Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 2 / A7

Let $\Lambda =h/\sqrt{2\pi m{k}_{\mathrm{B}}T}$ be the thermal de Broglie wavelength. The two free momentum components of one adsorbed atom give the phase-space integralEach momentum integral is a Gaussian integral, soThe second expression is the classical ideal-gas partition function for mobile, indistinguishable surface atoms. Its validity requires a dilute nondegenerate gas, $N{\Lambda }^2/\sigma \ll 1$, and neglects interactions and excited motion normal to the surface. The $N!$ appears here because particles move between positions, unlike the fully occupied fixed-site model of A6.

For the same atoms in a three-dimensional volume $V$, the one-atom translational partition function is $q_{\mathrm{trans}}^{(3)}=V/{\Lambda }^3$. Comparing the same $N$ atoms at the same temperature cancels the common $N!$ in their canonical partition functions. Therefore the molar adsorption confinement free energy, with the sign defined as adsorbed minus free, isThe logarithm is dimensionless: $V/\sigma$ and $\Lambda$ are both lengths. For a macroscopic gas depth $V/\sigma \gg \Lambda$, confinement raises the translational Helmholtz free energy. Equivalently, the translational state count is reduced by a factor $\sigma \Lambda /V$ per atom. The question supplies neither a volume nor an adsorption binding energy, so this answer is symbolic and describes the confinement cost, not a numerical net adsorption free energy.

For example, a binding energy $D>0$ per atom would supply a separate $-N_{\mathrm{A}}D$ to $\Delta F_m$, together with any normal-motion internal contribution retained by a more detailed model. Adsorption can then be favourable despite the loss of free translational phase space. Merely calling the change a loss must not reverse the sign of the explicitly defined $F_{\mathrm{ads}}-F_{\mathrm{gas}}$.

A planar rigid rotor has angular eigenfunctions $e^{im\varphi }$ and single-valuedness imposes $m=0,\pm 1,\pm 2,\dots$. Its energy depends on $m^2$, so if the level is labelled by $J=|m|$, thenThe two states at $J>0$ represent opposite directions of angular momentum. If the signed quantum number itself is called $J$, each signed state is counted once instead; one must not both sum over both signs and multiply by two again.

For an unrestricted three-dimensional linear rigid rotor, $J=0,1,2,\dots$ has $M=-J,\dots ,J$ and state degeneracy $2J+1$, with $E_J={\hslash }^{2}J(J+1)/(2I)$. Planar confinement removes one rotational coordinate and changes both the spectrum and degeneracy. These are the spatial counts before nuclear spin statistics. For homonuclear hydrogen, the allowed even/odd spatial states must also be paired with the appropriate parahydrogen or orthohydrogen nuclear states, as considered explicitly in A8.

Define ${\theta }_{\mathrm{rot}}={\hslash }^2/(2I{k}_{\mathrm{B}})$. Summing the spatial states counted in part (c),When $T\gg {\theta }_{\mathrm{rot}}$ the closely spaced terms can be replaced by a Gaussian integral:Here labelled means that the two ends of the molecular axis were distinguished in the angular count. In the usual high-temperature molecular convention that removes equivalent orientations and factors out nuclear-spin multiplicity, divide by the rotational symmetry number $s=2$ for homonuclear hydrogen:The full exact spin isomers of hydrogen count is instead the even-$m$ spatial sum plus three times the odd-$m$ sum; its high-temperature limit is $2\sqrt{\pi T/{\theta }_{\mathrm{rot}}}$, including the four nuclear-spin states. These conventions differ only by temperature-independent factors in this limit. All give $U_{\mathrm{rot},\mathrm{m}}=RT/2$ and $C_{V,m}^{\mathrm{rot}}=R/2$, consistent with one quadratic rotational degree of freedom. By comparison, the free linear rigid rotor has $q_{\mathrm{rot}}\propto T$ and $C_{V,m}^{\mathrm{rot}}=R$.