Question 5 of 5: Reactor Model and Activation Energy for Propylene Oxidation

Question 5: Reactor Model and Activation Energy for Propylene Oxidation (25 marks: a 3, b 7, c 9, d 6)

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. Gas-sparged slurry reactor, catalyst and liquid retained, product gases exit; first order in $C_3H_6$, zero order in O$_2$; $P = 1$ atm; space velocity $SV = 35\ \text{cc}\,\text{g}^{-1}\text{hr}^{-1}$; the four ($T$, $X$) pairs above.

Find. (a) the matching ideal reactor; (b) $k'$ at each $T$; (c) the activation energy over the three temperature intervals; (d) the physical reason for its trend.

Approach. Identify the flow pattern (vigorous sparging ⇒ back-mixed), apply the first-order CSTR relation to convert each conversion into $k'$, then take Arrhenius slopes between adjacent temperatures and interpret the falling $E_a$.

1. (a) Reactor model. The gas is bubbled (sparged) through a vigorously agitated slurry, so the reacting medium is essentially uniform in composition — every element of gas sees the same environment. This is the defining feature of a mixed-flow (CSTR / back-mixed) reactor, not plug flow or batch.
2. (b) Rate constants. For a first-order reaction with no volume change in a CSTR, $k'\tau = X/(1-X)$ with $\tau \propto 1/SV$, so per unit catalyst $$k' = SV\,\frac{X}{1-X}.$$ Evaluating with $SV = 35$: $$\boxed{k'(160) = 6.67,\ k'(180) = 15.0,\ k'(200) = 23.3,\ k'(230) = 27.5\ \ [\text{cc}\,\text{g}^{-1}\text{hr}^{-1}].}$$
3. (c) Interval activation energies. Between two temperatures, $E_a = R\,\dfrac{\ln(k'_2/k'_1)}{1/T_1 - 1/T_2}$ (T in K). Applying it to each interval: $$\boxed{E_a(160\text{--}180) = 66.2,\quad E_a(180\text{--}200) = 39.4,\quad E_a(200\text{--}230) = 10.8\ \ \text{kJ/mol}.}$$
4. (d) Interpretation. The activation energy falls steadily as temperature rises, which signals a change in the rate-controlling step. At low $T$ the intrinsic surface reaction is slowest and governs ($E_a \approx 66$ kJ/mol, a true chemical activation energy). As $T$ increases, the surface reaction speeds up faster than the transport steps that feed it (absorption of propylene from the gas bubbles into the dibutyl phthalate, then diffusion through the liquid to the catalyst particles). The regime becomes mixed and the apparent value falls to $\approx 39$ kJ/mol. Because the catalyst is a very fine powder, intraparticle (pore) diffusion is unlikely to be the limiting resistance, although it too would roughly halve the apparent $E_a$. At the highest temperatures the gas–liquid and liquid–solid mass-transfer steps, which depend only weakly on temperature, control, so the apparent $E_a$ collapses toward $\approx 11$ kJ/mol. A falling apparent $E_a$ with temperature is the classic diagnostic that transport, not chemistry, has become rate-limiting.