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For a reaction between an acid and a base

$$\ce{A1 + B1 <=> A2 + B2}$$

to be spontaneous $\ce{A1}$ must be a stronger acid than $\ce{A2}$. With this in mind, how can this reaction be understood in terms of spontaneity $$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$ Also sources on the internet say that $\ce{pK_a}$ of $\ce{H3O+}$ is $-1.74$ while that of acetic acid is $4.74$.

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    $\begingroup$ Note that the reaction should be written as an equilibrium, and, importantly, that the equilibrium lies well to the left (i.e. most of the acetic acid is not dissociated). $\endgroup$
    – airhuff
    Commented Apr 16, 2017 at 5:11
  • $\begingroup$ Then why do those few molecules dissociate anyway? Theoretically (considering the thermodynamics of the reaction) they shouldn't right? $\endgroup$
    – Azulene
    Commented Apr 16, 2017 at 12:27
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    $\begingroup$ @Azulene You now enter the world of entropy. Suppose you have $10^{23}$ molecules of acid. Will they all behave and not dissociate? That's unlikely. So you end up with a dynamic equilibrium with some molecules recombining, other dissociating etc. $\endgroup$
    – TAR86
    Commented Apr 16, 2017 at 12:36
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    $\begingroup$ I was thinking of the collection of molecules. In terms of pure energy ($U$), they should not dissociate. But there are so many of them - so some do, because it is not likely that they all stay associated. This is quantified by entropy, an expression of probability, which results in free enthalpy ($G$). $\endgroup$
    – TAR86
    Commented Apr 16, 2017 at 15:34
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    $\begingroup$ All chemical reactions are a balancing act between enthalpy and entropy. Strictly from an enthalpy standpoint, going to one side of the reaction completely should be favored. However, entropy is maximized by evenly mixing components. Combining these two factors, the equilibrium position is found by maximizing the free energy which has an enthalpic and entropic component. Due to this entropic contribution, all reactions are technically equilibria; they just might be shifted so far to one side that the reaction appears to have gone to completion. $\endgroup$
    – Tyberius ♦
    Commented Apr 16, 2017 at 18:13

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To expand on my comment above, I will reiterate that all chemical reactions are equilibria and this is due to the interplay of enthalpy and entropy. More specifically, it is due to enthalpy and entropy of the individual compounds of the reaction as well as an additional entropy component due to mixing.

Consider the general reaction $$\ce{A<=>B}$$ and let $\chi_\ce{i}$ denote the mole fraction of compound $\ce{i}$ in the solution or gas mixture. The enthalpy and the entropy of the compounds, combined together to form $\Delta G_{compounds}=\Delta H_{compounds}-T\Delta S_{compounds}$ or alternatively $$G=(\chi_\ce{B})\cdot H_{\ce{B}}+(1-\chi_\ce{B})\cdot H_{\ce{A}}-(\chi_\ce{B})\cdot T\Delta S_{\ce{B}}-(1-\chi_\ce{B})\cdot T\Delta S_{\ce{A}}$$ has a particular side of the reaction it favors. This component of the free energy leads to a complete preference; if this were the only factor, the solution would be all $\ce{A}$$(\chi_\ce{A}=1)$ or all $\ce{B}(\chi_\ce{B}=1)$. If we were to plot $G$ as a function of $\chi_\ce{B}$, it would either be linear increasing (reactants favored), linear decreasing (products favored), or flat (reactants and products equally favored).

However, there is an additional entropy component due to mixing, which leads to an additional free energy component. We can write the entropy of mixing as $$\Delta S_{mixing}=-\mathrm{R}[\chi_\ce{A}\ln(\chi_\ce{A})+\chi_\ce{B}\ln(\chi_\ce{B})]=-\mathrm{R}[(1-\chi_\ce{B})\ln(1-\chi_\ce{B})+(\chi_\ce{B})\ln(\chi_\ce{B})]$$ where we have used the fact that $\chi_\ce{B}+\chi_\ce{A}=1$. We can then write the free energy of mixing as $$\Delta G_{mixing}=\mathrm{RT}[(1-\chi_\ce{B})\ln(1-\chi_\ce{B})+(\chi_\ce{B})\ln(\chi_\ce{B})]$$ We can see graphically what these two mixing equations would look likeentropy and free energy of mixing

Combining all this together, we can write $$G_{total}=(\chi_\ce{B})\cdot H_{\ce{B}}+(1-\chi_\ce{B})\cdot H_{\ce{A}}-(\chi_\ce{B})\cdot T\Delta S_{\ce{B}}-(1-\chi_\ce{B})\cdot T\Delta S_{\ce{A}}+\mathrm{RT}[(1-\chi_\ce{B})\ln(1-\chi_\ce{B})+(\chi_\ce{B})\ln(\chi_\ce{B})]$$ We can take the derivative of this function with respect to $\chi_\ce{B}$ and set it equal to zero to determine where the miniumum free energy is $$\frac{\delta G_{total}}{\delta\chi_\ce{B}}=0= H_{\ce{B}} - H_{\ce{A}}- T\Delta S_{\ce{B}}- T\Delta S_{\ce{A}}+\mathrm{RT}[-\ln(1-\chi_\ce{B})+\ln(\chi_\ce{B})]$$ $$H_{\ce{B}} - H_{\ce{A}}- T\Delta S_{\ce{B}}- T\Delta S_{\ce{A}}=-\mathrm{RT}[-\ln(1-\chi_\ce{B})+\ln(\chi_\ce{B})]$$

This second equation is the crux of all this. While we cannot solve this for an exact value without knowing the enthalpy and entropy of the compounds, we can definitively say that $\chi_\ce{B}\ne1$ and $\chi_\ce{B}\ne0$. Why? Plugging either of the values into the equation will result in the right side being $\infty$, meaning the differences in enthalpy and entropy of the components would also have to be $\infty$ for the equation to be satisfied. Thus, the minimum free energy will never be attained by letting the reaction go to completion; it will always be more favorable to allow at least some mixing even if one side of the reaction has a much lower enthalpy and higher entropy of its compounds.

Thus all reactions are technically equilibria. I say technically because the mixing free energy might be negligible compared to the overall free energy and so the reaction could appear to have gone to completion to the extent that we can observe experimentally.

So for your reaction, the reason it doesn't all remain as $\ce{CH3COOH}$ and water is that the entropy of mixing makes it favorable for some of the $\ce{CH3COOH}$ to dissociate.

(For a brief version of this argument and where I got the graph, see http://rkt.chem.ox.ac.uk/tutorials/equilibrium/entropy_mixGas.html)

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  • $\begingroup$ Wow! Thanks a lot. This really made sense, especially that last equation. $\endgroup$
    – Azulene
    Commented Apr 17, 2017 at 4:34
  • $\begingroup$ @Azulene Not a problem. I had just received a good recap on this in my Statistical Mechanics course so and this question seemed like a convenient place to put that to use. Consider accepting the answer unless your holding out for something a little different. $\endgroup$
    – Tyberius ♦
    Commented Apr 17, 2017 at 4:52

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