GATE Surface Chemistry and Catalysis — Isotherms, Rate Laws and Worked Numericals

By Aniket Bhardwaj · 9 September 2026 · GATE Chemistry

Surface chemistry is one of the friendliest parts of the GATE Chemistry physical section, because almost every question reduces to one of three things: classify an adsorption process, linearise an isotherm and read off a slope and intercept, or write a rate law for a surface reaction. The concepts are few and the arithmetic is short. This guide covers the isotherms, the two surface rate-law mechanisms, and the catalysis ideas that keep reappearing — with every numerical worked out line by line.

Physisorption versus chemisorption

The first question type is pure classification. Learn this table once and you can answer it in ten seconds.

Property	Physisorption	Chemisorption
Force involved	van der Waals / dispersion	Chemical bond (covalent or ionic)
Enthalpy of adsorption (typical order)	roughly 20–40 kJ mol⁻¹	roughly 80–400 kJ mol⁻¹
Specificity	Non-specific — any gas on any solid	Highly specific to the adsorbate–adsorbent pair
Layers formed	Multilayer possible	Monolayer only
Activation energy	Essentially zero	Often appreciable (activated adsorption)
Reversibility	Readily reversible on lowering pressure	Often irreversible; desorption may give a new species
Effect of rising temperature	Decreases steadily	Rises, passes a maximum, then falls

One thermodynamic fact ties the table together and is itself a common one-mark question: adsorption is always exothermic. A gas molecule loses translational freedom when it sticks to a surface, so ΔS is negative; for the process to be spontaneous, ΔG = ΔH − TΔS must be negative, which forces ΔH to be negative. There is no exception to argue about.

The Langmuir isotherm

Langmuir's model assumes a fixed number of identical, independent sites, one molecule per site, and no interaction between adsorbed molecules. Setting the rate of adsorption equal to the rate of desorption gives the fractional coverage θ:

θ = KP / (1 + KP) and V = V_mKP / (1 + KP)

Here P is the equilibrium gas pressure, K is the adsorption equilibrium constant (units of pressure⁻¹), V is the volume adsorbed and V_m is the volume needed for a complete monolayer. GATE almost never asks you to use the curved form. It asks for the linearised form, because that is where the slope and intercept live:

P/V = 1/(V_mK) + P/V_m → plot P/V against P: slope = 1/V_m, intercept = 1/(V_mK)

Worked example 1 — extracting V_m and K from Langmuir data

A gas adsorbs on a catalyst at 298 K with the following data:

P / kPa	V adsorbed / cm³	P/V / kPa cm⁻³
10	33.33	10 ÷ 33.33 = 0.300
20	50.00	20 ÷ 50.00 = 0.400
40	66.67	40 ÷ 66.67 = 0.600
80	80.00	80 ÷ 80.00 = 1.000

Step 1 — slope. Using the first and last points:
slope = (1.000 − 0.300) ÷ (80 − 10) = 0.700 ÷ 70 = 0.0100 cm⁻³

Step 2 — monolayer volume.
V_m = 1 ÷ slope = 1 ÷ 0.0100 = 100 cm³

Step 3 — intercept. Using the point (10, 0.300):
intercept = 0.300 − (0.0100 × 10) = 0.300 − 0.100 = 0.200 kPa cm⁻³

Step 4 — adsorption constant.
K = 1 ÷ (V_m × intercept) = 1 ÷ (100 × 0.200) = 0.0500 kPa⁻¹

Cross-check by a second route. At P = 20 kPa, KP = 0.0500 × 20 = 1.00, so θ = 1.00 ÷ (1 + 1.00) = 0.500. Independently, θ = V/V_m = 50.00 ÷ 100 = 0.500. The two agree, so the fit is right.

Note also the middle point: at 40 kPa the calculated P/V is 0.600, and slope × P + intercept = 0.0100 × 40 + 0.200 = 0.600. All four points sit exactly on the line, which is what a "the data obey the Langmuir isotherm" question is telling you.

Freundlich and BET — knowing which one applies

The Freundlich isotherm is empirical and describes adsorption on heterogeneous surfaces where the sites are not all equal:

x/m = k P^1/n (n > 1) → log(x/m) = log k + (1/n) log P

Plot log(x/m) against log P: slope = 1/n, intercept = log k. Freundlich has no saturation term, so it cannot describe the plateau at high pressure — that is its known limitation and a favourite "which statement is incorrect" option.

The BET (Brunauer–Emmett–Teller) isotherm extends Langmuir to multilayer adsorption and is the basis of surface-area measurement by N₂ adsorption at 77 K:

P / [V(P₀ − P)] = 1/(V_mC) + [(C − 1)/(V_mC)] × (P/P₀)

Plot the left-hand side against P/P₀. Slope + intercept = 1/V_m, and C = 1 + (slope ÷ intercept). C is related to the difference between the heat of adsorption of the first layer and the heat of liquefaction of the adsorbate — a large C means the first layer binds much more strongly than subsequent ones.

Isotherm	Surface assumed	Layers	Linear plot
Langmuir	Uniform, identical sites	Monolayer	P/V vs P
Freundlich	Heterogeneous	Not specified; no saturation	log(x/m) vs log P
BET	Uniform, layers stack	Multilayer	P/[V(P₀−P)] vs P/P₀

Surface reaction rate laws — Langmuir–Hinshelwood and Eley–Rideal

Once you can write θ, you can write the rate law. Both mechanisms assume adsorption is at equilibrium and the surface step is rate determining.

Langmuir–Hinshelwood: both reactants adsorb, then react with each other on the surface. The rate is proportional to the product of the two coverages:

r = k θ_A θ_B = k K_AP_A K_BP_B / (1 + K_AP_A + K_BP_B)²

Eley–Rideal: only A adsorbs; B reacts straight out of the gas phase:

r = k θ_A P_B = k K_AP_A P_B / (1 + K_AP_A)

The diagnostic that GATE tests: in Langmuir–Hinshelwood, raising P_A too far lowers the rate, because A crowds B off the surface. The rate passes through a maximum. Eley–Rideal shows no such maximum — the rate rises monotonically with P_B at fixed P_A and saturates in P_A. If a question describes a rate that peaks and then falls as one partial pressure is increased, the answer is Langmuir–Hinshelwood.

Worked example 2 — how much a catalyst actually speeds a reaction up

A reaction has an activation energy of 75 kJ mol⁻¹ uncatalysed. A catalyst provides a path with E_a = 50 kJ mol⁻¹. Assuming the pre-exponential factor is unchanged, by what factor does the rate increase at 298 K?

Step 1 — the ratio from the Arrhenius equation. With k = A e^−Ea/RT and the same A for both paths:
k_cat/k_uncat = e^{(E_a,uncat − E_a,cat)/RT}

Step 2 — the exponent.
ΔE_a = 75 − 50 = 25 kJ mol⁻¹ = 25 000 J mol⁻¹
RT = 8.314 × 298 = 2477.57 J mol⁻¹
ΔE_a/RT = 25 000 ÷ 2477.57 = 10.090

Step 3 — evaluate.
e^10.090 = e¹⁰ × e^0.090 = 22 026.5 × 1.0946 = 24 110

Answer: about 2.4 × 10⁴ times faster.

The point to carry into the exam: a 25 kJ mol⁻¹ drop in E_a at room temperature is worth four orders of magnitude in rate. That is why heterogeneous catalysts matter industrially even when the surface area is small.

What a catalyst does and does not do

This is a guaranteed conceptual question in some form.

A catalyst lowers the activation energy by providing a different mechanism.
It speeds the forward and reverse reactions by the same factor, so it changes the time taken to reach equilibrium but never the position of equilibrium.
It does not change ΔG°, ΔH°, ΔS° or the equilibrium constant K. Those are state functions of the reactants and products only.
It is regenerated, so it does not appear in the overall stoichiometry — but it very often appears in the rate law.

Practical vocabulary that turns up in one-liners: a promoter increases catalyst activity without being catalytic itself (K₂O and Al₂O₃ on iron in the Haber process); a poison binds to active sites irreversibly (sulfur on nickel or platinum); turnover frequency is the number of reactant molecules converted per active site per second; selectivity is the fraction of converted reactant that becomes the desired product. Zeolites are the classic shape-selective catalysts — the pore dimensions decide which molecules can enter, react or leave.

The Sabatier principle explains the volcano-shaped plot of activity against binding strength: if the substrate binds too weakly it never adsorbs, and if it binds too strongly the product cannot desorb. The best catalyst sits at the top of the volcano, binding intermediately. Questions phrase this as "why is the activity of metals for a given reaction not monotonic across the transition series".

Common mistakes that cost marks

Saying a catalyst shifts the equilibrium. It never does. It only shortens the time needed to get there.
Mixing up the two Langmuir plots. P/V versus P gives slope 1/V_m; 1/V versus 1/P gives slope 1/(V_mK). Both are valid, but the slope means something different in each — read what the question actually plotted.
Using Freundlich where saturation is described. If the data plateau at high pressure, Freundlich cannot fit them; use Langmuir.
Forgetting the square in Langmuir–Hinshelwood. The denominator is squared because two independent coverages multiply. Dropping it changes the whole pressure dependence and removes the rate maximum.
Claiming adsorption can be endothermic. ΔS is negative for adsorption, so a spontaneous process requires a negative ΔH.
Assuming chemisorption always increases with temperature. It increases first because it is activated, then falls once desorption dominates. The curve has a maximum.

Where this appears in GATE Chemistry

Surface chemistry sits in the physical chemistry portion of the GATE Chemistry (CY) syllabus and connects outward to kinetics, thermodynamics and organometallic catalysis. Question types you should be able to handle are listed below — for the number of questions, mark split and the exact syllabus wording for your year, always read the current official GATE information brochure and syllabus PDF rather than any secondary source.

Question type	What you must do
Classification MCQ	Decide physisorption or chemisorption from ΔH, specificity or layer count
Isotherm linearisation (NAT)	Convert data to the linear form, read slope and intercept, get V_m and K
Surface area from BET	Get V_m, convert to number of molecules, multiply by cross-sectional area
Rate-law mechanism	Identify Langmuir–Hinshelwood vs Eley–Rideal from the pressure dependence
Catalyst effect on E_a	Arrhenius ratio; also state what is unchanged (ΔG°, K)
Industrial catalysis one-liners	Promoters, poisons, zeolite shape selectivity, Sabatier principle

Check the catalysis numerical instantly. The Arrhenius calculator takes E_a, temperature and the pre-exponential factor and returns the rate constant, so you can compare the catalysed and uncatalysed paths without touching a scientific calculator.

Open the Arrhenius Equation Calculator →

Preparing for GATE, IIT-JAM, CSIR-NET or CUET-PG chemistry? ABC Chemistry runs dedicated competitive-exam batches at the Gurugram coaching centre and online across India — details at abcchemistry.in.