In order to calculate the equilibrium constant we must know the value of $\Delta_\mathrm{r}G^{\circ}$ (which is a function of $T$) so we can calculate $K$ as:

$$K = \exp\left(-\frac{\Delta_\mathrm{r}G^{\circ}}{RT}\right)$$

Now if the standard state is changed (for example from $\pu{1 atm}$ to $\pu{3 atm}$) would the equilibrium constant change? My intuition says no because we can experimentally measure $K$ which varies only with temperature. So the value of $\Delta_\mathrm{r}G^{\circ}$ must stay constant irrespective of the standard state.

What is the point then to specify the standard state?

Edit I would like to add a specific example that troubles me. Suppose we have the reaction:

$$\ce{A (g) -> B (g)}$$

that takes place in a closed container (fixed volume). We fill the container with $\ce{A}$ and after some time has passed equilibrium is established. We can predict the ratio of partial pressures of the two gases (omitting fugacity for simplicity) from the equilibrium constant $K$ which equals $\exp(-\Delta_\mathrm{r}G^{\circ}/RT)$.

What I don't understand is why if we change the standard state (picking a different pressure) must change the equilibrium constant. At least for gases it doesn't make sense because in equilibrium is their partial pressure that matters.

But for constant temperature the Gibbs free energy of the compounds must be:

$$G_i(P_i) = G(P^{\circ}) + nRT \ln \left( \frac{P_i}{P^{\circ}} \right) $$

which, as orthocresol said in the comments, shows that $\Delta_\mathrm{r}G^{\circ}$ depends on pressure. I want to find solution to that contradiction and understand the significance of the standard state.

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    $\begingroup$ The expression for K in terms of partial pressures inherently assumes that the standard state is 1 bar. If the number of moles changes, and the standard state is changed to 3 bars, the value for K will change. $\endgroup$ Commented Jun 5, 2021 at 12:32
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    $\begingroup$ Your example probably won't be affected much. Now think of the reactions which have gas on one side and no gas on another. $\endgroup$ Commented Jun 5, 2021 at 12:59
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    $\begingroup$ The standard state is selected once and for all. When selecting it, you don't care about one particular reaction. $\endgroup$ Commented Jun 5, 2021 at 13:18
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    $\begingroup$ Ever heard about the planet Earth? Well, that's why. $\endgroup$ Commented Jun 5, 2021 at 15:09
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    $\begingroup$ @Anton Please note that, since 1982, standard pressure is $1\ \mathrm{bar}$ and not $1\ \mathrm{atm}$. Also note that the use of the unit $\mathrm{atm}$ is deprecated. $\endgroup$
    – Loong
    Commented Jun 7, 2021 at 17:36

3 Answers 3


We all are familiar with the standard state as 1 bar pressure and 273 K.The standard state can be any well defined state. You can define it in any way that is convenient for you. There is no universal definition. We may want to choose different references depending on the calculation we are doing, so we can choose different standard states.

In thermodynamics we are usually interested in changes in quantities like internal energy, entropy etc, and to measure a change we need to choose a reference that we can measure the change relative to. That reference is the standard state.

What happens in situations where Pressure and temp keeps on changing ?

In situations where a property like pressure or temperature changes we generally have to calculate things by integrating from some initial state to some final state which can be tough , a lot of calculations .

Therefore , you can define standard state as the most simplest state.   Also , I discussed regarding this Q with John Rennie sir.


Let's say you have a reaction that has reached equilibrium. The reaction quotient $Q$ will be equal to the equilibrium constant $K$, and the Gibbs energy of reaction will be zero. Now, you decide to change the standard state (without touching your experiment, which is still at equilibrium). For example, if you are a biochemist, you could say that the pH at standard state is 7 instead of 0, or define a magnesium concentration smaller than $\pu{1 mol L-1}$ as standard concentration because that is closer to the situation in the cell.

The reaction will still be at equilibrium, the Gibbs energy of reaction will still be zero, the equilibrium concentrations will still be the same. However, $Q$, $K$, and the standard Gibbs energy of reaction will be different.

To understand why this happens, you have to be exact when defining $Q$ and $K$. These quantities depend on the activity (or fugacity) of the species, which are dimensionless quantities and have the value of 1 at standard state. This means that the activity or fugacity (or "concentration relative to the standard state" if you want to use an approximation and neglect non-ideality) depends on the choice of the standard state. The activity of hydrogen ions in a neutral solution is about $\pu{1e-7}$ for the conventional standard state (standard pH = 0) but is 1 when we define the standard state pH to be neutral.

Your specific example

For $$\ce{A(g)->B(g)},$$ changes in the standard pressure don't affect $K$. However, if the reaction is not a 1:1 stoichiometry (or more generally, the amount of gas changes), i.e. for




you would see a change in $K$ if you choose a different standard state. The most famous example is often labeled $K_\mathrm{P}$ vs $K_\mathrm{c}$.

My intuition says no because we can experimentally measure K which varies only with temperature.

The thermodynamic $K$ varies with the definition of the standard state (the equilibrium concentrations do not).

So the value of $Δ_r G^\circ$ must stay constant irrespective of the standard state.

No, $Δ_r G^\circ$ depends on the choice of standard state, whereas $Δ_r G$ for a given set of concentrations (and temperature) does not.

What is the point then to specify the standard state?

You define it such that the standard state is close to what you are interested in, or it becomes easy to calculate the activity of a species. For example, you choose the standard state of the solvent to be pure solvent because that is close to what you have in a dilute solution. You choose the standard state of a solute to be approximately $\pu{1 mol L-1}$ (extrapolated from infinite dilution) because it makes it easier to calculate $Δ_r G$ from $Δ_r G^\circ$ for dilute solutes. Another way to say it: you want the difference between $Δ_r G$ and $Δ_r G^\circ$ to be small so that in many cases, you can make arguments about the direction of reactions looking at $Δ_r G^\circ$ that carry over to states other than the standard state.


As Chet Miller and Karsten pointed out, the culprit is the simple definition of K which we use in textbooks.

As in your last equation $$Gi(Pi)=G(P∘)+niRTln(Pi/P∘)$$

(similar equation can be used for dilute solutions in liquid phase)

This equation (differential of this equation in the form of chemical potential) is the origin point of the relation $$ \Delta G = G∘ + RTln(Q) $$ 1

In this relation, we should not forget that all the concentration terms of Q are activities, ratios with respect to standard state (whatever that may be). It is only when we take 1 bar or 1 atm or 1 molar as a standard state we neglect these denominators in every concentration term.

So what will happen if the standard state will be 2 bar rather than 1 bar, every concentration term will get divided by 2 instead of 1 and K will be different.

What about concentrations of real-world experiment which are related to K, of course they won't change, but now we can't use previous definition of K, now every term in K will be concentration/2atm. And again it will lead to the same concentration values at equilibrium. So there is no contradiction.

We have defined 1atm or 1 bar as a standard state because our partial pressures are of the order of 1 atm or 1 bar. I can define X pascal or X molar as a standard state but then we will need to divide each term by X pascals or X molars.

Can we define 1 atm for one reactant and 2 atm for another reactant in the same reaction as standard states? Yes, Here we can define each standard state independently, require only that the concentrations are so dilute that your last equation holds true. Here $G∘$ is not standard Gibbs free energy of formation but Gibbs free energy of any specific state wrt which we are defining our activity. The value of which can be calculated from Gibbs free energy of formation $Gf$. The $G∘$ here shows how $G$ changes with concentration, $Gf$ shows how it is related with other substances (specifically, how it is related with free element's Gibbs free energy) through reaction.


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