Parallel Gas-Phase Reactions — Temperature for Selectivity, PFR Volume for Yield: 23-Chem-A4 Chemical Reactor Engineering, December 2015, Q1

Question 1: Parallel Gas-Phase Reactions — Temperature for Selectivity, PFR Volume for Yield (25 marks: a 8, b 17)

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. Three parallel, irreversible, first-order-in-$P_A$ gas-phase reactions with different Arrhenius constants; isothermal isobaric PFR at $P=1$ bar; pure A fed at $F_{A0}=2$ mol/s.

Quantity	Value
$k_1=5.4\times10^{2}\,e^{-20{,}000/RT}$	mol·s⁻¹·m⁻³·bar⁻¹
$k_2=6.5\times10^{3}\,e^{-40{,}000/RT}$	mol·s⁻¹·m⁻³·bar⁻¹
$k_3=2.1\times10^{2}\,e^{-20{,}000/RT}$	mol·s⁻¹·m⁻³·bar⁻¹
Feed / pressure	pure A, $F_{A0}=2$ mol/s, $P=1$ bar
Selectivity target (a) / yield target (b)	$S_D=0.20$ / $Y_D=0.10$

Find. (a) The operating temperature $T$ that gives 20% instantaneous selectivity to D. (b) At that $T$, the PFR volume $V$ that delivers a 10% overall yield of D from the fed A.

Figure 1 — Isothermal, isobaric PFR carrying three parallel first-order reactions of A; the product distribution is set purely by the rate-constant ratios, hence by temperature alone.

Approach. Because all three rates are first order in the same reactant partial pressure $P_A$, every selectivity is a pure ratio of rate constants (independent of $P_A$ and of position), so part (a) is a single algebraic equation in $T$; part (b) fixes the required conversion from the yield, then sizes the variable-volume gas PFR with the correctly weighted expansion factor.

Reduce the rate constants to one temperature group. Since $k_1$ and $k_3$ share the same activation energy (20 kJ/mol) and $k_2$ has twice that (40 kJ/mol), let $u=e^{-20{,}000/RT}$. Then $k_1=540\,u$, $k_2=6500\,u^{2}$, $k_3=210\,u$ (mol·s⁻¹·m⁻³·bar⁻¹).
Write the selectivity to D and solve for $u$. Each reaction consumes A once, so the fraction of reacted A going through path 3 is $S_D=\dfrac{k_3}{k_1+k_2+k_3}=\dfrac{210u}{750u+6500u^{2}}=\dfrac{210}{750+6500u}=0.20.$ Hence $210=150+1300u\Rightarrow u=\dfrac{60}{1300}=0.04615.$
Recover the temperature. From $u=e^{-20{,}000/RT}$, $T=\dfrac{-20{,}000}{R\ln u}=\dfrac{-20{,}000}{8.314\ln(0.04615)}.$ $$T=\boxed{782\ \text{K}\ (\approx 509\ ^\circ\text{C})}$$ At this $T$: $k_1=24.92$, $k_2=13.85$, $k_3=9.69$, so $k_T=k_1+k_2+k_3=48.46$ and the path fractions are $f_1=0.514,\ f_2=0.286,\ f_3=0.200$ — confirming $S_D=0.20$.
Convert the yield target to a required conversion. Yield of D on fed A is $Y_D=f_3\,X$ where $X$ is the overall conversion of A. Setting $Y_D=0.10$ with $f_3=0.20$ gives $$X=\frac{0.10}{0.20}=\boxed{0.50}.$$
Set the expansion factor for the gas PFR. The three paths change the mole count by $\Delta n=0,\ +1,\ +2$ per mole A reacted (A→B; A→2C; A→3D). The mean change per mole A reacted is $\delta=f_1(0)+f_2(1)+f_3(2)=0.286+0.400=0.686.$ For pure A feed, $\varepsilon=y_{A0}\delta=0.686.$
Size the variable-volume PFR. With $-r_A=k_T P_A$ and $P_A=P\dfrac{1-X}{1+\varepsilon X}$, the design equation $F_{A0}\,dX=-r_A\,dV$ integrates to $$V=\frac{F_{A0}}{k_T P}\int_0^{X}\frac{1+\varepsilon X}{1-X}\,dX=\frac{F_{A0}}{k_T P}\Big[(1+\varepsilon)\ln\tfrac{1}{1-X}-\varepsilon X\Big].$$ Substituting $F_{A0}=2$ mol/s, $k_T=48.46$, $P=1$ bar, $\varepsilon=0.686$, $X=0.5$: the bracket $=(1.686)(0.6931)-(0.686)(0.5)=0.826$, so $$V=\frac{2}{48.46}(0.826)=0.0341\ \text{m}^3=\boxed{34.1\ \text{L}}.$$

Quantity	Result
(a) Operating temperature for 20% selectivity to D	782 K (509 °C)
Path fractions at 782 K ($f_1/f_2/f_3$)	0.514 / 0.286 / 0.200
(b) Conversion for 10% yield of D	$X=0.50$
(b) Expansion factor $\varepsilon$	0.686
(b) Required PFR volume	34.1 L

Concept & Theory

Topic. Selectivity and yield of parallel reactions of equal order; variable-volume (mole-changing) gas PFR sizing.

Key relations. Instantaneous selectivity $S_D=r_3/\sum r_i$; overall yield $Y_D=S_D\,X$ when the selectivity is composition-independent; expansion factor $\varepsilon=y_{A0}\,\delta$; isothermal isobaric PFR $V=\frac{F_{A0}}{k_TP}\big[(1+\varepsilon)\ln\frac{1}{1-X}-\varepsilon X\big]$.

Why this works. All three rates are first order in the same $P_A$, so $r_i/r_j=k_i/k_j$ everywhere in the reactor — the product split is fixed by temperature and never changes with conversion. That decouples the selectivity problem (algebra in $T$) from the sizing problem (integration in $X$). The mole-weighted $\delta$ captures that different paths expand the gas differently, which dilutes $P_A$ and enlarges the volume.

Common pitfall. Using the raw stoichiometric $\delta$ of only reaction 3 (or ignoring expansion entirely). The correct $\varepsilon$ is the selectivity-weighted mole change over all active paths; ignoring $\varepsilon$ here understates $V$ by roughly 15%.

Source. Fogler, Elements of Chemical Reaction Engineering §8 (parallel reactions, selectivity/yield) and §4–5 (stoichiometric table, $\varepsilon$; isothermal PFR design).

Question 1 of 5: Parallel Gas-Phase Reactions — Temperature for Selectivity, PFR Volume for Yield

Question 1: Parallel Gas-Phase Reactions — Temperature for Selectivity, PFR Volume for Yield (25 marks: a 8, b 17)