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Alternatively, for the same reaction, I thought that maybe this equation would help:ΔG = -RT ln KΔG = -RT ln ([B']/[A]) Using B' since the forum turns it into the bold command.Since we're using 1.00 mol in the beginning, thenΔG = -RT ln ((1-[A])/[A])Since we're using x = [A] to make it look neater,ΔG = -RT ln ((1-x)/x)Differentiating w.r.t. x,δΔG/δx = - RT x/(1-x) * δ((1-x)/x)/δxδΔG/δx = - RT x/(1-x) * - x^{-2}δΔG/δx = RT /[x(1-x)]But this expression does not have a δΔG/δx = 0 so obviously I'm missing something...

Hmm, that's right but once re-expressed isn't it that the Gibbs Free Energy at the state before equilibrium can be found this way?As in, when I read the equation, I see it as Delta G depends on the logarithm of the ratio of B to A. So in a sense, it does vary but I may have made that doubtful by using K since K is the equilibrium constant.Any suggestions how we could then evaluate this?