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?