# Total entropy change for reversible reactions and its relationship to equilibrium constant

I'm in year 12 of school doing A2 chemistry learning about entropy and its relationship to reversible equilibria. I learned about the following three ideas from my course book (verbatim), for a reaction at equilibrium:

1. $$\Delta S_{\text{total}} = 0$$
2. $$\Delta S_{\text{total}}$$[forward reaction] = $$\Delta S_{\text{total}}$$[backward reaction].
3. $$\Delta S_{\text{total}} = R\ln K$$, where $$R$$ is the gas constant and $$K$$ is either $$K_c$$ or $$K_p$$.

I'm confused about points 1 and 3 and would like to know if my understanding of 2 is correct.

For point 2, does it imply that at equilibrium the total entropy change from both the forward and backward reactions are zero, but if the forward reaction is say $$\Delta S_{\text{sys}} + \Delta S_{\text{surr}} = 22 + (-22)$$, then the backward reaction has to be $$\Delta S_{\text{sys}} + \Delta S_{\text{surr}} = (-22) + 22$$ (as opposed to something like $$(-60) + 60$$ which is also zero but the individual magnitudes are different)?

For points 1 and 3, if they're both agreed to be true then $$\Delta S_{\text{total}} = 0=R\ln K \implies K = 1$$, no matter the reaction. This result is just absurd because it's implying all reversible reactions have a position of equilibrium close to the middle. What exactly is going wrong? Later on in the book, there is another contradiction where it provides the following reaction at equilibrium: $$\ce{SO2(g) + 1/2O2(g) <=> SO3(g)}\quad\Delta S^\circ_{\text{total}}=238.3 JK^{-1}mol^{-1}$$ How come the total entropy change is nonzero during equilibrium? My guess is that it has something to with the fact that this entropy change is standard whereas the one in point 1 is not. Similarly, I think the entropy change in point 3 should also be standard. If that's the case, what exactly is the difference between $$\Delta S_{\text{total}}$$ and $$\Delta S^\circ_{\text{total}}$$?

Note, I don't know what Gibbs free energy is but from the expression I've seen online, it seems to be equal to $$-T\Delta S_{\text{total}}$$ so I'd appreciate if the explanations of my questions are in terms of entropy changes.

Edit: the book is Edexcel IAL Chemistry Student Book 2 (ISBN 13: 9781292244723), page 72 as requested in the comments.

• What is the book? Could you edit the question to include the title and author, or the ISBN?
– Karsten
Sep 4, 2022 at 17:32
• @Karsten I've added it at the bottom.
– Typo
Sep 4, 2022 at 17:38
• When you say that $Delta S=0$ or $Delta S=R\lnK$, please precisely state the initial and final states for these changes, in terms of the temperatures and pressures of the reactants and products (and if the reactants and products are pure, or not). Sep 4, 2022 at 21:17

For points 1 and 3, if they're both agreed to be true then $$\Delta S_{\text{total}} = 0=R\ln K \implies K = 1$$, no matter the reaction.

The entropy change of a reaction is concentration-dependent: $$\Delta S = \Delta S^\circ - R \ln Q\tag{1}$$

For the standard state (indicated by the $$^\circ$$), the reaction quotient $$Q$$ is one. At equilibrium, $$Q = K$$. So the entropy change is different at standard state and a equilibrium. Point 3 should indicate standard state to make sense, like this:

$$\Delta S^\circ_\mathrm{total} = R \ln K\tag 2$$In fact, you can derive (2) by setting $$\Delta S = 0$$ and $$Q = K$$ (true for equilibrium) and substitute into (1).

Later on in the book, there is another contradiction where it provides the following reaction at equilibrium: $$\ce{SO2(g) + 1/2O2(g) <=> SO3(g)}\quad\Delta S^\circ_{\text{total}}=\pu{238.3 J K^-1 mol^-1}$$

The double-harpoon in the chemical equation does not indicate that we are at equilibrium, it just says this reaction shows some forward and reverse reaction at the same time. The entropy change at standard state is non-zero, meaning there will be a net change in the forward direction until we reach equilibrium. At this point, the entropy change of a net reaction will be virtually zero, so there will be no further net reaction (i.e. forward and reverse rate will be equal). So this statement is fine once you figure out the meaning of the chemical equation.

How come the total entropy change is nonzero during equilibrium? My guess is that it has something to with the fact that this entropy change is standard whereas the one in point 1 is not. Similarly, I think the entropy change in point 3 should also be standard. If that's the case, what exactly is the difference between $$\Delta S_{\text{total}}$$ and $$\Delta S^\circ_{\text{total}}$$?

Without the $$^\circ$$, concentrations could be anything. With the $$^\circ$$, concentrations are all at standard state, and $$Q = 1$$.

$$\Delta S_{\text{total}}$$[forward reaction] = $$\Delta S_{\text{total}}$$[backward reaction]

Point 2 is written in a way that confuses me. "Total" implies it covers everything that goes on, so separating it into forward and backward reaction is surprising. I asked the OP for the source of this statement in the comments. Maybe the context in the book would shed some light on the intention of the textbook author.

• I'm starting to understand but I just need some clarification on somethings if you don't mind. My understanding of standard state is that temperature and pressure get specified and are kept constant (298K and 1 atm). What does "concentrations are all at standard state" mean? Why does the reaction quotient = 1?
– Typo
Sep 4, 2022 at 18:00
• "At this point, the entropy change of a net reaction will be virtually zero". So if the standard change of entropy keeps decreasing as the reaction reaches equilibrium, then don't we again arrive the problem of K equaling 1?
– Typo
Sep 5, 2022 at 5:06
• The standard entropy change is constant, it refers to a specific set of concentrations. The actual entropy change becomes smaller and smaller, approaching zero as equilibrium is approached. For the definition of standard state (rather than STP), see en.wikipedia.org/wiki/…
– Karsten
Sep 5, 2022 at 11:40
• Sorry if I sound like a broken record but my confusions just aren't clearing up... If at standard state, the standard entropy change is constant, doesn't that contradict "The entropy change at standard state is non-zero, meaning there will be a net change in the forward direction"? As in, if it is always constant, even during equilibrium, it should be a constant so it doesn't actually indicate whether the forward reaction currently has a greater rate? It appears to me that standard entropy change is actually just a number attributed to a reaction derived from K, is that correct?
– Typo
Sep 5, 2022 at 16:01
• Sorry, comments might not be the right space for this. Anyway, the standard state values only apply to the standard state. Even if you start at the standard state, concentrations will change, so most of the reaction will have a state different from the standard state, and thermodynamic quantities different from the standard state. You are welcome to ask a question directly about this.
– Karsten
Sep 5, 2022 at 16:43

(1) The entropy reaches a maximum at equilibrium, thus any change will depart from equilibrium, but this cannot happen if we are at equilibrium unless some other parameter changes such a temperature or pressure thus any change in entropy is zero.

If the reaction is $$A+B\rightleftharpoons C$$ if we start with one mole of A and one of B the formation of C can occur only for as long as the entropy can continue to increase. If we start with just C then A+B will be produced until equilibrium is again formed, i.e. some composition of A,B and C that maximises the entropy.

(2) I'm not clear what is meant here, possibly entropy of mixing?

(3) The expression for the Gibbs free energy is $$\Delta G^0 = -RT\ln(K_P)$$ where $$\Delta G^0$$ is the free energy at standard conditions ( 1 mol of vapour at 1 atm. pressure for a gas and 1 mol for a liquid independent of pressure) and $$K_p$$ is the equilibrium constant. The free energy is defined as $$\Delta G=\Delta H-T\Delta S$$ so (3) assumes that any change in $$H$$ is zero which is not generally true, i.e only accidentally so. The free energy $$G$$ is a minimum at equilibrium.

(That $$\Delta G^0$$ is not zero at equilibrium is because it is the slope of free energy $$G$$ (not standard /reference value $$G^0$$) wrt. extent of reaction $$\xi$$ that is zero, i.e $$dG/d\xi=0$$ at constant $$T,P$$.)

• Thanks for the response. I understand your (1) but (3) is completely flying over my head unfortunately. I mentioned in the OP I have yet to study free energy G and I don't know what $\xi$ is. It is not part of my current syllabus of A levels. If it's not too much to ask, could you address my questions directly: why the result $K=1$ is wrong and why $\Delta S^\circ_{\text{total}} > 0$ at equilibrium? If it must require the knowledge of free energy then maybe can you give a conceptual explanation like (1) (to get me through until I study the required topics)?
– Typo
Sep 4, 2022 at 17:17
• At equilibrium there is no change in entropy then $\Delta S=0$, if this is true then $K=1$ but this cannot be general. $\xi$ is simple, it is the extent of reaction and varies from 0 (no product) to 1 (all product) . So the energy is a minimum (at equilibrium) when plotted against $xi$, i.e. at some composition of product and reactant. continued. Sep 5, 2022 at 7:22
• continued. You cannot avoid free energy, it is simple anyway. In a reaction $\Delta G=\Delta G^0+RT\ln(P_a/P_b)$ where $P$ is pressure of species a and b, this ratio is also called the quotient $Q$ and then $K_p$ at equilibrium. Divide by $T$ then $\Delta S=\Delta S^0+R\ln(K)$ assuming $\Delta S=\Delta G/T$ and at equilib $\Delta S^0=-R\ln(K)$ where the superscript indicates standard state. Sep 5, 2022 at 7:26