Ideal PFR with Product-Inhibited Kinetics: 23-Chem-A4 Chemical Reactor Engineering, Undated paper, Q1

Question 1: Ideal PFR with Product-Inhibited Kinetics (25 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. Liquid-phase (constant-density), isothermal, ideal plug-flow reactor; rate law $-r_A=kC_A/(1+KC_B)$.

Quantity	Value
Rate constant $k$ (50 °C)	$4.08\ \text{min}^{-1}$
Inhibition constant $K$	$10\ \text{L/mol}$
Feed $C_{A0}$, $C_{B0}$	$1\ \text{mol/L}$, 0
Reactor volume $V$	$200\ \text{L}$
Volumetric flow $v_0$	$1200\ \text{L/h}=20\ \text{L/min}$

Find. The fractional conversion $X$ of A at the reactor outlet.

Figure 1 — Single ideal, isothermal liquid-phase PFR: 200 L fed at 1200 L/h (space time 10 min).

Approach. Write $C_A$ and $C_B$ in terms of conversion (constant density), substitute into the PFR design equation ($V=F_{A0}\int_0^X dX/(-r_A)$, Fogler Ch. 2), integrate in closed form and solve the resulting transcendental equation for $X$. The only trap is the time unit: $k$ is per minute while the flow is per hour.

Space time in consistent units. $v_0=1200/60=20\ \text{L/min}$, so $\tau=V/v_0=200/20=10.0\ \text{min}$ and $k\tau=4.08\times10.0=40.8$.
Rate in terms of conversion. With no B in the feed and 1:1 stoichiometry at constant density, $C_A=C_{A0}(1-X)$ and $C_B=C_{A0}X$, so $-r_A=\dfrac{kC_{A0}(1-X)}{1+KC_{A0}X}$.
PFR design equation. $\displaystyle \tau=C_{A0}\int_0^X\frac{dX}{-r_A}=\frac1k\int_0^X\frac{1+aX}{1-X}\,dX,\qquad a\equiv KC_{A0}=10.$
Integrate. Since $\dfrac{1+aX}{1-X}=-a+\dfrac{1+a}{1-X}$, $$k\tau=-aX-(1+a)\ln(1-X)\quad\Rightarrow\quad 40.8=-10X-11\ln(1-X).$$
Solve for X. The right-hand side increases monotonically with $X$. Near complete conversion put $X\approx1$ in the linear term: $\ln(1-X)=-(40.8+10)/11=-4.618$, so $1-X=0.0099$; one more pass with $X=0.990$ changes nothing at three figures: $$\boxed{X=0.990\ \ (99.0\%)}$$ Check: $-10(0.990)-11\ln(0.00996)=-9.90+50.70=40.8$ ✓.

The inhibition term matters even at this long space time. Without the $(1+KC_B)$ denominator the same reactor would leave only $e^{-40.8}$ of A unconverted (essentially 100% conversion). With inhibition, about 1.0% of A leaves unreacted, because once B builds up to about $1\ \text{mol/L}$ the rate falls to one-eleventh of its first-order value.

Quantity	Result
Space time $\tau$ / $k\tau$	10.0 min / 40.8
Inhibition group $a=KC_{A0}$	10
Fractional conversion of A	0.990 (99.0%)

Concept & Theory

Topic. Isothermal PFR design with a product-inhibited (self-retarding) rate law.

Key relations. $\tau=C_{A0}\int_0^X dX/(-r_A)$ (Fogler Ch. 2, with $\tau=V/v_0$); constant density $C_A=C_{A0}(1-X)$, $C_B=C_{A0}X$; closed form $k\tau=-aX-(1+a)\ln(1-X)$, $a=KC_{A0}$.

Why this works. For a liquid the volumetric flow does not change along the reactor, so each fluid element behaves like a small batch reactor held for exactly $\tau$. Because conversion sets both concentrations, the rate is a function of $X$ alone and the design integral can be done by hand. Rearranged as $k\tau=-\ln(1-X)+a[-\ln(1-X)-X]$, the second term is the extra space time the product inhibition costs.

Common pitfall. Mixing time units ($k$ in min⁻¹, $v_0$ in L/h). Using $v_0=1200$ with $k$ per minute gives $k\tau=0.68$ and a false conversion of about 25%. A second trap is dropping the $1+KC_B$ term, which predicts complete conversion.

Source. Fogler, Elements of Chemical Reaction Engineering, Ch. 2 and 5 (PFR sizing, isothermal design).

Question 1 of 5: Ideal PFR with Product-Inhibited Kinetics

Question 1: Ideal PFR with Product-Inhibited Kinetics (25 marks)