galpinj

Hey everyone, Why do we include ΔG° in this equation? Why does the Gibbs Free energy at standard conditions matter when finding the Gibbs Free energy of a certain reaction? I know that it is important for comparison and can help to determine which way a reaction will go, but I do not see why it is a fundamental part of the equation. Why does the free energy of a reaction have to rely on the free energy from the standard form? That seems very strange to me, and you don't see that kind of dependence in ΔG = ΔH - TΔS. Is the equation ΔG = ΔG° + RT ln(Q) somehow derived from ΔG = ΔH - TΔS? Thank you for any help!