Question 5 of 5: First-Order Fit and External Mass-Transfer Diagnosis in a Packed PFR

Question 5: First-Order Fit and External Mass-Transfer Diagnosis in a Packed PFR (25 marks: a 13, b 5, c 7)

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 ideal packed-bed PFR, 3.9 mm spheres, $\Delta H_r\approx0$, $C_{A0}=1.1\times10^{-4}$ mol/cm³.

Find. (a) Whether first order fits experiments 1–4 and the apparent rate constant; (b) whether external mass transfer affects performance; (c) whether the 425 °C run is consistent with (b).

Approach. (a) For a first-order reaction in an ideal PFR at constant density ($\Delta n=0$ for A → R), $k\tau=-\ln(1-X)$ (Fogler Ch. 5, on a catalyst-weight basis), so a constant $k=-\ln(1-X)/\tau$ confirms the order. (b) Experiments 5–6 repeat runs 1–2 at the same space time but three times the mass velocity. Intrinsic kinetics cannot depend on flow, but the film mass-transfer coefficient does (Fogler Ch. 14). (c) Transport and chemical kinetics respond very differently to temperature, so the 425 °C run is a second, independent test.

1. (a) Rate constant for each run. $k=-\ln(1-X)/\tau$: run 1 $=-\ln0.50/0.18=3.85$; run 2 $=-\ln0.25/0.36=3.85$; run 3 $=-\ln0.06/0.72=3.91$; run 4 $=-\ln0.02/1.08=3.62$ cm³/(g·s).
2. (a) Is it first order? Yes. A plot of $-\ln(1-X)$ against $\tau$ is a straight line through the origin (least-squares slope 3.72, $R^2=0.995$; Figure 5b). Runs 1–3 agree within 1.5%. Run 4 is the only one lower, and its conversion is quoted to the nearest percent at 98%: 97.5–98.5% spans $k=3.42$–$3.89$, which covers the others. Averaging the four runs, $$\boxed{k_{app}\approx3.8\ \text{cm}^3/(\text{g cat}\cdot\text{s})}$$
3. (b) Effect of mass velocity. At the same space times, tripling $G$ from 0.19 to 0.57 raises conversion from 50% to 75% (exp. 5) and from 75% to 94% (exp. 6). The apparent constants become $k=-\ln0.25/0.18=7.70$ and $-\ln0.06/0.36=7.82$: twice the 400 °C value at $G=0.19$. If the catalyst were kinetically controlled, conversion at fixed $\tau$ would not change with velocity. External mass transfer therefore has a strong influence; at $G=0.19$ it is the controlling resistance.
4. (b) Quantitative. For a film-controlled first-order process $k_{app}=k_ca\propto G^n$. Here $n=\ln(7.70/3.85)/\ln3=0.63$ (0.64 from exp. 6). Packed-bed correlations give $k_c\propto G^{0.5}$ (Thoenes–Kramers, $Sh\propto Re^{1/2}$) to $G^{0.6}$ (Colburn $j_D\propto Re^{-0.4}$), so the whole of the velocity dependence is explained by the film. Treating the resistances in series, $1/k_{app}=1/k_r+1/(k_ca)$ with $k_c\propto G^{0.5}$, would require a negative reaction resistance. The surface reaction is therefore much faster than film transport, and the $k_{app}=3.8$ of part (a) is really $k_ca$, not an intrinsic rate constant. The first-order fit in (a) was expected anyway, because film diffusion is also first order in $C_A$.
5. (c) Temperature response. Exp. 7 repeats exp. 1 (same $\tau$ and $G$) at 425 °C: $k=-\ln0.46/0.18=4.31$, only 12% higher than 3.85. The apparent activation energy is $$E_{a,app}=\frac{R\ln(k_{698}/k_{673})}{1/673.15-1/698.15}=\frac{8.314\ln1.12}{5.32\times10^{-5}}=\boxed{\approx18\ \text{kJ/mol}}$$
6. (c) Interpretation. Surface kinetics typically have $E_a$ of 40–200 kJ/mol. Even at 40 or 80 kJ/mol, a 25 °C rise would multiply $k$ by 1.29 or 1.67 and push exp. 7 to 59% or 69% conversion, not 54%. Film mass transfer depends only weakly on temperature: $k_c\propto D_{AB}^{2/3}$ with $D_{AB}\propto T^{1.75}$ gives about +4% for this step, before small velocity and viscosity effects at fixed $G$. The observed +12% ($E_a\approx18$ kJ/mol) is in the diffusion-controlled range and far below chemical control. Yes, exp. 7 is consistent with (b): external mass transfer controls the rate. Because $\Delta H_r\approx0$ there are no pellet heating effects to confuse the result.