3.667948e+1
2.445299e+1
2.445299e+1
—
3.667948e+1
2.445299e+1
2.445299e+1
—
The Kp Calculator converts the concentration-based equilibrium constant (Kc) to the pressure-based equilibrium constant (Kp) for gas-phase reactions. The relationship between Kp and Kc depends on the change in moles of gaseous species (Δn) and the temperature. Kp is essential when working with gaseous equilibria where partial pressures are more convenient than concentrations. This calculator uses the ideal gas constant R = 0.08206 L·atm/(mol·K) to bridge the two expressions. Understanding the Kp–Kc relationship is critical for industrial chemistry, atmospheric chemistry, and any system involving gaseous reactants and products.
The conversion between Kc and Kp is given by:
$$K_p = K_c \times (RT)^{\Delta n}$$
where:
When Δn = 0 (equal moles of gas on both sides), Kp = Kc. When Δn > 0, Kp > Kc (assuming RT > 1). When Δn < 0, Kp < Kc. The (RT)Δn factor accounts for the conversion between concentration (mol/L) and pressure (atm) units through the ideal gas law PV = nRT.
The Kp value tells you the equilibrium position in terms of partial pressures. If Kp > 1, products are favored. The (RT)Δn factor shows how much the conversion scales Kc. If Δn = 0, the factor is 1 and Kp = Kc. For reactions that produce more gas molecules (Δn > 0), Kp will be larger than Kc. The log(Kp) value helps compare equilibria across different orders of magnitude.
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Results
Δn = 2 − 1 = 1. Kp = 0.00466 × (0.08206 × 298)¹ = 0.00466 × 24.45 = 0.1139. Since Δn > 0, Kp > Kc.
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Results
Δn = 2 − 4 = −2. Kp = 0.5 × (0.08206 × 500)^(−2) = 0.5 × (41.03)^(−2) = 0.5 × 0.000594 = 0.000297. Since Δn < 0, Kp << Kc.
Kc uses molar concentrations (mol/L) in the equilibrium expression, while Kp uses partial pressures (typically in atm). They describe the same equilibrium but in different units.
Kp = Kc when Δn = 0, meaning the total moles of gaseous products equal the total moles of gaseous reactants. The (RT)0 factor equals 1.
Use R = 0.08206 L·atm/(mol·K) when pressures are in atmospheres. If using SI units (Pa), use R = 8.314 J/(mol·K), but the Kp value and units will differ accordingly.
Temperature appears in the ideal gas law (PV = nRT). Higher temperatures increase pressure for a given concentration, so the (RT)Δn conversion factor changes with temperature even if Kc itself also changes.
Yes. If the reaction produces fewer moles of gas than it consumes (e.g., synthesis reactions like N₂ + 3H₂ → 2NH₃, Δn = −2), then Kp will be smaller than Kc.
The Kp = Kc(RT)Δn relationship assumes ideal gas behavior. For real gases at high pressures, fugacity coefficients should be used instead of partial pressures.
Count the stoichiometric coefficients of all gaseous species on the product side, then subtract the total for the reactant side. Only include species in the gas phase.
No. Kp is only meaningful for gas-phase equilibria. For reactions involving only liquids or solids, Kc (or activity-based K) is used.
As temperature increases, the (RT)Δn factor grows for positive Δn, making Kp much larger than Kc. However, Kc itself also changes with temperature, so the net effect depends on the reaction's enthalpy.
Thermodynamically, Kp is defined using activities (P/P°) and is dimensionless. In practice, if standard state P° = 1 atm, Kp carries units of atmΔn when computed from raw pressures.
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