CSTR Kinetic Determination for a Mole-Changing Gas Reaction: 23-Chem-A4 Chemical Reactor Engineering, December 2014, Q5

Question 5: CSTR Kinetic Determination for a Mole-Changing Gas Reaction (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. Isothermal 1-L CSTR, pure gaseous A feed $C_{A0}=120$ mmol/L, reaction A → 3R (moles triple). Pure A feed so $y_{A0}=1$ and the expansion factor is $\varepsilon_A=y_{A0}\delta=1\cdot(3-1)/1=2$. Four steady-state $(v_0,C_A)$ points.

$v_0$ (L/min)	0.06	0.48	1.5	8.1
$C_A$ (mmol/L)	30	60	80	105

Find. The order $n$ and rate constant $k$ in $-r_A=kC_A^{n}$.

Figure 5 — Rate (back-calculated from each CSTR run with the expansion-corrected conversion) versus exit $C_A$ on log-log axes. The slope is $2.00\approx2$, so the reaction is second order with $k=0.0040$ L mmol⁻¹ min⁻¹.

Approach. For each run convert the measured exit $C_A$ to conversion via the gas-expansion relation, back-calculate $-r_A$ from the CSTR mole balance, then fit $-r_A=kC_A^{n}$ on log-log axes.

Conversion with expansion. For a mole-changing gas, $C_A=C_{A0}\dfrac{1-X}{1+\varepsilon_AX}$, which inverts to $X=\dfrac{C_{A0}-C_A}{C_{A0}+\varepsilon_AC_A}$. With $\varepsilon_A=2$, e.g. at $C_A=30$: $X=\dfrac{120-30}{120+2(30)}=\dfrac{90}{180}=0.500$.
Rate from the CSTR balance. $-r_A=\dfrac{F_{A0}X}{V}=\dfrac{C_{A0}v_0X}{V}$ ($V=1$ L). At $C_A=30$: $-r_A=\dfrac{120(0.06)(0.500)}{1}=3.6$ mmol/L·min. Repeating for all four gives $-r_A=\{3.6,\,14.4,\,25.7,\,44.2\}$ mmol/L·min at $C_A=\{30,60,80,105\}$.
Order from a log-log fit. Plotting $\ln(-r_A)$ against $\ln C_A$ (Figure 5) is straight with slope$$n=2.00\approx2,$$so the decomposition is second order in A.
Rate constant. With $n=2$, $k=-r_A/C_A^{2}$ is constant across the runs ($3.6/30^2=0.0040$, …, $44.2/105^2=0.0040$), giving$$\boxed{-r_A=0.0040\,C_A^{2}\ \text{mmol/L}\cdot\text{min}}\quad(k=0.0040\ \text{L}\,\text{mmol}^{-1}\text{min}^{-1}).$$

Quantity	Result
Expansion factor $\varepsilon_A$	2 (A → 3R, pure feed)
Reaction order $n$	2.00 (second order)
Rate constant $k$	0.0040 L·mmol⁻¹·min⁻¹
Rate equation	$-r_A=0.0040\,C_A^{2}$

Concept & Theory

Topic. Differential kinetic analysis in a CSTR for a gas reaction with change in number of moles.

Key relations. Expansion factor $\varepsilon_A=y_{A0}\delta$; $C_A=C_{A0}(1-X)/(1+\varepsilon_AX)$; CSTR mole balance $-r_A=C_{A0}v_0X/V=F_{A0}X/V$; power-law fit $\ln(-r_A)=\ln k+n\ln C_A$.

Why this works. A CSTR is a differential-rate instrument: each steady state gives one $(-r_A,\,C_A)$ pair directly, with no integration, so scanning the flow rate maps the rate law. The expansion term is essential — A→3R triples the moles, so ignoring $\varepsilon_A$ would misread the conversion (giving $X=0.75$ instead of 0.50 at $C_A=30$) and distort the apparent order.

Common pitfall. Using $X=(C_{A0}-C_A)/C_{A0}$ (valid only at constant density); for this expanding gas that overstates conversion and returns a wrong order and $k$.

Source. Levenspiel, Chemical Reaction Engineering §5 (search for a rate equation using a mixed reactor); Fogler §3–4 (stoichiometric table with $\varepsilon$; CSTR data analysis).

Question 5 of 5: CSTR Kinetic Determination for a Mole-Changing Gas Reaction

Question 5: CSTR Kinetic Determination for a Mole-Changing Gas Reaction (20 marks)