Acetaldehyde Decomposition — Batch Kinetics and a PFR: 23-Chem-A4 Chemical Reactor Engineering, May 2014, Q3

Question 3: Acetaldehyde Decomposition — Batch Kinetics and a PFR (20 marks: a 10, b 10)

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. $\text{CH}_3\text{CHO}\rightarrow\text{CH}_4+\text{CO}$ (mole change $\delta=+1$ per acetaldehyde). Constant-volume batch, 450 °C = 723.15 K, initially 50% A / 50% N₂ at $P_0=1$ atm ($P_{A0}=0.5$, inert $P_{N_2}=0.5$). Total-pressure data as tabulated. (b) PFR $V=15$ L, feed 40% A / 60% N₂ at 5 atm, 723.15 K, target $X=0.5$.

Find. (a) confirm first order and report $k$; (b) the feed flowrate (volumetric and molar) giving 50% decomposition in the 15-L PFR.

Figure 3a — Acetaldehyde partial pressure recovered from total pressure via $P_A=1.5-P_{tot}$; the straight semilog plot of $\ln P_A$ vs $t$ confirms first order.

Approach. Convert total pressure to acetaldehyde partial pressure via a mole-change (stoichiometric) table, test the first-order integrated law, then size the PFR flow with the first-order gas design equation including expansion.

Total pressure → acetaldehyde partial pressure (part a). Each mole of A that reacts adds one net mole (CH₄+CO replace one A), so $P_{tot}=P_0+(P_{A0}-P_A)$ and, with $P_0=1,\ P_{A0}=0.5$, $P_A=1.5-P_{tot}$. The data map to $P_A=\{0.5,0.4,0.3,0.2,0.1\}$ atm at $t=\{0,2.2,5.1,9.2,16.1\}$ min.
First-order test. For first order $k=\dfrac{1}{t}\ln\dfrac{P_{A0}}{P_A}$ should be constant. Evaluating at each point gives 0.1014, 0.1002, 0.0996, 0.1000 min⁻¹ — constant to within reading error, and the semilog plot (Figure 3a) is straight. Hence the reaction is first order with$$\boxed{k=0.100\ \text{min}^{-1}}\quad(450\,{}^{\circ}\text{C}).$$
PFR expansion factor (part b). With 40% A in the feed and $\delta=+1$, the volume-change parameter is $\varepsilon=y_{A0}\delta=0.4(1)=0.4$. The isothermal first-order gas PFR design equation is$$k\tau=(1+\varepsilon)\ln\frac{1}{1-X}-\varepsilon X.$$
Required space time. At $X=0.5$: $(1.4)\ln 2-0.4(0.5)=0.970-0.200=0.770$, so $\tau=0.770/0.100=7.70$ min.
Feed flowrate (part b result). The inlet acetaldehyde concentration is $C_{A0}=y_{A0}P/RT=0.4(5)/(0.08206\cdot723.15)=0.0337$ mol/L. From $\tau=V/v_0$,$$v_0=\frac{V}{\tau}=\frac{15}{7.70}=\boxed{1.95\ \text{L/min}},\qquad F_{A0}=C_{A0}v_0=0.0337(1.95)=0.0656\ \text{mol/min}.$$

Quantity	Result
(a) Reaction order	first order (constant $k$, straight $\ln P_A$)
(a) Rate constant	0.100 min⁻¹
(b) Required space time $\tau$	7.70 min
(b) Volumetric feed $v_0$	1.95 L/min
(b) Molar acetaldehyde feed $F_{A0}$	0.0656 mol/min

Concept & Theory

Topic. Constant-volume batch kinetics from total pressure; gas-phase PFR with change in number of moles.

Key relations. Stoichiometric pressure bookkeeping $P_A=P_{A0}-(P_{tot}-P_0)$; first-order integral $\ln(P_{A0}/P_A)=kt$; expansion factor $\varepsilon=y_{A0}\delta$; PFR design $k\tau=(1+\varepsilon)\ln[1/(1-X)]-\varepsilon X$.

Why this works. In a rigid vessel, partial pressure is proportional to moles, so measuring total pressure and subtracting the (constant) inert plus the net moles produced recovers $P_A$ directly. In the flow reactor the gas expands as one mole becomes two, lowering concentration and lengthening the required residence; $\varepsilon$ carries that correction.

Common pitfall. Forgetting the inert nitrogen when writing $P_A(P_{tot})$, or mishandling $\varepsilon$ in the PFR equation: dropping only the $-\varepsilon X$ term under-states the allowable flow by ~21%, while ignoring $\varepsilon$ altogether over-states it by ~11%.

Source. Fogler, Elements of Chemical Reaction Engineering §3 (stoichiometry, $\varepsilon$) and §5 (batch data analysis; gas-phase PFR).

Question 3 of 5: Acetaldehyde Decomposition — Batch Kinetics and a PFR

Question 3: Acetaldehyde Decomposition — Batch Kinetics and a PFR (20 marks: a 10, b 10)