Kp Calculator — Equilibrium Constant in Partial Pressures
Convert the concentration equilibrium constant (Kc) to the pressure equilibrium constant (Kp) using the ideal-gas relationship Kp = Kc × (RT)^Δn, and derive the standard Gibbs free energy change ΔG° = −RT ln Kp at any temperature.
K
Kp expressed in terms of partial pressures (atm units)
- 1
RT (L·atm/mol)
0.082057 × 298 = 24,4531 - 2
(RT)^Δn
24,4531^1 = 24,453087 - 3
Kp = Kc × (RT)^Δn
0,0821 × 24,453087 = 2,0076Kp relates partial pressures; Kc relates molar concentrations. The (RT)^Δn factor bridges the two.
Kaip veikia ši skaičiuoklė?
Kp = Kc × (RT)^Δn, where R = 0.082057 L·atm/(mol·K), T is temperature in kelvin, and Δn = moles of gaseous products minus moles of gaseous reactants. If Δn = 0, Kp = Kc. Standard Gibbs energy ΔG° = −RT ln Kp: negative ΔG° means products are favoured.
Formulė
How this is calculated
At chemical equilibrium the reaction quotient equals the equilibrium constant. Two equivalent forms exist: Kc uses molar concentrations [mol/L] while Kp uses partial pressures [atm]. For an ideal gas, the partial pressure of component i equals [i]RT, so the ratio of Kp to Kc carries a factor of (RT)^Δn, where Δn is the net change in the number of moles of gaseous species: Δn = Σn(gaseous products) − Σn(gaseous reactants). If Δn = 0 (equal moles of gas on both sides), Kp = Kc exactly.
The gas constant R = 0.082057 L·atm/(mol·K) is used here because Kp is expressed in atmospheres. If you need Kp in SI pascals you would use R = 8.314 J/(mol·K) with pressure in Pa — the numerical value of Kp changes, but the equilibrium position does not.
The standard Gibbs free energy change ΔG° = −RT ln Kp (R = 8.314 J/mol·K) quantifies whether the reaction is thermodynamically spontaneous under standard conditions: ΔG° < 0 means products are favoured (Kp > 1); ΔG° > 0 means reactants are favoured (Kp < 1); ΔG° = 0 means Kp = 1. This calculator assumes ideal-gas behaviour and does not account for fugacity corrections at high pressure.
Dažnai užduodami klausimai
Kp = Kc whenever Δn = 0, i.e. when the number of moles of gaseous products equals the number of moles of gaseous reactants. Example: H₂(g) + I₂(g) ⇌ 2 HI(g) has Δn = 2 − 2 = 0, so Kp = Kc at any temperature.
A large Kp (≫ 1) indicates the equilibrium lies strongly toward products — the reaction proceeds nearly to completion under those conditions. A small Kp (≪ 1) means the equilibrium favours reactants and very little product forms. Kp = 1 means products and reactants coexist in roughly equal concentrations at equilibrium.
No. At constant temperature, Kp is a true constant and does not change when total pressure is altered — only the position of equilibrium shifts (Le Chatelier's principle). Kp changes only when temperature changes, which is described by the van't Hoff equation.
Taip pat žinomas kaip
TG we-Calculate Editorial Team. (2026). Kp Calculator — Equilibrium Constant in Partial Pressures [Online calculator]. TG we-Calculate. https://we-calculate.com/lt/calculator/kp-calculator
TG we-Calculate Editorial Team. "Kp Calculator — Equilibrium Constant in Partial Pressures." TG we-Calculate. 2026. https://we-calculate.com/lt/calculator/kp-calculator.
TG we-Calculate Editorial Team, "Kp Calculator — Equilibrium Constant in Partial Pressures," TG we-Calculate, 2026. [Online]. Available: https://we-calculate.com/lt/calculator/kp-calculator
@misc{wecalculate_kp_calculator, title = {Kp Calculator — Equilibrium Constant in Partial Pressures}, author = {{TG we-Calculate Editorial Team}}, howpublished = {\url{https://we-calculate.com/lt/calculator/kp-calculator}}, year = {2026}, note = {TG we-Calculate} }
Ar ši skaičiuoklė jums padėjo?