KJ/mol
    1 moles CO2: -393.519 KJ/mol
    = 965.269 KJ

    801.44 KJ ≠ 965.269 KJ

    Where did I go wrong?
    Or my understanding of the concept of heat combustion and enthalpy is wrong.
    Thanks for any answer.
  2. jcsd
  3. Sep 29, 2015 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Check the "phase" of water (liq/vap).
  4. Sep 29, 2015 #3
    Of course. Thanks. The numbers are close now.
  5. Sep 29, 2015 #4
    Liquid water vs water vapor accounts for only about half the difference. You also forgot to subtract the heat of formation of methane, which is -75 kJ/mole.

  6. Oct 3, 2015 #5

    James Pelezo

    User Avatar
    Gold Member

    You might also read about Hess's Law Equation. That is, if you haven't already done so. Heat of Rxn (or, Heat of Combustion in this case) = (Sum of Enthalpies of Formation of Products) - (Sum of Enthalpies of formation of Reactants) Google 'Enthalpy of Formation Tables' You'll get many 'energy of formation' values to apply.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook