Calculating ΔG at the extremes of reaction extent

Question

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:

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 $\Delta_\mathrm{r} G$?

Answer 1

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:

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

or:

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

$\Delta_\mathrm{r} G^{\circ}$ 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.

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:

2. 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
3. $\Delta_\mathrm{r} G^{\circ}$ 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

Answer 2

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

Answer 3

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

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

where

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

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

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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:

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

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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):