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Heat phosphorus pentachloride vapour and it partly splits into PCl₃ and Cl₂. One density measurement is enough to tell you how much has split, and from there you can get Kp. The method is short, but students often mix up which density goes where, or use it on reactions where it cannot work. This post derives the formula in four lines, works three examples, checks one against an NCERT problem and lists the mistakes to avoid.
Is this on the syllabus?
The NCERT Class 11 unit Equilibrium (reprint 2026-27) treats PCl₅ ⇌ PCl₃ + Cl₂ in its worked problems on Kc, but it does not use the vapour density method. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list "equilibrium constants (Kp and Kc) and their significance" and Le Chatelier's principle, and the JEE (Advanced) 2026 brochure lists Kp, Kc and Le Chatelier's principle. The method is an application of those lines rather than a named topic. Check the current year's syllabus.
Why does the vapour density fall?
Vapour density is the density of a gas divided by the density of hydrogen at the same temperature and pressure, which is about M/2 for a gas of molar mass M.
Start with 1 mol of PCl₅ and let a fraction α dissociate:
PCl₅: 1 − α
PCl₃: α
Cl₂: α
Total: 1 + α
The mass has not changed, but there are now (1 + α) moles of gas. At fixed temperature and pressure, density is mass divided by volume, and volume grows with the number of moles. So the observed vapour density d is smaller than the theoretical value D by the same factor:
D ÷ d = 1 + α, so α = (D − d) ÷ d.
For a reaction where one molecule gives n molecules, the same reasoning gives α = (D − d) ÷ [(n − 1)d]. For PCl₅, n = 2.
Worked example 1: α from one reading
Molar mass of PCl₅ = 30.974 + 5 × 35.453 = 208.24 g/mol, so D = 104.12.
Sample numbers: at a certain temperature the vapour density of the equilibrium mixture is 70.0.
α = (104.12 − 70.0) ÷ 70.0 = 34.12 ÷ 70.0 ≈ 0.487
So about 49% of the PCl₅ has dissociated.
Worked example 2: from α to Kp
Suppose that reading was taken at a total pressure of 1.0 bar. The mole fractions are (1 − α)/(1 + α) for PCl₅ and α/(1 + α) each for PCl₃ and Cl₂. Partial pressure is mole fraction times total pressure P, so
Kp = [α/(1 + α)]²P ÷ [(1 − α)/(1 + α)] = α²P ÷ (1 − α²).
With α = 0.487 and P = 1.0 bar:
Kp = 0.237 ÷ 0.763 ≈ 0.31 bar.
Now double the pressure to 2.0 bar at the same temperature. Kp stays 0.31, so solve α²(2.0) ÷ (1 − α²) = 0.31: α² = 0.31 ÷ 2.31, giving α ≈ 0.37. The new vapour density is 104.12 ÷ 1.37 ≈ 76. Higher pressure pushes the equilibrium towards fewer moles, exactly as Le Chatelier's principle predicts, and the density reading shows it.
Worked example 3: checking against NCERT
NCERT's Problem 6.9 puts 3.00 mol of PCl₅ in a 1 L vessel at 380 K with Kc = 1.80, and finds that 1.59 mol/L dissociates. That makes α = 1.59 ÷ 3.00 = 0.53.
What vapour density would that mixture show? The mean molar mass is 208.24 ÷ 1.53 ≈ 136, so the vapour density is about 68. Run the formula backwards: (104.12 − 68.0) ÷ 68.0 ≈ 0.53. The two routes agree. Our post on ICE tables for equilibrium constants shows the concentration route in full.
When does the method fail?
It needs the number of moles to change. For 2HI ⇌ H₂ + I₂, two moles of gas give two moles of gas, so n − 1 = 0 and the vapour density is the same at every degree of dissociation. A density reading tells you nothing there.
It also assumes ideal gas behaviour, since it uses volume proportional to moles at fixed temperature and pressure.
Which mistakes cost marks?
Using M (208.24) instead of M/2 (104.12) for D.
Putting d and D the wrong way round, which gives a negative α.
Leaving α as a percentage in the Kp formula.
Treating Kp as pressure-dependent. When pressure changes at constant temperature, α changes and Kp does not.
Applying the method to a reaction with no change in moles.
For the related idea in solutions, where α is found from conductivity instead, see our post on degree of dissociation and Ostwald's dilution law.
If you keep setting up the mole table wrongly, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
A practice routine
Derive α = (D − d)/d from a mole table without notes.
Redo Worked example 1 with your own sample readings of 60, 80 and 95. Check that α falls as d rises.
For N₂O₄ ⇌ 2NO₂ (M = 92.0), find α from a sample vapour density of 30.0, then Kp at 1.0 bar.
Explain in two sentences why the method fails for HI.
Today, convert NCERT Problem 6.9 into a vapour density question and solve it both ways.
