Why is DeltaGo Temperature Dependent in Standard State?

[Conservation]

I see why, numerically speaking, deltaGo, defined as deltaGo = - RT ln K or deltaGo = deltaHo - TdeltaSo would be temperature dependent. But why is deltaGo temperature dependent, when it is simply Gibbs Free Energy at Standard State?

Is this essentially saying that the definition of standard state differs per temperature given? As in deltaGo=A for the "standard" state of 298K and 1 bar, but deltaGo=B for the "standard" state of 312K and 1 bar?

Thank you.

 

[Yanick]

Standard state does not take into account the temperature of a system. This is confusing because you read and hear about STP (standard temperature and pressure) which is not the same as saying standard state from the perspective of thermodynamics.

Ref: http://goldbook.iupac.org/S05925.html