Isothermal PFR — Second-Order Gas Reaction $2A\rightarrow A_2$: 23-Chem-A4 Chemical Reactor Engineering, December 2013, Q5

Question 5: Isothermal PFR — Second-Order Gas Reaction $2A\rightarrow A_2$ (20 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. Pure A, so the stoichiometry $2A\rightarrow A_2$ (i.e. $A\rightarrow\tfrac12 A_2$ per mole of A) shrinks the gas: $\varepsilon=y_{A0}\delta=(1)(-\tfrac12)=-0.5$. Ideal gas throughout.

Quantity	Value
Tube: $D$, $L$	0.025 m, 3.2 m
Feed $F_{A0}$ (pure A)	1.5 mol/h
Inlet $T$, $P$	593.15 K, 101.3 kPa
Conversion $X$ / expansion $\varepsilon$	0.58 / −0.5

Find. (a) the space time $\tau=V/v_0$; (b) the second-order rate constant $k$.

Figure 5 — Isothermal PFR. Pure A enters at 320 °C, 101.3 kPa; the gas contracts ($\varepsilon=-0.5$) as $2A\rightarrow A_2$, reaching 58% conversion at the outlet.

Approach. (a) reactor volume from the tube geometry and inlet volumetric flow from the ideal-gas law give $\tau$. (b) integrate the PFR design equation for a second-order gas reaction with volume change using Fogler’s closed form.

Reactor volume. A cylinder of $D=0.025$ m, $L=3.2$ m:$$V=\frac{\pi}{4}D^2 L=\frac{\pi}{4}(0.025)^2(3.2)=1.571\times10^{-3}\ \text{m}^3=1.571\ \text{L}.$$
Inlet volumetric flow. Ideal gas at inlet, $F_{A0}=1.5/3600=4.17\times10^{-4}$ mol/s:$$v_0=\frac{F_{A0}RT}{P}=\frac{(4.17\times10^{-4})(8.314)(593.15)}{101300}=2.03\times10^{-5}\ \text{m}^3/\text{s}.$$
Part (a) — space time. Dividing volume by inlet flow:$$\tau=\frac{V}{v_0}=\frac{1.571\times10^{-3}}{2.03\times10^{-5}}=\boxed{77.4\ \text{s}}.$$
Inlet concentration of A. Pure A, so $C_{A0}=P/RT$:$$C_{A0}=\frac{101300}{(8.314)(593.15)}=20.54\ \text{mol/m}^3=0.02054\ \text{mol/L}.$$
Part (b) — integrate the PFR design equation. For $(-r_A)=kC_A^2$ with $C_A=C_{A0}\dfrac{1-X}{1+\varepsilon X}$, Fogler’s integrated form is$$k\,\tau\,C_{A0}=2\varepsilon(1+\varepsilon)\ln(1-X)+\varepsilon^2X+(1+\varepsilon)^2\frac{X}{1-X}.$$With $\varepsilon=-0.5$, $X=0.58$ the right side is $0.434+0.145+0.345=0.924$.
Solve for $k$. Rearranging with $\tau=77.4$ s and $C_{A0}=0.02054$ mol/L:$$k=\frac{0.924}{\tau\,C_{A0}}=\frac{0.924}{(77.4)(0.02054)}=\boxed{0.581\ \text{L}\,\text{mol}^{-1}\text{s}^{-1}}.$$

Quantity	Result
Reactor volume $V$	1.571 L
(a) Space time $\tau$	77.4 s
Inlet $C_{A0}$	0.02054 mol/L
(b) Rate constant $k$	0.581 L·mol⁻¹·s⁻¹

Concept & Theory

Topic. Isothermal PFR design for a second-order gas-phase reaction with a change in total moles ($\varepsilon\neq0$).

Key relations. $\varepsilon=y_{A0}\delta$; $C_A=C_{A0}\dfrac{1-X}{1+\varepsilon X}$; PFR $\tau=C_{A0}\displaystyle\int_0^X\frac{dX}{(-r_A)}$; the closed integral above.

Why this works. In a variable-volume gas reactor the volumetric flow changes down the tube, so concentration is not simply $C_{A0}(1-X)$. Substituting $C_A(X)$ and integrating yields an algebraic expression relating $k$, $\tau$, and $X$ without a numerical solver.

Common pitfall. Setting $\varepsilon=0$ (would apply only to constant-density liquids or equimolar gas reactions), or getting $\delta$ wrong: base it on the rate species A, $A\rightarrow\tfrac12 A_2$, giving $\delta=-\tfrac12$. Space time uses the inlet volumetric flow.

Source. Fogler §5.3–5.4 (isothermal reactor design; variable-volume gas systems, integrated second-order form); Levenspiel Ch. 5.

Question 5 of 5: Isothermal PFR — Second-Order Gas Reaction $2A\rightarrow A_2$

Question 5: Isothermal PFR — Second-Order Gas Reaction $2A\rightarrow A_2$ (20 marks)