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My understanding of standard free energy is that it is only valid at the specified temperature, which is typically 25$^\circ$C. Thus, if T changes, so does $\Delta$G$^\circ$$_{rxn}$ according to this equation:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}$$

(assuming $\Delta$H$^\circ_{rxn}$ and $\Delta$S$^\circ_{rxn}$ are constant with temperature, which is a common assumption I'm finding in texts)

For example, $\Delta$G$^\circ_{rxn}$ at 25$^\circ$C would be different than at 250$^\circ$C. My confusion is with regards to $\Delta G^\circ_{rxn}$'s relationship to chemical equilibrium (or reactions) as:

$$\Delta G^\circ_{rxn}=-RT~ln~K$$

I've seen resources on the internet treat this oppositely, which has caused me great confusion. Let's take the Haber Bosch as an example, where we want to find K at 250$^\circ$C: $$\ce{N2 + 3H2 <=> 2NH3}$$ From standard tables, at 25$^\circ$C (all kJ, mole, and K):

$${H_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1306$$

$${N_2}: \Delta G^\circ=0, \Delta H^\circ=0, \Delta S^\circ=0.1915$$

$${NH_3}: \Delta G^\circ=-16.48, \Delta H^\circ=-46.1, \Delta S^\circ=0.192$$

For this reaction:

$$\Delta G^\circ_{rxn}=2(-16.48)-(0+0)= -32.96~kJ/mole$$

$$\Delta H^\circ_{rxn}=2(-46.1)-(0+0)= -92.2~kJ/mole$$

$$\Delta S^\circ_{rxn}=2(0.192)-[(0.1915)+3(0.1306)= -0.1993~kJ/mole~K$$

First approach (how I think it is):

1. Calculate $\Delta$G$^\circ$ at 250$^\circ$C:

$$\Delta G^\circ_{rxn}=\Delta H^\circ_{rxn}-T\Delta S^\circ_{rxn}=-92.2-(523.15 K)(-0.1993)=12.06~kJ/mole$$

2. Calculate K (assuming kJ, mole, and K for R and other values):

$$K=e^{-\Delta G/RT}=e^{-12.06/[(8.314E-3)(523.15 K)]}=0.062$$

Second approach (also commonly seen):

1. Calculate K using $\Delta$G$^\circ$ from tables, i.e., 25$^\circ$C, but inserting a new T:

$$K=e^{-\Delta G/RT}=e^{32.96/[(8.314E-3)(523.15)]}=1954.6$$

Edit: I want to also add that this reaction is more favorable at lower temperatures, so Approach 1 makes sense and Approach 2 does not, even though many treat problems this way. Also, I get $K=10^{5.77}$ at 25$^\circ$C, so the decreasing K matches up.

Question: Which one do you consider correct and why is there so much confusion and discrepancy in online tutorials? If this first approach is correct, is the error in the second common because of the misinterpretation of "standard"?

And I'd like to add: do the rules follow for $\Delta G_{rxn}$ and non-equilibrium as well? That is:

$$\Delta G_{rxn}=\Delta G^\circ_{rxn}+RT~ln~Q$$

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    $\begingroup$ you're right that the error in the second approach is due to the misinterpretation of the "standard" term... Although the second method seems right at first glance, the error is that the $\Delta G^o$ is not the value to be used at this temperature $\endgroup$ – AbhigyanC Aug 31 '17 at 17:30
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    $\begingroup$ To expand on what @Abhigyan has said, the standard enthalpy and entropy of formation (and thus free energy of formation) are functions of temperature. This certainly has to be taken into account. One approach for including the temperature dependence uses the Van't Hopf equation, and the other approach is to determine the changes in H and S with temperature. $\endgroup$ – Chet Miller Aug 31 '17 at 17:52
  • $\begingroup$ @Chester Miller. Thank you. So, why do some texts claim that std change in enthalpy and entropy dependence on temperature can be ignored for calculations of standard delG at different temperatures? What is their basis for this? $\endgroup$ – prof.kvothe Aug 31 '17 at 18:27
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    $\begingroup$ I guess it's OK if the temperature change is not very large. But we do know the enthalpy and entropy are functions of temperature. $\endgroup$ – Chet Miller Aug 31 '17 at 18:45
  • $\begingroup$ Here is an example where someone took the second approach: thoughtco.com/free-energy-and-pressure-example-problem-609491 $\endgroup$ – prof.kvothe Aug 31 '17 at 21:06

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