# Is standard change in activation Gibbs free energy indepedent of pressure?

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.

• 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. Mar 13, 2021 at 22:01