Speed both directions without moving equilibrium

Change the pathway, then compare the same temperature

A catalyst participates in a reaction pathway and is regenerated overall. It can change how quickly a system approaches equilibrium without changing the equilibrium condition at fixed temperature. That does not mean every catalyst survives forever: deactivation, poisoning or physical loss can reduce performance. Those effects require their own evidence.

The OpenStax discussion of catalysis explains alternative pathways and regenerated catalysts. Real mechanisms can have multiple intermediates and barriers. Our widget replaces those details with a single adjustable effective peak so that the relationship between forward and reverse changes can be inspected.

At 300 K and zero barrier lowering, the assigned forward and reverse activation energies are 60 and 70 kJ/mol. Increase lowering to 20 kJ/mol. Both become smaller: 40 kJ/mol forward and 50 kJ/mol reverse. The endpoints stay at 0 and −10 kJ/mol. The control does not make B a different substance or lower only the product energy.

Explain the shared multiplication factor

With the same fixed Arrhenius prefactor, lowering a barrier by an amount d multiplies its rate constant by exp(d/RT), with d expressed in J/mol. At 300 K and d = 20000 J/mol, that factor is approximately 3036.93. It applies to both directions in this assigned model.

The forward constant changes from about 0.003570 to 10.84251 s⁻¹. The reverse changes from about 0.00006479 to 0.196749 s⁻¹. The time to halve displacement from equilibrium decreases from about 190.69 s to 0.06279 s. These large factors come from the chosen synthetic parameters; they are not performance claims for a named catalyst.

At an unchanged composition, both forward and reverse reaction rates are multiplied by that same factor. If they were unequal, the net approach becomes faster. If they were equal, the net rate remains zero: speeding both balanced directions does not create a net conversion. Microscopic turnover can increase while the macroscopic composition stays constant.

Show why the equilibrium ratio does not change

For A ⇌ B with the specified first-order rates, equilibrium requires k_forward[A] = k_reverse[B]. Dividing gives [B]/[A] = k_forward/k_reverse. Multiplying both constants by the same positive factor cancels in that ratio. At 300 K the ratio is approximately 55.108 for every barrier-lowering setting.

If total A+B is conserved, equilibrium B fraction is ratio/(1 + ratio). This quantity stays fixed when you move the barrier control at fixed temperature. The display reports a fraction of the two-state inventory, not an experimental yield after an arbitrary duration. Neither the equation nor the control removes reactants from the mixture.

A tempting but inconsistent edit would lower only the forward barrier, keep the reverse barrier unchanged and still call the model a catalyst for the same overall transformation. That would change k_forward/k_reverse and therefore its predicted equilibrium ratio. For the same endpoints, temperature and thermodynamically consistent model, a catalyst cannot create a new equilibrium constant merely by providing a faster pathway.

Real catalysts need not multiply every individual elementary rate constant by one common factor. This equal-factor construction is a teaching simplification for two effective rates. More detailed mechanisms must still respect the same overall equilibrium constraints.

Contrast catalysis with heating

Keep the barrier lowering fixed and change temperature. Both constants rise as temperature rises, but their ratio changes in this model because the forward and reverse barriers differ. Heating and adding a catalyst are different interventions; neither should be described simply as “making particles react more.”

The model’s equilibrium ratio is exp(10000/RT), from its equal prefactors and 10 kJ/mol barrier difference. It decreases as temperature rises. This result follows from these assigned parameters. It is not a universal prediction that every reaction has a smaller equilibrium product fraction at higher temperature. More general thermodynamic treatment requires the reaction’s actual energetic and entropic properties.

Compare fixed-time observations honestly

Suppose two mixtures begin away from equilibrium at the same temperature. One has the baseline pathway; the other has the assigned lowered barrier. At a short observation time, the second can have more B because it is closer to the common equilibrium composition. That observation alone does not show a different equilibrium mixture.

A useful investigation compares initial conditions, temperature, concentrations over time and a justified equilibrium criterion. A single endpoint measurement taken too early can confuse kinetic differences with equilibrium differences. In this widget the equilibrium relation is known because it is stipulated; physical measurements would need independent evidence for that model.

Transfer: A hypothetical reversible first-order system has k_forward = 0.12 s⁻¹ and k_reverse = 0.03 s⁻¹. An assigned catalyst multiplies both by five. Determine both new constants, the equilibrium B/A ratio, and the equilibrium B fraction. Explain what happens if the mixture starts at equilibrium.

Check the catalyst transfer

The new constants are 0.60 and 0.15 s⁻¹. The ratio is still 4, so equilibrium B fraction remains 4/(1+4) = 0.80. Displacement from equilibrium relaxes five times faster. A mixture already at equilibrium retains zero net conversion even though both opposing rates can be higher.