Forgive me if this is not the correct place to ask this. When I encounter a molar entropy, is it possible to rearrange the Boltzmann entropy formula $$S=k\log W$$ and meaningfully interpret the number of micro-states thus calculated?

In particular, I found the molar entropy of quartz $(\ce{SiO2})$ in a table: $$S=41.36\ \text{J K}^{-1} \text{mol}^{-1}.$$

Using the Boltzmann formula for a single formula unit of $\ce{SiO2}$ (rather than a mole), the number of micro-states $W$ is: $$W=\exp\left(\frac{S}{kN_A}\right)\approx145.$$

Is there a sense in which a single formula unit of $\ce{SiO2}$ within a quartz crystal can be arranged into one of $145$ different micro-states? Is there any way of describing what those states are? What else am I missing?

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    $\begingroup$ Welcome! Great question - one thing that would be helpful (via a comment here) would be to know a bit about your background. Have you had thermodynamics or statistical mechanics? Trying to figure out the best level to frame the answer... $\endgroup$ Commented Dec 19, 2022 at 21:08
  • $\begingroup$ I'm actually a geology student; I found the table in my textbook but couldn't understand the meaning of entropy other than a tool to construct phase diagrams for interpreting rock formation conditions. I haven't taken either of the courses you mentioned. I found the Boltzmann equation on Wikipedia as I was trying to better understand entropy. $\endgroup$
    – Henry
    Commented Dec 19, 2022 at 21:21

1 Answer 1


As you note, entropy can be a measure of the number of possible accessible microstates. In other words, it's a measure of disorder.

One common example I give in class is if I toss 2-3 cards on the floor there are only a few ways they could be arranged and it's not a big mess. If I throw a whole deck of cards randomly on the floor, there are a lot of possible microstates (arrangements).

The third law of thermodynamics indicates that at absolute zero, for a perfect crystal, there is only one microstate:

$S=k \ln W=k \ln 1=0$

So how do we get the standard molar entropies? Well, there must be some disorder. In a molecule, you get some disorder because the molecule can rotate, translate, vibrate, and in many cases flop around (i.e., have different conformations).

For a real solid, you can have some entropy due to defects (i.e., the positions of the defects could be different = microstates).

For an ideal solid at finite temperature, without defects, the entropy derives from the vibrations / movements of the atoms. That is, the atoms will move around based on the vibrations of the solid. We often consider these as little harmonic oscillator springs. As a rule of thumb, weaker bonds tend to have weaker spring constants, so vibrations (and the corresponding entropy) tend to be higher.

In solids, we often compare different ionic compounds to see that the stronger ionic bonds have lower molar entropy (e.g., $\ce{LiF}$ vs. $\ce{NaCl}$ vs. $\ce{KCl}$):

From Chem LibreTexts:

  • $\ce{LiF}$ - 35.7 J/(mol•K) vs. 577 kJ/mol $\Delta H_f$
  • $\ce{NaCl}$ - 72.1 J/(mol•K) vs. 410 kJ/mol $\Delta H_f$

Note that the entropy is smaller for the more ionic $\ce{LiF}$. It's not a perfect correlation because it's about the shape of the vibrational energies rather than the heat of formation, but it's a useful heuristic.

In other words, for this type of entropy, it's not a function of equivalent microstates like throwing cards on the floor. It's a function of the vibrational entropy at finite temperature.

This implies not the Boltzmann formula (which requires equal weightings of the different microstates) but the Gibbs entropy formula:

$S=-k_{\mathrm{B}} \sum_i p_i \ln \left(p_i\right)$

in which $p_i$ is the probability of a particular micro state $i$ due to different energies $E_i$ and $k_{\mathrm{B}}$ is the Boltzmann constant.

Note this is essentially just a modification of the Boltzmann form, accounting that different microstates have different probabilities due to different relative energies.

So it's not that there really are 145 (or whatever) equivalent microstates at finite temperature. Some vibrations in a solid are lower energy and will contribute more to the entropy through what's called the zero-point energy. Some vibrations will be higher energy and will contribute less:

$p_i \propto e^{-\varepsilon_i /(k_B T)}$

  • $\begingroup$ I am still curious about the dimensionless number $W$ that would emerge from this calculation even for a defect-free crystal at non-zero temperature (if I understand correctly). If not a measurement of equivalent micro-states, is there another interpretation for $W$ as a measurement of the vibrational states you describe? Is the Boltzmann equation simply not appropriate for this situation? $\endgroup$
    – Henry
    Commented Dec 19, 2022 at 22:37
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    $\begingroup$ Boltzmann would be fine if all the vibrations were equally likely. They're not, because some will be higher energy than others. So you need to use a somewhat different formula to calculate entropies. $\endgroup$ Commented Dec 20, 2022 at 1:39
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    $\begingroup$ @Henry, to add on to Geoff's answer, the Boltzmann entropy formula is applicable for a system with, among other things, a fixed energy E. In most (and your) circumstances, we are interested in the entropy at fixed temperature T instead, so the Boltzmann formula does not apply. $\endgroup$ Commented Dec 20, 2022 at 17:40

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