Past exam of the chemistry A course of the University of Cambridge / 2023 / Paper 2 / A6 / a / Solution

Let ${\epsilon }_s$ be the energy of one adsorbed atom, using the same energy reference as the gas-phase value ${\epsilon }_0$. For equivalent, noninteracting adsorption sites, choosing the occupied positions givesEvery configuration has energy $N_1{\epsilon }_s$, so the canonical lattice-gas partition function isThere is no further $1/N_1!$ factor: the combinatorial choice already counts indistinguishable adsorbates correctly. The Helmholtz free energy, using Stirling's approximation for a macroscopic number of occupied and empty sites, isHence the adsorbed-phase chemical potential at fixed $T,M$ isThe logarithm comes from configurational entropy: adding an atom is less favorable when vacancies are scarce. In the gas, the stipulated expression can be writtenwhere $\Lambda$ is the thermal de Broglie wavelength. Equality of chemical potentials at chemical equilibrium givesThereforeThe binding energy $E_b={\epsilon }_0-{\epsilon }_s$ is positive for favorable binding. The units of $b$ are inverse pressure, making $bp$ dimensionless. Choosing the adsorbed energy as zero simply sets ${\epsilon }_s=0$ and leaves the same physics; it must not remove the relative binding energy. This is the Langmuir isotherm, with low-pressure linear uptake and high-pressure monolayer saturation.