Equilibrium constant calculator
K or Q from a reaction, Kc to Kp, combined reactions, and K from ΔG°.
Reactions to combine (leave K blank to skip one)
How do you write the equilibrium constant expression?
For aA + bB ⇌ cC + dD, K = [C]^c [D]^d ÷ ([A]^a [B]^b): products over reactants, each raised to its coefficient. Use concentrations for Kc or partial pressures for Kp. Pure solids, pure liquids and the solvent are left out, because their amounts don’t change the ratio (OpenStax Chemistry 2e, Section 13.2). Type the reaction and mark solids and liquids with (s) or (l), like CaCO3(s) = CaO(s) + CO2(g), and the calculator leaves them out and writes the expression for you.
How can I tell which way a reaction will shift?
Work out the reaction quotient Q with the same expression but the amounts you have now, and compare it with K. Q < K: the reaction runs forward (more products). Q > K: it runs in reverse. Q = K: it is at equilibrium. OpenStax Example 13.3 uses CO + H₂O ⇌ CO₂ + H₂ with Kc = 0.64: with 0.020 M of each reactant and 0.0040 M of each product Q = 0.040, so it goes forward; with [CO] = 0.011, [H₂O] = 0.0011, [CO₂] = 0.037 and [H₂] = 0.046 M, Q = 1.4 × 10² and it goes in reverse.
How do you convert Kc to Kp?
Kp = Kc × (RT)^Δn, where Δn is the moles of gas on the product side minus the moles of gas on the reactant side and T is in kelvins. R = 0.08206 L·atm/(mol·K), which is the exact SI gas constant 8.314 462 618 J/(mol·K) divided by 101.325. OpenStax Example 13.4: CS₂ + 4H₂ ⇌ CH₄ + 2H₂S has Kc = 0.28 at 900 °C, Δn = 3 − 5 = −2, so Kp = 0.28 × (0.08206 × 1173)⁻² = 3.0 × 10⁻⁵. If your Kp is in bar, switch the pressure unit: R then becomes 0.08314 L·bar/(mol·K).
What happens to K when I reverse, multiply or add reactions?
Reverse a reaction and K becomes 1/K. Multiply all the coefficients by n and K becomes Kⁿ. Add two reactions and their K values multiply (OpenStax Section 13.2). Example 13.5: for 2NH₃ + 3I₂ ⇌ N₂ + 6HI, reverse N₂ + 3H₂ ⇌ 2NH₃ (Kc = 0.50) to get 2.0, triple H₂ + I₂ ⇌ 2HI (Kc = 50) to get 50³, and multiply: 2 × 125,000 = 2.5 × 10⁵. OpenStax prints 2.4 × 10⁵ because it rounds 1.25 × 10⁵ down to 1.2 × 10⁵ before multiplying; the calculator keeps full precision.
How is the equilibrium constant related to ΔG°?
ΔG° = −RT ln K, or K = e^(−ΔG°/RT), with R = 8.314 J/(mol·K) and T in kelvins (OpenStax Section 16.4). K > 1 goes with a negative ΔG° (products favored), K < 1 with a positive ΔG°. OpenStax Example 16.13: dissolving AgCl has ΔG° = 55.7 kJ/mol at 25 °C, which gives K = 1.74 × 10⁻¹⁰ (1.745 × 10⁻¹⁰ with the full-precision R). Use the ΔG° ⇄ K tab for this; to get ΔG° from ΔH and ΔS, see the Gibbs free energy calculator. As in textbooks, K is treated as a plain number without units.