# How many micro states exist for a molecule?

Aug 24, 2017

A lot! For example, a single molecule of methane could have over ${10}^{9}$ microstates. That is a bit on the high side, but usually a single molecule will have larger than a million microstates.

And thus, a usual sample of $\text{1 mol}$ of molecules would have around ${10}^{30}$ microstates or more.

Also, a note about ${q}_{e l e c}$: it can be not $1$ if the ground state of a molecule is not a singlet. For example, ${q}_{e l e c} = 3$ for ${\text{O}}_{2}$ in its ground state because it is a ""^3 Sigma_g^-, a triplet-sigma-gerade-minus state.

For most molecules, namely those that are in a scenario where the number of occupied states is much less than the number of available states, one can find the $\ln$ of the number of microstates as:

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

where:

• ${N}_{i}$ is the number of particles in state $i$ with energy ${\epsilon}_{i}$.
• ${g}_{i}$ is the degeneracy of the energy ${\epsilon}_{i}$, i.e. how many ${\epsilon}_{i}$ there are for a given state $i$.
• $\Omega$ is the number of microstates.

One can then use this in Boltzmann's expression for the entropy to find:

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

where ${k}_{B}$ is the Boltzmann constant. Often the form $\text{0.695 cm"^(-1)"/K}$ of ${k}_{B}$ is useful in Statistical Mechanics.

The above form of $\ln \Omega$ is not all that useful. With a lot of manipulation, one would eventually obtain (Statistical Mechanics by Norman Davidson):

$S = {k}_{B} {\overbrace{\left[\ln \left({q}_{t o t} / N\right) + \frac{E}{N {k}_{B} T} + 1\right]}}^{\ln \Omega}$

where:

• For simple systems, ${q}_{t o t} \approx {q}_{t r} {q}_{r o t} {q}_{v i b} {q}_{e l e c}$ is the partition function for a given molecule, and is based on its translational, rotational, vibrational, and electronic degrees of freedom.
• $N$ is the number of particles.
• $E$ is the internal energy of a set of $N$ particles in a given system.
• $T$ is the temperature in $\text{K}$.

And so, the number of microstates is given by:

$\textcolor{b l u e}{\overline{\underline{|}} \stackrel{\text{ ")(" "Omega = "exp"(ln(q_(t ot)/N) + E/(Nk_BT) + 1)" }}{|}}$

Take methane, ${\text{CH}}_{4}$, for example (because I've already done this).

At $\text{298.15 K}$ and $\text{1 atm}$, it has:

• ${q}_{t r} = 0.02560 {M}^{3 / 2} {T}^{5 / 2} \approx \underline{2.525 \times {10}^{6}}$,

$M = 16.0426$ is the molar mass in $\text{amu}$

• ${q}_{r o t} = \frac{1}{12} \left(0.014837 {I}^{3 / 2} {T}^{3 / 2}\right) \approx \underline{36.72}$,

$I \approx 3.216$ is moment of inertia in "amu" cdot Å^2

• ${q}_{v i b} = {\prod}_{i = 1}^{4} {q}_{v i b , i} \approx \underline{6.373 \times {10}^{- 10}}$,

${q}_{v i b , i} = {e}^{- {\Theta}_{v i b , i} / 2 T} / \left(1 - {e}^{- {\Theta}_{v i b , i} / T}\right)$

${\Theta}_{v i b , i} = {\omega}_{i} / {k}_{B}$ is the vibrational temperature in $\text{K}$, where ${\omega}_{i}$ is the fundamental vibrational frequency in ${\text{cm}}^{- 1}$ for a given molecular motion, and ${k}_{B} = \text{0.695 cm"^(-1)"/K}$.

• $\underline{{q}_{e l e c} \approx 1}$ for any closed-shell molecule (zero unpaired electrons), since the next electronic energy level is usually naturally inaccessible except at temperatures in the tens of thousands of $\text{K}$.

However, if the molecule has unpaired electrons, that will make ${q}_{e l e c} = n + 1$, where $n$ is the number of unpaired electrons.

As a result, we get that:

${q}_{t o t} = {\overbrace{\left(2.525 \times {10}^{6}\right)}}^{\text{translational"overbrace((36.72))^"rotational"overbrace((6.373 xx 10^(-10)))^"vibrational"overbrace((1))^"electronic}} \approx 0.0591$

To make this simple, I have already calculated that the internal energy $E$ of methane at $\text{298.15 K}$ is $\text{5012.18 cm"^(-1)"/molecule}$ (or about $\text{60 kJ/mol}$).

As a result, for $1$ molecule ($N = 1$), we have:

ln Omega = ln(0.0591) + ("5012.18 cm"^(-1)"/molecule")/(1("0.695 cm"^(-1)"/molecule"cdot"K")("298.15 K")) + 1

$= 22.36$

And so, the number of microstates a single methane molecule has is:

$\textcolor{b l u e}{\Omega} = {e}^{22.36}$

$\approx$ $\underline{\textcolor{b l u e}{5.14 \times {10}^{9} {\text{molecule}}^{- 1}}}$

A more usual quantity is for a mol of methane molecules, which would then possess about $\underline{3.10 \times {10}^{33} {\text{mol}}^{- 1}}$ microstates!

SHOWING THAT THE NUMBER OF MICROSTATES IS CORRECT

From here one could also calculate the standard molar entropy ${S}^{\circ}$ to check that $\Omega$ is correct, since $\Omega$ was defined for $\text{298.15 K}$ and $\text{1 atm}$ (the same conditions needed for ${S}^{\circ}$):

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

$= \text{0.695 cm"^(-1)"/K} \times 22.36$

$= \text{15.54 cm"^(-1)"/molecule"cdot"K}$

Or, in more familiar units, multiply by $h c {N}_{A}$ to get:

${S}^{\circ} = \left(6.626 \times {10}^{- 34} \text{J"cdotcancel"s") cdot (2.998 xx 10^(10) cancel"cm""/"cancel"s") cdot 6.0221413 xx 10^(23) cancel"molecules""/mol" cdot (15.54 cancel("cm"^(-1)))/(cancel"molecule" cdot "K}\right)$

$=$ $\underline{\text{185.93 J/mol"cdot"K}}$

whereas the literature value was $\text{186.25 J/mol"cdot"K}$ (0.17% error).