How many micro states exist for a molecule?

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.

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).