Question

How many micro states exist for a molecule?

Answer 1

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 `"1 mol"` of molecules would have around `10^30` microstates or more.

Also, a note about `q_(el ec)`: it can be not `1` if the ground state of a molecule is not a singlet. For example, `q_(el ec) = 3` for `"O"_2` in its ground state because it is a `""^3 Sigma_g^-`, a triplet-sigma-gerade-minus state.

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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 [N_i ln(g_i/N_i) + N_i]`

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 `"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_Boverbrace([ln(q_(t ot)/N) + E/(Nk_BT) + 1])^(ln Omega)`

where:

• For simple systems, `q_(t ot) ~~ q_(tr)q_(rot)q_(vib)q_(el ec)` 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 `"K"`.

And so, the number of microstates is given by:

`color(blue)(barul|stackrel(" ")(" "Omega = "exp"(ln(q_(t ot)/N) + E/(Nk_BT) + 1)" ")|)`

Take methane, `"CH"_4`, for example (because I've already done this).

At `"298.15 K"` and `"1 atm"`, it has:

• `q_(tr) = 0.02560M^(3//2)T^(5//2) ~~ ul(2.525 xx 10^6)`,

  `M = 16.0426` is the molar mass in `"amu"`

• `q_(rot) = 1/12 (0.014837I^(3//2)T^(3//2)) ~~ ul(36.72)`,

  `I ~~ 3.216` is moment of inertia in `"amu" cdot Å^2`

• `q_(vib) = prod_(i=1)^(4) q_(vib,i) ~~ ul(6.373 xx 10^(-10))`,

`q_(vib,i) = e^(-Theta_(vib,i)//2T)/(1 - e^(-Theta_(vib,i)//T))`

`Theta_(vib,i) = omega_i/k_B` is the vibrational temperature in `"K"`, where `omega_i` is the fundamental vibrational frequency in `"cm"^(-1)` for a given molecular motion, and `k_B = "0.695 cm"^(-1)"/K"`.

• `ul(q_(el ec) ~~ 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 `"K"`.

However, if the molecule has unpaired electrons, that will make `q_(el ec) = n + 1`, where `n` is the number of unpaired electrons.

As a result, we get that:

`q_(t ot) = overbrace((2.525 xx 10^6))^"translational"overbrace((36.72))^"rotational"overbrace((6.373 xx 10^(-10)))^"vibrational"overbrace((1))^"electronic" ~~ 0.0591`

To make this simple, I have already calculated that the internal energy `E` of methane at `"298.15 K"` is `"5012.18 cm"^(-1)"/molecule"` (or about `"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:

`color(blue)(Omega) = e^(22.36)`

`~~` `ulcolor(blue)(5.14 xx 10^9 "molecule"^(-1))`

A more usual quantity is for a mol of methane molecules, which would then possess about `ul(3.10 xx 10^33 "mol"^(-1))` microstates!

SHOWING THAT THE NUMBER OF MICROSTATES IS CORRECT

From here one could also calculate the standard molar entropy `S^@` to check that `Omega` is correct, since `Omega` was defined for `"298.15 K"` and `"1 atm"` (the same conditions needed for `S^@`):

`S^@ = k_BlnOmega`

`= "0.695 cm"^(-1)"/K" xx 22.36`

`= "15.54 cm"^(-1)"/molecule"cdot"K"`

Or, in more familiar units, multiply by `hcN_A` to get:

`S^@ = (6.626 xx 10^(-34) "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")`

`=` `ul("185.93 J/mol"cdot"K")`

whereas the literature value was `"186.25 J/mol"cdot"K"` (`0.17%` error).