SO₂ Adsorption Isotherm and Surface Area: 23-Chem-A2 Unit Operations and Separation Processes, Undated paper, Q6

Question B3: SO₂ Adsorption Isotherm and Surface Area (25 marks: a 10, b 6)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

$p$ (mm Hg)	10	20	30	40	50	60	70	80
$V$ (cm³ STP)	2.20	2.40	2.62	2.75	2.88	3.00	3.05	3.12

Constants: $N_A=6.023\times10^{23}$ mol⁻¹; $\rho_{\text{liq,SO}_2}=1430$ kg/m³; $M_{SO_2}=64.07$ g/mol; molar gas volume at STP = 22 414 cm³/mol.

Find. (a) best-fit isotherm and its constants; (b) total surface area of the adsorbent.

Plotting $p/V$ against $p$ gives a straight line ($R^2=0.997$), confirming the Langmuir isotherm; slope $=1/V_m$ and intercept $=1/(V_m b)$ deliver the constants.

Approach. Test the Langmuir linearization $p/V$ vs $p$ (and compare a Freundlich log–log fit); read $V_m$ and $b$ from the line; then convert the monolayer volume $V_m$ into a molecule count and multiply by the projected area per SO₂ molecule to get the surface area.

(a) Identify the isotherm. Tabulating $p/V$ against $p$ gives a straight line (least squares $R^2=0.997$), whereas a Freundlich $\ln V$ vs $\ln p$ fit is poorer ($R^2=0.993$) and, crucially, has no saturation limit. The data follow the Langmuir isotherm $$V=\frac{V_m\,b\,p}{1+b\,p}\ \Longleftrightarrow\ \frac{p}{V}=\frac{1}{V_m b}+\frac{p}{V_m}.$$
Evaluate the constants. The regression line is $p/V=0.2967\,p+2.252$, so $$V_m=\frac{1}{\text{slope}}=\frac{1}{0.2967}=3.37\ \mathrm{cm^3\,(STP)},\qquad b=\frac{\text{slope}}{\text{intercept}}=\frac{0.2967}{2.252}=0.132\ \mathrm{mmHg^{-1}}.$$ Langmuir: $V_m=3.37\ \mathrm{cm^3(STP)}$, $b=0.132\ \mathrm{mmHg^{-1}}$.
(b) Molecules in the monolayer. The monolayer volume corresponds to $$n_m=\frac{V_m}{22\,414}=\frac{3.37}{22\,414}=1.504\times10^{-4}\ \mathrm{mol}\ \Rightarrow\ 1.504\times10^{-4}\times N_A=9.06\times10^{19}\ \text{molecules}.$$
Projected area per molecule from the liquid density: $$a_m=1.091\left(\frac{M}{\rho_\text{liq}N_A}\right)^{2/3}=1.091\left(\frac{64.07}{1.430\times6.023\times10^{23}}\right)^{2/3}=1.93\times10^{-19}\ \mathrm{m^2}.$$
Total surface area. $$S=(\text{molecules})\times a_m=(9.06\times10^{19})(1.93\times10^{-19})=\boxed{17.4\ \mathrm{m^2}}.$$ Total surface area of the adsorbent sample $\approx 17\ \mathrm{m^2}$.

Quantity	Value
Best-fit isotherm	Langmuir ($R^2=0.997$ > Freundlich 0.993)
Monolayer volume $V_m$	3.37 cm³ (STP)
Langmuir constant $b$	0.132 mmHg⁻¹
Molecules in monolayer	$9.06\times10^{19}$
Projected area per molecule $a_m$	$1.93\times10^{-19}$ m²
Total surface area $S$	17.4 m²

Check. The surface area is the total for the adsorbent charge used in the experiment (the sample mass is not printed, so a per-gram value cannot be quoted). The projected-area prefactor 1.091 assumes hexagonal close packing of spherical molecules in the monolayer, evaluated from the liquid density — the conventional BET/Emmett–Brunauer choice.

Concept & Theory

Topic. Physical adsorption — Langmuir monolayer isotherm and the molecular-area route to surface area.

Key relations. Langmuir $V=V_mbp/(1+bp)$, linearized $p/V=1/(V_mb)+p/V_m$; monolayer moles $n_m=V_m/22\,414$; projected area $a_m=1.091(M/\rho_\text{liq}N_A)^{2/3}$; $S=n_mN_Aa_m$.

Why this works. The Langmuir model treats adsorption as a reversible reaction filling a fixed number of identical sites, giving a saturation volume $V_m$ = one complete monolayer. Because $V_m$ counts the molecules needed to tile the surface once, multiplying that count by the area each molecule occupies recovers the surface area directly — the same idea underlies the BET method, with Langmuir standing in when the data plateau at a single layer (as here, where $V$ levels off toward $\sim3.1$ cm³).

Common pitfall. Choosing Freundlich (a good empirical fit at low coverage) and then being unable to extract a monolayer capacity — only a saturating isotherm yields $V_m$. Also, forgetting to convert the adsorbed volume from STP gas volume to moles before counting molecules.

Source. Geankoplis §12.3 (Adsorption); Treybal Ch. 11; Smith, Chemical Engineering Kinetics (surface-area determination).

Question 6 of 6: SO₂ Adsorption Isotherm and Surface Area

Question B3: SO₂ Adsorption Isotherm and Surface Area (25 marks: a 10, b 6)