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

May 16, 2017

The vibrational temperature of ${\text{H}}_{2}$ (the temperature at which it accesses its vibrational states) is ${\Theta}_{v i b} = \text{6333 K}$.

(This was derived from the fundamental vibrational frequency of ${\text{4401.21"_3 "cm}}^{- 1}$ found on NIST, using

${\Theta}_{v i b} = \frac{t i l \mathrm{de} {\omega}_{e}}{k} _ B$ in terms of ${\text{cm}}^{- 1}$ units, resulting in units of $\text{K}$.)

This is way higher than $\text{298 K}$, so it means that at room temperature, ${\text{H}}_{2}$ will have minimal if any vibrational contribution to its standard molar entropy.

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

${\Theta}_{r o t} = \frac{t i l \mathrm{de} {B}_{e}}{k} _ B = \text{60.8530 cm"^(-1)/("0.695 cm"^(-1)"/K") = "87.56 K}$,

derived from the rotational constant $t i l \mathrm{de} {B}_{e}$ from NIST in terms of ${\text{cm}}^{- 1}$.

Therefore, if we assume ${S}_{a \mathrm{ds}}^{\circ} \equiv {S}_{v i b}^{\circ} + {S}_{r o t}^{\circ}$, then we expect ${S}_{a \mathrm{ds}}^{\circ} \approx {S}_{r o t}^{\circ}$.

Since

• the absolute entropy for the initial state (before the adsorption) is described by ${S}^{\circ} = \text{130 J/mol"cdot"K}$...
• the absolute entropy for the final state is only described by a vibrational contribution and a rotational contribution.

Let's check how big ${S}_{a \mathrm{ds}}^{\circ}$ actually is.

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

$\frac{S}{N} = {k}_{B} \ln \left(\frac{q}{N}\right) + \frac{\left\langle\epsilon\right\rangle}{T} + {k}_{B}$

where:

• $N$ is the number of molecules. Dividing by $N$ means this is the absolute entropy per molecule.
• ${k}_{B} = \text{0.695 cm"^(-1)"/K}$ is the molecular Boltzmann constant.
• $\frac{q}{N}$ is the total partition function for the molecule. If including adsorption potential energy $U$, $\frac{q}{N} \approx {q}_{t r a n s} {q}_{r o t} {q}_{v i b} {e}^{- U \text{/} {k}_{B} T}$.
• $\left\langle\epsilon\right\rangle = \frac{E}{N}$ 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}_{a \mathrm{ds}} / N = {k}_{B} \ln \left({q}_{v i b} {q}_{r o t} {e}^{- U \text{/} {k}_{B} T}\right) + \frac{{\left\langle\epsilon\right\rangle}_{a \mathrm{ds}}}{T}$$\text{ "" } \boldsymbol{\left(1\right)}$

To evaluate the absolute entropy formula, we will need ${q}_{v i b}$, ${q}_{r o t}$, and ${\left\langle\epsilon\right\rangle}_{a \mathrm{ds}}$ first.

The vibrational molecular partition function, treating the zero of energy as $0 h \nu$, is:

${q}_{v i b} = \left({e}^{- {\Theta}_{v i b} \text{/"2T))/(1-e^(-Theta_(vib)"/} T}\right)$ $\text{ "" } \boldsymbol{\left(2\right)}$

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

${q}_{r o t} \approx \frac{T}{\sigma {\Theta}_{r o t}}$,

with $\sigma = 2$ for ${\text{H}}_{2}$ 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 $U$) is related to ${q}_{v i b}$ and ${q}_{r o t}$ according to

${\left\langle\epsilon\right\rangle}_{a \mathrm{ds}} = {k}_{B} {T}^{2} {\left(\frac{\partial \left(\ln {q}_{v i b} {q}_{r o t} - \frac{U}{{k}_{B} T}\right)}{\partial T}\right)}_{V}$,

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")$\text{ "" } \boldsymbol{\left(3\right)}$

Therefore, from plugging $\left(2\right)$ and $\left(3\right)$ into $\left(1\right)$, the absolute ro-vibrational molecular entropy is:

${S}_{a \mathrm{ds}} / N = {k}_{B} \ln \left(\left({e}^{- {\Theta}_{v i b} \text{/"2T))/(1-e^(-Theta_(vib)"/"T))) + k_B ln(T/(2Theta_(rot))) - cancel(k_BcdotU/(k_BT)) + k_B/T[Theta_(vib)/2 + (Theta_(vib))/(e^(Theta_(vib)"/} T} - 1\right) + \cancel{\frac{U}{{k}_{B}}}\right]$

= "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)]

$\approx \underline{\text{0.3695 cm"^(-1)"/K"cdot"molecule}}$

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"

$= \textcolor{b l u e}{\underline{\text{4.420 J/mol"cdot"K}}}$

You can see that this is much less than $\text{130 J/mol"cdot"K}$ from the initial state. Therefore:

$\textcolor{b l u e}{\Delta {S}^{\circ}} = {S}_{a \mathrm{ds}}^{\circ} - {S}_{f r e e}^{\circ}$

$= \text{4.420 J/mol"cdot"K" - "130 J/mol"cdot"K}$

$= \textcolor{b l u e}{\underline{- \text{125.58 J/mol"cdot"K}}}$

This says that almost all entropy for the system was lost during the adsorption, even if ${\text{H}}_{2}$ was rotating on (and presumably not translating across) the surface.

(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.)