# How do you calculate microstates in chemistry?

May 25, 2018

Well, you can calculate the NUMBER of microstates at $\text{298.15 K}$ with tabulated standard molar entropies.

• I go over what microstates are, here.
• An example that should be easy to follow is here.
• A much more detailed calculation example with methane is gone into here.

Boltzmann's formulation of entropy states:

$S = {k}_{B} \ln \Omega$

where $\Omega$ is the number of microstates, and ${k}_{B} = 1.38065 \times {10}^{- 23} \text{J/K}$ is the Boltzmann constant.

There is a more complicated formula for $\Omega$ that works for systems where the number of available states is much larger than the number occupied:

$\Omega = \text{exp} \left\{{\sum}_{i = 1}^{N} \left[{N}_{i} \ln \left({g}_{i} / {N}_{i}\right) + {N}_{i}\right]\right\}$

where ${N}_{i}$ is the number of particles with energy ${\epsilon}_{i}$ in a state with degeneracy ${g}_{i}$.

For simplicity, we can calculate the value of $\Omega$ indirectly.

$\Omega = {e}^{S / {k}_{B}} = \text{exp} \left(S / {k}_{B}\right)$

As an example, ${\text{O}}_{2}$ has ${S}^{\circ} = \text{205.15 J/mol"cdot"K}$ at $\text{298.15 K}$ and $\text{1 bar}$. So, in those conditions:

color(blue)(Omega) = "exp"((205.15 cancel"J""/"cancel"mol"cdotcancel"K")/(1.38065 xx 10^(-23) cancel"J""/"cancel"molecule"cdotcancel"K" xx (6.0221413 xx 10^(23) cancel"molecules")/(cancel"1 mol")))

$= \textcolor{b l u e}{5.197 \times {10}^{10}}$

So, there are $5.197 \times {10}^{10}$ ways for ${\text{O}}_{2}$ to exist such that it has an entropy of $\text{205.15 J/mol"cdot"K}$.