I have learnt that the standard free energy change is related to the equilibrium constant of a reaction by, $$\Delta G^\circ = -RT \ln K$$

Here, does $K$ refer to $K_p$ or $K_c$?

Also, please give me the derivation of this formula.

On the net, I saw that we can use either $K_p$ or $K_c$. But doesn't that give two different free energy values?

Moreover, if the formula was derived for gaseous reactions (using $K_p$), how can we just extend this to other reactions saying that we can use $K_c$ as well?


2 Answers 2


As noted in this previous question, the correct definition of the equilibrium constant $K$ depends on activities. If you are interested in the derivation of the equation $\Delta G^\circ = -RT \ln K$ (which requires "proper" thermodynamics), read Philipp's answer to that question. For a reaction

$$0 \longrightarrow \sum_i \nu_i \ce{J}_i$$

(this is a fancy way of writing the reaction which makes sure that the stoichiometric coefficients $\nu_i$ of the reactants are negative), the equilibrium constant is defined as follows:

$$K = \prod_i a_i^{\nu_i}$$

where $a_i$ is the activity of the species $\ce{J}_i$ at equilibrium. (Note that since $\nu_i$ for reactants is negative, the terms for the reactants will appear in the denominator of a fraction.)

The question is, why are $K_c$ and $K_p$ taught at an introductory level despite them not actually being the equilibrium constant $K$? The answer is likely partly because it is difficult to introduce the idea of an activity, but also because $K_c$ and $K_p$ are good approximations to the actual equilibrium constant, $K$. To see why, we have to look at the mysterious quantity called the activity and see how it is related to the concentration of a solution, or the partial pressure of a gas.

If we assume certain ideality conditions, then we can write the activity of a gas as:

$$a_i = \frac{p_i}{p^\circ}$$

where $p_i$ is the partial pressure of the gas and $p^\circ$ is the standard pressure, defined by IUPAC to be equal to $1~\mathrm{bar}$. Therefore, $a_i$ is nothing but the partial pressure of the gas in bars, but without the units. If we therefore define $K_p$ as follows:

$$K_p = \prod_i p_i^{\nu_i}$$

then, as long as the pressures you use are in bars, the numerical value of $K_p$ will be the same as that of $K$, the true equilibrium constant. As such, if you ignore the fact that you are taking the natural logarithm of a quantity with units, the value of $\Delta G^\circ$ that you obtain will be "correct".

Similarly, for a solution, we can write:

$$a_i = \frac{c_i}{c^\circ}$$

where $c_i$ is the concentration of the species and $c^\circ$ is the standard concentration, again defined by IUPAC to be equal to $1~\mathrm{mol~dm^{-3}}$. So, the same idea applies. If you keep all concentrations in units of $\mathrm{mol~dm^{-3}}$, then the numerical value of $K_c$ will match that of $K$.

An example

Consider the Haber process (all species gaseous):

$$\ce{N2 + 3H2 <=> 2NH3}$$

We would define the "true" equilibrium constant as:

$$K = \frac{(a_{\ce{NH3}})^2}{(a_{\ce{N2}})(a_{\ce{H2}})^3}$$

and you would define $K_p$ as:

$$K = \frac{(p_{\ce{NH3}})^2}{(p_{\ce{N2}})(p_{\ce{H2}})^3}$$

Let's say (just as an example - the numbers are not correct!) that at a certain temperature $T$, we have the equilibrium partial pressures:

$$\begin{array}{c|c|c} p_{\ce{NH3}} & p_{\ce{N2}} & p_{\ce{H2}} \\ \hline 20~\mathrm{bar} & 50~\mathrm{bar} & 50~\mathrm{bar} \end{array}$$

Then you would have $K_p = 6.4 \times 10^{-5}~\mathrm{bar^{-2}}$. Now you can see why it is not correct to use $K_p$ in the equation above: you would have

$$\Delta G^\circ = -RT \ln(6.4 \times 10^{-5}~\mathrm{bar^{-2}})$$

which makes absolutely no sense since you can't take the logarithm of a unit. In a simplistic treatment you would just ignore the fact that a unit existed and just take the logarithm of the number.

However, you will not run into that problem if you use the "true" equilibrium constant, as you really should. If we assume ideality, i.e. $a_i = p_i/p^\circ$, then we find that $K = 6.4 \times 10^{-5}$. Since activities are dimensionless, you will find that $K$ is dimensionless and you can now take the logarithm. As I mentioned earlier, the numerical quantity of $K$ is exactly equivalent to $K_p$ as long as you keep your pressures in units of bars.

I have seen some questions where students are requested to determine $K_c$ for a gas-phase reaction using the ideal gas law and hence $\Delta G$:

$$\begin{align} pV &= nRT \\ c = \frac{n}{V} &= \frac{p}{RT} \end{align}$$

The value of $c_i$ and $p_i$ will therefore differ by a factor of $RT$. Therefore, the numerical values of $K_c$ and $K_p$ in gas-phase reactions will, in general, differ. (If you have the same number of moles of gas on both sides, then the factors of $RT$ will cancel out and numerically $K_c = K_p$, but that is not always the case.) As established earlier, because of the equation for the activity of an ideal gas, only the use of $K_p$ will lead to the correct numerical value for $\Delta G^\circ$.

If a question requires you to do so, then do it. Just bear in mind that it is not only wrong in terms of taking the natural logarithm of a quantity with units; it will also give you a wrong numerical result for $\Delta G^\circ$.

  • 1
    $\begingroup$ Does this mean that in any (ideal) equilibria (including heterogeneous equilibria also) the activities can be approximated - to partial pressures in case of gases, to concentrations in case of liquids and to 1 in case of solids. These 'so called activities' can give us K (equilibrium constant) using the equation mentioned in your answer and hence this K when substituted in the equation Δ G=RTln(K) gives the standard Δ G - 'standard' in the way it is defined by IUPAC. So we are infact complicating things using Kc and Kp (atleast in my opinion). Is my understanding right? $\endgroup$
    – vamsi3
    Commented May 9, 2016 at 2:56
  • $\begingroup$ @user104014 yes that is right. $\endgroup$ Commented May 9, 2016 at 7:19
  • 1
    $\begingroup$ "why are Kc and Kp taught at an introductory level despite them being wrong? " They are not wrong any more than the ideal gas law is wrong! They are approximate and converge on exactness for dilute systems (eg gases at low P). $\endgroup$
    – Buck Thorn
    Commented Nov 7, 2018 at 11:33
  • $\begingroup$ @TryHard, if you re-read my answer, I did write that they are approximations which are numerically correct for ideal gases/solutions. However, they are dimensionally inconsistent and should not be used as-is in the relation ΔG = -RT ln K. If your only objection is my choice of phrasing, then I am happy to reword it later when I am off work. I agree it was an exaggeration. $\endgroup$ Commented Nov 7, 2018 at 12:10
  • $\begingroup$ @orthocresol For a gas phase reaction or a heterogeneous solid-gas phase reaction, if $K_c$ is given, then it has not considered the standard state of reactants and products. So, by using $K_p = K_c(RT)^{\Delta n}$, we cannot get the value of $K_p$ at standard state? Then, what's the use of this expression? $\endgroup$
    – Apurvium
    Commented Sep 8, 2021 at 5:48

You are correct that you can use either $K_c$ or $K_p$; the trick is to carefully choose your values of $\Delta G_f$ to reflect whether you are working with gaseous compounds or compounds dissolved in solution.

For example, if you are interested in the equilibrium constant for the reaction of $\ce{CO2}$ with $\ce{H2O}$, you can imagine this happening in the gas phase or in solution:

$$\ce{CO2(g) + H2O(g) <=> H2CO3(g)}$$


$$\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)}$$

In both cases, we have

$$K = \exp\left(\frac{-\Delta G_\mathrm{rxn}}{RT}\right)$$

However, in the first case, we use

$\Delta G_\mathrm{rxn} = \Delta G_\mathrm{f}\ce{[H2CO3(g)]} - \Delta G_\mathrm{f}\ce{[CO2(g)]} + \Delta G_\mathrm{f}\ce{[H2O(g)]}$

corresponding to $K_p$ and in the second case, we must use

$\Delta G_\mathrm{rxn} = \Delta G_\mathrm{f}\ce{[H2CO3(aq)]} - \Delta G_\mathrm{f}\ce{[CO2(aq)]} + \Delta G_\mathrm{f}\ce{[H2O(aq)]}$

corresponding to $K_c$.

A good table of thermodynamic values will list several different values for the same compound, corresponding to various possible phases, including (aq). Unfortunately, the tables in most chemistry textbooks are quite truncated, so you won't always find both (g) and (aq) Gibbs energies of formation.

Contrary to the simplified definition given in most introductory texts, the equilibrium constant is defined in terms of activities. For gases, this is often approximated as the partial pressure divided by 1 atm (maybe it's 1 bar these days? I learned it as 1 atm…). For liquids, it's approximated as the molar concentration divided by 1 M. This is what you actually end up using in more complex systems where you have both gases and liquids (and solids) all in equilibrium - you no longer have the simple distinction between $K_p$ and $K_c$, you simply use K with the appropriate choices for activities and the appropriate choices for $\Delta G_\mathrm{f}$.

  • $\begingroup$ For a liquid-gas phase heterogeneous equilibrium, $K_{eq} \neq K_c \neq K_p$. For calculating $K_{eq}$, the substances in aqueous form must be in 1 M concentration and substances in gaseous form must be in 1 bar pressure? $\endgroup$
    – Apurvium
    Commented Jun 7, 2023 at 18:37

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