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When we write for an equilibrium that $$\Delta G^\circ = -RT \ln(K)$$ Which type of $K$ , ie is it $K_\text{concentration}, K_\text{pressure}$ or $K_{\chi}$ because all these have different numeric values which will give different values of $\Delta G^\circ$ which is obviously not possible.

Edit: In the question marked as being duplicate of, they still don't tell which $K$ it is. Even if we divide by the appropriate standard value, and make it unitless, $G$ will still give different values based on $K_\text{concentration}$, $K_\text{pressure}$ or $K_{\chi}$. Also can $K_{\chi}$ also be used as it is dimensionless less too?

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  • $\begingroup$ You may want to see chemistry.stackexchange.com/questions/43531/… and links therein to additional posts. $\endgroup$ – Buck Thorn May 23 at 9:44
  • $\begingroup$ The answer / question there does not include the mole fraction part. Because mole fraction is dimensionless is it used to for activity? Also they both answers don't tell what to do ,because in highschool atleast , we have to deal with this only. Even if we take activity , then too ,eg in formation of NH³ , we will get different answers as such for concentration and pressure $\endgroup$ – RandomAspirant May 23 at 9:57
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    $\begingroup$ $\Delta G^\circ$ is defined for a certain standard state. No matter whether you use mole fraction, concentration or pressure for a given species, as long as you divide these by the standard mole fraction, concentration or pressure of that species, you will obtain the correct (dimensionless) value for K. Another way of saying this: as long as Q = 1 for the standard state, you are good. This means the numerical value of K will change only if you switch to a different standard state (in this case, the value of $\Delta G^\circ$ will also change). $\endgroup$ – Karsten Theis May 23 at 14:44
  • $\begingroup$ @KarstenTheis But how is that possible, the Gibbs free energy has a definite value.How can it change based upon the different values of K we take. $\endgroup$ – RandomAspirant May 23 at 17:45
  • $\begingroup$ @NightWriter This question has been marked as duplicate but the other answers don't fully answer it. How to reopen this question.? $\endgroup$ – RandomAspirant May 23 at 17:46
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$K$, the standard thermodynamic equilibrium constant, is computed from $\Delta G^\circ$, using

$$\Delta G^\circ = -RT\log K \tag1$$

Generally speaking, $K$ in equation (1) is unitless. Its value depends on the specified reference standard states and $T$ (and obviously on the equilibrium activities of reactants and products). Both $K$ and $\Delta G^\circ$ change (and are certainly allowed to) if you change the choice of standard states for any species.

You can also compute an equilibrium constant from a value of $\Delta G^\circ$. To do this reverse operation you need to know the standard state ($p^\circ=\pu{1 bar}$, $c^\circ=\pu{1 M}$, or $m^\circ=\pu{1 molal}$, $\chi^\circ=\pu{1}$) associated with each species in the equilibrium equation. At equilibrium each product/reactant is at the same temperature and in the same phase as in its associated reference standard state (if not at the same partial pressure, concentration, or mole fraction).

When you use either $K$ or $\Delta G^\circ$ in practice, you need to know the standard state of each species (this is often apparent from additional information that is provided with these parameters). If you are working with a gas-phase reaction then you can refer to $K$ as $K_p$. If you are using standard states then all reference states are $\pu{1 bar}$ $^\dagger$, and partial pressures computed directly from $K_p$ are in $\pu{bar}$ units (these can obviously be converted later).

To compute $K_c$ from say $K_p$ you have to convert units with an appropriate conversion factor. This is described in the linked post. Same applies to problems where $K_\chi$ is provided, where concentrations are described in mole fraction units. Note that if you change units, for instance to compute $K_c$, then you have (perhaps implicitly) changed the reference state (perhaps to a nonstandard state), and in that case $\Delta G^\circ$ also changes. This is also explained in an answer to the linked post.

$\dagger$ Components whose activity is constant can usually be ignored

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