2
$\begingroup$

My broad question is how do you measure entropy change? I was doing a bit of digging into how thermodynamic tables are developed at I got this from NIST. Basically it seems like you need to indirectly determine the entropy of a substance from other experimental parameters. What NIST says is

  1. You use heat capacities and phase transition enthalpies to do a ground up calculation
  2. Statistical mechanics methods once you have a partition function
  3. Measure temperature dependence of equilibrium constants from van't hoft plot
  4. Calculate it from previously determined enthalpies and free energies.

The thing that's bothering me is that all of these measurements depend on other experimental parameters and it just seems to be a raw calculation of entropy after determining these other quantities.

My primary question then is it possible to do a calorimetry experiment to calculate $\Delta S_{reaction}$? After all, how did people measure entropy experimentally in the olden days before we had copious information on heat capacities?

I tried to think about this for a constant pressure apparatus. A reaction happens and I have a thermometer to measure $\Delta T$. I can correlate this to $\Delta H$ using the heat capacity of the calorimeter. If I have a resistive circuit running through the calorimeter, I can supply heat externally to lower $\Delta T$ so now I can say that $\Delta H$ occurred at the same initial and final temperatures

However, this only tells me about the $\Delta S_{surroundings} = -\frac{\Delta H}{T}$. How do I calculate the change in the reaction system? I tried to imagine the reversible path but if I start and end at the same point, then isn't $\Delta S_{system} = 0$? Where is the flaw in my reasoning or is it just something we cannot end up measuring?

$\endgroup$
4
  • $\begingroup$ You have the right idea in devising a reversible path and measuring the reversible heat from the surroundings for that path. But, it isn't that easy to devise a reversible path for a chemical reaction, starting with pure reactants in stoichiometric proportions at T and P and going to pure products in corresponding stoichiometric proportions at T and P. Are you familiar with the concept of the van't Hoff equilibrium box? $\endgroup$ – Chet Miller Mar 31 at 10:54
  • $\begingroup$ (1) I don't understand what you mean by a "raw" calculation of entropy. I would have just written "a calculation of entropy". Raw means "unprocessed", as in raw data. Given that the methods you list determine entropy by going through the process of a calculation, rather than by direct measurement, it's the opposite of raw. (2) It sounds like you are trying to measure entropy directly, rather that by calculating it from other measurable quantities and, in so doing, you are presenting it by contrast with $\Delta H$, which you seem to think can be directly measured... $\endgroup$ – theorist Apr 1 at 1:24
  • $\begingroup$ ....but no state function can be directly measured. They all need to be calculated from measurable quantities. E.g., we can't directly measure $\Delta H$. Rather, we can only determine $\Delta H$ by measuring the heat flow (a path function) at constant pressure. So I don't see a significant qualitative difference between calculating $\Delta H$ from $q_p$ at a single temp., vs. calculating $\Delta S$ from the temp-dependence of $q_p$ via the van't Hoff eqn.... $\endgroup$ – theorist Apr 1 at 1:33
  • $\begingroup$ ...I.e., we can determine $\Delta S$ calorimetrically the same way we determine $\Delta H$ calorimetrically—by measuring $q_p$; the only diff. is that, for $\Delta S$, we need to measure at multiple temps. (3) You wrote "in the olden days before we had copious information on heat capacities". You don't need copious info. on heat capacities to determine $\Delta H$. You just use a small quantity of reactants, such that the heat capacity of the rxn mixture is small (and can thus be ignored) relative to that of your calorimeter. Thus the only heat capacity you need to know is your calorimeter's. $\endgroup$ – theorist Apr 1 at 1:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.