Change in standard Gibbs energy is always zero?

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?

• Welcome to chemistry.SE!. Please visit this page, this page and this ‎one on how to format your future posts better with MathJax and Markdown. If that is really the statement given, then it is wrong indeed. The main thing is that $^\circ$ points to standard state which does not include a rule that activities must all be one. See this. Jan 29, 2017 at 19:32

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.