For a chemical reaction the activation volume is defined as:
$$Δ^{\ddagger}V={V^\ddagger}-V$$
where both $V^\ddagger$ and $V$ refer to standard states (I have omitted the symbol for simplicity). One can calculate activation volume by taking the partial derivative with respect to pressure of $$Δ^{\ddagger}G={G^\ddagger}-G$$
(again for simplicity the standard state symbol is omitted). I can't understand why $Δ^{\ddagger}G$ must be a function of pressure if it happens at a specific pressure (standard state). I was expecting that it would be only a function of temperature like ${Δ_rG}^\circ$ is. The motivation to ask this question was how rate constant varies with pressure:
$$\left(\frac{\partial{\ln k}}{\partial{P}}\right)_T=-Δ^{\ddagger}V$$
The only thing that comes to my mind is if the standard state that is used to derive the above equation it means a state with fixed concentrations for example and not pressure.
