Diffusion of Acetic Acid through a Stagnant Water Film: 23-Chem-A3 Heat and Mass Transfer, Undated paper, Q4

Question 4: Diffusion of Acetic Acid through a Stagnant Water Film (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.

Quantity	Value
Film thickness $z$	0.001 m
Boundary 1	9 wt% acetic acid, $\rho=1012$ kg/m³
Boundary 2	4 wt% acetic acid, $\rho=1003.2$ kg/m³
$D_{AB}$ at 25 °C	$1.11\times10^{-9}$ m²/s
$\mu$ at 25 / 17 °C	$1.1336\times10^{-3}$ / $1.2883\times10^{-3}$ Pa·s
$M_A$ (acetic acid) / $M_B$ (water)	60.05 / 18.02 g/mol

Find. The molar (and mass) diffusion flux $N_A$ of acetic acid through the stagnant water film at 17 °C.

Figure 4 — Acetic acid ($A$) diffusing through the stationary water film ($B$). The acid mole fraction falls from 0.0288 to 0.0123 over 1 mm; because water does not diffuse, the flux carries a small log-mean drift correction $x_{BM}$.

Approach. Correct the diffusivity from 25 °C to 17 °C with the Stokes–Einstein group $D\mu/T=\text{const}$, convert the two wall compositions from weight to mole fraction and to molar concentration ($c=\rho/M_{\text{avg}}$), then apply the “$A$ through stagnant $B$” flux relation.

Diffusivity at 17 °C. Stokes–Einstein gives $D\propto T/\mu$, so$$D_{17}=D_{25}\frac{T_{17}}{T_{25}}\frac{\mu_{25}}{\mu_{17}}=1.11\times10^{-9}\frac{290.15}{298.15}\frac{1.1336}{1.2883}=\boxed{9.51\times10^{-10}\ \text{m}^2/\text{s}}.$$
Wall mole fractions. Per 100 g of solution: at boundary 1, $n_A=9/60.05=0.1499$, $n_B=91/18.02=5.050$, so $x_{A1}=0.1499/5.200=0.0288$; at boundary 2, $n_A=4/60.05=0.0666$, $n_B=96/18.02=5.327$, so $x_{A2}=0.0666/5.394=0.0123$.
Molar concentration. With $M_{\text{avg}}=100/(n_A+n_B)$ and $c=\rho/M_{\text{avg}}$: $c_1=1012/19.23=52.6$ and $c_2=1003.2/18.54=54.1$ kmol/m³; mean $c=53.4\ \text{kmol/m}^3$ (from the liquid density, not $P/RT$).
Flux, $A$ through stagnant $B$. Using $x_{B1}=0.9712$, $x_{B2}=0.9877$, the compact form $N_A=\dfrac{D_{AB}\,c}{z}\ln\dfrac{x_{B2}}{x_{B1}}$ (equivalently $\dfrac{D_{AB}c}{z}\dfrac{x_{A1}-x_{A2}}{x_{BM}}$ with $x_{BM}=0.979$) gives$$N_A=\frac{(9.51\times10^{-10})(53.4)}{0.001}\ln\frac{0.9877}{0.9712}=\boxed{8.53\times10^{-7}\ \text{kmol/m}^2\text{s}}.$$In mass terms $N_A M_A=8.53\times10^{-7}(60.05)=5.12\times10^{-5}\ \text{kg/m}^2\text{s}$.

Quantity	Result
$D_{AB}$ at 17 °C	$9.51\times10^{-10}$ m²/s
Mole fractions $x_{A1}$ / $x_{A2}$	0.0288 / 0.0123
Mean concentration $c$	53.4 kmol/m³
Diffusion flux $N_A$	$8.53\times10^{-7}$ kmol/m²·s ($5.12\times10^{-5}$ kg/m²·s)

Concept & Theory

Topic. Steady molecular diffusion of a solute through a stagnant liquid solvent film.

Key relations. $N_A=\dfrac{D_{AB}c}{z}\ln\dfrac{x_{B2}}{x_{B1}}=\dfrac{D_{AB}c}{z}\dfrac{x_{A1}-x_{A2}}{x_{BM}}$; liquid $c=\rho/M_{\text{avg}}$; Stokes–Einstein $D\mu/T=$ const.

Why this works. When only $A$ crosses the film and $B$ is stationary, the net molar flux induces a convective drag on $A$ captured by dividing by the log-mean stagnant fraction $x_{BM}$. In a dilute liquid $x_{BM}\to1$, so the answer is close to the simple Fick value, but the correction is retained for rigour. The Stokes–Einstein scaling transfers the tabulated 25 °C diffusivity to the required 17 °C using the measured viscosities.

Common pitfall. Using the ideal-gas $c=P/RT$ for a liquid (it must be $\rho/M_{\text{avg}}$), leaving compositions in weight fraction, or ignoring the temperature/viscosity correction to $D$.

Source. Geankoplis, Transport Processes §6.2–6.3 (diffusion of $A$ through stagnant $B$; liquid diffusivity temperature correction); Treybal Ch. 2.

Question 4 of 6: Diffusion of Acetic Acid through a Stagnant Water Film

Part A — Heat Transfer

Part B — Mass Transfer

Question 4: Diffusion of Acetic Acid through a Stagnant Water Film (25 points)