23-Chem-A4 Chemical Reactor Engineering · December 2015
Question 2 of 5: Semibatch Reactor — Ether Concentration as a Function of Time
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — December 2015 — 04-Chem-A4 Chemical Reactor Engineering. Three-hour open-book exam; the designated Fogler textbook (any edition), unit-conversion / mathematical tables (CRC Handbook) and a non-communicating programmable calculator are permitted. Five questions are printed and any four constitute a complete paper (each worth 25 marks); all five are solved below for completeness. No credit is given for re-deriving standard rate expressions, so the batch / CSTR / PFR / packed-bed design equations are quoted and applied. Property look-ups not printed on the paper (the gas constant, molar volumes) are stated explicitly in each Given block as permitted open-book references.
Reference texts: H. S. Fogler, Elements of Chemical Reaction Engineering (4th/5th ed., Prentice Hall) — the designated open-book text: batch/CSTR/PFR design equations, the stoichiometric table with expansion factor $\varepsilon$ for gas-phase reactions with a change in moles, parallel-reaction selectivity/yield, the semibatch mole balances, and the adiabatic energy balance; O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley) — reactor sequencing for >1-order kinetics and packed-bed (catalyst-weight) design; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 2: Semibatch Reactor — Ether Concentration as a Function of Time (25 marks)
The paper prints $r=0.263\,C_A^{2}C_B$ with A = methanol, so the rate is second order in methanol, first order in trityl chloride, and $k=0.263$ L²·mol⁻²·min⁻¹ (third order overall). It is solved as printed. In Fogler’s version of this trityl-chloride problem, the squared term belongs to trityl chloride. With the orders swapped, $r=0.263\,C_AC_B^{2}$, the same integration gives a peak of about 0.0022 mol/L at about 334 min. That result is shown only for comparison. The question asks for the concentration “in moles”, so both $N_C(t)$ (mol) and $C_C(t)$ (mol/L) are tabulated.
Given. Isothermal semibatch operation: B (40 mol) is charged in $V_0=378$ L of benzene; a methanol solution ($C_{Af}=0.054$ mol/L) is fed at $v_0=3.78$ L/min, so the liquid volume grows as $V(t)=V_0+v_0t$. Rate as printed, $-r_A=-r_B=r_C=kC_A^{2}C_B$ (1:1:1 stoichiometry).
Quantity
Value
Initial charge of B, $N_{B0}$
40 mol
Initial volume, $V_0$
378 L
Feed conc. / rate, $C_{Af}$ / $v_0$
0.054 mol/L / 3.78 L/min
Molar feed of A, $F_{A0}=C_{Af}v_0$
0.204 mol/min
Rate constant, $k$
0.263 L²·mol⁻²·min⁻¹
Find. The moles of ether $N_C(t)$ and its concentration $C_C(t)=N_C(t)/V(t)$ versus time.
Figure 2 — Semibatch reactor: B is charged initially, A is fed continuously, nothing leaves. The volume grows with time, so the ether concentration reflects a competition between generation and dilution.
Approach. Write unsteady mole balances on A, B and C for a semibatch reactor with a growing volume, integrate the three coupled ODEs numerically (RK4), and report both $N_C$ and $C_C=N_C/V$.
Semibatch mole balances. Only A is fed; the reaction rate acting on the whole liquid is $(-r_A)V=kC_A^{2}C_BV=k\dfrac{N_A^{2}N_B}{V^{2}}$. Hence $$\frac{dN_A}{dt}=F_{A0}-k\frac{N_A^{2}N_B}{V^{2}},\quad \frac{dN_B}{dt}=-k\frac{N_A^{2}N_B}{V^{2}},\quad \frac{dN_C}{dt}=+k\frac{N_A^{2}N_B}{V^{2}},$$ with $V(t)=378+3.78\,t$, $F_{A0}=0.204$ mol/min, and initial state $N_A=0$, $N_B=40$, $N_C=0$. By stoichiometry $N_C=N_{B0}-N_B$ at all times.
Why no closed form, and what the system does. $V$ is time-dependent and the rate is non-linear in $N_A$, so the balances are non-separable; a fourth-order Runge–Kutta march (step 0.05 min) integrates them. Because the rate is second order in the dilute methanol, the reaction starts very slowly: fed methanol accumulates ($C_A\approx0.018$ mol/L at 50 min and $0.046$ mol/L at 1000 min, approaching the feed value 0.054), and only then does the ether build up appreciably.
Ether history. Evaluating $N_C$ and $C_C=N_C/V$ along the integration:
$t$ (min)
100
200
300
500
750
1000
1500
2000
3000
$N_C$ (mol)
0.34
1.35
2.72
5.85
9.87
13.66
20.21
25.34
32.19
$C_C$ (mol/L)
0.00045
0.00119
0.00180
0.00258
0.00307
0.00329
0.00334
0.00319
0.00275
The moles of ether rise monotonically toward the 40 mol of B charged (80% of B converted by 3000 min). The long times involve very large liquid volumes ($V\approx4160$ L at 1000 min), so a real vessel would have to be sized for this, or the feed stopped earlier.
Locate and explain the maximum. The ether concentration rises, passes through a broad peak, then declines: $$C_{C,\max}\approx\boxed{0.00336\ \text{mol/L at }t\approx1320\ \text{min}}$$ ($N_C\approx18$ mol, $V\approx5370$ L). Early on, generation outpaces the volume growth, so $C_C$ climbs. Once B is being depleted, the generation term falls while the feed keeps enlarging $V$, so the accumulated ether is diluted and $C_C$ falls even though $N_C$ keeps increasing.
