Does the equation G=-RTlnK work only if G is a standard gibbs free energy?

[m25]

Does the equation
G=-RTlnK
work only if G is a STANDARD gibbs free energy?(ie, under standard condition)

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[amy_k]

Yes that is correct it is dG0 = - RTlnK

This comes from the equation:
dG = dG0 + RTlnK

Since dG = 0 at equilibrium,

dG0 = -RTlnK

 

[m25]

Yes that is correct it is dG0 = - RTlnK

This comes from the equation:
dG = dG0 + RTlnK

Since dG = 0 at equilibrium,

dG0 = -RTlnK

Wait are you saying that dG=0 when under standard condition?

 

[Doctor Dream]

Wait are you saying that dG=0 when under standard condition?

By definition delta G is going to be the same as delta Go under standard conditions. dG isn't going to equal zero under standard conditions (because dGo would be zero too), but the equation dGo = -RTlnK assumes the reaction is in equilibrium I think (note we are using K and not Q), which is why dG=O.

 

[amy_k]

Yes, since we have K in this equation, we are at equilibrium and dG = 0.

 

[m25]

Yes, since we have K in this equation, we are at equilibrium and dG = 0.

By definition delta G is going to be the same as delta Go under standard conditions. dG isn't going to equal zero under standard conditions (because dGo would be zero too), but the equation dGo = -RTlnK assumes the reaction is in equilibrium I think (note we are using K and not Q), which is why dG=O.

Wait, so are you guys saying that the equation
dG = dG0 + RTlnK or dG = dG0 + RTlnQ
is the general equation that can be applied in ANY condition, and the equation
dG0 = -RTlnK
works only when it's under standard condition AND the reaction is in equilibrium?
And being under standard condition is a completely different matter than from being in equilibrium?
Actually, the standard condition means that it's occuring at 25C and at 1M, having having 1M will always make K=1, so wouldn't that make all standard gibbs free energy equal to 0? Or am I missing some sort of big concept here?