Gas-Phase Trimerization $3A\rightarrow B$ in a CSTR — Fractional Conversion: 23-Chem-A4 Chemical Reactor Engineering, May 2016, Q5

Question 5: Gas-Phase Trimerization $3A\rightarrow B$ in a CSTR — Fractional Conversion (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. First-order irreversible gas reaction $3A\rightarrow B$ (mole-changing, $\Delta n = -2$ per 3 A) in a CSTR; cold feed / hot reactor, so molar flows are set at feed conditions but concentrations at reactor conditions.

Quantity	Value
Reactor volume $V$	10,000 L
Feed	1:1 A:N$_2$, 8000 L/hr at 50 °C, 5 atm
Reactor conditions	350 °C, 5 atm
Rate law / constant	$-r_A = kC_A$; $k = 4\times10^{-5}$ hr$^{-1}$ at 100 °C
Activation energy	$E = 90$ kJ/mol

Find. The fractional conversion $X_A$ of A in the exit stream.

Figure 5.1 — Cold feed (50 °C) sets the molar flows; the CSTR at 350 °C sets the reacting concentration. N$_2$ is inert but shares the volume, and the total moles fall as $3A\rightarrow B$ proceeds.

Approach. Bring $k$ to reactor temperature with Arrhenius, compute the molar feed of A from the ideal-gas feed state, express the reactor concentration $C_A$ (mole-changing gas) at reactor $T,P$, and close the CSTR balance into a quadratic in $X$.

Scale the rate constant to 350 °C. $$k = k_{ref}\exp\!\Big[-\tfrac{E}{R}\Big(\tfrac{1}{T}-\tfrac{1}{T_{ref}}\Big)\Big] = 4\times10^{-5}\exp\!\Big[-\tfrac{90000}{8.314}\Big(\tfrac{1}{623.15}-\tfrac{1}{373.15}\Big)\Big] = 4.54\ \text{hr}^{-1}.$$
Molar feed of A from the feed state. Total molar feed $F_{T0} = \dfrac{P\,v_0}{RT_{feed}} = \dfrac{(5)(8000)}{(0.082057)(323.15)} = 1508$ mol/hr; being 1:1 with N$_2$, $F_{A0} = 754$ mol/hr.
Reactor concentration of A (mole-changing gas). For $3A\rightarrow B$ with equal A:N$_2$ feed, the total molar flow at conversion $X$ is $F_T = F_{A0}\big(2 - \tfrac{2X}{3}\big)$, and at reactor $T,P$, $$C_A = \frac{F_{A0}(1-X)}{F_T}\cdot\frac{P}{RT_r} = \frac{1-X}{2-\tfrac{2X}{3}}\cdot\frac{1}{c}, \qquad c = \frac{RT_r}{P} = 10.23\ \text{L/mol}.$$
Close the CSTR balance. $F_{A0}X = (-r_A)V = kC_A V$. Substituting $C_A$ and letting $K = \dfrac{kV}{F_{A0}\,c} = \dfrac{(4.54)(10000)}{(754)(10.23)} = 5.88$ gives $X\big(2-\tfrac{2X}{3}\big) = K(1-X)$, i.e. $$\tfrac{2}{3}X^2 - (2+K)X + K = 0.$$
Solve the quadratic. With $K = 5.88$: $\tfrac{2}{3}X^2 - 7.88X + 5.88 = 0$, whose physical root is $$X_A = \frac{(2+K) - \sqrt{(2+K)^2 - \tfrac{8}{3}K}}{4/3} = \boxed{0.80}.$$

Quantity	Value
$k$ at 350 °C	4.54 hr$^{-1}$
Molar feed of A, $F_{A0}$	754 mol/hr
Dimensionless group $K = kV/(F_{A0}c)$	5.88
Fractional conversion $X_A$	0.80

Concept & Theory

Topic. First-order gas reaction in a CSTR with mole change and non-isothermal feed vs. reactor states.

Key relations. Arrhenius $k(T) = k_{ref}e^{-\frac{E}{R}(1/T - 1/T_{ref})}$; molar feed from ideal gas $F_{T0} = Pv_0/(RT_{feed})$; mole-changing concentration $C_A = \tfrac{F_{A0}(1-X)}{F_T}\tfrac{P}{RT_r}$; CSTR balance $F_{A0}X = kC_AV$ (Fogler, Ch. 3–5).

Why this works. Molar flows are conserved quantities fixed by the feed metering (feed $T,P$), whereas the reacting concentration depends on the local reactor $T,P$ and the changing total moles. Keeping these two states separate — feed for $F_{A0}$, reactor for $C_A$ — is the crux; conflating them is the usual error.

Common pitfall. Computing $C_A$ or $F_{A0}$ at reactor temperature (they are metered cold), or ignoring the $2 - 2X/3$ mole-change factor and treating the gas as constant-density — both distort the balance and miss the clean $X = 0.80$.

Source. Fogler, Elements of Chemical Reaction Engineering, Ch. 5 (CSTR sizing) and Ch. 3 (variable-density gas stoichiometry).

Question 5 of 5: Gas-Phase Trimerization $3A\rightarrow B$ in a CSTR — Fractional Conversion

Question 5: Gas-Phase Trimerization $3A\rightarrow B$ in a CSTR — Fractional Conversion (25 marks)