Equilibrium Constant Calculator

K and the standard free energy change are the same fact written two ways — a factor of ten is worth 5.7 kJ/mol.

Kc5 × 10⁻¹

Kp1.24225 × 10⁻⁴in atmospheres, at 500 °C

Ratio Kp to Kc0.00025(RT) raised to the power -2

Direction of reactionforwards, towards productsQ is 1 × 10⁻¹ against a K of 5 × 10⁻¹

Standard free energy change4.4558 kJ/molfrom -RT ln K

Worth per factor of ten in K14.8018 kJ/molat this temperature

The reaction consumes gas molecules, so raising the pressure pushes it forwards. That is the whole reason the Haber process runs at two hundred atmospheres.

The formula

Kp = Kc (RT)^dn ; dG = -RT ln K ; Q vs K sets the direction

Kc, Kp and the gas count

The exponent that converts between them is the change in the number of moles of gas, products minus reactants. Solids and liquids do not appear. When that change is zero the two constants are numerically identical at every temperature, which is worth checking before reaching for the conversion at all.

Q against K

The reaction quotient has exactly the same form as the equilibrium constant but is evaluated at whatever concentrations happen to be present. Comparing the two says which way the reaction will move: Q below K means too few products, so it runs forwards; Q above K means too many, so it runs back. Q equal to K means nothing further happens.

The constant is a free energy in disguise

K and the standard free energy change are two ways of writing the same fact. A K of one corresponds to zero free energy change; every factor of ten in K is worth about 5.7 kJ/mol at room temperature. This is why equilibrium constants span such enormous ranges from quite ordinary energies.

Balanced first, always

Every quantity on this page depends on a balanced equation. The coefficients set the molar ratio, the ratio sets the theoretical yield, and the yield is what the percentage is measured against. An unbalanced equation does not give a slightly wrong answer — it gives an answer to a different question.