Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

5.2.2Physical Chemistry and Transition Elements

Kc and Kp

5 min read 5 parts Lesson 11 of 40

0%read

Section 1 of 5Writing the Expressions

5.2.2 Kc and Kp

For aA + bB ⇌ cC + dD, the equilibrium constant compares products with reactants.

Kc
  • Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ), using equilibrium concentrations in mol dm⁻³ with powers equal to the stoichiometric coefficients.
Kp
  • Kp = p(C)ᶜ p(D)ᵈ / (p(A)ᵃ p(B)ᵇ), using equilibrium partial pressures of gases only.
    What the symbols mean: Kp = equilibrium constant in terms of partial pressures · p = partial pressure of each gas at equilibrium (Pa or kPa) · powers come from the balanced equation
Heterogeneous systems
  • Pure solids and liquids are left out, so CaCO₃(s) ⇌ CaO(s) + CO₂(g) gives Kc = [CO₂] and Kp = p(CO₂).
Temperature
  • The value of K is fixed at a given temperature, and the temperature should always be stated with it.
Size of K
  • A large K means the position of equilibrium lies to the right, while a small K means it lies to the left.

Section 2 of 5Calculating K and Its Units

Follow the same sequence for any Kc or Kp calculation.

1
Write the balanced equation with state symbols and then the expression for K from it.
2
Work out the equilibrium amounts from the initial moles and the change, using the stoichiometric ratios.
3
Convert to concentrations by dividing by the volume in dm³, or to partial pressures using mole fractions and total pressure.
4
Substitute into the expression and evaluate the numerical value, keeping sensible significant figures.
5
Find the units by substituting the units into the expression and cancelling; if the powers on top and bottom match, K has no units.

Section 3 of 5Typical Examples

Typical Examples
N₂O₄(g) ⇌ 2NO₂(g)
  • Kc = [NO₂]²/[N₂O₄] has units mol dm⁻³,
    • because two concentration terms are divided by one.
      What the symbols mean: Kc = equilibrium constant in terms of concentration · [ ] = concentration at equilibrium (mol dm−3) · each power is the number in front of that substance in the balanced equation
H₂(g) + I₂(g) ⇌ 2HI(g)
  • Kc = [HI]²/([H₂][I₂]) has no units, as the powers cancel completely.
N₂ + 3H₂ ⇌ 2NH₃
  • Kp = p(NH₃)²/(p(N₂)·p(H₂)³) has units of pressure⁻², for example kPa⁻².
2SO₂ + O₂ ⇌ 2SO₃
  • Kc = [SO₃]²/([SO₂]²[O₂]) has units dm³ mol⁻¹, found by cancelling the units term by term.

Section 4 of 5Worked Example

Question

  • 1.00 mol of N₂O₄ is placed in a 5.0 dm³ vessel and decomposes: N₂O₄(g) ⇌ 2NO₂(g).
  • At equilibrium 0.80 mol of N₂O₄ remains.
  • Calculate Kc.
  • Then, for N₂ + 3H₂ ⇌ 2NH₃ with p(N₂) = 40 kPa, p(H₂) = 120 kPa and p(NH₃) = 40 kPa, calculate Kp.

  1. N₂O₄ reacted = 0.20 mol, so NO₂ formed = 2 × 0.20 = 0.40 mol.
  2. [N₂O₄] = 0.80/5.0 = 0.16 mol dm⁻³; [NO₂] = 0.40/5.0 = 0.080 mol dm⁻³.
  3. Kc = (0.080)²/0.16 = 0.0064/0.16 = 0.040 mol dm⁻³.
  4. Kp = 40²/(40 × 120³) = 1600/(6.912 × 10⁷) = 2.3 × 10⁻⁵ kPa⁻².

Answer Kc = 0.040 mol dm⁻³ and Kp = 2.3 × 10⁻⁵ kPa⁻².

Section 5 of 5Exam Checklist & Tips

5.2.2 – Key Points for the Exam

  • Kc The equilibrium constant expressed using equilibrium concentrations.
  • Kp The equilibrium constant expressed using equilibrium partial pressures of gases.
  • Heterogeneous equilibrium An equilibrium involving species in more than one phase.
  • Position of equilibrium The relative amounts of reactants and products present once equilibrium has been reached.

Common Mistakes

  • Including solids and pure liquids in the K expression, when their concentrations are constant and so omitted.
  • Using initial instead of equilibrium amounts when substituting into the expression.
  • Forgetting to divide moles by volume, which matters when the numbers of moles on each side differ.
  • Raising terms to the wrong power, such as ignoring the coefficient 3 in front of H₂.
\n