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"Argon is added to the equilibrium mixture. Which way does it shift?" The honest answer is "it depends on what is held constant", and exam setters know many students give one answer for both cases. This post explains both cases using partial pressures, works an example with numbers and shows how to spot which case a question describes.
Where is this in the syllabus?
The NTA Syllabus for JEE (Main) 2026 lists "factors affecting equilibrium concentration, pressure, temperature, the effect of catalyst, Le Chatelier's principle", and the NEET (UG) 2026 syllabus has the same items. The JEE (Advanced) 2026 brochure lists Kp, Kc, the reaction quotient and Le Chatelier's principle. The NCERT Class 11 chapter Equilibrium (Reprint 2026-27), section 6.8.3, covers the constant-volume case only. The constant-pressure case is not in the NCERT text, but it follows directly from the definition of Kp, as shown below.
Why does nothing happen at constant volume?
NCERT puts it plainly: if an inert gas is added at constant volume, "the equilibrium remains undisturbed. It is because the addition of an inert gas at constant volume does not change the partial pressures or the molar concentrations of the substance involved in the reaction." The total pressure rises, but Kp and Kc are written only in terms of the reacting species, so Q still equals K.
Why does it shift at constant pressure?
To keep the total pressure fixed, the container has to expand. Each gas's partial pressure is pᵢ = (nᵢ/n_total) × P. Adding inert moles raises n_total, so every reacting partial pressure falls by the same factor, exactly as if you had lowered the pressure. Writing it out:
Qp = (product of nᵢ terms) × (P/n_total)^Δn, where Δn is moles of gaseous products minus moles of gaseous reactants.
Δn > 0: Qp falls below Kp, so the reaction moves forward.
Δn < 0: Qp rises above Kp, so the reaction moves backward.
Δn = 0, as in H₂ + I₂ ⇌ 2HI: no shift in either case.
Worked example: N₂O₄ ⇌ 2NO₂
Use a sample value Kp = 0.50 bar at a fixed temperature, chosen for practice. Start with 1.0 mol N₂O₄ at a total pressure of 1.0 bar. Let α be the fraction that dissociates.
Before adding argon: moles are N₂O₄ (1 − α) and NO₂ 2α, total 1 + α. Then Kp = 4α²P/(1 − α²), so α = √(Kp/(4P + Kp)) = √(0.50/4.50) = 1/3. Both partial pressures are 0.50 bar. Check: 0.50²/0.50 = 0.50 bar.
Add 1.0 mol argon at constant pressure (1.0 bar):
Total moles just after: 4/3 + 1 = 7/3.
Partial pressures: N₂O₄ = (2/3)/(7/3) = 0.286 bar; NO₂ = (2/3)/(7/3) = 0.286 bar.
Qp = 0.286²/0.286 = 0.286 bar, less than 0.50, so more N₂O₄ dissociates.
New α from 4α²P/[(1 − α)(2 + α)] = 0.50 gives α ≈ 0.42. Dissociation rises from 33% to about 42%.
Add the same 1.0 mol argon at constant volume instead:
Total pressure rises from 1.0 bar to 1.0 × (7/3)/(4/3) = 1.75 bar.
Partial pressures of N₂O₄ and NO₂ stay at 0.50 bar, so Q = K and α stays 1/3.
For N₂ + 3H₂ ⇌ 2NH₃, where Δn = −2, argon at constant pressure would push the reaction backward and lower the ammonia fraction.
How do you tell which case a question means?
"Rigid vessel", "sealed flask", "fixed volume": constant volume, no shift.
"At constant pressure", "piston", "open to a fixed pressure": constant pressure, shift towards more gas moles.
If the question says only "inert gas is added", look for the other words before you answer.
Which mistakes do students make?
Seeing higher total pressure and shifting to fewer moles in a rigid vessel. Only partial pressures of reacting gases matter.
Changing K. It is fixed at a given temperature.
Leaving the inert gas out of n_total when working out mole fractions at constant pressure.
Forgetting Δn = 0 reactions, which do not shift either way.
A practice routine
Write NCERT's constant-volume reason in your own words.
Redo the example with 3.0 mol argon at constant pressure. (Check: α ≈ 0.52.)
Predict the direction for PCl₅ ⇌ PCl₃ + Cl₂ and for N₂ + 3H₂ ⇌ 2NH₃, both at constant pressure.
Work two more from our post on equilibrium constant calculations with ICE tables, then add an inert gas to each.
Our post on Le Chatelier's principle and Kp and Kc covers the other factors. If setting up the α equation is where you stall, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning at each step.
Do practice step 2 today and check your α against 0.52.
