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The contact process looks like a list to memorise: three steps, a catalyst, a temperature and a pressure. Exam questions rarely ask for the list. They ask why the conditions are what they are, and that is an equilibrium question. This post gives the steps, explains each condition with Le Chatelier's principle, and works three calculations that show what pressure and oxygen actually do.
Is the contact process still in the syllabus?
It depends on your exam. The JEE (Advanced) 2026 information brochure lists, under group 16, the "preparation/manufacture, properties, and uses of" sulphur dioxide and sulfuric acid, so JEE Advanced aspirants should know the process itself.
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus cover groups 13 to 18 only as a general introduction: trends and the unique behaviour of the first element. Both, however, list Le Chatelier's principle and the effect of a catalyst under equilibrium. The current NCERT Class 12 Part I (2026-27 reprint) has no p-block unit. The full description of the process was in the p-Block Elements unit of the earlier edition, which you can still read in an archived 2022-23 copy.
What survives in today's books is in the NCERT Class 11 chapter Equilibrium (2026-27 reprint). It notes that K for 2SO2 + O2 ⇌ 2SO3 is 1.7 × 10²⁶, yet the oxidation is very slow, so platinum or V2O5 is used as a catalyst. Its exercises 6.2 and 6.10 use the same reaction. So for JEE Main and NEET, learn this reaction as an equilibrium example. Check the current year's documents before you plan.
What are the three steps?
From the earlier NCERT unit:
Make SO2. Burn sulphur or sulphide ores in air: S + O2 → SO2. The gas is purified by removing dust and impurities such as arsenic compounds.
Oxidise SO2 to SO3. 2SO2(g) + O2(g) ⇌ 2SO3(g), with V2O5 as catalyst and ΔrH° = −196.6 kJ mol⁻¹. NCERT calls this the key step.
Absorb SO3. SO3 + H2SO4 → H2S2O7 (oleum). Diluting oleum with water gives acid of the desired concentration, and NCERT says the product is 96–98% pure.
Why 720 K and 2 bar?
Read the key step like an equilibrium problem.
It is exothermic. NCERT Class 11 states that K for an exothermic reaction decreases as temperature rises, so a lower temperature gives more SO3 at equilibrium.
It reduces gas moles: three on the left, two on the right, so Δn = −1. Higher pressure pushes it forward.
But the old NCERT adds that the temperature should not be too low, or the reaction becomes too slow.
In practice, it says, the plant runs at 2 bar and 720 K. The temperature is a compromise between yield and rate. The catalyst handles the rate problem without touching the yield: NCERT says a catalyst speeds up the forward and reverse reactions and "does not affect the equilibrium composition".
Worked example 1: what does compression do?
Sample numbers for practice: at equilibrium at some fixed temperature, p(SO2) = 0.20 bar, p(O2) = 0.50 bar and p(SO3) = 1.30 bar.
Kp = p(SO3)² / [p(SO2)² × p(O2)] = 1.69 / (0.04 × 0.50) = 84.5 bar⁻¹.
Now halve the volume. Every partial pressure doubles at once:
Q = 2.60² / (0.40² × 1.00) = 6.76 / 0.16 = 42.25 bar⁻¹.
Q is exactly half of Kp, so Q < Kp and the reaction moves forward, making more SO3. The factor of a half is no accident: doubling all pressures multiplies Q by 2^Δn = 2⁻¹.
Worked example 2: why extra oxygen helps
Rearrange the Kp expression: p(SO3)/p(SO2) = √(Kp × p(O2)).
With the same sample numbers, √(84.5 × 0.50) = 6.5, which matches 1.30/0.20. If p(O2) is raised to 2.0 bar, the ratio becomes √(84.5 × 2.0) = 13. Four times the oxygen doubles the ratio of SO3 to SO2, because of the square root. More of one reactant shifts the balance, but not in proportion.
Worked example 3: from Kp to Kc
Use Kp = Kc(RT)^Δn with Δn = −1, so Kc = Kp × RT. With R = 0.08314 L bar K⁻¹ mol⁻¹ and T = 720 K, RT = 59.86 L bar mol⁻¹. If the sample Kp of 84.5 bar⁻¹ applied at 720 K, Kc = 84.5 × 59.86 ≈ 5.06 × 10³ L mol⁻¹. Use Δn = +1 by mistake and you get 1.41, which is off by a factor of about 3,600.
One quick mass check: 1.00 kg of sulphur is 1000/32.06 = 31.2 mol, and each sulphur atom ends up in one H2SO4, so the most acid you can make is 31.2 × 98.08 g ≈ 3.06 kg. The key step alone releases about 31.2 × 98.3 ≈ 3.07 × 10³ kJ, since −196.6 kJ is per 2 mol of SO2.
Which mistakes cost marks?
Saying V2O5 increases the yield. It increases the rate only.
Getting Δn wrong by forgetting the coefficient 2 on SO3.
Writing "low temperature is best" without the rate caveat.
Absorbing SO3 directly in water. NCERT's route absorbs it in concentrated sulphuric acid to form oleum, then dilutes.
Reading a huge K as "fast".
A practice routine
Write the three steps, with equations, catalyst and conditions, from memory.
Solve NCERT Class 11 exercises 6.2 and 6.10.
Redo example 1 for doubling the volume. You should find Q = 2Kp, so the reaction moves backward.
Compare the Haber process, which NCERT Class 11 also uses as an example. Our chemical equilibrium problems guide has more practice, and the story of Henry Le Chatelier shows where the principle came from.
If the reasoning about Q and K feels shaky, the Fermi Concept Library covers every concept in chemistry with its own lesson, so you can fix that idea before returning to industrial processes.
Tonight, write the three steps from memory, then rework example 2 with p(O2) = 1.0 bar.
