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$\Delta_\text r G=RT\ln \frac{Q}{K}$

If we switch $K_p$ from $K_c$, accordingly $Q_p$ will change to $Q_c$. Therefore, value of $\Delta_\text r G$ remains the same.

$ΔG^\circ=−RT\ln K=ΔH^\circ−TΔS^\circ$

If we switch $K_p$ from $K_c$, accordingly $\Delta H^\circ$ and $\Delta S^\circ$, which were defined at 1 atm will change their values according to $\pu{1 mol/L}$. Therefore, value of $\Delta G^\circ$ remains the same.

Now consider, $K_p=K_c(RT)^{\Delta n}$

$\implies K_p=K_c(0.082\times T)^{\Delta n}$ $\pu{(atm)}^{\Delta n}$

$\implies K_p=K_c(8.314\times T)^{\Delta n}$ $\pu{(pascal)}^{\Delta n}$

Which is fine, but consider:

$ΔG^\circ=−RT\ln K_p$

$\implies ΔG^\circ=−0.082\times T\ln (K_c\times0.082\times T)^{\Delta n}$ $\pu{L.atm}$ $=−8.314\times T\ln (K_c\times0.082\times T)^{\Delta n}$ $\pu{Joule}$

and also as $\ln K_p$ is dimension-less,

$ΔG^\circ=−8.314\times T\ln (K_c\times8.314\times T)^{\Delta n}$ $\pu{Joule}$

Why are there two different values of change in standard Gibbs energy? What did I miss?

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    $\begingroup$ The relationship between $\Delta G^\circ$ and $K$ is only defined for the dimensionless value $K_{eq}$. $\endgroup$
    – Andrew
    Commented Sep 6, 2021 at 13:10
  • $\begingroup$ @Andrew $K_{eq}$ will be one of these $K_p$, $K_c$, $K_x$, etc.? $\endgroup$
    – Apurvium
    Commented Sep 6, 2021 at 15:07
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    $\begingroup$ short version: if all of your reactants and products are dissolved species, $K_{eq}=K_c$, and if all are gases, $K_{eq}=K_p$, but if you have any gases in your reaction, $K_{eq}\neq K_c$! Look up the defintion of $K_{eq}$ and "activity" in any general chem text for a full explanation. $\endgroup$
    – Andrew
    Commented Sep 6, 2021 at 18:12
  • $\begingroup$ @Andrew Suppose the reaction is a heterogeneous solid-gas phase and the value of $K_c$ is given. Then we will convert $K_c$ to $K_p$ by taking the value of $R$ as $\pu{0.082 L atm mol-1 K-1}$ and then we can use, $\Delta G^\circ =-RT \ln K_p$ $\pu{L atm}$? $\endgroup$
    – Apurvium
    Commented Sep 7, 2021 at 7:33
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    $\begingroup$ There is no reason to change the units and value of R that you are using, since the standard units for $\Delta G^\circ$ are still J/mol, and the logarithm is dimensionless. So R should always be in J/(mol K). $\endgroup$
    – Andrew
    Commented Sep 7, 2021 at 12:41

1 Answer 1

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Recall that $\Delta G^\circ$ represents the difference in Gibbs free energy between an equilibrium state and a "standard state." This standard state is completely arbitrary, and the definition has changed slightly over time. Currently, IUPAC defines it as having all dissolved substances at a concentration of 1 mol/L and all gases at a partial pressure of 1 bar.

When we calculate the magnitude of $K_{eq}$, we use dimensionless quantities known as "activities", which are approximately equal to the amount of substance relative to the reference state. The deviation of the magnitude of the activity from the magnitude of the concentration is beyond the scope of this answer. For now, just assume they are always equal if the units are correctly used as described below.

Thus, the activity of a dissolved substance is the concentration in mol/L divided by the reference state of 1 M, which simply removes the units. But it requires that the units are M. For gases, the same process applies, except that the reference state is 1 bar, so units must be bar (or you could convert the reference state of 1 bar to whatever units you are using, but that's a bit silly).

For solids and bulk liquids, the activity is always 1, because they cannot differ in concentration, so they are always the same as the reference state.

In contrast to $K_{eq}$, $K_c$ and $K_p$ are not defined with respect to a reference state. They are simply ratios of either concentrations or pressures of products and reactants at equilibrium. If all of our reactants and products are dissolved species and we express their concentrations in mol/L, then the magnitude of $K_c$ is equal to that of $K_{eq}$. Likewise, if reactants and products are all gases, the magnitude of $K_p$ is equal to that of $K_{eq}$. But if we have gases in the reaction, the magnitude of $K_c$ is not equal to that of $K_{eq}$ because we haven't accounted for the reference state of 1 bar.

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  • $\begingroup$ I am a bit confused about your first sentence: At equilibrium, to define $\Delta G^\circ$, all the reactants and products must be in standard state, say 1 mol/L. That means in any other instance they were not in that "standard state". $\endgroup$
    – Apurvium
    Commented Sep 10, 2021 at 1:37

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