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Enthalpy of formation

  1. Mar 23, 2005 #1
    How would you calculate the standard enthalpy of formation of N2O5(g) in kJ/mol from the following data (all species are gases):
    2 NO + O2-----2 NO2 ∆H° = –114.1 kJ
    4 NO2 + O2-----2 N2O5 ∆H° = –110.2 kJ
    N2 + O2------2 NO ∆H° = +180.5 kJ

    Any help is greatly appreciated!
     
  2. jcsd
  3. Mar 23, 2005 #2

    DH

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    ok, here is your answer. this one is the easy one!
    time first and third one with 2, then add three of them up, we have:
    4 NO + 2O2 + 4NO2 + O2 + 2N2 + 2O2 -> 4NO2 +2N2O5 + 4NO

    5O2 + 2N2 -> 2N2O5 ∆H° = 2*(-114.1) -110.2 + 2*180.5 = 22.6
    -> standard enthalpy of formation of N2O5 = 22.6/2 = 11.3 kJ
     
  4. Mar 12, 2011 #3
    heat of formation for lactic acid
     
  5. Mar 12, 2011 #4
    what is the standard value of heat of formation for lactic acid
     
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