The equilibrium constant of a chemical reaction can be calculated if the $\Delta_\mathrm rG^\circ$ is known which is a function of only temperature. So if a reaction happens inside a container subjected to external pressure $p$, varying $p$ would have no effect on the equilibrium constant. I was reading this book which states:

$\mathrm d\Delta_\mathrm rG^\circ = -\Delta_\mathrm rS^\circ\,\mathrm dT + \Delta_\mathrm rV^\circ\,\mathrm dp$

It uses that relation to derive the pressure dependence of the equilibrium constant as:

$$\left( \frac{\partial \ln K}{\partial p} \right)_T = \frac{-\Delta_\mathrm rV^\circ}{RT} $$

What I can't understand is that the standard state subscript denotes a specific pressure. What is the meaning then differentiating standard change in free energy of a reaction with respect to pressure? As I said $\Delta_\mathrm rG^\circ = f(T)$

Can someone shed some light on what is going on the differential and why the equilibrium constant depends on pressure?


1 Answer 1


You need to define what type of $K$ you are using. By definition $\Delta G^\text{O}=-RT\ln(K_p)$ where $K_p$ is defined by the standard states of the gases, i.e. 1 atm. pressure so cannot depend on pressure.

If $K$ is measured wrt. mole fraction, for example, with partial pressure $P_i=x_iP$ for species $i$ then

$$K_x = K_pP^{-\Delta n}$$

where $\Delta n$ is the difference in moles product-reactants and $P$ total pressure. Taking the log of both sides of this last expression and differentiating (at const. $T$) wrt. $P$ gives

$$\left(\frac{\partial \ln(K_x)}{\partial P} \right)_T=\left(\frac{\partial \ln(K_p)}{\partial P} \right)_T-\Delta n\left(\frac{\partial \ln(P)}{\partial P} \right)_T$$

The first term on the right is zero because $K_p$ is independent of pressure, which leaves

$$\left(\frac{\partial \ln(K_x)}{\partial P} \right)_T=-\frac{\Delta n}{P}=-\frac{\Delta V}{RT}$$

assuming an ideal gas.

( When you use $K_p$ and the pressure is changed the degree of dissociation changes to compensate.)

  • $\begingroup$ Kp is the pressures at equilibrium not standard states. $\endgroup$
    – jimchmst
    Dec 5, 2021 at 1:16
  • $\begingroup$ @jmchmst, No, it is defined wrt to free energies of species in the standard states of 1 atm. pressure. Thermodynamics always examines the equilibrium situation. $\endgroup$
    – porphyrin
    Dec 5, 2021 at 9:45
  • $\begingroup$ @porphyrin I am using the definition of K which includes activities. And also in the book as I said it differentiates the standard free energy change with respect to pressure. That is what it troubles me. How a derivative of a standard quantity with respect to pressure makes sense, if by definition, standard state refers to a fixed pressure? $\endgroup$
    – Anton
    Dec 5, 2021 at 20:48
  • $\begingroup$ It should all work out ok, the activity coeffs are constants so differentate wrt anything to zero. $\endgroup$
    – porphyrin
    Dec 7, 2021 at 18:49

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