Lesson 6 · 25 min

Catalysis

Hydrogen peroxide keeps for months in the bottle but fizzes on a cut. Catalysts lower the barrier without moving the equilibrium — and enzymes and metal surfaces share one more trait: saturation.

In this lesson

Explain how a catalyst speeds a reaction by providing a lower-barrier pathway without changing ΔG° or K
Compare homogeneous, heterogeneous and enzyme catalysis, with an example of each
Describe enzyme saturation with the Michaelis–Menten equation and obtain KM, Vmax and kcat from data
Use Langmuir adsorption to explain why surface reactions are first order at low pressure and zero order at high pressure

A bottle of 3 % hydrogen peroxide keeps in the medicine cabinet for months, yet poured on a cut it foams at once. The decomposition $\ce{2H2O2} \mathrel{\ce{->}} \ce{2H2O} + \ce{O2}$ is downhill all along; the barrier is simply too high. Uncatalysed, the activation energy is about 75 kJ/mol; iodide ions lower it to about 57 kJ/mol, and catalase in blood takes it below 10 kJ/mol. At room temperature, every 5.7 kJ/mol or so knocked off the barrier makes the reaction about ten times faster.

Predict first

At low substrate concentration an enzyme-catalysed rate is proportional to [S]. What happens when [S] is raised 100-fold?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

It rises 100-fold It rises, but levels off towards a maximum It falls It rises 10 000-fold

A catalyst changes the path, not the destination

A catalyst takes part in the reaction but is regenerated at the end of each cycle, so a small amount works over and over. It provides a pathway with a lower activation energy — for instance $\ce{I-}$ turns $\ce{H2O2}$ into $\ce{IO-}$ and is then restored by a second $\ce{H2O2}$ (see reaction mechanisms). The energies of reactants and products are untouched, so ΔH°, ΔG° and the equilibrium constant K are unchanged: a catalyst speeds forward and reverse reactions by the same factor and simply brings equilibrium sooner.

Homogeneous catalysis: the catalyst is in the same phase as the reactants, like $\ce{I-}$ or acid in solution.
Heterogeneous catalysis: usually a solid surface catalysing a gas or liquid reaction, like the iron in ammonia synthesis or the platinum, palladium and rhodium in a car’s catalytic converter.
Enzyme catalysis: biological catalysts made of protein, whose active sites are exquisitely selective.

v = \frac{V_\mathrm{max}[\ce{S}]}{K_\mathrm{M} + [\ce{S}]},\qquad V_\mathrm{max} = k_\mathrm{cat}[\ce{E}]_\mathrm{t},\qquad \frac{1}{v} = \frac{K_\mathrm{M}}{V_\mathrm{max}}\cdot\frac{1}{[\ce{S}]} + \frac{1}{V_\mathrm{max}}

The Michaelis–Menten equation (1913) and its double-reciprocal (Lineweaver–Burk) form. At [S] = KM, v = Vmax/2.

$K_\mathrm{M}$: Michaelis constant: the substrate concentration at half the maximum rate
$k_\mathrm{cat}$: turnover number: substrate molecules converted per enzyme per second
$[\ce{E}]_\mathrm{t}$: total enzyme concentration

Worked example

How fast is carbonic anhydrase?

Carbonic anhydrase catalyses $\ce{CO2} + \ce{H2O} \mathrel{\ce{<=>}} \ce{HCO3-} + \ce{H+}$ with $K_\mathrm{M}(\ce{CO2}) = 8.0$ mM and $k_\mathrm{cat} = 6.0\times10^5\ \mathrm{s^{-1}}$ . With 10 nM enzyme and $[\ce{CO2}] = 4.0$ mM, what is the initial rate?

How three kinds of reversible inhibitor change the apparent parameters
Inhibition	Inhibitor binds	Apparent KM	Apparent Vmax
Competitive	the free enzyme’s active site	increases	unchanged
Uncompetitive	the enzyme–substrate complex	decreases	decreases
Non-competitive	both, equally	unchanged	decreases

Michaelis–Menten · Lineweaver–Burk · Langmuir–Hinshelwood

Saturation kinetics: enzymes and surfaces

Adsorption constant K · 2 bar⁻¹ Gas pressure · 1.00 bar

\theta = \dfrac{Kp}{1 + Kp},\qquad r = k\theta = \dfrac{kKp}{1 + Kp}

Surface rate and coverage

Catalyst surface: 27 of 40 sites occupied Coverage θ = 0.67 Apparent order at this pressure: 0.33

Surface catalysis: gas molecules must first adsorb on a limited number of active sites. At low pressure the coverage θ ≈ Kp and the rate is proportional to p (first order); at high pressure the surface is full and the rate stops rising (zero order). Drag the pressure and watch the apparent order fall from 1 to 0. Switch back to “Enzyme” and compare the two curves.

Concept check

What does a competitive inhibitor do to the Michaelis–Menten parameters?

KM increases, Vmax unchanged KM unchanged, Vmax decreases Both decrease Neither changes

Try the calculation

Iodide lowers the activation energy for $\ce{H2O2}$ decomposition from 75.3 kJ/mol to 56.5 kJ/mol. If the pre-exponential factor stays the same, by what factor does the rate increase at 300 K?

Try the calculation

When [S] is a quarter of KM, what percentage of Vmax is the rate?

Key takeaways

A catalyst opens a lower-barrier pathway, speeds forward and reverse equally, and leaves ΔG° and K unchanged.
Homogeneous, heterogeneous and enzyme catalysis: same phase, a surface, or a biological macromolecule.
Michaelis–Menten, v = Vmax[S]/(KM + [S]): about first order at low [S], saturating to zero order at high [S].
Surface reactions saturate the same way: coverage θ = Kp/(1 + Kp).

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Data and references

J. M. Berg, J. L. Tymoczko, L. Stryer, Biochemistry, 5th ed., W. H. Freeman (2002), Ch. 8, Tables 8.5 and 8.6.
L. Michaelis, M. L. Menten, “Die Kinetik der Invertinwirkung”, Biochem. Z. 49, 333–369 (1913).
Chemistry LibreTexts, “Ea and catalysts” (H₂O₂ activation energies); Purdue University demonstration sheet 19.6, “The catalytic decomposition of hydrogen peroxide”.
I. Langmuir, “The adsorption of gases on plane surfaces of glass, mica and platinum”, J. Am. Chem. Soc. 40, 1361–1403 (1918).