Kp and Kc are equilibrium constants expressed in different units. Kc uses molar concentrations, while Kp uses partial pressures. For gas-phase reactions, they're related by Kp = Kc × (RT)^Δn, where Δn is the change in moles of gas. This relationship comes from converting between concentration and pressure using the ideal gas law (PV = nRT).
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Formula: Kp = Kc × (RT)^Δn, where R is the ideal gas constant (0.08206 L⋅atm/mol⋅K).
Δn = (Moles of gaseous products) - (Moles of gaseous reactants).
Temperature MUST be in Kelvin (K = °C + 273.15).
R = 0.08206 is used when pressure is in atmospheres (atm). If using kPa, R = 8.314.
If the reaction is H₂(g) + Cl₂(g) ⇌ 2HCl(g), then Δn = 2 - (1+1) = 0, so Kp = Kc.
Remember: K tells you WHERE the equilibrium lies, not how fast it gets there (that's kinetics).
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Fun Facts
"For reactions where the number of gas moles doesn't change (Δn = 0), Kp is exactly equal to Kc."
"Pressure only shifts the position of equilibrium (Le Chatelier's Principle) if Δn ≠ 0; it doesn't change the value of Kp itself."
"The Haber Process for ammonia used in fertilizers depends on high pressure to shift equilibrium toward the product side."
"The value of Kp can be enormous (e.g., 10³⁰) for complete reactions or tiny (e.g., 10⁻⁵⁰) for reactions that barely occur."
"Kp is always temperature-dependent; if the reaction is exothermic, Kp decreases as temperature increases."
"Units for Kp depend on the Δn value (e.g., atm, atm⁻¹, or unitless if Δn=0)."
"Liquid and solid species are ignored in the Kp expression because their concentrations are effectively constant."
Formula: Kp = Kc · (RT)^Δn
Kp vs. Temperature Complexity
Illustrating how the pressure-based equilibrium constant shifts across typical reaction temperatures.
