C1 — Film-model dissolution with an instantaneous reaction: 23-Chem-B1 Transport Phenomena, December 2019, Q5

Question 5: C1 — Film-model dissolution with an instantaneous reaction (25 marks: 7 + 18)

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. A stagnant film of thickness $\delta$ over the dissolving solid; saturation (interfacial) concentration of A at the solid surface $c_{A0}$; diffusivities $D_{AS}$ and $D_{BS}$ in the film; well-mixed bulk. Find. (a) the molar dissolution flux $N_A$ with negligible bulk A; (b) the flux when B is present and consumes A instantaneously at the reaction plane $z=\kappa\delta$, and the location $\kappa$ of that plane.

Fig. C1: Film model. In case (b) A diffuses from the solid ($c_{A0}$ at $z=0$) inward and B diffuses from the bulk inward; they annihilate at the reaction plane $z=\kappa\delta$ where both concentrations vanish. The instantaneous reaction shortens A’s diffusion path from $\delta$ to $\kappa\delta$, raising the dissolution rate.

Approach. With no homogeneous reaction inside the film and a dilute solute, the steady one-dimensional flux is constant and the concentration profile is linear (Fick’s law). For part (b) the reaction plane splits the film into an A-side and a B-side; a 1:1 stoichiometric flux match locates the plane and gives the enhanced rate.

(a) Diffusion across the whole film. Steady state with no reaction in the film means $dN_{Az}/dz=0$, so $N_{Az}$ is constant and, for dilute A, Fick’s law $N_{Az}=-D_{AS}\,dc_A/dz$ integrates to a linear profile between $c_A(0)=c_{A0}$ (saturation) and $c_A(\delta)\approx0$ (negligible bulk A): $$\boxed{N_{A}=\frac{D_{AS}\,c_{A0}}{\delta}.}$$ The dissolution rate is diffusion-limited across the full film thickness $\delta$.
(b) Set up the two sub-regions. The instantaneous, irreversible reaction confines A to $0\le z\le\kappa\delta$ (A cannot coexist with B), and B to $\kappa\delta\le z\le\delta$; both concentrations reach zero at the reaction plane. Each sub-region is source-free, so both profiles are linear with constant fluxes: $$N_A=\frac{D_{AS}\,c_{A0}}{\kappa\delta},\qquad N_B=\frac{D_{BS}\,c_{B0}}{(1-\kappa)\delta},$$ where $c_{B0}=c\,x_B$ is the bulk B concentration ($c$ = total molar concentration).
Locate the reaction plane by stoichiometry. Because $A+B\rightarrow$ product consumes A and B in a 1:1 ratio, the two fluxes arriving at the plane must be equal, $N_A=N_B$: $$\frac{D_{AS}c_{A0}}{\kappa\delta}=\frac{D_{BS}c_{B0}}{(1-\kappa)\delta}\ \Longrightarrow\ \boxed{\kappa=\frac{D_{AS}c_{A0}}{D_{AS}c_{A0}+D_{BS}c_{B0}}.}$$ A more soluble/faster-diffusing A pushes the plane outward (larger $\kappa$); more B in the bulk pulls it toward the solid.
Dissolution rate with reaction. Substituting $\kappa$ back into $N_A=D_{AS}c_{A0}/(\kappa\delta)$, the $c_{A0}$ cancels and $$\boxed{N_A=\frac{D_{AS}c_{A0}+D_{BS}c_{B0}}{\delta}=\frac{c\,(D_{AS}x_{A0}+D_{BS}x_B)}{\delta},}$$ with $x_{A0}=c_{A0}/c$ the saturation mole fraction of A.
Interpret the enhancement. Comparing with part (a), the reaction multiplies the dissolution rate by $$\frac{N_A^{(b)}}{N_A^{(a)}}=1+\frac{D_{BS}c_{B0}}{D_{AS}c_{A0}}=\frac1\kappa\;(>1).$$ The instantaneous reaction holds $c_A=0$ at $z=\kappa\delta$ instead of at $z=\delta$, steepening A’s gradient and speeding dissolution — the classic reaction-enhancement of interphase mass transfer.

Part	Result
(a) Dissolution flux, no reaction	$N_A=D_{AS}c_{A0}/\delta$
(b) Reaction-plane location	$\kappa=\dfrac{D_{AS}c_{A0}}{D_{AS}c_{A0}+D_{BS}c_{B0}}$
(b) Dissolution flux with reaction	$N_A=\dfrac{D_{AS}c_{A0}+D_{BS}c_{B0}}{\delta}$
Enhancement factor	$N_A^{(b)}/N_A^{(a)}=1/\kappa=1+D_{BS}c_{B0}/(D_{AS}c_{A0})$

Check — dilute-solution (linear-profile) assumption

Both parts assume dilute solutions, so the convective (bulk-flow) contribution to the flux is dropped and Fick’s law gives linear profiles — appropriate for a sparingly-soluble solid. If A were concentrated, the surface dissolution would need the log-mean drift factor $\ln[(1-x_{A\delta})/(1-x_{A0})]$; the instantaneous-reaction geometry (path shortened to $\kappa\delta$) is unchanged. The reaction is taken fast enough that A and B never coexist, which is the definition of the “instantaneous” limit.

Concept & Theory

Topic. Two-film (stagnant-film) model for interphase mass transfer with an instantaneous liquid-phase reaction.

Key relations. Fick’s first law $N_A=-D_{AS}\,dc_A/dz$; steady source-free film $\Rightarrow$ linear profile; stoichiometric flux match $N_A=N_B$ for $A+B\rightarrow$ product; reaction-plane fraction $\kappa=D_{AS}c_{A0}/(D_{AS}c_{A0}+D_{BS}c_{B0})$.

Why this works. The film model replaces the real hydrodynamic boundary layer by an equivalent stagnant layer of thickness $\delta$ across which molecular diffusion controls. An instantaneous reaction cannot let reactants overlap, so it collapses to a sharp plane whose position is fixed by supply-and-demand of the two diffusing species. Shortening A’s diffusion path is exactly why reactive absorption/dissolution outperforms the purely physical case.

Common pitfall. Placing $c_A=0$ at the film edge $z=\delta$ rather than at the reaction plane $z=\kappa\delta$ in case (b); forgetting that B’s diffusion path is the complementary $(1-\kappa)\delta$; or mismatching stoichiometry (here 1:1, so fluxes are simply equal). Concentrations must be consistent — $c_{B0}=c\,x_B$.

Source. Bird, Stewart & Lightfoot, Transport Phenomena, 2nd ed., §18.1–18.4 (diffusion with heterogeneous/instantaneous reaction); Geankoplis, Transport Processes, Ch. 7 (film theory and reaction enhancement).

Question 5 of 6: C1 — Film-model dissolution with an instantaneous reaction

Question 5: C1 — Film-model dissolution with an instantaneous reaction (25 marks: 7 + 18)