$$\Delta G^\circ=-RT\ln K$$

I have always assumed it to be $K_p$ because $\Delta G^\circ$ represents standard free energy change and the standard state for a gas is 1 bar. So I thought we must express the activities of gases as ratios of partial pressures to $\pu{1 bar}$. But while solving sums, I found my answer to match only if I consider $K$ as $K_c$. Where am I wrong? Or am I missing something trivial?

  • $\begingroup$ Your question is indeed natural to arise, but I think it's way complicated than it seems. Read more here. $\endgroup$ Feb 27, 2018 at 4:55
  • $\begingroup$ Sure thing! I don’t understand a word in that link. $\endgroup$
    – MollyCooL
    Feb 27, 2018 at 5:11

2 Answers 2


Neither. It is exactly what is given there $K$, the standard equilibrium constant, and the equation given is its definition, $$\Delta G^\circ=-RT\ln K.\tag{1}\label{def}$$

A general equilibrium constant for any reaction $$\ce{\nu_A A + \nu_B B + \nu_C C + \cdots <=>\nu_{A'} A' + \nu_{B'} B' + \nu_{C'} C' + \cdots }$$ is given by $$K_x = \frac{ x_{\ce{A'}}^{\ce{\nu_{A'}}}\cdot x_{\ce{B'}}^{\ce{\nu_{B'}}}\cdot x_{\ce{C'}}^{\ce{\nu_{C'}}}\cdots}{ x_{\ce{A}}^{\ce{\nu_A}}\cdot x_{\ce{B}}^{\ce{\nu_B}}\cdot x_{\ce{C}}^{\ce{\nu_C}}\cdots},$$ or more general, gathering all components in $\mathbb{B}=\{\ce{A, B, C, \dots, A', B', C', \dots }\}$ $$K_x = \prod_{B~\in~\mathbb{B}} x_B^{\nu_B}.\tag2$$

If you insert that into \eqref{def}, then $x$ becomes the chemical potential of the component. You can derive the link to the activity of the component (see here), and you can link the activity to concentrations, pressures, etc.

Given the right boundary conditions, e.g. ideal dilution, $K$ essentially becomes $K_c$, etc.

$$\mu(i) \propto a(i) \propto c(i) \implies K \propto K_c\tag3$$

Obviously, this is a very sloppy generalisation, and you should study thermodynamics in more detail to fully understand the differences between the quantities used, and when to use which.

  • 4
    $\begingroup$ x is never equal to chemical potential. I think you meant activity (or perhaps fugacity). Chemical potential is partial molar free energy, and has units of energy per mile. $\endgroup$ Feb 27, 2018 at 15:48
  • $\begingroup$ "ideal dilution" Do you mean infinite dilution i.e. the Kohlaurcsh law limit of ionic conductivity in electrochemistry? (that's a very vague connection though) $\endgroup$ Feb 27, 2018 at 15:54
  • 2
    $\begingroup$ Although the OP has accepted the answer, I don't see how this answers the question. It does not tell when to use which type of equilibrium constant. $\endgroup$ Feb 28, 2018 at 3:52
  • $\begingroup$ @ApoorvPotnis Martin does say that "you should study thermodynamics in more detail to fully understand the differences between the quantities used, and when to use which" (my emphasis) i.e. without knowing the exact specific conditions it is difficult to predict if it is $K_\mathrm{c}$ or $K_\mathrm{p}$ or a hybrid of both. $\endgroup$ Mar 2, 2018 at 11:41
  • $\begingroup$ @GaurangTandon Then how are we expected to use the equilibrium constants in problems that we are expected to solve? $\endgroup$ Mar 2, 2018 at 11:53

With gas phase reactions the equation for equilibrium is $\Delta G^\mathrm{o} =-RT\ln(K_p)$ where because the equilibrium constant $K_p$ may be defined in terms of the free energies in their standard states (as you mention), i.e. at 1 bar (or atm.) it must be independent of pressure.

In terms of the partial pressures if the reaction has species $M_i$ the general reaction is $\ce{v_1M_1 + v_2M_2<=>v_3M_3 + v_4M_4}$ then, with $p_i$ being the partial pressure of species $i$, $K_p = \prod p_i^{v_i}$ where the stoichiometric coefficients are $v_i$, and are positive for products and negative for reactants, thus for the reaction $\ce{A_2<=>2A}$, $K_p= p_A^2/p_{A_2}$.

The equilibrium constant can also be defined in terms of concentration since for a perfect gas the partial pressure of the $i^{\text{ th}}$ species is $p_i=n_iRT/V=c_iRT$. The new equilibrium constant ($K_c = \prod c_i^{v_i}$) when written in terms of $K_p$ becomes $\displaystyle K_c=K_p(RT)^{-\Sigma v_i}$ which is often written as $\displaystyle K_c=K_p(RT)^{-\Delta n}$ where $\Delta n= \Sigma v_i$. Thus for the reaction $\ce{2A + 3B<=> 4C}$, $\Delta n\equiv \Sigma v_i= -2-3+4=-1$ and so $\displaystyle K_c=RTK_p$

(Incidentally, the temperature dependence if the equilibrium constant $K_p$ is given by the Van't Hoff equation, sometimes called the Van't Hoff Isochore).


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