So I have this statement from our professor :
We have $\Delta G^\circ = -RT\ln K$ and under standard conditions the activities of all substances in the reaction is $1$ ($a=1$), therefore $\Delta G^\circ > 0$.
I dont understand it.If every $a=1$, then $K = 1$ (as the division of activities) and thus $\ln K =0$ which would make $\Delta G^\circ = 0$ as well instead of $>0$,right?
No. $K$ is not a ratio of instantaneous activities! Let's say you hypothetically prepare a system where all components have unit activity. In that case you would have the reaction quotient
$$Q = \prod_i a_i^{\nu_i} = 1$$
but $K$ is the ratio of activities at equilibrium. If that hypothetical system is not at equilibrium, then it will go towards equilibrium. When it reaches equilibrium, the activities are no longer going to be equal to unity, and $K$ will not be equal to 1.
The complete equation is $$\Delta G=\Delta G^\circ+RT\ln{Q}$$If $Q=1$, then $$\Delta G=\Delta G^\circ$$Maybe your professor meant to conclude that $\Delta G \neq 0$ if all the activities are equal to 1. It's hard to know what he meant to say.
