Langmuir and Freundlich Isotherms for Papain Adsorption: 23-Chem-A2 Unit Operations and Separation Processes, May 2018, Q5

Question B2: Langmuir and Freundlich Isotherms for Papain Adsorption (25 points)

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.

$C$ (g papain/g solution)	$q$ (g papain/g charcoal)	$C/q$
9.295	0.3511	26.47
18.667	0.6109	30.56
28.754	0.6614	43.47
38.278	1.1953	32.02
48.355	1.2626	38.30

Find. (a) Langmuir constants $q_m,\ K$; (b) Freundlich constants $K_F,\ 1/n$; (c) the better-fitting isotherm.

Figure B2 — The loading keeps rising as a power of $C$ with no saturation plateau — the Freundlich (heterogeneous-surface) signature, which the fit statistics confirm.

Approach. Linearise each isotherm — Langmuir as $C/q$ vs $C$ (slope $1/q_m$, intercept $1/q_mK$), Freundlich as $\ln q$ vs $\ln C$ (slope $1/n$, intercept $\ln K_F$) — fit both by least squares and compare $R^2$.

(a) Langmuir fit. $q=\dfrac{q_mKC}{1+KC}\Rightarrow \dfrac{C}{q}=\dfrac{1}{q_m}C+\dfrac{1}{q_mK}$. Regressing $C/q$ on $C$ gives slope $0.2582$, intercept $26.76$: $$q_m=\frac{1}{0.2582}=\boxed{3.87\ \text{g/g}},\qquad K=\frac{1}{q_m(26.76)}=\boxed{9.65\times10^{-3}},\qquad R^2=0.35.$$
(b) Freundlich fit. $q=K_FC^{1/n}\Rightarrow \ln q=\ln K_F+\tfrac{1}{n}\ln C$. Regressing $\ln q$ on $\ln C$ gives slope $1/n=0.785$, intercept $\ln K_F=-2.826$: $$K_F=e^{-2.826}=\boxed{0.0592},\qquad \frac{1}{n}=0.785\ (n=1.27),\qquad R^2=0.93.$$
(c) Better fit. With $R^2_\text{Freundlich}=0.93\gg R^2_\text{Langmuir}=0.35$, $$\boxed{\text{the Freundlich isotherm fits the data far better.}}$$ Physically, $q$ keeps climbing as a power of $C$ with no saturation plateau over the tested range — heterogeneous-surface (Freundlich) behaviour rather than monolayer (Langmuir) saturation.

Quantity	Value
(a) Langmuir $q_m$ / $K$	3.87 g/g / 9.65×10⁻³ ($R^2=0.35$)
(b) Freundlich $K_F$ / $1/n$	0.0592 / 0.785 ($n=1.27$, $R^2=0.93$)
(c) Better fit	Freundlich

Check: the "g/g solution" concentrations exceed 1 (a source unit slip — likely mg/g or g/L); the isotherm constants are fit as printed and simply carry those units. The conclusion (Freundlich fits better) is unaffected by the unit label because both models see the same $C$ values.

Concept & Theory

Topic. Adsorption equilibrium — Langmuir vs Freundlich isotherms and their linearisations.

Key relations. Langmuir $q=\tfrac{q_mKC}{1+KC}$ (linear $C/q$ vs $C$); Freundlich $q=K_FC^{1/n}$ (linear $\ln q$ vs $\ln C$); goodness by $R^2$.

Why this works. The Langmuir model assumes identical sites and finite capacity (monolayer saturation, $q\to q_m$); the Freundlich model is empirical for energetically heterogeneous surfaces with no capacity limit and $1/n<1$ signalling favourable adsorption. Fitting the two linear forms and comparing $R^2$ shows which physical picture the data support — here, a non-plateauing power law favours Freundlich decisively.

Common pitfall. Comparing $R^2$ between two different linear transforms is only a rough guide (each transform reweights errors); trust it only when the gap is large, as it is here. Also, do not force Langmuir on clearly non-saturating data.

Source. McCabe, Smith & Harriott ch. 25 (adsorption); Geankoplis §12.3; Perry's §16.

Question 5 of 6: Langmuir and Freundlich Isotherms for Papain Adsorption

Question B2: Langmuir and Freundlich Isotherms for Papain Adsorption (25 points)