Kc, Kp and the unitless K

You will meet three equilibrium constants. At A-level, $K_\mathrm{c}$ and $K_\mathrm{p}$ carry units. At university and in olympiads, $K$ has no units, because it goes inside a logarithm in $\Delta G^\circ = -RT\ln K$. They are the same idea, written with different conventions.

Kc: concentrations, with units

For $a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$, the concentration equilibrium constant uses equilibrium concentrations:

Its units are $(\text{mol dm}^{-3})^{\Delta n}$, where $\Delta n = (c + d) - (a + b)$. For $\mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3}$, $\Delta n = -2$, so $K_\mathrm{c}$ is in mol⁻² dm⁶. When $\Delta n = 0$ the units cancel, and so does the volume.

Kp: partial pressures, with units

For gases, use partial pressures instead. The partial pressure of a gas is its mole fraction times the total pressure, $p_\mathrm{A} = x_\mathrm{A}\,p_\text{total}$, and the partial pressures add up to the total. Then

with units of $(\text{pressure})^{\Delta n}$. The number depends on the unit chosen: for $\Delta n = -2$, $K_\mathrm{p}$ in Pa⁻² is $10^6$ times smaller than in kPa⁻². Only gases appear in $K_\mathrm{p}$.

From Kc to Kp

For an ideal gas, $p = \frac{n}{V}RT = cRT$, so every partial pressure is its concentration times $RT$. Substituting into $K_\mathrm{p}$, each factor of $RT$ appears once per mole of gas:

Use SI units throughout: concentrations in mol m⁻³ and pressures in Pa (so $R = 8.314$ J K⁻¹ mol⁻¹ links them directly).

The unitless K: activities

The thermodynamic $K$ replaces each concentration or pressure with an activity, a pure number that says how far a substance is from its standard state:

So $K = \prod_i a_i^{\nu_i}$ has no units, and $\Delta G^\circ = -RT\ln K$ makes sense: you can only take the logarithm of a pure number. For real, non-ideal mixtures each activity also carries an activity coefficient, $a = \gamma c/c^\circ$; olympiad problems almost always take $\gamma = 1$.

Numerically, $K$ equals $K_\mathrm{c}$ with concentrations in mol dm⁻³, and $K$ equals $K_\mathrm{p}$ with pressures in bar. Dividing the ideal-gas link by the standard states gives the unitless version of the conversion, the Kp from Kc relation:

Which one to use