Calculate the equilibrium constant Kp from partial pressures. Convert between Kp and Kc using the ideal gas law. Find Δn and reaction quotient Qp.
The equilibrium constant Kp expresses the ratio of partial pressures of products to reactants at chemical equilibrium for gas-phase reactions. For a general reaction aA + bB ⇌ cC + dD, Kp = (P_C^c × P_D^d) / (P_A^a × P_B^b), where each P represents the partial pressure in atmospheres (or bar) and the exponents are stoichiometric coefficients.
Kp is related to the concentration-based equilibrium constant Kc through the equation Kp = Kc(RT)^Δn, where Δn is the change in moles of gas (products minus reactants), R is the gas constant, and T is the absolute temperature. When Δn = 0, Kp = Kc. This relationship is fundamental for converting between pressure and concentration expressions of equilibrium.
Understanding Kp is essential for predicting the direction of gas-phase reactions, calculating equilibrium compositions at different temperatures and pressures, and applying Le Chatelier's principle. Industrial processes like the Haber process (ammonia synthesis), sulfuric acid production, and methanol synthesis all depend on accurate Kp calculations for optimizing yield and operating conditions.
Essential for chemical engineering, gas-phase reaction design, and chemistry courses. Quickly calculate equilibrium constants, convert between Kp and Kc, predict reaction direction, and explore how conditions affect gas-phase equilibria. This kp calculator (equilibrium constant in pressure) helps you compare outcomes quickly and reduce avoidable mistakes when making day-to-day care decisions. Use the estimate as a planning baseline and confirm final decisions with a qualified professional when risk is high.
Kp = Π(P_products^coefficients) / Π(P_reactants^coefficients). Kp-Kc relationship: Kp = Kc × (RT)^Δn, where Δn = Σ(product moles) − Σ(reactant moles), R = 0.08206 L·atm/(mol·K), T in Kelvin. ΔG° = −RT ln Kp gives the standard Gibbs free energy.
Result: Kp = 9.26 × 10⁻³
For N₂ + 3H₂ ⇌ 2NH₃: Kp = (0.5)² / (1 × 3³) = 0.25/27 = 9.26 × 10⁻³. Since Δn = 2 − 4 = −2, increasing pressure shifts equilibrium toward products (fewer moles of gas).
The Haber-Bosch process for ammonia synthesis (N₂ + 3H₂ ⇌ 2NH₃) operates at 400-500°C and 150-300 atm. Despite the reaction being exothermic, high temperature is needed for kinetics while high pressure compensates for the unfavorable Kp at elevated temperatures. Understanding the interplay between Kp, temperature, and pressure is crucial for optimizing industrial processes. Similar considerations apply to sulfuric acid production (2SO₂ + O₂ ⇌ 2SO₃), methanol synthesis, and Fischer-Tropsch processes.
The van't Hoff equation relates Kp at two temperatures: ln(K₂/K₁) = −ΔH°/R × (1/T₂ − 1/T₁). Plotting ln(Kp) vs. 1/T gives a straight line with slope −ΔH°/R. This is a powerful tool for determining reaction enthalpies from equilibrium data and for extrapolating Kp to new temperatures. For reactions where ΔH° changes significantly with temperature, the integrated form with heat capacity corrections is needed.
Strictly, the thermodynamic equilibrium constant K° is dimensionless and defined in terms of activities rather than pressures. For ideal gases, activity equals P/P° where P° = 1 bar (or 1 atm in older conventions). The distinction matters when calculating ΔG° = −RT ln K°, which requires a dimensionless K. In practice, Kp calculated with pressures in the standard unit (bar or atm) equals K° for ideal gas mixtures.
Kp uses partial pressures (in atm or bar) while Kc uses molar concentrations (mol/L). They're related by Kp = Kc(RT)^Δn. When Δn = 0, they're equal.
No. Kp depends only on temperature. Changing total pressure shifts the equilibrium position (changes individual partial pressures), but Kp remains constant. This is why Kp is called a constant.
For exothermic reactions, Kp decreases with increasing temperature. For endothermic reactions, Kp increases. The van't Hoff equation quantifies this: ln(K₂/K₁) = −ΔH°/R × (1/T₂ − 1/T₁).
If Qp < Kp, the reaction proceeds forward (toward products). If Qp > Kp, the reaction proceeds in reverse. If Qp = Kp, the system is at equilibrium.
Δn determines how pressure affects equilibrium. If Δn < 0 (fewer product gas moles), increasing pressure favors products (Le Chatelier). If Δn = 0, pressure changes don't affect equilibrium composition.
No. Since Kp is a product of partial pressures (all positive) raised to positive powers, Kp is always positive. Very large Kp means the reaction strongly favors products; very small Kp means it favors reactants.