Can we calculate instantaneous difference between enthalpy and entropy of product and reactants of a reaction as the reaction proceeds towards equilibrium? Ex: $\ce{R<->P}$ reaction, at some reaction coordinate, moles of $R = 1-x , P = x$, Enthalpy difference between products and reactants? Enthalpy difference if it reached equilibrium? Values of $$\Delta G^0, \Delta H^0, \Delta S^0$$ are known. Or is it impossible and the only relationship available is $$\Delta H = T \Delta S$$ at equilibrium or are their values only temperature dependent and activity independent?

We can do this for Gibbs free energy by $$\Delta G=\Delta G^0+\mathcal RT\ln Q$$

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    $\begingroup$ In order to do this, you need to accept the idea that the thermodynamic functions and be defined and calculated even for non-equilibrium states of a system. I have no problem accepting this, but certainly some physicists would not be willing to make this leap. $\endgroup$ – Chet Miller Dec 16 '16 at 15:38
  • $\begingroup$ G is also a thermodynamic function right? $\endgroup$ – Mrigank Dec 17 '16 at 0:31
  • $\begingroup$ Sure G is a thermodynamic function. $\endgroup$ – Chet Miller Dec 17 '16 at 2:54
  • $\begingroup$ en.m.wikipedia.org/wiki/Partial_molar_property It seems that Chemical Potential is bridging the gap and it's only connection is to G. H needs entropy to be constant. Any thoughts? $\endgroup$ – Mrigank Dec 17 '16 at 5:23
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    $\begingroup$ I agree with everything in the reference you provided. G is the most useful and convenient function for defining the chemical potential. In fact, the chemical potential is just the partial molar Gibbs free energy. $\endgroup$ – Chet Miller Dec 17 '16 at 12:35

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