Equilibrium constants Kc and Kp

Kc (spec 13.1)

For aA + bB ⇌ cC + dD at equilibrium:

Kc = [C]^c [D]^d / ([A]^a [B]^b), with concentrations in mol dm^-3.

For a heterogeneous equilibrium, leave out pure solids (and pure liquids not acting as solvents), because their concentration is constant. For example, for CaCO3(s) ⇌ CaO(s) + CO2(g), Kc = [CO2].

Kp (spec 13.2)

For gases, use partial pressures:

• mole fraction of A = moles of A / total moles of gas
• partial pressure of A, pA = mole fraction x total pressure (in atm)

Kp = pC^c pD^d / (pA^a pB^b). Again, solids are left out: for CaCO3(s) ⇌ CaO(s) + CO2(g), Kp = pCO2.

Calculating K (spec 13.3)

1. Write the starting moles.

2. Use the equation ratio to find the change, and so the equilibrium moles.

3. Convert to concentrations (divide by volume) or partial pressures.

4. Substitute into the expression and work out the units by cancelling. If there are the same number of moles on each side, K has no units and the volume cancels.

Worked example

1.00 mol of ethanoic acid and 1.00 mol of ethanol reach equilibrium with 0.667 mol of ethyl ethanoate formed. Find Kc. CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

1. At equilibrium: ester 0.667, water 0.667, acid 0.333, alcohol 0.333 mol
2. Moles are equal on both sides, so the volume cancels
3. Kc = (0.667 x 0.667) / (0.333 x 0.333)

Answer: Kc = 4.0 (no units)

Check yourself

1. What is the Kc expression for N2(g) + 3H2(g) ⇌ 2NH3(g)?
   • [N2][H2]³ / [NH3]²
   • [NH3]² / ([N2][H2]³)
   • 2[NH3] / ([N2] x 3[H2])
   • [NH3] / ([N2][H2])
   Show answer

   [NH3]² / ([N2][H2]³). Products over reactants, each raised to its coefficient.

2. What are the units of Kc for N2(g) + 3H2(g) ⇌ 2NH3(g)?
   • mol² dm^-6
   • mol dm^-3
   • No units
   • dm⁶ mol^-2
   Show answer

   dm⁶ mol^-2. (mol dm^-3)² / (mol dm^-3)⁴ = (mol dm^-3)^-2 = dm⁶ mol^-2.

3. What is the Kp expression for CaCO3(s) ⇌ CaO(s) + CO2(g)?
   • Kp = pCO2 x pCaO / pCaCO3
   • Kp = pCO2
   • Kp = 1 / pCO2
   • Kp = pCaO / pCaCO3
   Show answer

   Kp = pCO2. Solids are left out of K expressions.

4. In a 2.00 dm³ flask at equilibrium there are 0.100 mol PCl5, 0.050 mol PCl3 and 0.050 mol Cl2. PCl5 ⇌ PCl3 + Cl2. Calculate Kc in mol dm^-3.
   Show answer

   0.0125 mol dm^-3. Concentrations: 0.0500, 0.0250, 0.0250 mol dm^-3. Kc = 0.0250 x 0.0250 / 0.0500 = 0.0125 mol dm^-3.

5. A gas mixture at 2.00 atm total pressure contains 0.30 mol N2, 0.90 mol H2 and 0.20 mol NH3. What is the partial pressure of NH3?
   Show answer

   0.286 atm. Mole fraction = 0.20 / 1.40 = 0.143; p = 0.143 x 2.00 = 0.286 atm.

6. At equilibrium, N2O4(g) ⇌ 2NO2(g) contains 0.400 mol N2O4 and 0.200 mol NO2 at a total pressure of 1.00 atm. Calculate Kp in atm.
   Show answer

   0.167 atm. Mole fractions: NO2 = 0.200 / 0.600 = 0.333, N2O4 = 0.667. Kp = 0.333² / 0.667 = 0.167 atm.

7. 0.500 mol of ethanoic acid and 0.500 mol of ethanol reach equilibrium with 0.300 mol of ethyl ethanoate formed. Calculate Kc.
   Show answer

   2.25. Equilibrium: ester 0.300, water 0.300, acid 0.200, alcohol 0.200 mol. Kc = 0.300² / 0.200² = 2.25 (no units, volume cancels).

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Sources

• Pearson Edexcel IAL Chemistry (2018) specification
• Pearson Edexcel IAL Chemistry (2018) scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.