Question 2 of 6: Internal Diffusion Limitation in Immobilized Enzyme Beads
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-Chem-B4, Biochemical Engineering — May 2014. 3 hours, Closed-Book Exam
(any non-communicating calculator permitted). Six questions are printed; per the exam notes any five (5)
constitute a complete paper (100 marks) and only the first five as they appear in the answer book are
marked. All six are solved below for completeness.
Find. (i) Whether internal (intraparticle) diffusion limits the observed reaction rate
in the spherical bead; (ii) how the same test changes when the particle shape is a thin rectangular strip.
Approach. Use the generalized (Aris-normalized) Thiele modulus
$\phi=(V_p/S_p)\sqrt{k_1/D_e}$, built from each particle's own volume-to-surface ratio as its characteristic
diffusion length. This normalization makes the same numeric criterion — φ ≪ 1
⇒ reaction-limited (η ≈ 1, substrate not limiting); φ ≫ 1
⇒ diffusion-limited (η ≈ 1/φ) — apply to any particle shape, sphere or
strip alike.
Fig. 2 — characteristic diffusion length Vp/Sp for the
sphere vs. the strip.
Common rate/diffusion ratio. Both geometries share the same intrinsic kinetics and
diffusivity, so compute $\sqrt{k_1/D_e}$ once:
$$\sqrt{\frac{k_1}{D_e}}=\sqrt{\frac{1.0\times10^{-3}\ \text{s}^{-1}}{1\times10^{-6}\ \text{m}^2/\text{s}}}
=\sqrt{1000\ \text{m}^{-2}}=\boxed{31.6\ \text{m}^{-1}}$$
(i) Thiele modulus for the sphere. A sphere's volume-to-surface ratio is
$V_p/S_p=(\tfrac{4}{3}\pi R^3)/(4\pi R^2)=R/3$:
$$\frac{V_p}{S_p}\Big|_{sphere}=\frac{6\times10^{-3}}{3}=2.0\times10^{-3}\ \text{m}$$
$$\phi_{sphere}=(2.0\times10^{-3})(31.6)=\boxed{0.0632}$$
Since $\phi_{sphere}=0.063\ll1$ (well below the ≈0.3 threshold for the onset of diffusion control),
the effectiveness factor $\eta\approx1$: the reaction, not substrate diffusion, is rate-limiting
— substrate availability is NOT a limiting factor for the 6 mm bead.
(ii) Thiele modulus for the strip. The strip's own $V_p/S_p$ replaces R/3 as the
characteristic length:
$$\frac{V_p}{S_p}\Big|_{strip}=\frac{1\times10^{-9}\ \text{m}^3}{6\times10^{-6}\ \text{m}^2}
=1.67\times10^{-4}\ \text{m}$$
$$\phi_{strip}=(1.67\times10^{-4})(31.6)=\boxed{5.27\times10^{-3}}$$
$\phi_{strip}/\phi_{sphere}=0.083$: the strip's characteristic diffusion path is about 12× shorter
than the bead's, so its Thiele modulus is proportionally smaller. It is still (even more clearly)
reaction-limited — substrate availability remains not limiting, and the strip geometry moves
the system further into the reaction-limited regime, not closer to diffusion control.
