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Gibbs Free energy vs Gibbs Free energy at standard state

  1. Sep 20, 2013 #1
    ΔG° is the measure of Gibbs Free energy change at 1 bar but no specified temperature and also the stoichiometric amounts depending on the equation of the chemical reaction.

    For example, if X ->2Y then the ΔG° would be equal to ΔG°f(2Y)-ΔG°f(X). While for ΔG it is a general term for Gibbs Free Energy that might not have the same properties as the ΔG° (1 bar, stoichiometric amounts)

    So I was wondering, for a equilibrium reaction the ΔG=0 as when it reaches the lowest point the gradient=0. At this point, there is a certain composition of reactants to products that would not be 0% reactants and 100% products as that would mean the reaction is irreversible and reaches a lowest point at 100% products.

    The graph for the reversible reaction would look like this: http://postimg.org/image/r9vtp0fh3/

    So now in most questions, such as in phase changes questions they would give us the ΔH° and ΔS° and they then would want us to find the temperature at which equilibrium takes place. So using the formula ΔG°=ΔH°-TΔS° we would let ΔG° be 0 and solve for T. The part that i don't get is that ΔG° is a non-zero value and that we shouldn't be able to use ΔH° or ΔS° to find ΔG because either ΔH° or ΔS° represents 100% complete reaction. So I'm not sure why we are allowed to do those thing actually.

    Thanks in advance for the help :)
     
  2. jcsd
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