Immobilized-Enzyme Batch Reactor with Enzyme Deactivation: 20-Bio-A5 Systems Analysis & Control, December 2017, Q3

Question 3: Immobilized-Enzyme Batch Reactor with Enzyme Deactivation (15 marks)

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. First-order surface rate law $-r_s=V_m'S_wA_T/K_m$ with $S_w=S_b=S$ (well mixed); first-order enzyme deactivation $dV_m'/dt=-k_1V_m'$; initial substrate concentration $S_0$ at $t=0$; the list of standard integrals printed on the exam's final page.

Find. Show that $S\to S_{min}=S_0\exp[-A_TV_{m,0}/(k_1K_m)]$ as $t\to\infty$.

Approach. Write the batch substrate balance using the (already per-unit-reactor-volume) first-order rate law, substitute the exponentially decaying $V_m'(t)$, separate variables, and integrate over all time using the supplied $\int e^{ax}dx$ formula.

Enzyme deactivation. Integrating $dV_m'/dt=-k_1V_m'$ from $V_m'(0)=V_{m,0}$ gives the standard first-order decay $$V_m'(t)=V_{m,0}e^{-k_1t}.$$
Batch substrate balance. The stated rate law carries no explicit reactor-volume term, so it is already expressed on a per-unit-reactor-volume basis and the batch balance is simply $$\frac{dS}{dt}=-(-r_s)=-\frac{V_m'(t)\,A_T\,S}{K_m}=-\frac{A_TV_{m,0}}{K_m}\,e^{-k_1t}\,S.$$
Separate variables. $$\frac{dS}{S}=-\frac{A_TV_{m,0}}{K_m}\,e^{-k_1t}\,dt.$$
Integrate from $t=0$ ($S=S_0$) to $t\to\infty$ ($S\to S_{min}$, the value approached once the enzyme has essentially fully deactivated). $$\ln\frac{S_{min}}{S_0}=-\frac{A_TV_{m,0}}{K_m}\int_0^\infty e^{-k_1t}\,dt.$$
Evaluate the integral with the supplied formula $\int_{x_1}^{x_2}e^{ax}dx=\tfrac1a(e^{ax_2}-e^{ax_1})$, taking $a=-k_1$: as $t\to\infty$, $e^{-k_1t}\to0$, so $$\int_0^\infty e^{-k_1t}\,dt=\frac{1}{k_1}.$$
Substitute back. $$\ln\frac{S_{min}}{S_0}=-\frac{A_TV_{m,0}}{k_1K_m} \quad\Rightarrow\quad \boxed{S_{min}=S_0\exp\!\left(-\dfrac{A_TV_{m,0}}{k_1K_m}\right)}$$ — exactly the target expression. A finite, nonzero $S_{min}$ (rather than complete conversion) is the direct physical consequence of the enzyme's activity dying out before all of the substrate can be consumed.

Result	Expression
Minimum substrate concentration $S_{min}$	$S_0\exp[-A_TV_{m,0}/(k_1K_m)]$ (confirmed, matches target)

Concept & Theory

Topic. Enzyme deactivation coupled to a first-order surface reaction in a batch reactor.

Key relations. First-order deactivation $V_m'(t)=V_{m,0}e^{-k_1t}$; batch balance $dS/dt=-r_s(S,t)$; total conversion capacity is bounded because $\int_0^\infty V_m'(t)\,dt=V_{m,0}/k_1$ is finite.

Why this works. Because the deactivation is first order, its integral over infinite time converges to a finite value ($V_{m,0}/k_1$) rather than diverging — that finite "total catalytic exposure" is exactly what caps how much substrate can ever be consumed, producing a nonzero asymptotic $S_{min}$ instead of eventual complete conversion.

Common pitfall. Treating the rate law as needing an explicit reactor volume (it does not, here, since no $V$ term appears anywhere in the given target formula) or forgetting that the integration limit is $t\to\infty$, not some finite batch time — the "minimum" substrate concentration is a long-time asymptote, not the value at any specific finite $t$.

Source. Bailey & Ollis, Biochemical Engineering Fundamentals, Ch. 3 and Ch. 8 (enzyme deactivation kinetics; immobilized-enzyme rate laws).

Question 3 of 6: Immobilized-Enzyme Batch Reactor with Enzyme Deactivation

Question 3: Immobilized-Enzyme Batch Reactor with Enzyme Deactivation (15 marks)