Tuesday, October 18, 2011

Blog 10 Due Monday Oct 24 10:00 pm

Ethanol, C2H5OH, has been proposed an an alternate fuel source. Calculate the standard of enthalpy of combustion per gram of liquid ethanol.
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14 comments:

  1. NayiriOctober 23, 2011 at 5:06 PM

    C2H5OH + 3O2 = 2CO2 + 3H20
    heat of formation C2H5OH= -277 kJ/mol
    heat of formation C02= -394 kJ/mol
    heat of formation 02= 0
    heat of formation H20= -286 kJ/mol

    enthalpy= 2(C02)+ 3(H20) - (C2H50H)
    = -788 -858 +277
    = -1369 kJ/mol

    MM= 46 g/mol

    -1369/ 46= -29.76 kJ/g

    ReplyDelete
  2. AlecOctober 23, 2011 at 5:59 PM

    C2H5Oh + 3O2 = 2CO2 + 3H20
    C2H5OH= -277 kJ/mol
    C02= -394 kJ/mol
    02= 0 (element)
    H20= -286 kJ/mol

    E=2(c02) + 3(h20) - (c2h50h)
    E=-1369 KJ/mol

    molar mass of ethanol=46 g/mol (about)

    therefore.... -1396/46=-29.76 kJ/g

    ReplyDelete
  3. Hasmik.October 23, 2011 at 8:04 PM

    C2H5OH + 3O2 = 2CO2 + 3H20
    C2H5OH= -277 kJ/mol
    C02= -394 kJ/mol
    H20= -286 kJ/mol

    delta H = 2 x (-394 kJ/mol) + 3 x (-286 kJ/mol) - (-277 kJ/mol)
    delta H = -788 kJ/mol - 858 kJ/mol + 277 kJ/mol
    delta H = -1,369 kJ/mol

    Ethanol MM = 46.08 g/mol

    -1,369 kJ/ 1 mol x 1 mol / 46.08 g = -29.71 kJ/g

    ReplyDelete
  4. AnonymousOctober 23, 2011 at 8:32 PM

    C2H5OH + 3O2 ---> 2CO2 + 3H20
    C2H5OH= -277 kJ/mol
    C02= -394 kJ/mol
    H20= -286 kJ/mol
    (delta)H= 2(-394 kJ/mol) + 3(-286 kJ/mol) -(-277 kJ/mol)= -1369 kJ/mol
    MM for C2H5OH= 46.08 g/mol
    SO...
    (-1369 kJ C2H5OH/mol) x (1 mol C2H5OH/46.08 g C2H5OH)= -29.71 kJ/g C2H5OH

    ReplyDelete
  5. AnonymousOctober 23, 2011 at 8:58 PM

    C2H5OH (l) + 3 O2 (g)  2 CO2 (g) + 3 H2O (g)
    Heat of Formation of C2H5OH = -277 kJ/mol
    Heat of Formation of O2 = 0 kJ/mol
    Heat of Formation of CO2 = -394 kJ/mol
    Heat of Formation of H2O = -286 kJ/mol
    Enthalpy = (2*CO2) + (3*H2O) – (C2H5OH)
    Enthalpy = (2*-394 kJ/mol) + (3*-286 kJ/mol) – (-277 kJ/mol)
    Enthalpy = (-788 kJ/mol) – (858 kJ/mol) + (277 kJ/mol)
    Enthalpy = -1369 kJ/mol
    Molar Mass of C2H5OH = 46.08 g/mol
    (-1369 kJ/mol)(mol C2H5OH/46.08 g C2H5OH) = -29.71 kJ/g C2H5OH

    ReplyDelete
  6. AnnieOctober 23, 2011 at 10:14 PM

    C2H5OH + 3O2 = 2CO2 +3H2O
    heat of formation= sum of delta H products- sum of delta H reactants
    (2*CO2 +3*H2O)-(C2H5OH + 3*O2)
    ((2*-394) + (3*-286) – (-277 kJ/mol)= -1369 kJ/mol
    Molar Mass of C2H5OH = 46.08 g/mol
    -1369/46.06=-29.7 kJ/g C2H5OH

    ReplyDelete
  7. varak ghazarianOctober 23, 2011 at 10:22 PM

    C2H5OH + 3O2 => 2CO2 + 3H2O
    enthalpy of formation C2H5OH= -277 kJ/mol
    enthalpy of formation CO2= -394 kJ/mol
    enthalpy of formation O2= 0 kJ/mol
    enthalpy of formation H2O= -286 kJ/mol

    enthalpy= 2 x -394 kJ/mol + 3 x -286 kJ/mol - -277 kJ/mol= -1369 kJ/mol
    C2H5OH MM= 46.08 g/mol
    -1369 kJ C2H5OH/mol x 1 mol C2H5OH/46.08 g C2H5OH= -29.71 kJ/g of C2H5OH

    ReplyDelete
  8. Varak ArmoudikianOctober 23, 2011 at 11:50 PM

    C2H50H+302->2CO2+3H2O
    (2*-394 kj/mol)+(3*-286 kj/mol)-(-277 kj/mol)=
    -1369kj/mol
    MM C2H5OH=46.08 g/mol
    -1369kj/46.08g=-29.71 kj/g

    ReplyDelete
  9. Mr. POctober 24, 2011 at 2:35 PM

    Nice job everyone, you got it!! Good explanation and thank you for showing your work.

    ReplyDelete
  10. AnonymousOctober 24, 2011 at 10:24 PM

    C2H5OH + 3O2 = 2CO2 + 3H20
    C2H5OH= -277 kJ/mol
    C02= -394 kJ/mol
    02= 0
    H20= -286 kJ/mol

    enthalpy= 2(C02)+ 3(H20) - (C2H50H)
    = -1369 kJ/mol

    MM= 46 g/mol

    ReplyDelete
  11. Garni TerterianOctober 25, 2011 at 6:52 PM

    C2H5OH + 3O2 = 2CO2 +3H2O
    C2H5OH= -277 kJ/mol
    C02= -394 kJ/mol
    H20= -286 kJ/mol

    E =2 x (-394 kJ/mol) + 3 x (-286 kJ/mol) - (-277 kJ/mol)
    E= -1,369 kJ/mol

    MM of C2H5OH = 46.08 g/mol
    (-1369 kJ/mol)(mol C2H5OH/46.08 g C2H5OH)
    = -29.71 kJ/g C2H5OH

    ReplyDelete
  12. Michael C.October 25, 2011 at 9:16 PM

    C2H5Oh + 3O2 = 2CO2 + 3H20
    C2H5OH= -277 kJ/mol
    C02= -394 kJ/mol
    02= 0 (element)
    H20= -286 kJ/mol

    E=2(c02) + 3(h20) - (c2h50h)
    E=-1369 KJ/mol
    -1369 kJ C2H5OH/mol x 1 mol C2H5OH/46.08 g C2H5OH= -29.71 kJ/g of C2H5OH

    ReplyDelete
  13. AnonymousOctober 26, 2011 at 9:45 PM

    C2H5Oh + 3O2 = 2CO2 + 3H20
    C2H5OH=-277kj/mol
    H2O=-286kj/mol
    CO2=-394kj/mol

    deltaE=(-788-858)-(-277)=-1369kj/mol
    -1369C2H5OHkj/molx1molC2H5OH/46.08g= -29.71kj/g

    ReplyDelete
  14. ShauntikOctober 30, 2011 at 7:37 PM

    C2H5OH + 3O2 = 2CO2 + 3H20
    C2H5OH= -277 kJ/mol
    CO2= -394 kJ/mol
    H2O= -286 kJ/mol

    deltaH = 2x(-394kJ/mol) + 3x(-286 kJ/mol) - (-277kJ/mol) = -1369 kJ/mol

    (-1369 kJ/mol)/(46.08 g)= 29.71 kJ/g of C2H5OH

    ReplyDelete

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