Degree of Dissociation at Equilibrium
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Direct answer
Heat PCl5 vapour in a closed flask and it partly breaks into PCl3 and Cl2; the fraction of the original molecules that have split at equilibrium is the degree of dissociation, alpha. For PCl5 ⇌ PCl3 + Cl2 starting from pure PCl5 at total pressure p, one mole becomes 1 + alpha moles, the mole fractions become (1 − alpha)/(1 + alpha) for PCl5 and alpha/(1 + alpha) for each product, and Kp works out to alpha^2 × p/(1 − alpha^2). Alpha is measured experimentally from vapour density — the observed molar mass is M(calc)/(1 + alpha) for a two-fold dissociation — and for weak electrolytes it obeys the Ostwald dilution relation Ka = C × alpha^2/(1 − alpha).
What you must remember
- PCl5 pattern: Kp = alpha^2 p/(1 − alpha^2); for N2O4 ⇌ 2NO2, the doubled product count gives Kp = 4 alpha^2 p/(1 − alpha^2).
- Partial pressures at equilibrium (PCl5): p_PCl5 = (1 − alpha)p/(1 + alpha), and each product gets alpha p/(1 + alpha) — derive rather than memorise.
- Vapour density method: alpha = (D − d)/d for one molecule giving two, where D is the theoretical and d the observed vapour density; equivalently alpha = (M_calc − M_obs)/M_calc.
- Ostwald dilution law: Ka = C alpha^2/(1 − alpha), so alpha = sqrt(Ka/C) when alpha is small — dilution increases dissociation.
- Le Chatelier on dissociation: raising total pressure suppresses alpha (fewer moles side favoured); raising temperature increases alpha because dissociation is endothermic.
- Kp-Kc bridge: Kp = Kc(RT)^dn, with dn = 1 for both PCl5 and N2O4, so Kp exceeds Kc by a factor RT at every temperature.
Walking through the equilibrium table
Start with 1 mole PCl5 at total equilibrium pressure 1 atm, and let alpha = 0.5. Moles: PCl5 = 0.5, PCl3 = 0.5, Cl2 = 0.5, total 1.5. Mole fractions: 1/3 each — a pleasing symmetry. Partial pressures: 1/3 atm each. So Kp = (1/3 × 1/3)/(1/3) = 0.33 atm, and the shortcut Kp = alpha^2 p/(1 − alpha^2) = 0.25 × 1/0.75 = 0.33 atm agrees. Always run the shortcut and the table once each; they must match.
Now the reverse question, which is how JEE Advanced phrases it: Kp = 1/3 atm at total pressure 3 atm. Substitute: alpha^2 × 3 = (1/3)(1 − alpha^2), which rearranges to 9 alpha^2 = 1 − alpha^2, so alpha^2 = 0.1 and alpha = 0.316. The same algebra solves N2O4 problems with the factor 4 inserted — note how the brown NO2 colour deepens when you drop the pressure, because alpha rises.
Where students slip
The recurring error is partial pressure versus total pressure: alpha formulas use p as the total equilibrium pressure, and each species' partial pressure still needs its mole fraction applied. The second error is mole-fraction arithmetic at equilibrium — after dissociation the mixture has 1 + alpha moles, not 1, and every denominator must carry the correction. Third, in vapour-density problems students quote alpha as a percentage when the question asked for the fraction, or use molar masses where vapour densities (half the molar masses) were given. A final trap belongs to N2O4: students read "50% dissociation" as alpha = 50 moles — write alpha = 0.50 and recompute total moles as 1.5 before touching Kp.
Frequently asked questions
What is the Kp expression for PCl5 dissociation at total pressure p?
Kp = alpha^2 p/(1 − alpha^2), derived by feeding the equilibrium partial pressures (1 − alpha)p/(1 + alpha) and alpha p/(1 + alpha) into the equilibrium expression.
How is degree of dissociation found from vapour density?
Through alpha = (D − d)/d for a two-fold dissociation, since observed vapour density falls below theoretical in exact proportion to dissociation.
What happens to alpha if the pressure on N2O4 is increased?
It decreases, because dissociation doubles the number of gas moles and Le Chatelier shifts the equilibrium back toward N2O4 at higher pressure.
What does the Ostwald dilution law state?
For a weak electrolyte, Ka = C alpha^2/(1 − alpha), so alpha grows as sqrt(1/C) on dilution — dissociation is stronger in dilute solutions.
Why is Kp numerically larger than Kc for PCl5 dissociation?
Because dn = +1 and Kp = Kc(RT), the reaction produces one extra mole of gas from the reactant's one mole.