enter image description here

My interpretation of this image is that $\Delta G^\circ$ is equal to the difference between $G^\circ$ of the products and $G^\circ$ of the reactants.

But, I've also read $\Delta G^\circ$ as being given for calculating the difference in free energy between a point when $Q=1$ and the equilibrium position. That is, $\Delta G^\circ=-RT\ln(K)$.

Which one of these is the accurate interpretation?

  • 1
    $\begingroup$ $\Delta_\mathrm{r} G^\circ = -RT\ln K$ is undoubtedly correct, but "the difference in free energy between a point when Q=1 and the equilibrium position" sounds suspect. $\endgroup$ Apr 28, 2018 at 23:30
  • 1
    $\begingroup$ hmm, won't ΔG = ΔG° when Q= 1? is it more precise to say that ΔG° is equal to the ΔG btw q=1 and k? furthermore, im trying to visualize these equations against the curve above ( i wish the y-axis had some sort of values, even if just throw away for the point of illustration). anyway, is it incorrect to thing of q=1 as being a point being XkJ higher on the curve relative to the equilibrium point if ΔG° = -XkJ ? $\endgroup$
    – gerry
    Apr 28, 2018 at 23:43
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    $\begingroup$ $\Delta_\mathrm{r}G$ and $\Delta_\mathrm{r}G^\circ$ represent slopes of the curve (specifically, $\Delta_\mathrm{r}G$ refers to the slope at any one specified point and $\Delta_\mathrm{r}G^\circ$ refers to the slope when $Q=1$) but the Gibbs energy $G$ refers to the actual value of the curve. The thing is, there is no reason why $\Delta_\mathrm{r}G^\circ$, which is a slope, should relate to the difference in the value of $G$ at two points - unless the graph is perfectly linear, which it isn't. Note also that the units of $\Delta_\mathrm{r}G^\circ$ are kJ/mol, not kJ. $\endgroup$ Apr 29, 2018 at 0:00
  • $\begingroup$ The main problem in this diagram, I suppose, is the conflation of two very different concepts: changes in the Gibbs free energy of the system (i.e. the difference between two values of the curve, which is correctly denoted $\Delta G$) with the reaction Gibbs free energy (which is a slope and is more properly denoted with $\Delta_\mathrm{r}G$). This might help: chemistry.stackexchange.com/q/50569/16683 $\endgroup$ Apr 29, 2018 at 0:13
  • $\begingroup$ if $\xi$ is the extent of reaction the slope of the graph is $\left(\partial G/\partial \xi\right)_{T,p}\equiv\Delta_rG=\Delta G_r^\mathrm{O}+RT\ln(Q)$, at equilibrium (point 3 on the plot) $\Delta_r G =\left(\partial G/\partial \xi\right)_{T,p}=0$ $\endgroup$
    – porphyrin
    Apr 29, 2018 at 6:13


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