Kp from Kc

Worked example

Convert Kc = 14 to Kp at 819 K for a reaction with Δn_gas = −1.

1. 1. Kp from Kc: solve for K_p

   $K_p = K_c\left(\dfrac{c^\circ RT}{p^\circ}\right)^{\Delta n}$

   = 0.2056

Answer: K_p = 0.206

Method

Kp = Kc(c°RT/p°)^Δn; with c° = 1 mol dm⁻³ and p° = 1 bar, c°R/p° = 0.08314 K⁻¹.

Common mistakes

• The exponent is upside down: Kp = Kc(c°RT/p°)^Δn with Δn = gaseous products − reactants.

Formulas used

• Kp from Kc

  $K_p = K_c\left(\dfrac{c^\circ RT}{p^\circ}\right)^{\Delta n}$
  Derivation: Kp from Kc

More equilibrium questions

• pH of a weak acid
• Buffer pH (Henderson–Hasselbalch)
• 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 from equilibrium amounts and total pressure
• 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