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I understand that $K$ will be dimensionless. Let's say an equilibrium involving $K_p$ and the reaction is $\ce{A -> B}$, $K_p$ will be dimensionless. If the reaction $\ce{A + B -> C}$, then based on the formula of $K_p$ will be:

$$ \frac{(\text{partial pressure of C})}{[(\text{partial pressure B})\times(\text{partial pressure of A})]}$$

Since partial pressure can be in $\mathrm{Pa}$, or $\mathrm{bar/atm}$, with the difference between them will be roughly 100000 times. So the $K_p$s will have 100000 difference. Putting the $K_p$ into $\Delta G=-RT\ln K$, there will be a huge difference of $\Delta G$. So what units of $K_p$ should we use in this case?

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marked as duplicate by Todd Minehardt, Mithoron, Gaurang Tandon, Tyberius, airhuff Apr 24 at 5:28

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What units of K should we use in ΔG = -RT lnK?

The full formula is:

$$\Delta_r G^\circ = - R T \ln K$$

The circle after the G refers to a standard state, where reactants and products are present at standard concentrations/pressures. The equilibrium constant has to be given with reference to the same standard state. The subscript r stands for reaction. The balanced chemical reaction has to match between the Gibbs energy of reaction and the equilibrium constant expression. Let's say our balanced reaction is:

$$\ce{ \nu_A A + \nu_B B + \nu_C C + ... <=> \nu_Y Y + \nu_Z Z + ...}$$

Here, the different $\nu$ are the stoichiometric coefficients.

The correct equilibrium expression for this reaction uses activities of species raised to their respective stoichiometric coefficient. Activities are dimensionless, and are equal to one at standard state. For dilute solutions, these can be estimated as the ratio of a concentration (or partial pressure) divided by the standard concentration (or partial pressure):

$$ a_\text{species} \approx \frac{c_\text{species}}{c^\circ_\text{species}} = \frac{p_\text{species}}{p^\circ_\text{species}}$$

As a consequence, the equilibrium constant expression will always be dimensionless, and will not depend on the concentration measure used, as long as the standard state concentration is expressed using the same measure.

Since partial pressure can be in Pa, or bar/atm, with the difference between them will be roughly 100000 times.

No, because you are dividing by the standard state, using the same concentration measure (and for ease of cancelling units, in the same units). For example, let's say the standard state of a species in the gas phase is defined as one molar. At ambient conditions, this would correspond to a partial pressure of about 0.04 atm. If the equilibrium concentration is - say - 2 M, that would correspond to a partial pressure of about 0.08 atm. Either measure of concentration gives the same result: The species is present at twice the concentration (partial pressure) as the standard state. No matter what concentration measure I choose, I get the same value for the equilibrium constant.

The same goes for switching pressure units. As you are always dividing by the same standard pressure, you always get the same result (either by converting the units of the partial pressure, or the units of the standard partial pressure, and then cancelling).

Switching to a different standard state

In biology and biochemistry, a different standard state is used. This affects both the values of the standard Gibbs energy of reaction, and the values of equilibrium constants. However, these changes are self-consistent, i.e. you still can calculate K from the Gibbs energy and vice versa, and you can convert easily from one convention of the standard state to the other.

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