Thermodynamics and Kinetics of Catalytic Steps

Introduction

A catalyst speeds up a reaction without changing its overall thermodynamics. Yet within a catalytic cycle, every elementary step has its own free-energy change and barrier. Some steps are downhill and effectively irreversible; others lie close to equilibrium. Drawing these on a free-energy profile shows at a glance which intermediate the catalyst sits in, which barrier limits the rate and which steps can run backwards. This page builds that picture using ideas from earlier work on catalytic cycles and elementary steps.

Core explanation

Catalysts and equilibrium. The overall reaction A + B → P has a fixed ΔG° that no catalyst can change. The catalyst provides a new pathway with lower barriers, so it speeds both forward and reverse reactions equally and does not alter the equilibrium constant.

Free-energy profiles for cycles. A catalytic cycle can be drawn as a sequence of intermediates and transition states. Because the cycle repeats, the profile after one turnover begins again at the starting catalyst but shifted down by ΔG of the overall reaction. Each intermediate is a local minimum; each transition state is a maximum.

The effective barrier. A common error is to take the highest single barrier as rate-controlling. What matters is the energetic span : the difference between the most stable intermediate (the likely resting state) and the highest transition state that follows it in the cycle. If the highest transition state comes before the lowest intermediate, the reaction energy of one turnover is added to the span, since the system must climb from the low intermediate to that transition state in the next cycle. In practice, a very stable intermediate raises the effective barrier even if the step leading out of it is modest.

Resting state. The resting state is the most abundant catalyst species under turnover conditions, often the lowest intermediate on the profile. A deep thermodynamic well acts like a trap: much of the catalyst sits there, and the rate falls.

Reversibility. A step is effectively reversible when its reverse is faster than the next forward step. Ligand dissociation, alkene coordination and β-hydride elimination/reinsertion are often reversible. Steps that release a stable product or form strong bonds, such as reductive elimination of C–C bonds, are usually irreversible. By microscopic reversibility, the forward and reverse reactions share the same transition state, so knowing one path defines the other.

Consequences for selectivity. When steps before the selectivity-determining step are reversible, product ratios reflect the difference in transition-state energies after equilibration (Curtin–Hammett conditions), not the populations of the preceding intermediates. This explains why a minor intermediate can give the major product in asymmetric hydrogenation.

Good catalysts avoid both extremes described by the Sabatier principle: intermediates bound too weakly form slowly; intermediates bound too strongly never leave.

Formulae

Eyring equation: k = (kB T / h) exp(−ΔG‡ / RT). At 298 K, kB T / h ≈ 6.2 × 10¹² s⁻¹. Energetic span: δE = T(TDTS) − I(TDI) if the TDTS follows the TDI; otherwise δE = T(TDTS) − I(TDI) + ΔGr, where ΔGr is the (negative) reaction free energy.

Step-by-step reasoning

To read a catalytic free-energy profile:

1. Locate the lowest intermediate; this is probably the resting state. 2. Find the highest transition state reached after it. 3. Take the difference as the effective barrier. 4. Check each step: is its reverse faster than the next step? 5. Use the effective barrier in the Eyring equation to estimate the turnover frequency.

Visual explanation

Draw a staircase of hills and valleys descending from left to right. Shade the deepest valley and the tallest peak beyond it. The vertical gap between them, not the tallest single hill measured from its own valley, is the true climb for each turnover.

Real-world analogy

Hiking a mountain route with a deep valley in the middle: what exhausts you is not the steepest short slope, but the long climb from the lowest valley floor up to the highest pass that follows it.

Real-world example

Computational chemists routinely build free-energy profiles for new cross-coupling and hydrogenation catalysts. Identifying an overly stable palladium or rhodium intermediate has guided ligand redesign that destabilises the trap and increases turnover frequency.

Why?

Why does a stable intermediate slow a cycle? Most of the catalyst accumulates there, so the concentration of the species that must cross the high barrier is small. The effective barrier therefore includes the energy needed to leave the stable well as well as the barrier itself.

Common misconception

"The step with the largest individual barrier is always rate-determining." A step with a large barrier starting from a high-energy intermediate may be easy overall, while a smaller barrier following a deep resting state may control the rate.

Worked example

Question: A cycle has intermediates at 0, −30 and −10 kJ mol⁻¹ and transition states at +40, +45 and +60 kJ mol⁻¹, in the order I₁ (0), TS₁ (+40), I₂ (−30), TS₂ (+45), I₃ (−10), TS₃ (+60). Estimate the effective barrier.

Reasoning: The lowest intermediate is I₂ at −30 kJ mol⁻¹. The highest transition state after it is TS₃ at +60 kJ mol⁻¹. Span = 60 − (−30) = 90 kJ mol⁻¹.

Answer: About 90 kJ mol⁻¹, far larger than any single step barrier measured from its own starting intermediate.

Quick check

1. Does a catalyst change the equilibrium constant of the overall reaction it catalyses? Answer: No; it lowers barriers for forward and reverse directions equally, leaving ΔG° and K unchanged.

Exam focus

Be able to sketch a free-energy profile, identify the resting state, and calculate the effective barrier from the lowest intermediate to the highest following transition state. Explain reversibility using relative rates and link reversible pre-equilibria to Curtin–Hammett selectivity.

Advanced insight

The energetic span model, developed by Kozuch and Shaik, shows that turnover frequency depends mainly on one intermediate and one transition state, but "degree of turnover-frequency control" analysis quantifies how much every state contributes. Several states can share control when their energies are close, so improving only one may give a smaller gain than expected.

Summary

Catalysts do not alter overall thermodynamics but provide cycles of elementary steps, each with its own ΔG and barrier. The effective barrier is the gap between the most stable intermediate and the highest following transition state. Deep wells create resting states that slow turnover. Reversible steps equilibrate before irreversible ones, which controls selectivity. Good catalysts bind intermediates neither too strongly nor too weakly.

Practice questions

1. Explain the Sabatier principle in terms of a free-energy profile. Answer: Intermediates that are too stable form deep wells that slow later steps, while too-weakly bound ones form slowly; optimal binding is intermediate in strength. 2. What makes an elementary step effectively reversible within a cycle? Answer: Its reverse reaction is faster than the next forward step, so the two species equilibrate before moving on. 3. Using the Eyring equation, how does lowering an effective barrier by 5.7 kJ mol⁻¹ affect the rate at 298 K? Answer: It increases the rate by about a factor of ten, since exp(5700 / (8.314 × 298)) ≈ 10. 4. Why can a minor intermediate lead to the major product? Answer: If intermediates interconvert rapidly, selectivity depends on the relative transition-state energies, so the less abundant intermediate may react much faster.