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

Homework Help: Per mol of what?

  1. Feb 8, 2007 #1
    1. The problem statement, all variables and given/known data
    The combustion of the fuel gas butane is given by:

    2(C4H10) +13(O2) --> 8(CO2) + 10(H2O) del(H)= –5760kJ/mol

    What mass of butane would be needed to produce 33440J of energy?



    2. Relevant equations



    3. The attempt at a solution
    First of all how do you interpret del(H)= –5760kJ/mol
    Per mol of what?

    I assume a mol is the complete reaction of
    [2(C4H10) +13(O2) --> 8(CO2) + 10(H2O)]

    33440J/(5760kJ/mol)=0.0058mol of the reaction

    So a total of 0.0058*2=0.0116 moles of butane?

    Which would be 0.012*(12.01*4 + 10.1)=0.675g of butane but the answers suggested 0.337g which is a half of what I worked out. Hence they didn’t multiply by 2. This confusion comes back to my first question. ‘Per mol of what?’ In del(H)= –5760kJ/mol
     
  2. jcsd
  3. Feb 9, 2007 #2
    No. of moles is the stoichiometric coeff. in a balanced eqn.

    oh BTW, how is the eqn represented in the text...i mean like
    2(C4H10) +13(O2) --> 8(CO2) + 10(H2O) + 5760 J
    or
    is the energy evolved given separately like 5760 J/mol
    In case 1, 2 mol give 5760 J while in case 2, 1 mol gives 5760 J
     
    Last edited: Feb 9, 2007
  4. Feb 9, 2007 #3

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If it is specified as [itex]\Delta H_c[/itex] (enthalpy of combustion), then it is per mole of the hydrocarbon being combusted. If it is specified as [itex]\Delta H_r[/itex] (enthalpy of reaction), then it would not come in units of kJ/mol, and would simply be in units of kJ. This would indicate the question is talking about an enthalpy of combustion, so it's for 1 mole of butane.
     
  5. Feb 10, 2007 #4
    That's strange, in another part of the book, it stated the equation
    2(C2H6) + 7(O2) --> 4(CO2) + 6(H2O) del(H)=-2856kJ/mol

    and explained, "This equation means that if 2 moles of C2H6 reacts with 7 moles of O2, then 2856 kJ of heat energy will be released." But the answers worked out didn't go by this convention.
     
  6. Feb 10, 2007 #5

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is definitely WRONG. Which text is this?

    Did the given answers use the correct definition?
    http://www.answers.com/topic/standard-enthalpy-change-of-combustion
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook