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Is this correct Hess' law, thermodynamics

  1. Feb 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate ∆rH° for the reaction
    c2h50h+3o2-->2co2+3h20
    Given that ∆rU° = - 1373 kJ mol-1 at 298K.

    2. Relevant equations
    c2h50h+3o2-->2co2+3h20

    3. The attempt at a solution
    delta H f CO2(g) = -393.5 kJ/mole
    delta H f H2O(l) = -241.8 kJ
    delta H f O2(g) = 0 these values are from my textbook


    delta H reaction = ((2 moles CO2)(-393.5 kJ/mole) + (3 moles H2O)(-241.8kJ/mole)) - ((1 mole C2H5OH)(delta H f C2H5OH(l))

    -1373 kJ = (-787.0 kJ - 725.4 kJ) - (delta H f C2H5OH(l)
    -139.4 kJ = delta H f C2H5OH(l)
     
  2. jcsd
  3. Feb 29, 2016 #2
    You are supposed to do this problem without using heats of formation. Any thoughts on how?
     
  4. Feb 29, 2016 #3
    no clue.
     
  5. Feb 29, 2016 #4
    What is the definition of ΔH in terms of ΔU?
     
  6. Feb 29, 2016 #5
    i do know DH=DU+PDV
     
  7. Feb 29, 2016 #6
    ΔH=ΔU+Δ(PV)
    At constant temperature, for ideal gases, Δ(PV)=RTΔn.

    What is the change in the number of moles of gases Δn between reactants and products for you chemical reaction? If you neglect the volume of the liquids relative to the gases, what is the change in the PV in going from your reactants to going to your products?
     
  8. Feb 29, 2016 #7
    WAIT ITS A COMBUSTION REACTION
     
  9. Feb 29, 2016 #8
    Well 1 mole of c2h5OH and 3 molews of 02 , 2 Moles of co2 and 3moles of h20 so the change is 1 mole 5- 4 moles
     
  10. Feb 29, 2016 #9
    I asked for only the change in the number of moles of gases. The changes in PV resulting from the changes in the liquids is negligible.
     
  11. Feb 29, 2016 #10
    Change will be 2. so the relationship between enthalpy and internal energy is DH=DU=DNgasRT
     
  12. Feb 29, 2016 #11
    What if I told you that I get a change in the number of moles of gas as -1, not 2?
     
  13. Feb 29, 2016 #12
    how? you didn't cater for h2o as a gas?
     
  14. Feb 29, 2016 #13
    because water isnt gas in its standard state ?
     
  15. Feb 29, 2016 #14
    I see (l)'s next to your H2O's for the reaction. Apparently, they are saying that the product water is in the liquid state.
     
  16. Feb 29, 2016 #15
    my mistake water is supposed to be a gas.
    so dh= -1373* 2 mol*8.314*298 k
     
  17. Feb 29, 2016 #16
    i got -3582.14 KJ thank you for your assistance. (-1373+ (2 MOLES * 8.314* 298))
     
  18. Feb 29, 2016 #17
    Not correct. Watch out for the units on that PV term.
     
  19. Feb 29, 2016 #18
    OH IT WILL BE JOULES EVERYTHING ELSE CANCELS ?
     
  20. Feb 29, 2016 #19
    But the Delta U is in kJ. The units of the two terms have to be consistent.

    Also, as a check, how does your answer compare with what you get using heats of formation? (I know you must have thought of doing this check)
     
  21. Feb 29, 2016 #20
    I did compare the enthalpy of formation against the enthalpy of reaction . so the unit is Joule per kelvin?
     
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