Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 4 / 64

For cubes of side $\ell$, total area divided by total mass is $S=6/(\rho \ell )$, independently of how many cubes share the $100\mathrm{g}$ sample. With $S=5000{\mathrm{m}}^2\mathrm{k}{\mathrm{g}}^{-1}$ and $\rho =2330\mathrm{kg}{\mathrm{m}}^{-3}$,This assumes accessible external faces, negligible contact-area losses and equally sized nonporous cubes; BET adsorption can also count accessible internal surface in a porous powder.

The coherent neutron scattering length density is atomic number density times scattering length:The density was converted to $\mathrm{g}{\mathrm{m}}^{-3}$ to match the molar mass in grams per mole, and $1{\mathrm{m}}^{-2}=10^{-20}{\mathring{\mathrm{A}}}^{-2}$.

Use the water data to determine the common oxygen contribution rather than assume identical molar volumes. Ordinary water has molecular number densityso $b_O={\rho }_b^{H_{2}O}/n_H-2b_H\simeq 5.81\times 10^{-15}\mathrm{m}$. Therefore $b_{D_{2}O}=2b_D+b_O\simeq 1.915\times 10^{-14}\mathrm{m}$ andThe assumptions are isotopically pure waters, additive nuclear scattering lengths, the same oxygen isotope contribution, and molar masses approximately $18$ and $20\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. Equal molar volumes would give the rougher estimate $20/18=1.11\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$.

Assuming additive liquid volumes, the scattering length density is linear in D${}_2$O volume fraction ${\varphi }_D$:where all scattering length densities are in $10^{-6}{\mathring{\mathrm{A}}}^{-2}$. The $100\mathrm{g}$ of heavy water occupies about $90.99\mathrm{c}{\mathrm{m}}^3$. The required ordinary-water volume isThus add approximately $147\mathrm{g}$ of H${}_2$O. Volume fraction, rather than mass fraction, enters this contrast-matching equation. Isotope exchange to form HDO does not change the total nuclear scattering-length balance in the additive-volume approximation.

Without the film, a sharp silicon-water boundary gives the usual Fresnel-like reflectivity governed by its scattering-length-density contrast. If the contrast-matched mixture from part (iii) is used, that ideal interface signal is strongly suppressed.

Polyethylene has a different scattering length density from both matched media. Its two interfaces generate interference oscillations, or Kiessig fringes, in reflectivity versus normal momentum transfer $q_z$. Their approximate spacing isfor $d=100\mathrm{nm}=1000\mathring{\mathrm{A}}$. Fit several fringes, with refraction, roughness and instrument resolution included if necessary, to determine the thickness accurately. Even with unmatched water the film changes the reflection envelope and adds thickness-dependent interference.

The Langmuir adsorption model assumes equivalent independent sites, a maximum of one adsorbate per site, no lateral adsorbate interactions, reversible adsorption-desorption equilibrium and non-dissociative adsorption here. Surface coverage obeys $\theta =bC/(1+bC)$ for the solution analogue.

BET adsorption allows multiple layers: occupation of one layer creates sites for further adsorption above it, and higher-layer binding is treated differently from first-layer binding, commonly using a condensation-like energy. Langmuir saturation is monolayer saturation; the BET model is not simply Langmuir with a different numerical equilibrium constant.

Invert the stated expression:Hence the line of $1/\Gamma$ against $1/C$ has slope $M=1/(b{\Gamma }_{\infty })$ and intercept $1/{\Gamma }_{\infty }$. Their ratio gives $b=(\mathrm{intercept})/(\mathrm{slope})$.

A least-squares straight-line fit in the requested reciprocal coordinates gives approximatelywhen the supplied numerical units are used. ThusThe slope and intercept units follow from the plotted reciprocal quantities. These parameters are the result of the requested reciprocal linear fit; fitting the original nonlinear isotherm with a different error model would give slightly different estimates.

The inferred saturation density is physically implausible for a single molecular layer in the printed units:That is far smaller than an atomic cross-section, let alone a butyric-acid molecule. The near-linear reciprocal plot therefore does not validate the physical monolayer interpretation. Possible issues include reported units or normalization, inaccessible/internal area not accounted for correctly, multilayer adsorption or other uptake processes. The numbers must not silently be relabelled as micromoles to obtain a more comfortable molecular area.

Electric-dipole sum-frequency generation spectroscopy is a coherent second-order optical probe. It is sensitive to appropriately oriented, symmetry-allowed molecular responses at the interface; randomly oriented populations can cancel in the orientational average, and centrosymmetric arrangements give no net electric-dipole second-order response.

Thus an ordered fraction of surface-bound carboxylate or hydrocarbon groups may dominate the measured spectrum, while disordered, differently oriented or weakly active adsorbates contribute little. The observed vibrational intensity is not a direct count of every molecule retained in an adsorption measurement. Selection rules, polarization and interference between contributions also matter.

For competing reactant and product species on shared equivalent sites, add each $b_j{C}_j$ term to the denominator:Rapid product desorption makes product coverages small, reducing competitive site blocking and simplifying the rate to reactant coverage alone. Without that assumption, product adsorption can inhibit conversion and make the apparent rate evolve as products accumulate.

$0.1\mathrm{M}=100\mathrm{mol}{\mathrm{m}}^{-3}=10^5\mathrm{mmol}{\mathrm{m}}^{-3}$. ThereforeThe surface is almost saturated, so the rate approaches the supplied limiting constant. This assumes rapid adsorption equilibrium, negligible product coverage, no competing species and unchanged substrate area/photochemical conditions.

For simultaneous competitive adsorption, use the butyric-acid constant inferred in part (c), $b_b\simeq 1.21{\mathrm{m}}^3\mathrm{mmo}{\mathrm{l}}^{-1}$, and $C_b=10^6\mathrm{mmol}{\mathrm{m}}^{-3}$:The propionic-acid rate is therefore approximatelyThe abundant, more strongly adsorbing butyric acid occupies most common sites, reducing propionic conversion by roughly a factor of $35$. This calculation follows the requested competitive Langmuir model despite the physical concern about the fitted saturation density noted earlier.