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Calculating ΔH°rxn for a Reaction

  1. Mar 28, 2015 #1

    Drakkith

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    Staff: Mentor

    1. The problem statement, all variables and given/known data

    Given the data in the table below, ΔH°rxn for the reaction

    4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (l)

    is ________ kJ.

    Substance ΔH∘f(kJ/mol)
    H2O (l) -286
    NO (g) 90
    NO2 (g) 34
    HNO3 (aq) -207
    NH3 (g) -46

    2. Relevant equations

    ΔH°rxn = ΔH°products - ΔH°reactants

    3. The attempt at a solution

    ΔH°rxn = [(4x90)+(6x-286)] - [(4x-46)+(X)]

    ΔH°rxn = (-1356) - (-184+X)

    ΔH°rxn = -1356 + 184 - X

    ΔH°rxn = -1172 + X

    At first I selected, D.) The ΔH°f of O2 (g) is needed for the calculation.
    However, apparently the answer is -1172, but I have no idea why. How can -1172 be the answer here? And why do they give me the enthalpy of formation for NO2 and HNO3 if neither of those compounds are in my reaction?
     
  2. jcsd
  3. Mar 28, 2015 #2

    Drakkith

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    Staff: Mentor

    Okay, I just realized the ΔH°f of O2 is defined to be zero...
    Now it all makes sense.
     
  4. Mar 29, 2015 #3

    Borek

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    Staff: Mentor

    And you answer - that it is needed - wasn't wrong in general. Defining ΔH°f of O2 as zero means you know its value ;)
     
  5. Apr 4, 2015 #4
    why?
     
  6. Apr 4, 2015 #5

    Borek

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    Staff: Mentor

    Either to confuse him, or because it was copy/pasted from another problem and someone didn't bother to leave just what is important for the question.
     
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