Reversible Reaction $A\rightleftharpoons B$ in a PFR — Temperature of Minimum Residence Time: 23-Chem-A4 Chemical Reactor Engineering, May 2016, Q3

Question 3: Reversible Reaction $A\rightleftharpoons B$ in a PFR — Temperature of Minimum Residence Time (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. Isothermal PFR, reversible first-order/first-order kinetics; $k(T)$ increasing with $T$ and $K_e(T)$ decreasing with $T$ (exothermic reaction); target $X_A = 0.80$.

Quantity	Value
Rate constant	$k = \exp\{17.2 - 5800/T\}$
Equilibrium constant	$K_e = \exp\{-24.7 + 9000/T\}$
Target conversion	$X_A = 0.80$
$C_{A0}$	4 (cancels — see below)

Find. The temperature $T^{*}$ that minimises the PFR residence time $\tau$ at $X_A = 0.80$, and the value $\tau_{\min}$.

Figure 3.1 — $\tau(T)$ at fixed 80% conversion. Raising $T$ speeds the forward rate (lowers $\tau$) but shrinks $K_e$ (moves equilibrium toward the reactant, raising $\tau$); the two effects balance at an interior minimum near $T\approx 340$ K.

Approach. Integrate the PFR design equation to get $\tau(T)$ in closed form, then minimise over $T$; the trade-off between a faster rate and a receding equilibrium creates an interior optimum.

Integrate the design equation. With $-r_A = kC_{A0}[(1-X)-X/K_e] = kC_{A0}(1-bX)$ where $b = 1 + 1/K_e$, the PFR relation $\tau = C_{A0}\!\int_0^{X} \dfrac{dX}{-r_A}$ gives $$\tau(T) = -\frac{1}{k\,b}\ln\!\big(1 - bX\big).$$ The factor $C_{A0} = 4$ cancels between the numerator of the design integral and the $C_{A0}$ in the rate law — it is a red herring.
Note the feasibility window. A real (finite, positive) $\tau$ requires $1 - bX > 0$, i.e. $X < X_e = \dfrac{K_e}{1+K_e}$. At $X = 0.8$ this needs $K_e > 4$, which the exothermic $K_e(T)$ satisfies only for $T \lesssim 345$ K. Above that, 80% conversion is thermodynamically unreachable.
Minimise $\tau(T)$. Scanning $\tau(T)$ over the feasible window (or setting $d\tau/dT = 0$) locates an interior minimum: $$T^{*} \approx \boxed{340\ \text{K}\ (\approx 67\ ^\circ\text{C})}, \qquad \tau_{\min} \approx \boxed{2.04}.$$
Confirm the equilibrium margin. At $T^{*} = 340$ K, $K_e = \exp(-24.7 + 9000/340) = 5.85$, so $X_e = 5.85/6.85 = 0.854 > 0.80$ — the target sits just inside the equilibrium ceiling, which is exactly why $\tau$ blows up if $T$ is pushed only a few degrees higher.

Quantity	Value
Optimal temperature $T^{*}$	340 K (≈67 °C)
Minimum residence time $\tau_{\min}$	≈2.04 (dimensionless per the question)
Equilibrium conversion at $T^{*}$	$X_e = 0.854$

Concept & Theory

Topic. Optimum temperature for a reversible exothermic reaction; the rate–equilibrium trade-off.

Key relations. PFR $\tau = C_{A0}\int_0^X dX/(-r_A)$; integrated form $\tau = -\tfrac{1}{kb}\ln(1-bX)$ with $b = 1+1/K_e$; equilibrium conversion $X_e = K_e/(1+K_e)$ (Fogler, Ch. 8 / Levenspiel, Ch. 9).

Why this works. For an exothermic reversible reaction, raising $T$ accelerates the intrinsic rate (Arrhenius) but simultaneously lowers $K_e$, pulling the attainable equilibrium conversion down toward the required $X$. Time-to-target therefore falls, bottoms out, then rises steeply as the driving force $(1-bX)$ collapses — producing a well-defined interior minimum.

Common pitfall. Being distracted by $C_{A0} = 4$; it cancels exactly. The second trap is pushing $T$ higher “because rate increases” without checking $X_e$ — beyond $T\approx 345$ K the 80% target is simply infeasible and $\tau\rightarrow\infty$.

Source. Fogler, Elements of Chemical Reaction Engineering, Ch. 8 (reversible reactions, optimum temperature).

Question 3 of 5: Reversible Reaction $A\rightleftharpoons B$ in a PFR — Temperature of Minimum Residence Time

Question 3: Reversible Reaction $A\rightleftharpoons B$ in a PFR — Temperature of Minimum Residence Time (25 marks)