# How to calculate the change in the entropy of hydrogen molecule before and after the adsorption?

##### 1 Answer

The vibrational temperature of

(This was derived from the fundamental vibrational frequency of

#Theta_(vib) = (tildeomega_e)/k_B# in terms of#"cm"^(-1)# units, resulting in units of#"K"# .)

This is way higher than

Another contribution is rotation on the surface, which means we should consider the rotational temperature of

#Theta_(rot) = (tildeB_e)/k_B = "60.8530 cm"^(-1)/("0.695 cm"^(-1)"/K") = "87.56 K"# ,

derived from the rotational constant

Therefore, if we assume

Since

- the absolute entropy for the initial state (before the adsorption) is described by
#S^@ = "130 J/mol"cdot"K"# ... - the absolute entropy for the final state is only described by a vibrational contribution and a rotational contribution.

Let's check how big

From my text (*Statistical Mechanics* by Norman Davidson), the expression for the **absolute total molecular entropy** at any temperature is:

#S/N = k_Bln(q/N) + (<< epsilon >>)/T + k_B# where:

#N# is the number of molecules. Dividing by#N# means this is the absolute entropy per.molecule#k_B = "0.695 cm"^(-1)"/K"# is the molecular Boltzmann constant.#q/N# is the total partition function for the molecule. If including adsorption potential energy#U# ,#q/N ~~ q_(trans)q_(rot)q_(vib)e^(-U"/"k_BT)# .#<< epsilon >> = E/N# is the molecular internal energy.

It turns out that the ro-vibrational contribution to the entropy, when including adsorption potential energy, is only this part:

#S_(ads)/N = k_Bln(q_(vib)q_(rot)e^(-U"/"k_BT)) + (<< epsilon >>_(ads))/T# #" "" "bb((1))#

To evaluate the absolute entropy formula, we will need

The **vibrational molecular partition function**, treating the zero of energy as

#q_(vib) = (e^(-Theta_(vib)"/"2T))/(1-e^(-Theta_(vib)"/"T))# #" "" "bb((2))#

The **rotational diatomic molecular partition function** at the high-temperature limit is:

#q_(rot) ~~ T/(sigmaTheta_(rot))# ,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 molecular adsorption energy (the ro-vibrational energy when including

#<< epsilon >>_(ads) = k_BT^2((del(lnq_(vib)q_(rot)-U/(k_BT)))/(delT))_V# ,

and after some derivation (we will skip some steps because it's just a lot of product rule), we obtain:

#<< epsilon >>_(ads) = overbrace(k_B[Theta_(vib)/2 + (Theta_(vib))/(e^(Theta_(vib)"/"T) - 1) + U/(k_B)])^("Vibrational") + overbrace(k_BT)^("Rotational")# #" "" "bb((3))#

Therefore, from plugging *absolute ro-vibrational molecular entropy* is:

#S_(ads)/N = k_B ln((e^(-Theta_(vib)"/"2T))/(1-e^(-Theta_(vib)"/"T))) + k_B ln(T/(2Theta_(rot))) - cancel(k_BcdotU/(k_BT)) + k_B/T[Theta_(vib)/2 + (Theta_(vib))/(e^(Theta_(vib)"/"T) - 1) + cancel(U/(k_B))]#

#= "0.695 cm"^(-1)"/K" cdot ln((e^(-"6333 K/2"cdot"298 K"))/(1-e^(-"6333 K/298 K"))) + ("0.695 cm"^(-1)"/K")cdot ln("298 K"/(2 cdot "87.56 K")) + ("0.695 cm"^(-1)"/K")/("298 K")["6333 K"/2 + ("6333 K")/(e^("6333 K/298 K") - 1)]#

#~~ ul("0.3695 cm"^(-1)"/K"cdot"molecule")#

Converting to molar units, we get:

#color(blue)(S_(ads)^@) = 0.3695 cancel("cm"^(-1))"/K"cdotcancel"molecule" xx 2.998 xx 10^(10) cancel("cm")/cancel"s" xx 6.626 xx 10^(-34) "J"cdotcancel"s" xx (6.0221413 xx 10^(23) cancel"molecules")/"mol"#

#= color(blue)ul("4.420 J/mol"cdot"K")#

You can see that this is much less than

#color(blue)(DeltaS^@) = S_(ads)^@ - S_(f re e)^@#

#= "4.420 J/mol"cdot"K" - "130 J/mol"cdot"K"#

#= color(blue)ul(-"125.58 J/mol"cdot"K")#

This says that almost all entropy for the system was lost during the adsorption, even if

(Furthermore, it turned out that the adsorption energy numerically cancels out and the entropy is independent of what it is because it is assumed roughly constant with respect to position.)