Catalytic Packed Bed — Rate Expression from Conversion Data: 23-Chem-A4 Chemical Reactor Engineering, May 2018, Q2

Question 2: Catalytic Packed Bed — Rate Expression from Conversion Data (25 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. A single-mole-change-free gas reaction ($1\rightarrow1$, so $\varepsilon=0$ and the volumetric flow is constant) run over catalyst; concentration falls as catalyst weight rises.

$W$ (g)	0.5	1.0	2.5
$C_A$ (mol/m$^3$)	30	20	10
$1/C_A$ (m$^3$/mol)	0.0333	0.0500	0.1000

Find. The reaction order and rate constant, i.e. the rate law $-r_A'$ (per gram of catalyst).

Figure 2 — A straight line of $1/C_A$ against catalyst weight $W$ (through the intercept $1/C_{A0}=0.0167$) confirms second-order kinetics; the slope equals $k/v$.

Approach. Write the packed-bed mole balance in terms of catalyst weight, test which integer order linearizes the data, and read the rate constant from the slope.

Packed-bed mole balance. With $\dot F_{A0}=vC_{A0}$ and $\varepsilon=0$ (so $v$ is constant), $\dot F_{A0}\,dX=-r_A'\,dW$ reduces to $-v\,\dfrac{dC_A}{dW}=-r_A'.$ For a trial law $-r_A'=kC_A^{\,n}$ the data should linearize under the corresponding integral.
Reject first order. First order predicts $\ln(C_{A0}/C_A)=\tfrac{k}{v}W$ (constant slope). The ratios $\ln(60/30)/0.5=1.39$, $\ln(60/20)/1.0=1.10$, $\ln(60/10)/2.5=0.72$ are not constant — first order fails.
Confirm second order. Second order predicts $\dfrac{1}{C_A}-\dfrac{1}{C_{A0}}=\dfrac{k}{v}W.$ Evaluating: $(1/30-1/60)/0.5=0.0333$, $(1/20-1/60)/1.0=0.0333$, $(1/10-1/60)/2.5=0.0333$ — a constant slope, so the reaction is $$\boxed{\text{second order: }-r_A'=kC_A^{2}.}$$
Extract the rate constant. The slope $k/v=\tfrac{1}{30}\ \text{m}^3\text{mol}^{-1}\text{g}^{-1}$; with $v=3\ \text{L/min}=3\times10^{-3}\ \text{m}^3/\text{min},$ $$k=\frac{1}{30}\times3\times10^{-3}=1.0\times10^{-4}\ \frac{\text{m}^6}{\text{mol}\cdot\text{g}\cdot\text{min}}.$$ Equivalently, with $C_A$ in mol/L, $-r_A'=100\,C_A^{2}\ \text{mol}\,\text{g}^{-1}\text{min}^{-1}.$

Quantity	Result
Reaction order	2 (in $C_A$)
Rate constant $k$	$1.0\times10^{-4}\ \text{m}^6\,\text{mol}^{-1}\text{g}^{-1}\text{min}^{-1}$
Rate expression	$-r_A'=1.0\times10^{-4}\,C_A^{2}$ (mol·g$^{-1}$·min$^{-1}$, $C_A$ in mol/m$^3$)

Concept & Theory

Topic: Determining catalytic reaction order and rate constant from integral packed-bed data.

Key relations: catalyst-weight mole balance $W/\dot F_{A0}=\int_0^X dX/(-r_A')$; integrated second-order form $1/C_A-1/C_{A0}=(k/v)W$; conversion–concentration link $C_A=C_{A0}(1-X)$ when $\varepsilon=0$ (Fogler Ch. 10; Levenspiel Ch. 18).

Why this works: For a heterogeneous PFR the natural design variable is the mass of catalyst $W$, not volume. When the volumetric flow is constant, the integrated rate law becomes linear in $W$; the order that produces the straight line is the correct one, and its slope carries $k$.

Common pitfall: Confusing volume-based ($-r_A$, mol·m$^{-3}$·s$^{-1}$) with mass-based ($-r_A'$, mol·g$^{-1}$·min$^{-1}$) rates; and applying $C_A=C_{A0}(1-X)$ when $\varepsilon\neq0$ — valid here only because moles are conserved.

Source: Fogler, Ch. 10 (catalysis and catalytic reactors); Levenspiel, Ch. 18 (solid-catalyzed reactions).

Question 2 of 5: Catalytic Packed Bed — Rate Expression from Conversion Data

Question 2: Catalytic Packed Bed — Rate Expression from Conversion Data (25 marks)