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Specifically, can $\Delta G=\Delta G^{\circ}+RT \ln Q$ be used to calculate reaction quotients if we choose T to be a different temperature than the one provided by ΔG°?

For example, suppose that we are given ΔG° at 25°C and we are tasked with finding the equilibrium constant at 30°C. Do these equations hold? $$0 \stackrel{?}{=} \Delta G^{\circ}+RT \ln K$$ $$K \stackrel{?}{=} e^\frac{-\Delta G^{\circ}}{RT}$$

I believe this is illegal, as it contradicts the fact that some equilibria are less favored at higher temperatures. However, I'm having trouble finding confirmation online, especially since the above equation has no easily searchable name.

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    $\begingroup$ Delta G is temperature dependent. See also van't Hoff equation : $\frac{\mathrm{d}(ln K)}{\mathrm{d}T} = \frac{\Delta_r H)}{RT^2}$ // Try to learn more flexible searching, like site:stackexchange.com OR site:libretexts.org equilibrium constant temperature dependance // I am not aware of any law that forbids writing false equality statements, so it is probably legal, at least in most countries. $\endgroup$
    – Poutnik
    May 18, 2023 at 7:56
  • $\begingroup$ Related: chemistry.stackexchange.com/q/127313/72973. In both cases, temperature is an explicit part of the mathematical relationship but other quantities in it are highly temperature-dependent as well. $\endgroup$
    – Karsten
    May 19, 2023 at 11:17

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As mentioned in the comments, $K$ is itself temperature dependent. That is to say, $K = K(T)$ is a function of temperature, so

$$\Delta_\mathrm{r} G^\circ = -RT\ln K(T)$$

and the temperature dependence in $\Delta_\mathrm{r} G^\circ$ not only arises from the $RT$ term but also the $K$ term.

The dependence of $K$ on temperature is described by the van 't Hoff equation. I've written about it here (a while ago!).

$$\frac{\mathrm d(\ln{K})}{\mathrm dT} = \frac{\Delta_\mathrm{r} H^\circ}{RT^2}$$

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