If (k_d) is the desorption rate constant then the mean desorption time (t_d) for the adsorbed molecule will be . Correspondingly, if the adsorption rate constant is (k_a), then the mean adsorption time for a free molecule in the mobile phase will be .

Consider a peak moving down a column. During this migration process, adsorption and desorption steps will constantly and frequently occur and each occurrence will be a random event. Now a desorption step will be a random movement forward as it releases a molecule into the mobile phase. Conversely, an adsorption step will be a random movementbackward, as it is a period of immobility for the molecule while it resides in the stationary phase and the rest of the zone moves forward. The total number of random steps, taken as the solute mean position moves a distance  along the column, is the number of forward steps plus the number of backward steps. Now the distribution of the solute is dynamic and is also an equilibrium system, consequently, each desorption step must be followed by an adsorption step. It follows that the total number of steps will be twice the number of adsorption steps that take place in the migration period.