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    If we have two equilibrium constants K1 and K2, measured at T1 and T2 respectively, for a certain reaction, should there not be a ΔG° relationship between them? If ΔG°=-R*T*ln(K) then it follows that ln(K2/K1)=-(ΔG°/R)*(1/T2-1/T1) as far as I can see (just divide K2 by K1). But then if you try replacing with ΔG°=ΔH°-T·ΔS°, ΔS° falls out and we end up with ln(K2/K1)=-(ΔH°/R)*(1/T2-1/T1). How can both formulae be correct - doesn't that suggest ΔG°=ΔH°, which is not necessarily (almost never) true?
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    (Original post by Big-Daddy)
    If we have two equilibrium constants K1 and K2, measured at T1 and T2 respectively, for a certain reaction, should there not be a ΔG° relationship between them? If ΔG°=-R*T*ln(K) then it follows that ln(K2/K1)=-(ΔG°/R)*(1/T2-1/T1) as far as I can see (just divide K2 by K1). But then if you try replacing with ΔG°=ΔH°-T·ΔS°, ΔS° falls out and we end up with ln(K2/K1)=-(ΔH°/R)*(1/T2-1/T1). How can both formulae be correct - doesn't that suggest ΔG°=ΔH°, which is not necessarily (almost never) true?
    I think this is because you are assuming dH to be T independent (Which it isn't)
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    (Original post by Big-Daddy)
    If we have two equilibrium constants K1 and K2, measured at T1 and T2 respectively, for a certain reaction, should there not be a ΔG° relationship between them? If ΔG°=-R*T*ln(K) then it follows that ln(K2/K1)=-(ΔG°/R)*(1/T2-1/T1) as far as I can see (just divide K2 by K1). But then if you try replacing with ΔG°=ΔH°-T·ΔS°, ΔS° falls out and we end up with ln(K2/K1)=-(ΔH°/R)*(1/T2-1/T1). How can both formulae be correct - doesn't that suggest ΔG°=ΔH°, which is not necessarily (almost never) true?
    Why does ΔS "fall out"?
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    (Original post by JMaydom)
    I think this is because you are assuming dH to be T independent (Which it isn't)
    Possibly ... but then both relations are wrong, no? I've seen the ΔH° relationship all over the Internet, but derived the ΔG° one for myself, which I've seen nowhere else but in which I can't find the flaw ...

    (Original post by charco)
    Why does ΔS "fall out"?
    Try the derivation, it naturally falls out. Go from ΔH°-T1*ΔS°=-R*T1*ln(K1) and ΔH°-T2*ΔS°=-R*T2*ln(K2) and try and find ln(K2/K1), you will find ΔS° is not needed. (Of course we are assuming that ΔH° and ΔS° are temperature independent as JMaydom said above.)

    If you'd rather just see the equation derived, here: http://www.chem.ufl.edu/~itl/4411/lectures/lec_v.html.

    It's not ln(K2/K1)=-(ΔH°/R)*(1/T2-1/T1) which is in doubt, it's definitely correct. ln(K2/K1)=-(ΔG°/R)*(1/T2-1/T1) is in doubt.
 
 
 
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