Catalysis — Changing the Path, Not the Equilibrium

By Aniket Bhardwaj · 16 September 2026 · Chemistry Concept

A catalyst is one of the few ideas in chemistry that students can define correctly and still misuse in the very next question. "A substance that increases the rate of a reaction without being consumed" is the textbook line — but the moment an equilibrium question appears, half the class writes that the catalyst "increases the yield". It does not. A catalyst changes the route a reaction takes; it never changes where the reaction ends up. This article separates the two ideas with numbers, so the distinction becomes something you can calculate rather than something you have to remember.

What a catalyst actually does

Reactants must climb an energy barrier before they can become products. The height of that barrier is the activation energy, E_a. A catalyst offers a different mechanism — usually involving an intermediate that includes the catalyst itself — whose barrier is lower. More colliding molecules then have enough energy to react, so the rate rises sharply.

Arrhenius equation: k = A e^−E_a/RT
k = rate constant · A = pre-exponential (frequency) factor · E_a = activation energy (J/mol) · R = 8.314 J K⁻¹ mol⁻¹ · T = temperature (K)

Because E_a sits inside an exponential, a modest drop in the barrier produces a very large jump in rate. That single fact is the whole commercial value of catalysis.

Worked example 1 — how much faster?

An uncatalysed reaction has E_a = 75 kJ/mol. A catalyst provides a route with E_a = 50 kJ/mol. Assuming the frequency factor A is unchanged, how many times faster is the catalysed reaction at 298 K?

k_cat/k_uncat = e^−50000/RT ÷ e^−75000/RT = e^{(75000 − 50000)/RT}

RT = 8.314 × 298 = 2477.57 J/mol
Exponent = 25000 ÷ 2477.57 = 10.0905
e^10.0905 = e¹⁰ × e^0.0905 = 22026.5 × 1.0947 = 2.41 × 10⁴

The catalysed reaction is about 24,000 times faster — a reaction that would take a day now finishes in about three seconds. And the catalyst is recovered at the end.

Worked example 2 — the equilibrium constant does not move

Here is the calculation that settles the "does a catalyst increase yield" argument. The equilibrium constant is fixed by thermodynamics alone:

ΔG° = −RT ln K, so K = e^−ΔG°/RT

Take a reaction with ΔG° = −20.0 kJ/mol at 298 K.

ΔG°/RT = −20000 ÷ 2477.57 = −8.0724
K = e^8.0724 = e⁸ × e^0.0724 = 2980.96 × 1.0751 = 3.20 × 10³

Now add a catalyst. Which symbol in that equation changed? None of them. ΔG° is a difference between the standard free energies of products and reactants; the catalyst appears in neither. T is the same, R is a constant. So K is still 3.20 × 10³. The mixture reaches that same composition — it simply gets there far sooner.

The same argument works from the kinetic side. For a reversible reaction, K = k_forward / k_reverse. The catalyst lowers the barrier for the forward step and the reverse step by exactly the same amount, because both directions cross the same transition state. If k_forward rises 24,000 times, so does k_reverse, and the ratio — the equilibrium constant — is untouched.

Worked example 3 — the temperature a catalyst saves

Using the same numbers as example 1: at what temperature would the uncatalysed reaction (E_a = 75 kJ/mol) run as fast as the catalysed one does at 298 K (E_a = 50 kJ/mol)?

Set the exponents equal: 75000 ÷ (8.314 T) = 50000 ÷ 2477.57

Right-hand side = 50000 ÷ 2477.57 = 20.1810
So 8.314 T = 75000 ÷ 20.1810 = 3716.4
T = 3716.4 ÷ 8.314 = 447 K ≈ 174 °C

Without the catalyst you would have to heat the reactor to about 174 °C to match a rate the catalyst delivers at 25 °C. In an exothermic industrial reaction, being able to run cooler does improve the equilibrium yield — but notice the cause: the temperature changed, not the catalyst's effect on K. This indirect route is the only honest way a catalyst helps yield, and it is worth writing exactly that way in an answer.

The three kinds of catalysis

Type	Catalyst phase	How it works	Standard example
Homogeneous	Same phase as reactants	Forms a soluble intermediate that is regenerated	Acid catalysis of ester hydrolysis; NO in the lead-chamber process
Heterogeneous	Different phase (usually a solid)	Reactants adsorb on the surface, bonds weaken, products desorb	Fe in the Haber process; V₂O₅ in the Contact process; Ni in hydrogenation of oils
Enzyme (biocatalysis)	Protein in solution or bound	Substrate binds a specific active site; extremely selective	Amylase splitting starch; catalase decomposing H₂O₂

Three supporting terms appear constantly in exam questions. A promoter increases a catalyst's activity without being a catalyst itself — molybdenum, or potassium and aluminium oxides, with iron in the Haber process. A poison destroys activity by binding irreversibly to the active sites; sulphur compounds poison many metal catalysts, which is why fuels are desulphurised. In autocatalysis a product of the reaction catalyses it, so the rate starts slow and then accelerates — the permanganate–oxalic acid titration is the standard school example, where Mn²⁺ formed in the reaction speeds up what follows.

Heterogeneous catalysis, step by step

The adsorption theory explains a solid catalyst in five stages, and this sequence is worth memorising as a list because it answers most "explain the mechanism" questions:

Reactant molecules diffuse to the catalyst surface.
They are adsorbed on active sites — often chemisorption, which forms real but weak bonds to the surface.
Those surface bonds weaken the bonds inside the reactant, and the molecules are held close together in the right orientation.
Reaction occurs on the surface, forming the product.
Products desorb, freeing the active site for the next cycle.

This is also why surface area matters so much. Finely divided metals, or metals dispersed on a porous support, expose far more active sites per gram than a solid lump.

Common mistakes that cost marks

"A catalyst increases the yield at equilibrium." It does not. It changes how fast equilibrium is reached. Only a change in temperature, pressure or concentration shifts the position of equilibrium.
"A catalyst lowers ΔH." No. ΔH, ΔG and ΔS for the reaction are properties of the reactants and products alone. Only E_a changes.
"A catalyst only speeds up the forward reaction." It speeds both directions by the same factor.
"A catalyst takes no part in the reaction." It takes part fully — it is simply regenerated. Saying "not consumed" is correct; saying "does not react" is not.
Drawing the energy profile wrongly. The catalysed curve must start and finish at exactly the same levels as the uncatalysed one; only the peak is lower, and it often has two smaller humps for the two-step catalysed mechanism.
Ignoring selectivity. A catalyst can change which product dominates, by accelerating one competing pathway more than another. That is a kinetic effect on a choice between reactions — it still does not alter K for any one of them.

Where catalysis appears in exams

Exam	Typical question
CBSE / ICSE Class 12	Surface chemistry: adsorption, types of catalysis, promoters and poisons, shape-selective catalysis by zeolites
JEE / NEET	Energy-profile diagrams; effect of catalyst on E_a, K and rate
IIT-JAM / CUET-PG	Arrhenius calculations with and without a catalyst; enzyme kinetics basics
GATE / CSIR-NET	Heterogeneous catalysis mechanisms, turnover number, organometallic catalytic cycles

Try the rate ratio yourself. The Arrhenius calculator takes E_a, temperature and the frequency factor and returns the rate constant, so you can put in 75 kJ/mol and 50 kJ/mol and watch the barrier drop turn into a four-figure speed increase.

Open the Arrhenius Equation Calculator →

Surface chemistry and kinetics carry steady marks in the Class 12 board paper and in entrance exams. ABC Chemistry teaches both at the Gurugram centre and in online classes across India — see abcchemistry.in.