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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.

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  • $\begingroup$ In the last equation I don't think it can be true that the V daggers refer to standard states. Thus the meaning of V dagger in your last equation is different than in your first one. $\endgroup$
    – Curt F.
    Mar 13, 2021 at 22:01

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