Figure 1: Reaction Space with Entropy Contours
It is often interesting to ask whether it is possible for a given transformation to proceed spontaneously – and if so, in which direction.
For starters, consider the reaction of carbon with oxygen to form carbon dioxide.
| C + O2 ⇒ CO2 (1) |
Under some conditions this reaction proceeds forward, as stated ... while under other conditions the reverse reaction occurs, i.e. the decomposition of carbon dioxide to form carbon and oxygen.
More generally, we need to consider other possibilities, such as the possible presence of carbon monoxide. This is now a multi-dimensional situation, as shown schematically in figure 1.
Consider the contrast:
| In one dimension, we can speak of a given transformation proceeding forward or backward. Forward means proceeding left-to-right as written in equation 1, while backwards means proceeding right-to-left. | In multiple dimensions, we cannot speak of forward, backward, right, or left. We need to specify in detail what sort of step is taken when the transformation proceeds. |
The term “transformation” is meant to be very general, including chemical reactions and phase transitions among other things. (We do not consider it useful, or even possible, to distinguish “chemical processes” from “physical processes”.)
Entropy and the second law of thermodynamics will play a central role in our analysis of spontaneity and reversibility.
Keep in mind that the second law is only one law among many. Other laws include conservation of energy (aka the first law of thermodynamics), other conservation laws, various symmetries, spectroscopic selection rules, et cetera.
A process can proceed only if it complies with all the laws. Therefore if a process is forbidden by one of the laws, it is unconditionally forbidden. In contrast, if the process is allowed by one of the laws, it is only conditionally allowed, conditioned on compliance with all the other laws.
Based on a second-law analysis alone, we can determine that a process absolutely will not proceed spontaneously. In contrast, a second-law analysis does not allow us to say that a process “will” proceed spontaneously. Until we do a more complete analysis, the most we can say is that it might proceed spontaneously.
We are on much firmer ground when it comes to reversibility. In the context of ordinary chemical reactions,1 if we know the reaction can proceed in one direction, it is reversible if and only if it is isentropic. That’s because of the reversibility of all the fundamental laws governing such reactions, except the second law. So if there is no barrier to the forward reaction, there should be no barrier to the reverse reaction, other than the second law. (If the reaction cannot proceed in either direction, it is pointless to ask whether it is reversible and/or isentropic.)
Terminology note: If the world were logical, an irreversible reaction would be called exentropic (in analogy with exergonic and similar terms) … but usually people just say "entropic" instead of "exentropic".
This is meant to be a review article, not a tutorial, so we will turn the normal pedagogical sequence on its head and start with the most general and most reliable formulation of the problem. Later we will discuss various restricted and simplified versions.
Here’s the fundamental, essential criterion: A transformation will not proceed in any direction that reduces the amount of total entropy. This is essentially a statement of the second law of thermodynamics. (The second law, and the definition of entropy, are discussed in reference 1.)
| Δ St ≥ 0 (2) |
To say the same thing the other way: Any step that produces a positive amount of total entropy is irreversible in the thermodynamic sense.
Suppose a transformation can proceed in a direction that increases the total entropy. Then the transformation in all likelihood will proceed in such a direction, in preference to directions that leave the total entropy unchanged.
In thermodynamic equilibrium, no such direction exists. The system might already be in a configuration of globally maximal entropy ... or it might be restricted by huge activation barriers, symmetries, selection rules, or other laws of physics such that it cannot move toward higher entropy at any non-negligible rate.
Note that the spontaneity and reversibility criteria are based on the total entropy (St), which consists of the entropy (S) inside the region of interest plus whatever entropy, if any, leaked out of the region across the boundaries during the transformation.
To quantify which directions correspond to increasing total entropy, it suffices to look at the exterior derivative, dSt. This is a vector; in particular it is a one-form, as opposed to a pointy vector. As such, it is best visualized as a set of contours, namely contours of constant St, as shown schematically in figure 1. (For details on one-forms and their application to thermodynamics, see reference 2.)
The system may sit at a point of maximal St (thermodynamic equilibrium), or it may skate along a contour of constant St (a reversible transformation) or it may step across the contours in any direction that increases St (an irreversible transformation).
This completes the general analysis. This is the whole story. This must be correct, since it is a direct consequence of the law of paraconservation of entropy. For an explanation of this law, see reference 1.
As our first simplification, we don’t need to differentiate St with respect to every imaginable direction, just the directions in which the system is free to move. That is, if there are constraints which forbid certain changes, we can neglect the corresponding components of dSt.
If we have enough constraints, the problem reduces to the one-dimensional case, as mentioned in section 1.1. In such a case, it suffices to look at the directional derivative of St, namely the derivative in the direction in which the reaction coordinate is free to move. This is a corollary of the statement in the previous paragraph. It is slightly less general, but has the advantage of being easier to explain to students who don’t know what an exterior derivative is.
As a less-general corollary, which is even easier to explain to the mathematically unwashed, we can approximate the directional derivative by a finite difference, ΔSt. As always, we define
| ΔSt := St(after) − St(before) (3) |
and as always, we must be completely clear as to what we mean by "before" and "after".
In this case, the only sensible choice is two points on the locus of the reaction coordinate, both very close to the "current" reaction coordinate, and ordered according to the "forward" direction of the reaction in question. This version is particularly easy to sketch, along the lines of figure 2. In this example, ΔSt is positive, so the reaction will proceed forward, left-to-right as written in equation 1.
We now rescind the simplifications mentioned in section 1.4, so that we can consider a different genre of simplifications.
Specifically, we now restrict attention to conditions of constant pressure and constant positive temperature. (Section 1.4 was free of such restrictions). We shall see that this allows us to shift attention from the total entropy St (the truly general idea) to the free enthalpy G (a less-general idea). Note that G is the free enthalpy inside the region of interest. Similarly S is the entropy inside the region of interest (in contrast to St, which includes the entropy outside the region).
Here’s an outline of the usual calculation that shows why dG is interesting:
| (4) |
On the next-to-last line we expressed the change in total entropy (dSt) in terms of the change in system entropy (dS) plus the entropy that leaks across the boundary (−dH/T) as the heat-bath struggles to maintain constant temperature T. So under the stated constraints (constant P and constant positive T), anything we could predict based on dSt we can equally well predict based on −dG.
The minus sign is important; minimizing G corresponds to maximizing St.
Beware: Under other conditions (other than constant P and constant positive T), you cannot reliably predict the direction of spontaneity based on G or derivatives of G. Indeed there are plenty of cases where S is well defined but G is not even definable.
We now restrict attention to conditions of constant volume and constant positive temperature. This is just like section 1.5, but with constant volume instead of constant pressure. We shall see that this allows us to shift attention from the total entropy St (the truly general idea) to the free energy F (a less-general idea). Note that F is the free energy inside the region of interest. Similarly S is the entropy inside the region of interest (in contrast to St, which includes the entropy outside the region).
Here’s an outline of the usual calculation that shows why dF is interesting:
| (5) |
On the next-to-last line we expressed the change in total entropy (dSt) in terms of the change in system entropy (dS) plus the entropy that leaks across the boundary (−dE/T) as the heat-bath struggles to maintain constant temperature T. So under the stated constraints (constant V and constant positive T), anything we could predict based on dSt we can equally well predict based on −dF.
The minus sign is important; minimizing F corresponds to maximizing St.
Beware: Under other conditions (other than constant V and constant positive T), you cannot reliably predict the direction of spontaneity based on F or derivatives of F. Indeed there are plenty of cases where S is well defined but F is not even definable.
Suppose we are interested in the following reaction:
| (6) |
which we carry out under conditions of constant pressure and constant temperature, so that when analyzing spontaneity, we can use G as a valid proxy for St, as discussed in section 1.5.
Equation 6 serves some purposes but not all.
Let x represent some notion of reaction coordinate, proceeding in the direction specified by equation 6, such that x=0 corresponds to 100% reactants, and x=1 corresponds to 100% products.
Equation 6 is often interpreted as representing the largest possible Δ x, namely leaping from x=0 to x=1 in one step.
That’s OK for some purposes, but when we are trying to figure out whether a reaction will go to completion, we usually need a more nuanced notion of what a reaction is. We need to consider small steps in reaction-coordinate space. One way of expressing this is in terms of the following equation:
| a N2 + b O2 + c NO → (a−є) N2 + (b−є) O2 + (c+2є) NO (7) |
where the parameters a, b, and c specify the “current conditions” and the RHS of this equation has been displaced from the LHS by an amount є in the direction specified by equation 6.
The amount of free enthalpy liberated by equation 7 will be denoted δG. It will be infinitesimal, in proportion to є. If divide δG by є, we get the directional derivative ∇x G.
This tells us what we need to know. If ∇x G is positive, the reaction is allowed proceed spontaneously by (at least) an infinitesimal amount in the +x direction. We allow it to do so, then re-evaluate ∇x G at the new “current” conditions. If ∇x G is still positive, we take another step. We iterate until we come to a set of conditions where ∇x G is no longer positive. At this point we have found the equilibrium conditions (subject of course to the the initial conditions, and the contraint that the reaction equation 6 is the only allowed transformation).
Naturally, if we ever find that ∇x G is negative, we take a small step in the −x direction and iterate.
If the equilibrium conditions are near x=1, we say that the reaction goes to completion as written. By the same token, if the equilibrium conditions are near x=0, we say that the reaction goes to completion in the opposite direction, opposite to equation 6.
Let’s consider the synthesis of ammonia:
| (8) |
We carry out this reaction under conditions of constant P and T. We let the reaction reach equilibrium. We arrange the conditions so that the equilibrium is nontrivial, i.e. the reaction does not go to completion in either direction.
The question for today is, what happens if we increase the pressure? Will the reaction, formerly at equilibrium, now proceed to the left or the right?
We can analyze this using the tools developed in the previous sections. At constant P and T, subject to mild restrictions, the reaction will proceed in whichever direction minimizes the free enthalpy:
| (9) |
where x is the reaction coordinate, i.e. the fraction of the mass that is in the form of NH3, on the RHS of equation 8. See section 1.5 for an explanation of where equation 9 comes from.
Note that P and T do not need to be constant "for all time", just constant while the d/dx equilibration is taking place.
As usual, the free enthalpy is defined to be:
| G = E + PV − TS (10) |
so the gradient of the free enthalpy is:
| dG = dE + PdV − TdS (11) |
There could have been terms involving VdP and SdT, but these are not interesting since we are working at constant P and T.
In more detail: If (P1,T1) corresponds to equilibrium, then we can combine equation 11 with equation 9 to obtain:
| (12) |
To investigate the effect of changing the pressure, we need to compute
| (13) |
where P2 is slightly different from the equilibrium pressure:
| P2 = (1+δ)P1 (14) |
We now argue that E and dE are insensitive to pressure. The potential energy of a given molecule depends on the bonds in the given molecule, independent of other molecules, hence independent of density (for an ideal gas). Similarly the kinetic energy of a given molecule depends on temperature, not on pressure or molar volume. Therefore:
| ⎪ ⎪ ⎪ ⎪ |
| = |
| ⎪ ⎪ ⎪ ⎪ |
| (15) |
Having examined the first term on the RHS of equation 13, we now examine the next term, namely the P dV/dx term. It turns out that this term is insensitive to pressure, just as the previous term was. We can understand this as follows: Let N denote the number of gas molecules on hand. Let’s say there are N1 molecules when x=1. Then for general x we have:
| N = N1 (2−x) (16) |
That means dN/dx = −N1, independent of pressure. Then the ideal gas law tells us that
| (P V) | ⎪ ⎪ ⎪ ⎪ |
| = |
| (N kT) | ⎪ ⎪ ⎪ ⎪ |
| (17) |
Since the RHS is independent of P, the LHS must also be independent of P, which in turn means that P dV/dx is independent of P. Note that the V dP/dx term is automatically zero.
Another way of reaching the same conclusion is to recall that PV is proportional to the kinetic energy of the gas molecules: PV = N kT = (γ−1) E, as discussed in reference 1. So when the reaction proceeds left to right, for each mole of gas that we get rid of, we have to account for RT/(γ−1) of energy, independent of pressure.
This is an interesting result, because you might have thought that by applying pressure to the system, you could simply “push” the reaction to the right, since the RHS has a smaller volume. But it doesn’t work that way. Pressurizing the system decreases the volume on both sides of the equation by the same factor. In the PdV term, the P is larger but the dV is smaller.
Also note that by combining the pressure-independence of the dE/dx term with the pressure-independence of the P dV/dx term, we find that dH/dx is pressure-independent also, where H is the enthalpy.
Now we come to the −TdS/dx term. Entropy depends on volume. Consider what would happen if we were to run the reaction backwards (i.e. right to left), in the hypothetical situation where we had N1 molecules on each side of the equation. Then as we went from x=1 to x=0, the system volume would double, and we would pick up N1 kln(2) units of entropy ... independent of pressure. It is the volume ratio that enters into this calculation. The ratio is independent of pressure, and therefore cannot contribute to explaining any pressure-related shift in equilibrium.
The real story is that we have not N1 but 2N1 molecules when x=0. So when we run the real reaction backwards, in addition to simply letting N1 molecules expand, we have to create another N1 molecules from scratch.
For that we need the full-blown Sackur-Tetrode equation. For a pure monatomic nondegenerate ideal gas in three dimensions:
| (18) |
which gives the entropy per particle in terms of the volume per particle, in an easy-to-remember form.
For the problem at hand, we can re-express this as:
| (19) |
where the index i runs over the three types of molecules present (N2, H2, and NH3). We have also used the ideal gas law to eliminate the V dependence inside the logarithm in favor of our preferred variable P. We have (finally!) identified a contribution that depends on pressure and also depends on x.
We can understand the qualitative effect of this term as follows: The −TS term always contributes a drive to the left. According to equation 18, at higher pressure this drive will be less. So if we are in equilibrium at pressure P1 and move to a higher pressure P2, there will be a net drive to the right.
We can quantify all this as follows: It might be tempting to just differentiate equation 19 with respect to x and examine the pressure-dependence of the result. However, it is easier if we start by subtracting equation 12 from equation 13, and then plug in equation 19 before differentiating. A lot of terms are unaffected by the change from P1 to P2, and it is helpful if we can get such terms to drop out of the calculation sooner rather than later:
| (20) |
This quantitative result reinforces the previous qualitative analysis: If P2 is greater than P1, the reaction will proceed in the +x direction, since that is what will minimize G.
The calculation involved many steps, but each step was reasonably easy.
Remark: The result is surprisingly simple. Whenever a complicated calculation produces a simple result, I take it as a sign that I don’t really understand what’s going on. I suspect there is an easier way to obtain this result. In particular, since we have figured out that the entropy term is running the show, I conjecture that it may be possible to start from first principles, all the way back to the entropy in equation 2, and just keep track of the entropy.
Remark: In equation 20, by a first order expansion of the logarithm on the last line, you can verify that when the reaction is pushed toward equilibrium, the amount of push is proportional to δ, which makes sense.
Exercise: Use a similar argument to show that increasing the temperature will shift the equilibrium of equation 8 to the left. Hint: In a gas-phase reaction such as this, the side with the more moles will have more entropy.
One sometimes sees equilibrium (and or the shift of equilibrium) “explained” by reference to Le Chatelier’s principle. This is highly problematic.
LeChatelier in his lifetime gave two inconsistent statements of the "principle”. One is trivially tautological, and the other is false.
As a familiar, important example of a system that exhibits zero stability, consider ice in equilibrium with water. Let x represent the reaction coordinate, i.e. the percent of the mass of the system that is in the form of ice. If you perturb the system by changing x – perhaps by adding water, adding ice, or adding energy – the system will exhibit zero tendency to return to its previous x-value.
