GATE Chemical Equilibrium — Kp/Kc Numericals
Chemical equilibrium numericals in GATE Chemistry come in a small number of recurring shapes: finding Kc from equilibrium concentrations, converting between Kp and Kc, deciding which way a reaction will shift by comparing Q to K, and working out how much of a gas has dissociated at a given pressure. This guide works through each shape completely, with the algebra shown at every step.
The two equilibrium constants and how they relate
Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ Kp = PCᶜPDᵈ / PAᵃPBᵇ
Kp = Kc (RT)^Δn where Δn = (c + d) − (a + b), and R = 0.0821 L atm mol⁻¹ K⁻¹
Only gas-phase species with genuinely variable concentration enter K — pure solids and pure liquids are omitted, because their "concentration" (activity) is fixed at 1 regardless of how much is present.
Worked example 1 — Kc from an ICE table. 1.0 mol of PCl₅ is placed in a
1 L flask. At equilibrium, 60 % has dissociated: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Find Kc.
Initial: [PCl₅] = 1.0 M, [PCl₃] = 0, [Cl₂] = 0
Change: −0.6, +0.6, +0.6 (α = 0.60, so 0.6 mol dissociates)
Equilibrium: [PCl₅] = 0.4 M, [PCl₃] = 0.6 M, [Cl₂] = 0.6 M
Kc = (0.6 × 0.6) / 0.4 = 0.36 / 0.4 = 0.90 (dimensionally mol/L here, since Δn = 1)
Worked example 2 — Kp from Kc. For the same reaction at 500 K, with
Kc = 0.90, find Kp.
Δn = (1 + 1) − 1 = +1
Kp = Kc (RT)¹ = 0.90 × (0.0821 × 500) = 0.90 × 41.05 = 36.9 atm
Worked example 3 — predicting direction with Q. For
N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = 0.50 at a given temperature. At some moment, [N₂] = 1.0 M,
[H₂] = 1.0 M and [NH₃] = 0.50 M. Which way will the reaction shift?
Q = [NH₃]² / ([N₂][H₂]³) = (0.50)² / (1.0 × 1.0³) = 0.25 / 1.0 = 0.25
Q (0.25) < Kc (0.50), so the system has not yet made enough product. The reaction
proceeds forward (toward more NH₃) until Q rises to meet Kc.
Degree of dissociation and pressure — the formula GATE actually wants
For a single gas dissociating into two product gases, A(g) ⇌ B(g) + C(g), starting with 1 mole of pure A and letting α be the fraction dissociated at equilibrium, the total moles at equilibrium are 1 + α. Working through the mole fractions and multiplying each by the total pressure P gives a clean closed-form result:
Worked example 4 — finding α from Kp and P. For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g),
Kp = 1.00 atm at a given temperature, and the total pressure is held at 3.0 atm. Find the
degree of dissociation α.
1.00 = α²(3.0) / (1 − α²)
1.00 (1 − α²) = 3.0 α² → 1 − α² = 3.0 α² → 1 = 4.0 α² → α² = 0.25 → α = 0.50
(50 % dissociated)
Check by substitution: Kp = (0.5)²(3.0) / (1 − 0.25) = 0.75 / 0.75 = 1.00 ✓ —
matches the given Kp exactly.
This is also the quantitative face of Le Chatelier's principle: since Kp is fixed at a given temperature, raising P must be compensated by α falling (less dissociation), because the right-hand side has more moles of gas and the system resists the pressure increase by favouring the side with fewer moles.
Common mistakes that cost marks
- Forgetting Δn is gas-phase products minus gas-phase reactants only. Solids, pure liquids and solvent are excluded from the equilibrium expression entirely, so they never contribute to Δn either.
- Using Kp = Kc(RT)^Δn when Δn = 0 and expecting a different number. When the mole count of gas is unchanged, Kp = Kc numerically — there is nothing to convert.
- Believing K changes with the initial amounts or the container's pressure. K depends only on temperature. Initial concentrations and total pressure change the position of equilibrium (how far the reaction proceeds, i.e. α), never the value of K itself.
- Comparing Q and K backwards. Q < K means not enough product yet, so the reaction goes forward; Q > K means too much product, so it goes in reverse. The direction is always the one that moves Q toward K.
- Using R = 8.314 with pressure in atm. The value 0.0821 L atm mol⁻¹ K⁻¹ is paired specifically with pressure in atmospheres and volume in litres; using the SI value 8.314 J mol⁻¹ K⁻¹ needs pressure in Pa and volume in m³ instead — mixing the two gives an answer wrong by many orders of magnitude.
Quick reference
| You are asked for | Use | Watch for |
|---|---|---|
| Kc from equilibrium amounts | Build an ICE table, then the mass-action expression | Divide moles by volume to get concentration first |
| Kp from Kc (or the reverse) | Kp = Kc(RT)^Δn | Δn sign; only gas-phase species count |
| Direction of shift | Compute Q, compare to K | Q < K → forward; Q > K → reverse |
| α from Kp and total pressure | Kp = α²P/(1 − α²) for A ⇌ B + C | Formula is specific to a 1-mole-into-2-mole dissociation |
Cross-check your gas-law arithmetic. The ABC Chemistry Calculator Suite's ideal gas calculator is a useful companion for the partial-pressure bookkeeping behind Kp problems.
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