[Big-Daddy]

I'm getting some conflicting ideas from different textbooks and Internet sites regarding the equations of entropy and Gibbs free energy. Please help me see where I'm wrong because the way I see it so far one of my sources must be making a mistake.

ΔS(total)=ΔS(system)+(-ΔH/T)

The definition of Gibbs free energy is:

ΔG=ΔH-(T*ΔS(system))

Therefore:

ΔG=-T*ΔS(total)

ΔS(total)=R*ln(K)
ΔG=-RT*ln(K)

All consistent with each other.

This leads to us being able to solve for the equilibrium constant at any T with ΔG or for any T with one equilibrium constant at another T already known and ΔH. ΔS(total) provides the role of telling us whether the reaction can proceed (if positive) or not (if negative), which ΔG can also do (if negative, it proceeds; if positive it does not), and in both cases if ΔS(total) or ΔG is known, as well as temperature T, you can calculate K at that temperature. In the case of ΔS(total) you need only know the value of ΔS(total) to get K (though admittedly you wouldn't know what temperature that K relates to if you didn't work out ΔS(total) using T in the first place), whereas in the case of ΔG you need to know the temperature and can then work out K at that temperature. ΔS(total), I assume, plays no more role, and nor does ΔS(system)? (Except, as I put above, in the derivation of ΔG itself, but it carries no more important meaning now.)

Please let me know if this is all correct. I'm extremely anxious to know if my ideas are accurate.

My issue is that many websites I look on are confusing the issue. For example, this one makes life difficult for me: http://www.chem1.com/acad/webtext/thermeq/TE5.html because no matter how hard I try I can't get the right answer on the calculation ("Problem Example 1").

[Yggdrasil]

This leads to us being able to solve for the equilibrium constant at any T with ΔG or for any T with one equilibrium constant at another T already known and ΔH.

This is not true.  In general, ΔG and ΔS(total) vary with temperature, so the ΔG value at one temperature will not work at another temperature (the same applies to ΔS(total)).

My issue is that many websites I look on are confusing the issue. For example, this one makes life difficult for me: http://www.chem1.com/acad/webtext/thermeq/TE5.html because no matter how hard I try I can't get the right answer on the calculation ("Problem Example 1").

Perhaps if you show your approach to solving the problem, we can help you determine where you made an error.

[Big-Daddy]

This leads to us being able to solve for the equilibrium constant at any T with ΔG or for any T with one equilibrium constant at another T already known and ΔH.

This is not true.  In general, ΔG and ΔS(total) vary with temperature, so the ΔG value at one temperature will not work at another temperature (the same applies to ΔS(total)).

In the case of ΔS(total) you need to be given the entropy at a certain temperature and can then work out K at that temperature, but you don't necessarily need to be told what the temperature is to perform the calculation.

In the case of ΔG, I was under the impression that ΔG itself cancels out to leave behind just ΔG^O=-RT*ln(K), which would indicate the standard ΔG of reaction is the one taken in the equation (standard meaning 1 bar pressure, 1 mol/dm³ concentrations and 298 K) and thus, with whatever T you have known, you can work out K at that T. Or would it be necessary for this ΔG^O to nonetheless refer to the "standard" Gibbs' free energy at whichever temperature you're at? This depends on the definition of "standard", which I'm unclear about - does it include a temperature of 298, or can it refer to any temperature (and thus its standardization is based on pressures of 1 bar)?

[Yggdrasil]

In the case of ΔS(total) you need to be given the entropy at a certain temperature and can then work out K at that temperature, but you don't necessarily need to be told what the temperature is to perform the calculation.

That is correct, if you know ΔS(total) at an unknown temperature, you can calculate the K for that unknown temperature.  However, it is rarely the case that you would know ΔS(total).  At best, you could calculate ΔS(system) from the standard entropies.

In the case of ΔG, I was under the impression that ΔG itself cancels out to leave behind just ΔG^O=-RT*ln(K), which would indicate the standard ΔG of reaction is the one taken in the equation (standard meaning 1 bar pressure, 1 mol/dm³ concentrations and 298 K) and thus, with whatever T you have known, you can work out K at that T. Or would it be necessary for this ΔG^O to nonetheless refer to the "standard" Gibbs' free energy at whichever temperature you're at? This depends on the definition of "standard", which I'm unclear about - does it include a temperature of 298, or can it refer to any temperature (and thus its standardization is based on pressures of 1 bar)?

The ΔG^o has to have been measured/calculated for whatever temperature you wish to calculate K.  For example, consider H₂O_(l) H₂O_(s).  K > 1 for T < 273 K, but K < 1 for T > 273 K.  This would not be possible if ΔG^o were invariant with temperature.