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Enthalpy of Combustion - Alkanes

  1. Oct 20, 2007 #1
    Enthalpy of Combustion -- Alkanes

    Write a balanced equation for the combustion of CH4(g) (methane)
    (its reaction with O2(g) forming the products CO2(g) and H2O(l)).

    Given the following standard heats of formation:
    [tex]\Delta[/tex]Hf° of CO2(g) is -393.5 kJ/mol
    [tex]\Delta[/tex]Hf° of H2O(l) is -286 kJ/mol
    [tex]\Delta[/tex]Hf° of CH4(g) is -74.8 kJ/mol

    Calculate the difference, [tex]\Delta[/tex]H-[tex]\Delta[/tex]E=[tex]\Delta[/tex](PV) for the combustion reaction of 1 mole of methane.
    (Assume standard state conditions and 298 K for all reactants and products.)


    -----------------------------------------------------------------------------------
    The ideal gas law must be used.
    The incremental change in volume due to liquids is neglected (cf the volume of 1 mole of H2O(l) is 18 mL, whereas the volume of 1 mol of a gas at 298 K is about 24,450 mL.
    Be careful to account for the total change in number of moles of gas molecules (be sure the signs and units are correct too).
    Note that certain alkanes are gases in the standard state.

    -----------------------------------------------------------------------------------


    My balanced equation was CH4 (g) + 2O2 (g) ---> CO2 (g) + 2H2O (l)

    it says use PV=nRT but i was confused about the change in moles like is it just the change in moles of the GASES so molesfinal - molesininitial = 1 - 3 = -2 ????
    and then
    (-2)(0.08206)(298) = the difference....???

    but i think thats wrong!
     
    Last edited: Oct 20, 2007
  2. jcsd
  3. Oct 20, 2007 #2

    Astronuc

    User Avatar

    Staff: Mentor

    The number of moles doesn't change - for each mole of CH4 and two moles of O2, one mole of CO2 and 2 moles of H2O are produced. If the reactants start at STP, the products would certainly not, but the problem states
    but, think about the fact that one is given H2O (l) as opposed to H2O (g), which would occupy a greater volume. Compare the volume of liquid water to vapor.
     
  4. Oct 20, 2007 #3
    huh? sorry i dont really understand what i need to compare..

    'The incremental change in volume due to liquids is neglected (cf the volume of 1 mole of H2O(l) is 18 mL, whereas the volume of 1 mol of a gas at 298 K is about 24,450 mL. "

    i thought volume didnt matter since it was neglected :S
     
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