Chemistry · Lesson

Catalysis, selectivity and the distinction between rate and equilibrium

Follow catalyst regeneration through a cycle and compare kinetic barriers with unchanged endpoints. Distinguish rate, equilibrium, selectivity and catalyst lifetime.

Back to the Chemistry map

By the end, you can

Identify a catalyst that participates and is regenerated.
Explain why a catalyst does not change the same reaction’s equilibrium constant at fixed conditions.
Separate selectivity and transport effects from an intrinsic rate law.

Regeneration is a net-cycle statement

An original abstract catalytic cycle is C + A → I followed by I → C + P. Let A and P have the same conserved composition, C carry its own conserved composition and I contain both. Summing these neutral-species steps cancels C and I to give A → P. C is the catalyst; I is an intermediate.

The catalyst participates in the proposed steps and is regenerated in the complete cycle. It is not absent from the chemistry. Real catalysts can become poisoned, deactivate or be lost from the apparatus, so regeneration in an ideal reaction equation does not guarantee an unlimited operating lifetime.

An alternative pathway keeps the same endpoints

A catalyst can make an alternative pathway available and change the rate. Our separate original schematic marks reactant energy 10, product energy −10 and a barrier point 50 for one uncatalysed path or 30 for an alternative path, all in arbitrary units. Forward differences are 40 and 20; reverse differences are 60 and 40. The endpoint difference remains −20.

The graph shows only stipulated endpoint and barrier markers, not a measured continuous energy surface or a complete catalytic cycle. Its coordinate is not time, and its arbitrary energy scale is not a measured activation energy or Gibbs energy. A real catalytic mechanism can contain several barriers and intermediates.

Alternative barrier markers with unchanged endpoints

Schematic energy / arbitrary units

Reaction coordinate / arbitrary units

Uncatalysed schematic markers
Alternative schematic markers

Underlying data

Reaction coordinate / arbitrary units	Schematic energy / arbitrary units
0	10 · Uncatalysed schematic markers
0.5	50 · Uncatalysed schematic markers
1	-10 · Uncatalysed schematic markers
0	10 · Alternative schematic markers
0.5	30 · Alternative schematic markers
1	-10 · Alternative schematic markers

Original schematic only: reaction coordinate 0, 0.5, 1 marks reactant, barrier point and product. Arbitrary energies: uncatalysed 10,50,−10; alternative 10,30,−10. Circles and squares show discrete stipulated markers without connecting curves; neither series is measured data, a full mechanism, a time trace or Gibbs energy.

Faster approach does not change the same equilibrium

For the same overall reaction at fixed temperature and other thermodynamic conditions, a catalyst does not change the equilibrium constant. It changes kinetics and can help the system approach that equilibrium faster through its pathway. It does not change the reactants’ and products’ standard thermodynamic states simply by offering a different route.

In a network with competing products, a catalyst can alter selectivity by affecting pathways differently. That is a statement about kinetic allocation under the specified conditions. It is different from changing the equilibrium constant of one fixed reaction, and a maintained flow or externally driven process needs its own system description.

Observed process speed also depends on transport

A homogeneous catalyst operates in the same phase as the reacting mixture; a heterogeneous catalyst brings an interface between phases into the process. Surface access, mixing and heat or mass transport can limit an observed rate even when the intrinsic molecular steps are fast.

For an original irreversible parallel model A → P with k_P = 0.0500 s⁻¹ and A → Q with k_Q = 0.0100 s⁻¹, both first order in the same A, the branch ratio is 5 and the fraction allocated to P is 5/6 ≈ 0.833. These are exact chosen model inputs. This kinetic branching example does not determine a reversible mixture’s equilibrium composition.

Worked examples

Reference each barrier to its starting side

For the original schematic markers, reactants are at 10, products at −10 and barrier points at 50 or 30 arbitrary units. Calculate forward and reverse differences.

Uncatalysed forward difference: 50 − 10 = 40.
Alternative forward difference: 30 − 10 = 20.
Uncatalysed reverse difference: 50 − (−10) = 60.
Alternative reverse difference: 30 − (−10) = 40.
Both endpoint differences are −10 − 10 = −20.

Both barrier directions are lower in this stipulated alternative schematic, while the endpoint difference is unchanged.

Calculate a kinetic branch fraction

For the exact chosen irreversible parallel first-order model, k_P = 0.0500 s⁻¹ and k_Q = 0.0100 s⁻¹ act on the same A under fixed conditions. What fraction of consumed A goes to P?

Both branch rates share [A], so their ratio is k_P/k_Q = 5.
The total consumption rate is (k_P + k_Q)[A].
P’s fraction of that consumption is k_P/(k_P + k_Q) = 0.0500/0.0600 = 5/6.

About 0.833, or 83.3%, is allocated to P in this model. This is kinetic selectivity, not a changed equilibrium constant.

Try it yourself

In the stated C + A → I; I → C + P cycle, which symbol is the catalyst?
Show the answer and reasoning
C.
It participates first and is regenerated later, cancelling from the net change.
Does a catalyst change the same reaction’s equilibrium constant at fixed thermodynamic conditions?
Show the answer and reasoning
No.
It changes the route and kinetics while the reaction’s thermodynamic endpoint relation stays the same.
Does ideal regeneration guarantee that a real catalyst stays active forever?
Show the answer and reasoning
No.
Poisoning, deactivation and physical loss can limit its useful lifetime.
Can mass transport limit a measured process rate even if the intrinsic steps are fast?
Show the answer and reasoning
Yes.
Reactants must reach accessible reacting regions, and the measured process includes that transport.

What to carry forward

A catalyst participates and is regenerated in the net cycle.
Compare barriers relative to their starting states.
Keep kinetic selectivity distinct from a fixed reaction’s equilibrium constant.
Retain catalyst lifetime and transport limits when interpreting process rates.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.