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Thermodynamics - Heat of Reaction

  1. Feb 16, 2016 #1
    • Moved from a technical forum, so homework template missing
    Calculate the difference between heat of reaction at constant pressure and at constant volume for following reaction at 298 K.
    N2(g)+3H2(g)→2NH3(g)

    My attempt ~
    At constant pressure
    ΔHp=ΔUp+PΔV

    At constant volume
    ΔHv=ΔUv

    ∴ΔHp-ΔHv
    =ΔUp-ΔUv+ PΔV
    And PΔV = ΔnRT
    But what to do of ΔUp-ΔUv?
    In solution,
    For both , at constant pressure and volume
    ΔU is mentioned and
    ΔU-ΔU =0 is given..
    But how is it possible that
    ΔUp=ΔUv ?
     
  2. jcsd
  3. Feb 16, 2016 #2
    Well, if you move from one state to another, it doesen't matter how you move. ΔU is zero anyway. So yes, ΔUp = ΔUv.
     
  4. Feb 16, 2016 #3
    so, ΔHp should also be equal to ΔHv since it is a state function like internal energy U , why it is not so?
     
  5. Feb 16, 2016 #4

    TSny

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    What thermodynamic state variables determine the internal energy of an ideal gas?

    If you know this, then you should be able to compare the final internal energy for the constant pressure process with the final internal energy for the constant volume process.
     
  6. Feb 17, 2016 #5
    ΔH is defined as the enthalpy change of a reaction at constant pressure and is a state function.. There is no such thing as ΔHp and ΔHv. What you must mean is Qp and Qv.
     
  7. Feb 17, 2016 #6

    TSny

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    ΔH is defined for any process between any two equilibrium states. As you say, H is a state variable. The process need not be at constant pressure in order to calculate ΔH.
     
  8. Feb 17, 2016 #7
    To directly calculate ΔH, I agree the process need not be done at constant pressure. But ΔH is defined as such.
     
  9. Feb 17, 2016 #8

    Ygggdrasil

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    ΔH is not defined as the heat produced at constant pressure and temperature, though it is true that ΔH = q for such a process. H is most rigorously defined as H = U + PV.

    Also a helpful note. Performing the reaction at constant temperature does not lead to the same final state as performing the reaction at constant volume. In one case, the final volume will be different and in the other case the final pressure will be different.
     
  10. Feb 17, 2016 #9

    TSny

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    Is this true?

    ΔH = Δ(U + PV) = ΔU + Δ(PV).

    Does Δ(PV) equal 0 for the reaction at constant volume?
     
  11. Feb 20, 2016 #10
    Constant pressure and temprature @Ygggdrasil ?
     
  12. Feb 20, 2016 #11
    Yes. For tabulated heats of reaction, both the temperature and the pressure are specified to not change between the initial state of pure reactants and for the final state of pure products.
     
  13. Feb 21, 2016 #12
    So basically we start at some T,P... do the reaction with reactant A.... get product B... bring product B to the same state T,P... measure the heat exchanged... and label this as the heat of the reaction? OK, that actually makes sense.
     
  14. Feb 21, 2016 #13
    Yes. That's basically right.
     
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