Relationship between the Equilibrium Constants $K_p$ and $K_c$

For a purely gaseous equilibrium, concentration is not the only convenient way to describe how much

of each species is present — partial pressure works equally well, since for an ideal gas pressure is

directly proportional to concentration at a fixed temperature. This gives rise to a second

equilibrium constant, $K_p$, defined identically to $K_c$ but with partial pressures (usually in atm)

in place of molar concentrations. For the ammonia equilibrium,

Because $K_p$ and $K_c$ describe the same physical equilibrium, they must be related to one another,

and that relationship follows directly from the ideal gas equation, $PV = nRT$, which can be

rearranged to give the concentration of any gas as $\dfrac{n}{V} = \dfrac{P}{RT}$, i.e.

$[\text{gas}] = P/RT$. Substituting this relation for every species in the $K_c$ expression and

collecting the powers of $RT$ gives the general result

where $\Delta n_g$ is the change in the number of moles of gas on going from reactants to products

(moles of gaseous products minus moles of gaseous reactants), and $R$ must be expressed in units

consistent with pressure in atm, i.e. $R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$.

Three cases are worth noting explicitly. If $\Delta n_g = 0$ (equal moles of gas on both sides, as in

$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$), then $(RT)^0 = 1$ and $K_p = K_c$

exactly, with no unit conversion needed at all. If $\Delta n_g > 0$ (more moles of gaseous product

than reactant), $K_p$ is larger than $K_c$ by a positive power of $RT$. If $\Delta n_g < 0$, as in

ammonia synthesis where $\Delta n_g = 2 - 4 = -2$, then $K_p$ is smaller than $K_c$, since

$(RT)^{-2}$ is a fraction less than 1 (for $T$ well above $1/R$).

This relationship is essential in practice because chemists frequently measure equilibrium data as …