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?




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.


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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.


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