Reversible Gas-Phase Dimerization $2A\rightleftharpoons B$ in a Tubular Reactor — Reactor Length: 23-Chem-A4 Chemical Reactor Engineering, May 2016, Q4

Question 4: Reversible Gas-Phase Dimerization $2A\rightleftharpoons B$ in a Tubular Reactor — Reactor Length (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.

Check — rate-constant units

Taken literally, $k = 118$ kg·mol/(L·hr·atm²) yields a reactor barely 1 cm long — physically absurd for this classic problem. The sensible reading is $k = 118$ mol/(L·hr·atm²) $\equiv 118$ kmol/(m³·hr·atm²) (the “kg·mol/L” is a transcription error for mol/L). This gives $V = 90.4$ L and $L \approx 11.5$ m, the accepted textbook result.

Given. Isothermal isobaric tubular (plug-flow) reactor; feed 9 kmol/hr of butadiene + steam in ratio 1 : 0.5 (steam inert), so $F_{A0} = 9/1.5 = 6$ kmol/hr; $P = 1$ atm, $T = 640$ °C.

Quantity	Value
Total feed / butadiene feed	9 kmol/hr; $F_{A0} = 6$ kmol/hr
Steam (inert)	0.5 mol per mol A ⇒ 3 kmol/hr
Forward rate constant	$k = 118$ kmol/(m³·hr·atm²)
Equilibrium constant	$K_e = 1.27$
Diameter / target	$D = 0.10$ m; $X = 0.40$

Find. The reactor length $L$ for 40% conversion of butadiene.

Figure 4.1 — Isothermal tubular (plug-flow) dimerization. Steam rides through as an inert diluent, lowering the partial pressures of A and B and thus the net rate; the mole count falls from 3 to $3-X$ (per 2-mol A basis) as A dimerizes.

Approach. Express the partial pressures as functions of $X$ from a stoichiometric table (with inert steam), write the reversible rate law, and integrate the PFR design equation numerically to the target conversion; divide the volume by the tube cross-section for the length.

Build the mole table (basis: 2 mol A fed). With inert steam scaled to 1 mol on this basis, at conversion $X$: $n_A = 2(1-X)$, $n_B = X$, $n_{inert} = 1$, total $n_T = 3 - X$. At $P = 1$ atm the partial pressures are $p_A = \dfrac{2(1-X)}{3-X}$ and $p_B = \dfrac{X}{3-X}$ atm.
Write the reversible rate law. $$-r_A = k\Big(p_A^2 - \frac{p_B}{K_e}\Big) = 118\Big[\Big(\tfrac{2(1-X)}{3-X}\Big)^2 - \tfrac{1}{1.27}\tfrac{X}{3-X}\Big]\ \text{kmol/(m}^3\text{}\cdot\text{hr)}.$$ The equilibrium conversion (where $-r_A = 0$) is $X_e \approx 0.503$, so the requested 40% sits well into the region where the reverse term matters — the rate falls almost five-fold across the interval (from 52.4 to 10.8 kmol/(m³·hr)), so the integrand grows by the same factor.
Integrate the PFR design equation. $$V = F_{A0}\int_0^{0.4} \frac{dX}{-r_A}.$$ Numerical quadrature (Simpson / trapezoidal, 4×10$^5$ panels) with $F_{A0} = 6$ kmol/hr gives $$V = 0.0904\ \text{m}^3 = \boxed{90.4\ \text{L}}.$$
Convert volume to length. The tube cross-section is $A_c = \tfrac{\pi}{4}D^2 = \tfrac{\pi}{4}(0.10)^2 = 7.854\times10^{-3}$ m², so $$L = \frac{V}{A_c} = \frac{0.0904}{7.854\times10^{-3}} = \boxed{11.5\ \text{m}}.$$

Quantity	Value
Butadiene molar feed $F_{A0}$	6 kmol/hr
Equilibrium conversion $X_e$	0.503
Reactor volume $V$	90.4 L
Reactor length $L$ ($D = 10$ cm)	11.5 m

Concept & Theory

Topic. Variable-mole reversible gas reaction in a PFR with an inert diluent; partial-pressure rate law and numerical sizing.

Key relations. Stoichiometric table for $2A\rightleftharpoons B$ with inert; $p_i = (n_i/n_T)P$; reversible rate $-r_A = k(p_A^2 - p_B/K_e)$; PFR design $V = F_{A0}\int dX/(-r_A)$ (Fogler, Ch. 3&8 / Levenspiel, Ch. 5).

Why this works. Because the total moles change (2→1) and steam dilutes, the partial pressures are non-trivial functions of $X$; expressing them explicitly lets the design integral be evaluated numerically. The reverse term $p_B/K_e$ grows as B accumulates, slowing the rate and inflating the required volume as $X$ approaches $X_e$.

Common pitfall. The unit trap on $k$ (see the check note) — a literal “kg·mol/L” gives an absurd 1-cm reactor. A second trap is omitting the inert steam from $n_T$, which would overstate the partial pressures and understate the volume.

Source. Fogler, Elements of Chemical Reaction Engineering, Ch. 3 (stoichiometry with inerts) and Ch. 8 (reversible gas reactions).

Question 4 of 5: Reversible Gas-Phase Dimerization $2A\rightleftharpoons B$ in a Tubular Reactor — Reactor Length

Question 4: Reversible Gas-Phase Dimerization $2A\rightleftharpoons B$ in a Tubular Reactor — Reactor Length (25 marks)