When calculating the Gibbs energy of reaction for reaction mixtures at arbitrary concentrations, you use the following expression: $$\Delta_\mathrm r G = \Delta_\mathrm r G^\circ + RT\ln Q$$ The term $RT\ln Q$ has to do with the entropy of mixing, we are told.

How does that work for reactions that are continuously mixed? If I have a reaction like

$$\ce{A(aq) + B(aq) <=> C(aq) + D(aq)}$$


$$\ce{A(g) + B(g) <=> C(g) + D(g)},$$

the reactants themselves are already a mixture. In fact, when there is a solvent, even a single reactant is part of a solution, i.e. already in a mixture. If (hypothetically) the reaction goes to completion, you still have a mixture. So why do we have to consider the entropy of mixing if the reaction mixture is already mixed?

  • $\begingroup$ Do you mean that A+B are continuously added or that an equilibrium already exists as shown in your equation ? The calculation for entropy of mixing is given in most textbooks as well as many times on this site. $\endgroup$
    – porphyrin
    Aug 7 at 19:24
  • 1
    $\begingroup$ Will you accept an answer based on a reaction of ideal gases? $\endgroup$ Aug 7 at 22:42
  • $\begingroup$ @porphyrin One scenario is mixing A and B, and calculating $\Delta_r G$ at various times while the reaction reaches equilibrium. $\endgroup$
    – Karsten
    Aug 8 at 1:17
  • $\begingroup$ @ChetMiller Yes, ideal gases or ideal solutions would be fine. The simpler, the better, I would think. $\endgroup$
    – Karsten
    Aug 8 at 1:18

3 Answers 3


So, I am gonna use the most basic definition of entropy - the statistic one

S = kB ln $\Omega$

$\Omega$ is the total number of microstates possible.

In your case, different microstates can be formed by arranging momenta , energies and also by arranging POSITIONS as any molecule can be found at any possible "position" in a solution.

I will use basic principles of Permutations and Combinations to find $\Omega$.

Suppose I have 'a' and 'b' moles of A and B initially , while no C and D initially present. After a time 't', x moles of A and B have reacted to form x moles of C and D.

\begin{align} \ce{A(aq) + B(aq) <=> C(aq) + D(aq)}\\ \end{align} \begin{align} \ce{t=0..... . a. . . . . . b. . . ....0... . . 0 ....... . . . . }\\ \end{align} \begin{align} \ce{t=t..... (a-x). . . . .(b-x). . ... x... . . x ...... . . . . . . }\\ \end{align}

The total number of molecules remain same in your example , and is (a+b+w) N0 , where N0 is Avogadro's number , and w is the number of moles of water.

Let me simplify the model by assuming there are (a+b+w) N0 positions available. Any molecule can "sit" at any position and can have any of the available value of momentum and energy. Changing either of position or momentum of any one molecule will lead to a new microstate.

But there's a catch. Suppose molecule 1 (m1) sits at position r1 and has energy E1 , momentum p1. Another identical molecule (i.e. the same compound) m2 has position r2 , energy E2 , momentum p2. If I exchange all - their positions, energies and momenta - I will not get a new microstate. But if m2 was not identical, it would surely have resulted in a new microstate.

So, assume I haven't put any molecule at any position. First I assign every position a value of energy and momentum , and write down each possible arrangement in my notepad (my notepad has infinite pages...so don't worry , it won't run out ; ) ). Then I count them. There turn out to be Z possible arrangements of energy and momentum.

Now , I start placing the molecules randomly ( there are just A , B and water present for now). How many unique arrangements can I have only due to virtue of positioning? That I can easily calculate. It will be

\begin{equation} V_{1} = \frac{[ (a+b+w) N_{0} ]!}{( aN_{0} )! * ( bN_{0} )! * ( wN_{0} )!} \end{equation}

So, total number of microstates ($\Omega$1) = V1 * Z
Hence, S1 = kB ln( V1 Z )

Now, after time t , I again do the same.
The number of arrangements of energies and momenta would still be same , i.e. , Z .

But, the number of arrangements due to positioning will now be ,

\begin{equation} V_{2} = \frac{[(a+b+w) N_{0}]!} {[(a-x) N_{0}]! * [(b-x) N_{0}]! * (xN_{0})! * (xN_{0})! * (wN_{0})! } \end{equation}

$\Omega$2 = V2 * Z
Hence, S2 = kB ln(V2 Z)

So, Change in Entropy $\Delta$S = S2 - S1 = kB ln (V2 / V1 )

Dividing V2 by V1 , one can easily see it is equal to \begin{equation} \frac{V_{2}}{V_{1}} = \binom{aN_{0}}{xN_{0}} * \binom{bN_{0}}{xN_{0}} \end{equation}

\begin{equation} Hence, \Delta S = k_{B} * \ln\left[ \binom{aN_{0}}{xN_{0}} * \binom{bN_{0}}{xN_{0}} \right] \end{equation}

where\begin{equation} \binom{n}{r}\end{equation} represents binomial term nCr

This is the increase in entropy due to "mixing".

I even plotted a graph to verify. Here's the graph.

I didn't know how to approximate its derivative (as the graph is broken when viewed closer) So I instead compared the graph with another one made by integrating Rln(K/Q). Both come out to be same at some value of K. This way, we have also calculated the Equilibrium Constant :)

NOTE: In the graph, I have also considered change in "Internal Entropies" apart from Entropy of Mixing. This "internal" entropy arise due to arrangements possible within a molecule. It is characteristic of a given compound.

The Black graph is the one made by integrating Rln(K/Q) and the Red haze visible behind it is the graph made by the formula we just derived.

I took a and b = 1
SA , SB , SC , SD (the molar "internal entropies" of A , B , C and D respectively) equal to 1 , 1.5 , 2.3 , 2 kJ/mol*K (not to scale....just example)
And K came out to be nearly 1.2

Y axis = Entropy change due to mixing , X axis = Moles of A and B reacted

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    $\begingroup$ Sorry. I don't understand the meaning of $\pu{ \frac{V_2}{V_1}}$ with your strange parenthesis. $\endgroup$
    – Maurice
    Aug 7 at 19:55
  • $\begingroup$ It is binomial term... (aNo C xNo) $\endgroup$ Aug 7 at 19:57
  • $\begingroup$ I took No (Avogadro's No.) equal to 500 in graphing calculator because its system broke down at higher values. But it doesn't matter as the values reach a limit on increasing No....So, graph is same for No = 500 and No = 6*10^23 $\endgroup$ Aug 7 at 20:08
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    $\begingroup$ minor formatting suggestions: using "\ln" will keep ln from being italicized, and "\left[" and "\right]" instead of just "[" and "]" will cause the size (ie height) to adjust properly to match the contents within the brackets. $\endgroup$
    – Andrew
    Aug 7 at 21:30
  • $\begingroup$ What a tour de force! But you showed nicely that for Q = 1, the microstates are maximal. To calculate K, you would have to know the standard entropy and enthalpy of reaction. If you just consider the entropy of mixing, your are setting these other quantities to zero, in which case you would expect K = 1 indeed. $\endgroup$
    – Karsten
    Aug 8 at 1:57

I'm a little unsure exactly what you are looking for but I try to show the connection between the Gibbs energy and mixing energy as reaction proceeds. We can start with the chemical potential $\mu$ and calculate the Gibbs function. Thus,

$$\displaystyle \mu_i=\mu_i^0+RT\ln(P/P^0)+RT\ln(x_i)$$

where $P$ is the total pressure, ($P^0$ is unit pressure to make the log dimensionless and is assumed to be included from now on), $x_i$ the mole fraction of species $i$ and $\mu_i^0$ is a function of temperature only, i.e independent of composition so is unchanged as the mole fraction changes $x_i =0\to 1$.

In the reaction $A+B=C+D$ the Gibbs function is

$$\displaystyle G=n_a\mu_a+n_b\mu_b+n_c\mu_c+n_d\mu_d$$

and if we assume a perfect gas mixture, using the first equation produces

$$\displaystyle G=[n_a\mu_a^0+n_b\mu_b^0+n_c\mu_c^0+n_d\mu_d^0+RT(n_a+n_b+n_c+n_d)\ln(P)]+RT\big(n_a\ln(x_a)+n_b\ln(x_b)+n_c\ln(x_c)+n_d\ln(x_d)\big)$$

The first two terms (in square brackets) is the total Gibbs function for the four gases as if each is in a separate container at pressure $P$, the term on the right is therefore the free energy of mixing, $RT\sum_i n_i\ln(x_i)$ and if divided by temperature is minus the entropy of mixing.

If the system initially consists of only one mole each of A and B then

$$\displaystyle n_b=n_a,\quad n_c=n_d=1-n_a,\quad x_a=x_b=n_a/2,\quad x_c=x_d=(1-n_a)/2$$

and now $G$ only depends on $n_a$ which can vary between $0\to 1$

$$\displaystyle G= n_a(\mu_a^0+\mu_b^0)+(1-n_a)(\mu_c^0+\mu_d^0)+2RT\ln(P)+2RT(n_a\ln(n_a/2)+(1-n_a)\ln((1-n_a)/2)$$

and without loss of generality choose the pressure to be unity, then with some rearranging,

$$\displaystyle G= n_a(\mu_a^0+\mu_b^0)+2RTn_a\ln(n_a/2)+(1-n_a)(\mu_c^0+\mu_d^0)+2RT(1-n_a)\ln((1-n_a)/2)$$

When $n_a=1$ so that only reactants are present we find that $G=\mu_a^0+\mu_b^0+2RT\ln(1/2)$ where $2RT\ln(1/2)$ has to be the free energy of mixing before any reaction has occurred and similarly when $n_a=0$ the same energy of mixing is present for the products there being only two species present in either case. Elsewhere both terms contribute to the mixing free energy.

The shape of the plot of $G$ vs. $n_a$ is due primarily to the log terms, if this term were absent the plot would follow the line W-V, see figure. The lower part of $G$ shows that the free energy at equilibrium becomes even smaller than that if there were complete conversion to products, because here there are four species in various proportions not just two. Entropy of mixing depends on the mole fractions is maximal at equilibrium.

The position of the minimum, and so the equilibrium, is determined by $\mu_a^0+\mu_b^0-\mu_c^0-\mu_d^0$ which is the difference in free energies between reactants and product in their pure states. This is the case because the mixing terms are symmetrical in $n_a$ but the sloping energy terms biases this. The minimum can be calculated by differentiating $(\partial G/\partial n_a)_{T,P}=0$ or more easily and just as generally via

$$\displaystyle dG=-SdT+VdP+(\mu_a+\mu_b-\mu_c-\mu_d)dn_a$$

where the stoichiometry of the reaction was used ($-dn_a = -dn_b=dn_c=dn_d$) and at constant $T$ and $P$ when $\partial G/\partial n_a=0$, we find that

$$\displaystyle \mu_a+\mu_b=\mu_c+\mu_d$$

and substituting for $\mu_i$ and mole fractions leads to the solution for $n_a$ at equilibrium just as if $G$ had been differentiated.

What this shows is that the same relationship exists at equilibrium as in the chemical equation $A+B=C+D$ from which the equilibrium constant is normally obtained and shows the connection to the entropy of mixing as the reaction proceeds. The reaction is therefore seen as a competition between attractive forces causing the reactants to form products on the one hand and increasing entropy forming a mixture on the other.

mixing energy Gibbs function of a reacting system


For a reaction of ideal gases, such as, say, aA+bB-->cC+dD, the $\Delta G_r$ in your equation is the change in Gibbs free energy in going at constant temperature T from a moles of pure A and b moles of pure B at arbitrary pressures $p_A$ and $p_B$, respectively, to c moles of pure C and d moles of pure D at arbitrary pressures $p_C$ and $p_D$, respectively. If the pressures were not arbitrary, but, instead happened to coincide with those for an equilibrium reaction mixture of A, B, C, and D, then $\Delta G_r$ would be equal to zero, and Q would be equal to the equilibrium constant for the reaction. Under these circumstances, the pure gas pressures would be equal to the partial pressures of the corresponding components in the equilibrium reaction mixture, and could be placed in equilibrium with the components in the mixture through 4 semipermeable membranes which allowed passage of only one of the corresponding components.

The challenge in all this is to figure out how to interpret all this in terms of a reversible process to go from the initial state of pure A and B to the final state of pure C and D so that $\Delta G_r$ can be expressed in terms of the initial and final pressures. This typically is facilitated by the use of a so-called van't Hopf equilibrium box.

  • $\begingroup$ I did not know about the van't Hoff equilibrium box, chemistry.stackexchange.com/a/127915 $\endgroup$
    – Karsten
    Aug 8 at 12:54
  • $\begingroup$ So for ideal gases, the entropy gain of mixing is the same as the entropy gain of expansion? It does not matter whether it is a pure gas or a mixture, the entropy depends only on the (partial) pressure. $\endgroup$
    – Karsten
    Aug 8 at 14:32
  • $\begingroup$ Basically, yes. The partial molar entropy of a component in an ideal gas mixture is the same as the molar entropy of the pure component at the same temperature and at a pressure equal to the partial pressure of the component in the mixture. This follows from Gibbs Theorem, $\endgroup$ Aug 8 at 18:11

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