1
$\begingroup$

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?

Improve this question
$\endgroup$
7
  • 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. $\endgroup$
    – orthocresol
    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
  • 1
    $\begingroup$ $\Delta_\mathrm{r}G$ 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. $\endgroup$
    – orthocresol
    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 $\endgroup$
    – orthocresol
    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

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.