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Spontaneity of reactions and free energy

  1. Dec 2, 2008 #1
    Does anyone know of a clean proof that a reaction will occur at constant temperature if the change in Helmholtz free energy is negative, or at constant temperature and pressure if the change in Gibbs free energy is negative?

    The only `proofs' I've found rely on the fact that the entropy change of the system is given by [itex]\Delta Q /T[/itex]. It seems to me as though this assumption is unjustified since the relation [itex]\Delta S = \Delta Q /T[/itex] holds only for reversibly exchanged heat, but chemical reactions are irreversible processes in general.
     
  2. jcsd
  3. Dec 2, 2008 #2

    Mapes

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    It's true that all real-life processes are irreversible. But as we narrow our focus from a laboratory to a calorimeter to a micro-scale region of gas to individual atoms, things look more and more reversible. After all, irreversibility is caused by smoothing of energy gradients over spatial regions, so let's consider only the smallest regions. Reactions are reversible at the molecular level; it's the dispersion of heat, momentum, mass, etc. to other areas that is irreversible. That's why the [itex]\Delta H=0[/itex] and [itex]\Delta G=0[/itex] criteria are still useful.

    Another way of looking at it is that the energy equation [itex]dE=T\,dS-p\,dV+\sum\mu\,dN[/itex] holds for a system in equilibrium. As before, the closer we look, the more things look like they're at equilibrium. Now do a Legendre transform to get [itex]dG=-S\,dT+V\,dp+\sum\mu\,dN=0[/itex] at constant temperature, pressure, and mass, and we have your Gibbs free energy spontaneity criterion again.
     
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