Equilibrium Constant Calculator (Kc and Kp)
For aA + bB ⇌ cC + dD, enter the coefficients and concentrations. Set a coefficient to 0 for a species that is not in the reaction. Kp uses pressures in atm.
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Formula
For aA + bB ⇌ cC + dD:
Kc = [C]^c [D]^d / ([A]^a [B]^b)with equilibrium concentrations in mol/LΔn = (c + d) - (a + b)for gasesKp = Kc × (R × T)^Δn, withR = 0.0820573661 L atm/(mol K)andTin kelvin- The reaction quotient
Qhas the same form asKcbut uses the current concentrations:Q < Kshifts forward,Q > Kshifts in reverse,Q = Kis at equilibrium.
Worked example
N2 + 3 H2 ⇌ 2 NH3 with [N2] = 0.5, [H2] = 0.6, [NH3] = 0.2 mol/L at 298.15 K:
Kc = 0.2² / (0.5 × 0.6³) = 0.04 / 0.108 =0.37037Δn = 2 - (1 + 3) = -2RT = 0.0820573661 × 298.15 = 24.4654, soKp = 0.37037 × 24.4654^-2 =6.1877 × 10^-4
Frequently asked questions
How do I calculate Kc?
Kc = [C]^c [D]^d / ([A]^a [B]^b), using equilibrium concentrations in mol/L. For N2 + 3 H2 ⇌ 2 NH3 with 0.5, 0.6 and 0.2 mol/L, Kc = 0.2² / (0.5 × 0.6³) = 0.04 / 0.108 = 0.370.
How do I convert Kc to Kp?
Kp = Kc × (RT)^Δn, with R = 0.08206 L atm/(mol K), T in kelvin and Δn = moles of gaseous products minus gaseous reactants. For Kc = 0.370, T = 298.15 K and Δn = -2, Kp = 0.370 × (24.465)^-2 = 6.19 × 10^-4.
What does the reaction quotient Q tell me?
Q has the same form as Kc but uses the concentrations right now. If Q < K the reaction proceeds forward, if Q > K it proceeds in reverse, and if Q = K it is at equilibrium. For N2O4 ⇌ 2 NO2 with Kc = 0.36, Q = 0.2² / 0.1 = 0.4 is above K, so it shifts in reverse.
Are solids and liquids included?
Pure solids and pure liquids have activity 1 and are left out of K, so give them coefficient 0. Concentrations here are treated as activities; this is the textbook approximation and Kc depends on temperature only.