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Equilibrium Constants Kc, Kp and Kx Explained

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The equilibrium constants Kc, Kp and Kx describe the same chemical equilibrium using concentrations, partial pressures, and mole fractions respectively. For aA + bB ⇌ cC + dD: Kc uses mol/L, Kp uses partial pressures, and Kx uses mole fractions.

They matter because exam and lab data come in different units — gas equilibria are measured in pressures, solution equilibria in concentrations — and you must convert cleanly between them and compare the reaction quotient Q against the right K to predict the direction of change.

Key Concepts

  • Kc = [C]^c[D]^d / [A]^a[B]^b, with concentrations in mol/L; pure solids and liquids are omitted.
  • Kp has the same form with partial pressures (atm or bar); Kx uses mole fractions and is dimensionless.
  • Δn = (moles of gaseous products) − (moles of gaseous reactants); only gases count.
  • K depends only on temperature — never on concentration, pressure, or catalyst.
  • Q has the same formula as K but with non-equilibrium values: Q < K means the reaction shifts right, Q > K shifts left, Q = K is equilibrium.

Formulas

  • Kp = Kc(RT)^Δn, with R = 0.0821 L·atm/mol·K (if Kp is in atm) and T in kelvin.
  • Kx = Kp / (P_total)^Δn, since each partial pressure p(i) = x(i) × P_total.
  • Assumptions: ideal-gas behaviour, so that pV = nRT connects pressure and concentration.

Worked Example

For N2O4(g) ⇌ 2NO2(g) at 100 °C (373 K), Kc = 0.211. Convert to Kp:

  1. Count gases: Δn = 2 − 1 = 1.
  2. Kp = Kc(RT)^Δn = 0.211 × (0.0821 × 373)^1.
  3. 0.0821 × 373 = 30.62; Kp = 0.211 × 30.62 = 6.46.

So Kp = 6.46 (in atm) while Kc = 0.211 (in mol/L) — same equilibrium, different number because the units differ. For the reverse reaction 2NO2 ⇌ N2O4, K would be the reciprocal, 1/6.46 = 0.155.

Common Misconceptions

  • "K is always unitless." Only Kx is. Kc and Kp carry units derived from their concentration or pressure terms (strict thermodynamics uses a dimensionless K defined against standard states).
  • "K changes with pressure or concentration." Changing those shifts the position of equilibrium (Q changes), but K itself moves only with temperature.
  • "A catalyst changes K." A catalyst speeds up both forward and reverse reactions equally, so K is untouched — equilibrium is reached faster, not shifted.

Real-World Applications

  • Haber process: Kp guides the choice of high pressure to favour ammonia, since Δn = −2.
  • Esterification: Kc tells manufacturers how much excess reactant is needed for a good yield.
  • Solubility equilibria: Ksp is simply Kc written for a dissolving sparingly soluble salt.

Kc, Kp and Kx are three dialects of one language. Master the conversions Kp = Kc(RT)^Δn and Kx = Kp/(P_total)^Δn, remember that only temperature changes K, and use Q versus K as your compass for which way any mixture will shift.

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