This question follows on from a great post on comparing $\Delta_\mathrm{r} G^{\circ}$ to $\Delta_\mathrm{r} G$ answered here

The G vs ξ graph that is used in the answer is an excellent tool to understand how $\Delta_\mathrm{r} G$ changes with the extent of reaction: enter image description here

If $\Delta_\mathrm{r} G$ is found on this graph by the slope of the line, what happens at the two extremes, where we have pure reactants on the left ($Q = 0$) and pure products on the right ($Q ⟶ ∞$)?

The equation to calculate $\Delta_\mathrm{r} G$ is: $$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^{\circ} + RT\ln Q$$

This would give $\Delta_\mathrm{r} G ⟶ -∞$ (vertical slope asymptotic to the y axis) when $Q=0=ξ$ (i.e., when we have pure reactants). Logically, that would mean that absolutely pure reactants have infinite G (which doesn't seem right).

A similar problem occurs as we approach pure products: $\Delta_\mathrm{r} G ⟶ ∞$ when $Q⟶∞,ξ⟶max$ (vertical slope asymptotic to a line at ξ = max).

Can anyone reconcile this graph with the equation to calculate $\Delta_\mathrm{r} G$?


3 Answers 3


After a lot of help, I have the following to suggest as an answer:

Imagine a reaction with $\Delta_\mathrm{r} G^{\circ} = -1000Jmol^{-1}$ at 298K. Using the following equations: $$\Delta_\mathrm{r} G^{\circ} = - RT\ln K$$ $$e^{-\frac{\Delta_\mathrm{r} G^{\circ}}{RT}} = K$$ This would give a K value of 1.50, which indicates the Q value on the curve where G is at a minimum (favoring products slightly in the equilibrium mixture). Using another equation (valid e.g. for a reaction $\ce{A(aq) <=> B(aq)}$): $$ξ = \frac{Q}{1+Q}$$

Gives the ξ value (1.50/2.50 = 0.600) which is the x coordinate of the minimum on the graph, where the x axis runs from ξ = 0 (pure reactants) to ξ = 1 (pure products). This graph is pretty similar to the one posted in the question:

enter image description here

Using the equation:$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^{\circ} + RT\ln Q$$

And substituting in:$$Q = \frac{ξ}{1-ξ}$$

Gives us:$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^{\circ} + RT\ln \frac{ξ}{1-ξ} $$


$$\frac{\delta G}{\delta \xi} = \Delta_\mathrm{r} G^{\circ} + RT\ln \frac{ξ}{1-ξ} $$

$\Delta_\mathrm{r} G^{\circ}$ is a constant even as G changes as the value of ξ changes, so are R and T. This differential equation can be separated and integrated to gives us the original function of how G depends on ξ:

$$\begin{align} \int \delta G = \Delta_\mathrm{r} G^{\circ}ξ + RT \int \ln \frac{ξ}{1-ξ}\delta \xi \end{align}$$

$$\begin{align} G = \Delta_\mathrm{r} G^{\circ}ξ + RT~(ξ \ln (ξ) + (1-ξ) \ln (1-ξ) - 1 + C)\end{align}$$

Graphing this on desmos gives us the correct graph for how G varies with ξ. enter image description here

The graph shows the correct ξ value for the equilibrium point and also the difference in G between the "pure" reactants and products. All of this is as predicted by the theory.

If you trace the line along the domain of x towards zero (or 1) the y values increase until the value becomes undefined, but I have zoomed in until my brain hurt and the value never seems to shoot upwards (as I thought the maths predicted). L'Hopital's rule seems to prevent this from happening.

I can't pretend that I fully understand the maths (I don't), maybe others' comments will help here. If anyone wants to use the desmos calculator to produce their own plots (you can change T or $\Delta_\mathrm{r} G^{\circ}$ to other values to see the effect) please do so, I may use this in teaching the material in the future.

I credit @orthocresol @Night Writer and @Karsten Theis with helping me. Also Darrell Bach and my students Nick Kleiderer and Sylver Riddell.

****** Addition to deal with limits ******

The limits of G at each end of the domain of ξ are below. (For ease of reading I will use x = ξ)

$$\begin{align} G = \Delta_\mathrm{r} G^{\circ}ξ + RT~(ξ \ln (ξ) + (1-ξ) \ln (1-ξ) - 1 + C)\end{align}$$ $$ G(x) = \Delta_\mathrm{r} G^{\circ}x + RT~(x \ln (x) + (1-x) \ln (1-x) - 1 + C)$$

Limit of G when $x \to 0 ^+$ :

$$ \lim_{x \to 0 ^+} G(x) = \lim_{x \to 0 ^+} [\Delta_\mathrm{r} G^{\circ}x + RT~(x \ln (x) + (1-x) \ln (1-x) - 1 + C)] $$

$$\lim_{x \to 0 ^+} G(x) = \lim_{x \to 0 ^+} [\Delta_\mathrm{r} G^{\circ}x - RT + RTC] + \lim_{x \to 0 ^+} [RT(x \ln (x) + (1-x) \ln (1-x))] $$

Note (1): $ \lim_{x \to 0 ^+}[(1-x) \ln (1-x)] = 0 $

Note (2): $\lim_{x \to 0 ^+} [x \ln (x)] = 0$ (by l'Hopital's rule)

$$\lim_{x \to 0 ^+} G(x) = - RT + RTC + RT(0+0) $$

$$\lim_{x \to 0 ^+} G(x) = - RT + RTC $$

$$\lim_{x \to 0 ^+} G(x) = RT(C-1) $$

Limit of G when $x \to 1 ^-$ :

$$ \lim_{x \to 1 ^-} G(x) = \lim_{x \to 1 ^-} [\Delta_\mathrm{r} G^{\circ}x + RT~(x \ln (x) + (1-x) \ln (1-x) - 1 + C)] $$

$$\lim_{x \to 1 ^-} G(x) = \lim_{x \to 1 ^-} [\Delta_\mathrm{r} G^{\circ}x - RT + RTC] + \lim_{x \to 1 ^-} [RT(x \ln (x) + (1-x) \ln (1-x))] $$

Note (1): $ \lim_{x \to 1 ^-}[(1-x) \ln (1-x)] = 0 $ (by l'Hopital's rule)

Note (2): $\lim_{x \to 1 ^-} [x \ln (x)] = 0$

$$\lim_{x \to 1 ^-} G(x) = \Delta_\mathrm{r} G^{\circ} - RT + RTC + RT(0+0) $$

$$\lim_{x \to 1 ^-} G(x) = \Delta_\mathrm{r} G^{\circ} - RT + RTC $$

$$\lim_{x \to 1 ^-} G(x) = \Delta_\mathrm{r} G^{\circ} + RT(C-1) $$

These limits to G show that there is a value to G as you approach pure reactants and products along the curve of G vs ξ. The algebra shows some interesting points, which agree with other parts of the theory of equilibria and state functions:

  1. The curve of G vs ξ can be thought of as the sum of a simple line between the G of reactants and the G of products $$G = \Delta_\mathrm{r} G^{\circ}ξ + RT(C-1)$$

for $C = 0$

$$G = \Delta_\mathrm{r} G^{\circ}ξ - RT$$

and a second function (which is always negative) that deals with the Gibbs free energy of mixing $$ + RT~(ξ \ln (ξ) + (1-ξ) \ln (1-ξ) - 1)$$

This is seen as a dip from the line, reducing the Gibbs free energy in the mixed system. This disappears as ξ approaches 0 or 1, as the system is becoming pure reactants or products:

enter image description here

  1. For a reaction where K > 1 and $\Delta_\mathrm{r} G^{\circ} <0$ (such as the graphs in this example) the pure products are lower in G than the reactants, but the equilibrium mixture is lower still
  2. $\Delta_\mathrm{r} G^{\circ}$ refers to the tangent of the curve at standard conditions ($Q=1$ and $ξ = 0.5$) and also to the slope of the straight line that connects $G_{reactants}$ and $G_{products}$. As ξ changes by 1 from reactants to products, $\Delta_\mathrm{r} G^{\circ}$ is also the absolute $\Delta G$ between pure reactants and products
  • 1
    $\begingroup$ Two short comments, since I did not inspect your math thoroughly yet. (1) Again, it's the slope that diverges at 0 and 1, not $\Delta G$ which you obtained by integration. (2) There is a small error which I spot right away in the $\Delta G^\circ$ term: should be $\times (\xi_f - \xi_i)$ but this gets buried in the integration constant. $\endgroup$
    – Buck Thorn
    May 21, 2019 at 16:43
  • $\begingroup$ @Night Writer. Thanks, I'll inspect and make changes. $\endgroup$
    – Withnail
    May 21, 2019 at 17:04
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    $\begingroup$ I also zoomed and zoomed. The trick is to seeing the increasing slope is keep the aspect ratio constant, otherwise you don't see the larger and larger slope as you zoom in. If you express ξ as a negative power of 10, the slope is linear with the exponent as ξ approaches zero, so it is subtle. Another way of saying it is that in order to see extreme slopes, you have to zoom in $10^\text{extremely}$ and keep the aspect ratio or, when letting the graphing program choose the aspect ratio, pay attention to the distinct scale of the two axes. $\endgroup$
    – Karsten
    May 21, 2019 at 17:47
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    $\begingroup$ Note that from your own analysis you can conclude that the initial value of the (integral) Gibbs free energy is $G_i=RT(C−1)$ $\endgroup$
    – Buck Thorn
    May 22, 2019 at 19:59
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    $\begingroup$ Nice! Of course, now your answer is much more complete than mine... $\endgroup$
    – Karsten
    May 22, 2019 at 21:49

G is a finite quantity for ξ = 0 or max. The slope of the graph is vertical on the extremes, though. This is because the chemical potential for reagents approaches negative infinity on one side, and that of the product positive infinity on the other side. (A vertical or infinite slope does not mean that the function value has to be infinite - a half-circle or the letter U is an example to illustrate).

The interpretation is that (unless $\Delta_r G^\circ$ is of extremely large magnitude), there are always a least a couple molecules of solute or gas species at equilibrium.

The sketch of the graph could be made more accurate by adjusting the slope at the ends accordingly. For a graph of $\Delta_r G$, see here: https://chemistry.stackexchange.com/a/115544

  • $\begingroup$ Thanks for the link, I have seen that graph, it is a good one (thanks to @Night Writer for posting that) and it makes sense. It seems that the reason for there being an absolute value for G (rather than infinite) for a pure species is that the species needs to be in equilibrium with something (even a few molecules of products will do) in order for the the equations we are using to actually hold. So the graph essentially fails at the extreme values of ξ as this is no longer true (either no reactant, or no products exist). Would you agree? $\endgroup$
    – Withnail
    May 21, 2019 at 0:58
  • $\begingroup$ I would also like to thank @orthocresol for answering my initial comments. Hopefully they will get a chance to comment on the answers. $\endgroup$
    – Withnail
    May 21, 2019 at 1:18
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    $\begingroup$ @Withnail - The graph does not fail; x ln x approaches zero when x approaches zero. This is because x approaches zero "faster" than ln x approaches infinity. So there is no discontinuity (except for the usual one that substances are made of particles, and you go abruptly from zero to one particle, i.e. the x-axis is not continuous if you look too closely). $\endgroup$
    – Karsten
    May 21, 2019 at 3:18
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    $\begingroup$ @Withnail Thanks for sharing your insights! They are the same because of $\Delta_r G^\circ$s definition. We are imagining that the reaction proceeds while the concentrations remain at standard state, and it is per mole. So you can either see it as differential, or in this somewhat artificial reaction where you keep $\Delta_r G$ constant, either by having large vats of reactants, or by having very little react. I recommend Treptow (1996) J Chem Ed 79(1) 51-54. $\endgroup$
    – Karsten
    May 21, 2019 at 19:32
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    $\begingroup$ @Withnail It looks like we've attempted roughly the same thing. I also edited my answer and added what seems a satisfying if somewhat anticlimactic math explanation. It's been an interesting set of detailed observations you've made. $\endgroup$
    – Buck Thorn
    May 21, 2019 at 20:00

Logically, that would mean that absolutely pure reactants have infinite G (which doesn't seem right).

It's the slope (rate of change) that is infinite, not the actual property (Gibbs free energy).

$\Delta_r G$ (expressed sometimes as $\Delta G'$, sometimes you'll see just $\Delta G_m$ without subscripts indicating explicitly that this is the molar Gibbs free energy change for a reaction) is a partial molar quantity and describes the change in the Gibbs free energy per mole unit change in the reaction progress coordinate. It is a differential quantity (a slope).

If you want to know the value of the Gibbs free energy at any point during the reaction, what you actually want $^\dagger$ to compute at constant $T$ and $p$ is

$$G=\sum_i n_i \mu_i \tag{1}$$

where $n_i$ and $\mu_i$ are the number of moles and chemical potential of substance $i$.

In order to arrive at the expression for $\Delta_r G$ you need to take the derivative of equation (1) with respect to the reaction progress coordinate ($d\xi=\frac{dn_i}{\nu_i}$):

$$\Delta_r G =\frac{dG}{d\xi}=\sum_i \mu_i \left(\frac{\partial n_i}{\partial\xi}\right)_{n_j}=\sum_i \nu_i \mu_i$$

(where use was made of the Gibbs-Duhem equation).

With appropriate substitutions this last expression can be reformulated into the more familiar expression

$$\Delta_r G =\Delta_r G^\circ + RT \log(Q)$$


$$\Delta_r G^\circ = \sum_i \nu_i \mu_i^\circ $$ and $$Q = \prod_{i} a_{i} ^{\nu_i}$$

$\dagger$ Want but can't as written. You instead compute $\Delta G$, the value as a difference relative to a reference state.

Mathematical aside

if the slope is infinite as ξ approaches zero, the line will never hit the y axis, so the y intercept should be infinite too

Consider the inverse problem, determining the change in G from its slope. This involves evaluating integrals of the sort $\int_{c_i}^{c_f}\log(c)dc$:

$$\begin{align} \int_{c_i}^{c_f}\log(c)dc = \left[c\log(c)-c \right]^{c_f}_{c_i} \end{align}$$

Now if one of the limits ($c_i$ or $c_f$) equals zero, it would appear at first glance to result in a singularity, but it doesn't. To see why we can evaluate $0 \times \log(0)$ by invoking L'H$\hat{\text{o}}$pital's rule:

$$\begin{align} \lim_{c\rightarrow 0} [c\log(c)] &=\lim_{c\rightarrow 0} \left[\frac{\log(c)}{1/c}\right] \\ &= \lim_{c\rightarrow 0} \left[\frac{1/c}{-1/c^2}\right]\\ &= \lim_{c\rightarrow 0} \left[-c\right]\\&=0 \end{align}$$

In the poster's answer an expression analogous to the following is provided for the integral of $\Delta_r G$:

$$G = G_i + \Delta_r G^\circ \xi + RT[\xi\log(\xi) + (1-\xi)\log(1-\xi)] $$

If plotted and expanded around $\xi=0$, this provides little evidence of the presumed infinite slope, thanks to the log dependence on $\xi$. The slope (free energy of reaction) near $\xi=0$ can be written

$$ \Delta_r G = \Delta_r G^\circ + RT\log(\xi)$$

The logarithmic dependence on $\xi$ guarantees that the slope goes to $-\infty$, but that it does so slowly, as seen in the following plots (same parameters as in the poster's answer):

enter image description here

  • $\begingroup$ Thanks for the detailed reply. I know that the equation gives us change in G per mol of reaction proceeding (ξ) - i.e. the slope of the graph, but if the slope is infinite as ξ approaches zero, the line will never hit the y axis, so the y intercept should be infinite too. Please let me know what I am missing. $\endgroup$
    – Withnail
    May 21, 2019 at 0:47
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    $\begingroup$ @Withnail - I edited my answer to give examples of finite functions with infinite slope - a half circle or U shape. $\endgroup$
    – Karsten
    May 21, 2019 at 3:21
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    $\begingroup$ I searched for images that show the correct slope for no reactants or no products, and most "schematic" drawings have it wrong. Two exceptions: media.cheggcdn.com/… and commons.wikimedia.org/wiki/File:Entropy_of_Mixing.jpg @Withnail $\endgroup$
    – Karsten
    May 21, 2019 at 13:22
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    $\begingroup$ I am working on a maths solution, I have a colleague helping with the calculus, and will post as an answer when we are done. I appreciate your help, the function of G vs ξ is quite interesting as ξ approaches 0 or 1 @Karsten Theis $\endgroup$
    – Withnail
    May 21, 2019 at 13:53
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    $\begingroup$ Thanks for the graphs of G (and delta G) dependence on ξ. I will post the work on the limits when I get a sec. $\endgroup$
    – Withnail
    May 21, 2019 at 20:35

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