Catalysis and Rate Laws

Introduction

A catalyst speeds up a reaction without being consumed overall, but it is not a spectator. It enters the mechanism, forms intermediates and is regenerated at the end of each catalytic cycle . Because the number of catalyst molecules or surface sites is limited, catalysed reactions show a characteristic behaviour: the rate rises with substrate concentration at first and then levels off. This page shows how that behaviour follows from a simple cycle, and how to read measured orders and turnover numbers.

Core explanation

The catalyst balance. In any cycle, the total catalyst is conserved: [C]₀ = [C] + [CS] + any other catalyst-containing intermediates. This conservation equation, combined with steady-state or pre-equilibrium conditions, is the key to every catalytic rate law.

A two-step cycle. Let catalyst C bind substrate S, then convert it to product:

C + S ⇌ CS (k₁ forward, k₋₁ back); CS → C + P (k₂)

Applying the steady-state approximation to CS and using [C] = [C]₀ − [CS] gives

rate = k₂[C]₀[S] ÷ (K + [S]), with K = (k₋₁ + k₂)/k₁

This is a saturation rate law . At low [S] (much less than K) most catalyst is free, and rate ≈ (k₂/K)[C]₀[S], first order in S. At high [S] nearly every site is occupied, and rate ≈ k₂[C]₀, zero order in S. The same form appears for enzymes (Michaelis–Menten) and for surfaces (Langmuir adsorption).

Surface catalysis. On a solid, the fraction of sites covered by a gas A follows the Langmuir isotherm, θ = Kp/(1 + Kp). If adsorbed A decomposes, rate = kθ, giving first-order kinetics at low pressure and zero order at high pressure. The decomposition of ammonia on hot tungsten is zero order at typical pressures because the surface is saturated. In a Langmuir–Hinshelwood mechanism, two adsorbed species A and B react, so rate = kθ Aθ B. Because A and B compete for sites, increasing p A can eventually lower the rate by crowding B off the surface.

Measured rate and turnover. The rate is proportional to the amount of active catalyst, so it is often normalised as a turnover frequency (TOF): moles of product per mole of active sites per second. TOF measures intrinsic activity; turnover number (TON), the total product per site before deactivation, measures lifetime.

Apparent orders. Because the rate law is not a simple power law, a measured order depends on the range studied. An order between 0 and 1 in substrate usually indicates partial saturation, not a fractional stoichiometry.

Formulae

Catalyst conservation: [C]₀ = [C] + [CS].

Saturation law: rate = k₂[C]₀[S] ÷ (K + [S]).

Langmuir coverage: θ = Kp ÷ (1 + Kp). Langmuir–Hinshelwood: rate = kK AK Bp Ap B ÷ (1 + K Ap A + K Bp B)².

TOF = rate ÷ number of active sites.

Step-by-step reasoning

To derive a catalytic rate law:

1. Write the cycle as elementary steps and mark each catalyst-containing species. 2. Write the catalyst conservation equation. 3. Apply the steady-state (or pre-equilibrium) condition to each intermediate. 4. Solve for the intermediate concentrations in terms of [C]₀. 5. Substitute into the product-forming step and check limiting cases.

Visual explanation

Sketch rate against [S]: a straight line rising from the origin that bends over and flattens at a plateau of height k₂[C]₀. Beneath it, sketch the fraction of catalyst occupied, rising from 0 towards 1 along the same curve. The plateau is where every site is busy.

Real-world analogy

A car wash with five bays works like a catalyst. When few cars arrive, the output rises with the number of arrivals. When a queue forms, all five bays are busy and output is fixed by how fast each bay washes a car, however long the queue becomes.

Real-world example

In the Haber process, the iron catalyst surface becomes heavily covered with adsorbed nitrogen atoms, and adsorbed species compete for sites. Measured rates therefore show non-integer orders in N₂, H₂ and NH₃ that change with conditions, exactly as saturation and competition models predict.

Why?

Why does the rate become independent of substrate at high concentration? The rate is limited by how fast each occupied site converts substrate to product. Once all sites are occupied, adding substrate cannot occupy more sites, so the rate reaches its maximum value.

Common misconception

"A catalyst does not appear in the rate law because it is not consumed." The catalyst concentration usually appears directly: the rate is proportional to [C]₀. What the catalyst does not do is change the overall equilibrium constant.

Worked example

Question: A catalysed reaction has K = 2.0 mmol dm⁻³ and maximum rate k₂[C]₀ = 5.0 μmol dm⁻³ s⁻¹. Find the rate at [S] = 2.0 and 18 mmol dm⁻³.

Reasoning: rate = 5.0 × [S]/(2.0 + [S]). At 2.0: 5.0 × 2.0/4.0 = 2.5 μmol dm⁻³ s⁻¹. At 18: 5.0 × 18/20 = 4.5 μmol dm⁻³ s⁻¹.

Answer: 2.5 and 4.5 μmol dm⁻³ s⁻¹. A ninefold rise in [S] increases the rate by less than a factor of 2.

Quick check

1. In the saturation rate law, what is the rate when the substrate concentration equals K? Answer: Exactly half the maximum rate, because [S]/(K + [S]) = 1/2.

Exam focus

Show the catalyst conservation equation explicitly in derivations. Identify first-order and zero-order limits and explain them in terms of site occupancy. Distinguish TOF (activity) from TON (lifetime).

Advanced insight

The best catalysts balance binding strength: too weak and substrate rarely binds; too strong and product or intermediates block the sites. Plotting activity against adsorption energy gives a volcano plot , whose peak is the Sabatier optimum. This principle guides the search for electrocatalysts for hydrogen and oxygen evolution.

Summary

Catalysts act through closed cycles and are regenerated each turnover. Conservation of catalyst plus a steady-state or equilibrium condition gives saturation rate laws, first order at low substrate and zero order at high substrate. Measured orders therefore depend on conditions, and TOF normalises rate per active site.

Practice questions

1. Write the catalyst conservation equation for a cycle with intermediates CS and CP. Answer: [C]₀ = [C] + [CS] + [CP]. 2. Why is ammonia decomposition on tungsten zero order at moderate pressures? Answer: The surface is saturated with adsorbed ammonia, so the rate depends only on how fast occupied sites react. 3. A catalyst with 2.0 × 10⁻⁶ mol of active sites forms 3.6 × 10⁻³ mol of product in 60 s at constant rate. Calculate the TOF. Answer: Rate = 6.0 × 10⁻⁵ mol s⁻¹; TOF = 6.0 × 10⁻⁵ ÷ 2.0 × 10⁻⁶ = 30 s⁻¹. 4. In a Langmuir–Hinshelwood reaction, why can increasing p A at high values decrease the rate? Answer: A occupies most sites and displaces B, so θ B falls and the product θ Aθ B decreases.