Engineering & Technology · Reaction Engineering, Separations & Transport
Catalysis, selectivity and coupled reaction-transport effects
A more active catalyst does not necessarily give a proportional increase in the reactor’s output. Reactants must reach its active region, products must leave, and competing reactions may respond differently to local conditions.
Use Process Balances & Design for amount, flow, energy and boundary accounts. Review material structure and properties. Chemistry’s reaction rates and equilibrium explain the molecular and thermodynamic ideas further. Exponentials are evaluated explicitly; each model states its scope before calculation.
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Explain why catalyst activity, product selectivity and transport to an active surface are related but distinct performance questions.
Faster transformation is only one objective
A catalyst provides another reaction pathway and is regenerated in its reaction cycle. It can change the rates of competing pathways. At a fixed temperature it does not change the equilibrium constant of an otherwise unchanged reaction. A faster approach to equilibrium and a different equilibrium position are different claims.
Define selectivity explicitly. For two one-to-one parallel paths A → P and A → U, let instantaneous selectivity mean r_P/r_U, where P is desired and U is undesired. It describes rates at a particular local state, not automatically the product ratio accumulated over a changing batch or recovered after downstream separation. Conversion alone cannot tell how much useful product was made.
Transport can limit access to activity
A heterogeneous catalyst lies in another phase from its reactants, commonly a porous solid. Delivery from bulk fluid to its surface, diffusion through pores, adsorption, surface reaction and product departure can all matter. A rate measured from the entire particle can therefore incorporate transport effects. The concentration at an active site need not equal the value in a bulk sample.
Consider an original, isothermal, steady planar-surface model with no internal pore resistance. Supplied film coefficient h=0.004 m/s carries A from bulk concentration C_b=2.00 mol/m³ to surface concentration C_s. The surface consumes A with flux k_s C_s, where k_s=0.006 m/s. Equating delivery and consumption gives h(C_b−C_s)=k_s C_s. Thus C_s=h C_b/(h+k_s)=0.80 mol/m³ and flux J=0.0048 mol m⁻² s⁻¹.
A response to change can distinguish mechanisms
If h alone doubles in the supplied model, C_s rises to about 1.1429 mol/m³ and J to 0.006857 mol m⁻² s⁻¹. That is a42.9% increase, not a doubling. In the limit of extremely fast delivery, surface consumption still limits the flux to k_s C_b=0.012. The figure uses a normalised film coordinate so it does not claim a measured film thickness.
Real changes to flow or particle size can also change heat transfer, pressure drop, accessible area and material condition. Evidence for a transport limitation needs comparisons that identify those effects. Catalyst fouling, poisoning or structural change can alter activity over time; a fixed initial coefficient is not a lifetime guarantee. Conditions that improve one response may worsen another.
Separate selectivity from total speed
For a separate original competing-path exercise, supply r_P=k_P C_A with k_P=0.06 min⁻¹ and r_U=k_U C_A² with k_U=0.04 L mol⁻¹ min⁻¹. At C_A=1 mol/L, the rates are 0.06 and 0.04 mol L⁻¹ min⁻¹, so their ratio is 1.5 and the desired fraction of total consumption is 0.60.
At 0.5 mol/L the rates become 0.03 and 0.01; their ratio is 3 and the desired fraction is 0.75. Lower concentration improves this local fraction while reducing both absolute rates. These two assumed laws do not describe a named catalyst, and the local fractions do not determine whole-reactor yield without the concentration history and species balances.
Try the reasoning
Can a catalyst change the equilibrium constant merely by accelerating both directions?
Show the answer and reasoning
No. At the same temperature and unchanged chemistry, it changes the approach rate rather than the equilibrium constant.
Why does doubled film transport give less than doubled flux here?
Show the answer and reasoning
Surface consumption remains a second resistance, and its driving concentration also changes.
Does higher instantaneous selectivity establish more recovered product per hour?
Show the answer and reasoning
No. Throughput, conversion history, losses and downstream recovery also matter.
What to carry forward
Identify the local state, rate basis and selectivity definition before attributing process improvement to catalyst activity.
References
Teaching resources and research supporting this lesson’s explanations and examples.
H. Scott Fogler · University of Michigan2 sources