Kp from equilibrium amounts and total pressure

Worked example

At equilibrium, N₂O₄ ⇌ 2NO₂ has 1.10 mol N₂O₄ and 1.16 mol NO₂, with total pressure 213 kPa. Find the mole fractions, the partial pressures and hence Kp (with units).

1. 1. Add n(N₂O₄) + n(NO₂) → Total moles

   = 2.260 mol

2. 2. Divide n(N₂O₄) ÷ Total moles → x(N₂O₄)

   = 0.4867

3. 3. Divide n(NO₂) ÷ Total moles → x(NO₂)

   = 0.5133

4. 4. Partial pressure from mole fraction: solve for p(N₂O₄)

   $p_\mathrm{A} = x_\mathrm{A}\,p_\text{total}$

   = 103.7 kPa

5. 5. Partial pressure from mole fraction: solve for p(NO₂)

   $p_\mathrm{A} = x_\mathrm{A}\,p_\text{total}$

   = 109.3 kPa

6. 6. Kp from equilibrium partial pressures: solve for Kp = p(NO₂)² / p(N₂O₄)

   $K_\mathrm{p} = \prod_i p_{\mathrm{X}_i}^{\,\nu_i}$

   = 115.3 kPa

Answer: Kp = 115 kPa

Method

Mole fraction x = n ÷ n(total); partial pressure p = x × p(total). Then Kp = p(NO₂)² ÷ p(N₂O₄), with units kPa²/kPa = kPa.

Common mistakes

• N₂O₄ ⇌ 2NO₂: square the partial pressure of NO₂.

Formulas used

• Partial pressure from mole fraction

  $p_\mathrm{A} = x_\mathrm{A}\,p_\text{total}$

• Kp from equilibrium partial pressures

  $K_\mathrm{p} = \dfrac{p_\mathrm{C}^{\,c}\,p_\mathrm{D}^{\,d}}{p_\mathrm{A}^{\,a}\,p_\mathrm{B}^{\,b}}$

More equilibrium questions

• pH of a weak acid
• Buffer pH (Henderson–Hasselbalch)
• Kp from Kc
• Solubility from Ksp
• pH of a weak base
• Kc and its units for the Haber process
• Kc from equilibrium amounts and volume
• Kc from initial amounts and one equilibrium amount
• An equilibrium concentration from Kc
• Kp and its units for the Haber process
• Ksp from a solubility in g dm⁻³
• Solubility with a common ion
• pH of benzoic acid from a weighed sample
• H₂ + I₂ ⇌ 2HI: equilibrium amounts