# Why is the Standard Entropy of a Proton Zero?

It was my assumption that the standard entropy was the absolute entropy of a substance at standard state; however, my textbook states various standard entropies ranging from $$-144.77\ \mathrm{J/(mol\ K)}$$ for $$\ce{Al^3+}$$ to $$0\ \mathrm{J/(mol\ K)}$$ for $$\ce{H+}$$ to $$213\ \mathrm{J/(mol\ K)}$$ for $$\ce{CO2}$$. It is clear that these values are not absolute entropies, as the only substance that can have an absolute entropy of $$0$$ is a perfect crystal at $$0\ \mathrm K$$ according to the third law of thermodynamics. Why are standard entropies notated like this instead of just stating their absolute form? My assumption is that this is just a way to make calculations involving entropy easier, since they merely involve changes in standard entropies; however, why is the standard entropy still notated $$S^\circ$$ and not instead $$\Delta S^\circ$$ (since it is not absolute)?

Indeed, it is clear that these values are not... Wait! What "these values"?

You are putting two entirely different things in the same basket. (If your textbook does so, then it does a poor job.) It is much like treating dogs and cupboards similarly, on the basis of both having four legs. Then, however, many generalizations will fail quite miserably. You feed them meat, but the results are not always great. You call in the carpenter to fix them, but...

Same thing here. See, ions and molecules are very, very different things. You totally can have a mole of $$\ce{CO2}$$. It is a thing. You can work with it, can measure its heat capacity and stuff, and eventually work out the entropy. Indeed, the textbook value of $$\ce{CO2}$$ entropy is absolute.

On the other hand, you totally can't have a mole of $$\ce{Al^3+}$$. It is not a thing. Unless you have some counterions, that is, but in that case you can't tell how much of the entropy comes from the counterion and how much from $$\ce{Al^3+}$$.

How can we assign entropies to ions, then? Why, it is as simple as that: we assign one ion arbitrarily, then work out the entropies of all its counterions from there, and then turn to their counterions. Just what one ion could that be? Why, let's say it is $$\ce{H+}$$, because so much of our chemistry is dominated by the acid/base equilibria. It is kind of natural to consider its entropy $$0$$, much like the absolute elevation of the sea level is $$0$$. Now there you have them, the textbook entropies of ions, which are not quite absolute.

So it goes.

• So basically, it is impossible to determine the absolute entropy of an ion, so instead we determine its entropy relative to a H+. In the case of performing calculations (such as Gibbs free energy) it would not matter whether or not the value is absolute, just if there is a change in entropy values. This makes sense; however, if we ran a reaction in which ions were produced, such as the reaction between sodium and chlorine to form NaCl, these values would no longer be relative to each other (with one set absolute and the other relative to H+), so how do we compare these changes in entropy? – Sarah Brown Apr 24 '20 at 21:51
• Well, the values for cations can all be increased by the same arbitrary constant, but then the values for anions will be decreased by the same constant. See, these things are not quite absolute, but not quite arbitrary either. – Ivan Neretin Apr 24 '20 at 21:54