This says that almost all entropy for the system was lost during the adsorption, even if #"H"_2# was rotating on (and presumably not translating across) the surface. You can see that this is much less than #"130 J/mol"cdot"K"# from the initial state. Therefore, from plugging #(2)# and #(3)# into #(1)#, the absolute ro-vibrational molecular entropy is: #>_(ads) = k_BT^2((del(lnq_(vib)q_(rot)-U/(k_BT)))/(delT))_V#,Īnd after some derivation (we will skip some steps because it's just a lot of product rule), we obtain: The molecular adsorption energy (the ro-vibrational energy when including #U#) is related to #q_(vib)# and #q_(rot)# according to With #sigma = 2# for #"H"_2# because it has one principal axis of rotation that preserves its symmetry, and #sigma# is that plus #1#. The rotational diatomic molecular partition function at the high-temperature limit is: The vibrational molecular partition function, treating the zero of energy as #0hnu#, is: To evaluate the absolute entropy formula, we will need #q_(vib)#, #q_(rot)#, and #>_(ads)# first. It turns out that the ro-vibrational contribution to the entropy, when including adsorption potential energy, is only this part: #> = E/N# is the molecular internal energy.If including adsorption potential energy #U#, #q/N ~~ q_(trans)q_(rot)q_(vib)e^(-U"/"k_BT)#. #q/N# is the total partition function for the molecule.Dividing by #N# means this is the absolute entropy per molecule. the absolute entropy for the final state is only described by a vibrational contribution and a rotational contribution.įrom my text ( Statistical Mechanics by Norman Davidson), the expression for the absolute total molecular entropy at any temperature is:.Therefore, if we assume -= + then we expect ~~ the absolute entropy for the initial state (before the adsorption) is described by = "130 J/mol"cdot"K"#. This is way higher than #"298 K"#, so it means that at room temperature, #"H"_2# will have minimal if any vibrational contribution to its standard molar entropy.Īnother contribution is rotation on the surface, which means we should consider the rotational temperature of #Theta_(vib) = (tildeomega_e)/k_B# in terms of #"cm"^(-1)# units, resulting in units of #"K"#.) (This was derived from the fundamental vibrational frequency of #"4401.21"_3 "cm"^(-1)# found on NIST, using The vibrational temperature of #"H"_2# (the temperature at which it accesses its vibrational states) is #Theta_(vib) = "6333 K"#.
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