How do you calculate the number of microstates a compound has?

Jul 26, 2016

DISCLAIMER: The answer isn't that long, but has a link to extra info at the bottom for you extra, extra hard-working people.

A convenient equation is:

$\setminus m a t h b f \left(S = {k}_{B} \ln \Omega\right)$

where:

• $S$ is the statistical entropy at a specified temperature. For example, we can use ${S}^{\circ} = \text{191.6 J/mol"cdot"K}$ for ${\text{N}}_{2} \left(g\right)$ at $\text{298.15 K}$. As long as you shoot for $\text{298.15 K}$, you can look it up in the appendix of your textbook.
• ${k}_{B}$ is the Boltzmann constant, $1.3806 \times {10}^{- 23}$ $\text{J/K}$.
• $\Omega$ is the number of microstates consistent with a given macrostate.

So, if we can calculate or if we know the compound's entropy at a given temperature, we can calculate the number of microstates.

Note though, that we shouldn't be surprised if $\setminus m a t h b f \left(\Omega\right)$ is absurdly large, because ${k}_{B}$ is very small, and we ARE talking about quantum particle sizes here.

Using ${S}^{\circ}$, for ${\text{N}}_{2} \left(g\right)$ at $\text{298.15 K}$, we would get:

$\textcolor{b l u e}{\Omega} = {e}^{S \text{/} {k}_{B}}$

= "exp"(191.6 cancel("J/")"mol"cdotcancel("K")"/"1.3806xx10^(-23) cancel("J/molecule"cdot"K") xx cancel"1 mol"/(6.0221413 xx 10^23 cancel"molecules"))

$\approx \textcolor{b l u e}{1.019 \times {10}^{10} \text{ accessible microstates}}$

So really, the hard part is figuring out what the entropy for the system is if we don't have the number readily available. If you want to know how to do it, there's a very involved process, shown below.

Fortunately, if you want it at $\setminus m a t h b f \left(\text{298.15 K}\right)$, it's already been done for you: you can find ${S}^{\circ}$ in the appendix of any good chemistry textbook!

For you extra hard-working people:
Deriving Entropy at a Given Temperature