Phosphine Decomposition in a Plug-Flow Reactor: 23-Chem-A4 Chemical Reactor Engineering, May 2014, Q5

Question 5: Phosphine Decomposition in a Plug-Flow Reactor (20 marks)

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Given. $4\text{PH}_3\rightarrow\text{P}_4+6\text{H}_2$; first order $k=10$ hr⁻¹; $T=1200\,{}^{\circ}\text{F}=1659.67\,{}^{\circ}$R; $P=4.6$ atm; pure phosphine feed $F_{A0}=4$ lb-mol/hr; target $X=0.80$. Imperial gas constant $R=0.7302\ \text{ft}^3\!\cdot\text{atm}/(\text{lbmol}\cdot{}^{\circ}\text{R})$.

Find. (a) The PFR volume $V$ for 80% conversion of the pure phosphine feed.

Figure 5 — Pure phosphine enters an isothermal PFR; every 4 mol PH₃ become 7 mol of products, so the gas expands strongly ($\varepsilon=0.75$) as conversion proceeds.

Approach. Compute the expansion factor for the pure feed, evaluate the first-order gas PFR design integral, then convert the space time to a volume through the inlet concentration and feed rate.

Expansion factor. Per 4 mol PH₃ reacted, moles go $4\to1+6=7$, so $\delta=(7-4)/4=0.75$. With pure feed $y_{A0}=1$, $\varepsilon=y_{A0}\delta=0.75$.
Design integral. The isothermal first-order gas PFR relation gives $$k\tau=(1+\varepsilon)\ln\frac{1}{1-X}-\varepsilon X=(1.75)\ln 5-0.75(0.8)=2.816-0.600=2.217,$$ so $\tau=2.217/10=0.2217$ hr.
Inlet concentration and volumetric feed. For pure phosphine, $C_{A0}=P/RT=4.6/(0.7302\cdot1659.67)=3.80\times10^{-3}$ lbmol/ft³, and the inlet volumetric flow is $v_0=F_{A0}/C_{A0}=4/3.80\times10^{-3}=1054$ ft³/hr.
Reactor volume (part a). Combining $\tau=V/v_0$,$$V=\tau v_0=0.2217(1054)=\boxed{234\ \text{ft}^3\ (\approx6.6\ \text{m}^3)}.$$

Quantity	Result
Expansion factor $\varepsilon$	0.75
Space time $\tau$	0.222 hr
Inlet volumetric feed $v_0$	1054 ft³/hr
PFR volume $V$	234 ft³ (≈ 6.6 m³)

Concept & Theory

Topic. First-order gas-phase PFR with a large mole increase.

Key relations. $\delta=\sum\nu/|\nu_A|$; $\varepsilon=y_{A0}\delta$; first-order gas PFR $k\tau=(1+\varepsilon)\ln[1/(1-X)]-\varepsilon X$; ideal-gas $C_{A0}=P/RT$; $V=\tau v_0=\tau F_{A0}/C_{A0}$.

Why this works. As 4 moles become 7 the gas expands by up to 75%, diluting the reactant and slowing the volumetric conversion; the $(1+\varepsilon)\ln[1/(1-X)]-\varepsilon X$ form is the exact integral of the first-order rate over the expanding stream, so the required volume is larger than the constant-density estimate $-\ln(1-X)/k$.

Common pitfall. Using SI or the wrong $R$ with Fahrenheit/Rankine, or ignoring $\varepsilon$ (which would give $\tau=\ln5/10=0.161$ hr and undersize the reactor by ~27%).

Source. Fogler, Elements of Chemical Reaction Engineering §5 (gas-phase PFR sizing with $\varepsilon$); Levenspiel §5 (plug-flow reactor).

Question 5 of 5: Phosphine Decomposition in a Plug-Flow Reactor

Question 5: Phosphine Decomposition in a Plug-Flow Reactor (20 marks)