23-Chem-A4 Chemical Reactor Engineering · December 2014
Question 1 of 5: Enzyme Hydrolysis of Sucrose — Michaelis-Menten Constants
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — December 2014 — 04-Chem-A4 Chemical Reactor Engineering. Three-hour open-book exam; the designated Fogler textbook (any edition), unit-conversion / mathematical tables and a non-communicating programmable calculator are permitted. Five questions are printed and any four constitute a complete paper (each worth 20 marks); all five are solved below for completeness. No credit is given for re-deriving standard rate expressions, so the batch / PFR / CSTR design equations are quoted and applied. These are Levenspiel-style graphical / analytical reactor problems (enzyme kinetics, rate-data analysis, autocatalytic staging, parallel-reaction selectivity, and CSTR kinetic determination).
Reference texts: O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley) — batch/MFR/PFR design equations, the graphical “$1/(-r_A)$ vs $C_A$” area method, autocatalytic optimum staging, product distribution for parallel reactions, and constant-volume CSTR kinetic analysis; H. S. Fogler, Elements of Chemical Reaction Engineering (4th/5th ed., Prentice Hall) — the designated open-book text: stoichiometric tables with the expansion factor $\varepsilon$, gas-phase reactions with change in moles, and reactor sizing; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Given. Constant-volume batch reactor; Michaelis-Menten form $-dC_A/dt = k_3C_AC_{E0}/(C_A+M)$ with $C_{A0}=1.0$ mmol/L and $C_{E0}=1.0$ mmol/L (constant). Eleven $(t,C_A)$ pairs are supplied.
t (hr)
1
2
3
4
5
6
7
8
9
10
11
$C_A$ (mmol/L)
0.84
0.68
0.53
0.38
0.27
0.16
0.09
0.04
0.018
0.006
0.0025
Find. The kinetic constants $k_3$ (units hr⁻¹) and $M$ (mmol/L) that best fit the batch data.
Figure 1 — The integrated rate law plotted in linear form. The least-squares line (R² = 0.9989) has slope $k_3C_{E0}=0.196$ and intercept $-M=-0.188$.
Approach. Integrate the Michaelis-Menten rate law analytically, rearrange it into a straight-line form, and obtain $k_3$ and $M$ from the slope and intercept of a linear least-squares fit.
Integrate the rate law. Separating variables, $-\dfrac{C_A+M}{C_A}\,dC_A = k_3C_{E0}\,dt$. Integrating from $C_{A0}$ to $C_A$ and $0$ to $t$ gives the closed form$$(C_{A0}-C_A)+M\ln\frac{C_{A0}}{C_A}=k_3C_{E0}\,t.$$
Cast into straight-line form. Dividing through by $\ln(C_{A0}/C_A)$ isolates a line in two measured groups:$$\underbrace{\frac{C_{A0}-C_A}{\ln(C_{A0}/C_A)}}_{y}=k_3C_{E0}\underbrace{\frac{t}{\ln(C_{A0}/C_A)}}_{x}-M,$$so a plot of $y$ vs $x$ has slope $k_3C_{E0}$ and intercept $-M$.
Linear regression. Computing $x,y$ for all eleven points and fitting (Figure 1) gives slope $=0.196$ and intercept $=-0.188$, with $R^2=0.9989$ — an excellent straight line, confirming the Michaelis-Menten form.
Extract the constants. Since $C_{E0}=1.0$ mmol/L, the slope equals $k_3$ directly:$$\boxed{k_3=0.196\ \text{hr}^{-1}},\qquad \boxed{M=0.188\ \text{mmol/L}}.$$The small $M$ (comparable to the lowest concentrations) means the reaction is near zero-order at high $C_A$ and becomes first-order only once $C_A\lesssim M$ — exactly the accelerating decay the data show.
