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Enthalpy and Heat State Functions

  1. Feb 24, 2012 #1
    I'm sure this topic has been asked before so I'd be happy if you'd just direct me to the thread.
    I know U+PV=H and if the only work done is PV work then H=q-P∆V+P∆V=q. And I know that pressure must be constant. I also know that a state function is a function where the value of the function only depends on the change between the last and first states. I have 2 questions.
    1. If pressure were not constant wouldn't the PV work=0 and enthalpy would still equal q?
    2. I understand why enthalpy is a state function. The heat lost or gained only depends on the change between the first and last states. If enthalpy always equals q then why isn't q a state function? For example if 2 reactions happened one released 20 kJ of heat and one absorbed 10 kJ of heat the change in q or the heat lost or gained in the system is -20 kj + 10 kj =10 kJ lost.

    I feel like my reasoning is probably off somewhere. Can someone help me please?
     
  2. jcsd
  3. Feb 25, 2012 #2
    dH = dU + pdV + Vdp

    Only mechanical work (dU = δQ - pdV)

    dH = δQ + Vdp

    If p=constant

    dH|p = δQ|p
     
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