Why are equilibrium constants dimensionless?

2 Answers
Apr 14, 2015

An equilibrium constant for liquid-solid solutions is defined as:

K_(eq) = ([Y]^(nu_Y)[Z]^(nu_Z))/([A]^(nu_A)[B]^(nu_B))
Where A and B are reactants, Y and Z are products, and the nu_i are the stoichiometric coefficients.

If there are the same number of moles of products as there are for reactants, the stoichiometric coefficients add up to the same number. In other words:

nu_Y + nu_Z = nu_A + nu_B

As a result, your units are:

[(M^(nu_Y)*M^(nu_Z))/(M^(nu_A)*M^(nu_B))] = [M^(nu_Y+nu_Z)/M^(nu_A+nu_B)] = [M^(nu_A+nu_B)/M^(nu_A+nu_B)] = [no units]

So this is only true if the moles of products equal the moles of reactants.

Apr 14, 2015

Double check this Truong-Son. The number of moles of reactants does not always equal the moles of product.

Consider

N_2+3H_2rarr2NH_3

The equilibrium constant is

k=([NH_3]^2)/([N_2][H_2]^3) and the units of k would be

([moldm^"-3"]^2)/([moldm^"-3"][moldm^"-3"]^3)=1/(mol^2dm^"-6")