where ‹k›_{1} is some number. Now, if the concentration of one of the components, ‹e.g.› ‹A›, should be doubled, then the chances for collision and for action between molecules of ‹A› and ‹B› will be twice as great as before and the velocity of the action will be doubled. If only one-tenth of the concentration of ‹A› (one-tenth mole) is used, the velocity will only be one-tenth as great as originally, and, in general terms, if [‹A›] moles of ‹A› are used per liter, the ‹velocity of the change will be proportional to› [‹A›], ‹and equal to› ‹k›_{1} × [‹A›]. If the concentration of the other reacting component, ‹B›, is now doubled, the chances for action are again doubled, and, in general, the velocity of the action will be proportional also to the concentration [p093] [‹B›] of the second reacting substance. For the velocity, ‹v›_{1} of the action for any concentrations, [‹A›] and [‹B›], of ‹A› and ‹B› at any moment at a given temperature, we have ‹v›_{1} = ‹k›_{1} × [‹A›] × [‹B›]. Hence, if by the symbols [‹A›] and [‹B›] the concentrations at any given moment are represented, we may say that the velocity of the formation of ‹C› and ‹D› at that moment[169] ‹is proportional to the product of the concentrations of› ‹A› ‹and› ‹B›, ‹and to some constant›, which is characteristic of the interaction of ‹A› and ‹B›. The validity of this conclusion has been fully verified by ‹experiment›.[170] The case is an instance of the ‹law of mass action, which states that in chemical changes the velocity of the action is proportional at any moment to the molecular concentrations›[171] ‹of the reacting components, and to a constant, which is characteristic of the chemical nature of the reacting components› (and of the temperature). If we start with the reversed action ‹A› + ‹B› ← ‹C› + ‹D›, the relation may be developed in the same way. Thus the two substances ‹C› and ‹D› will react upon each other, at the given temperature, with a velocity proportional to a constant, ‹k›_{2}, and, at any given moment, proportional also to their respective concentrations at that moment: ‹v›_{2} = ‹k›_{2} × [‹C›] × [‹D›]. Equilibrium will be reached when the substances ‹A› and ‹B› are formed at any moment from ‹C› and ‹D› just as rapidly as they are used up to produce ‹C› and ‹D›, and ‹vice versa›. Such is the case, [p094] when ‹the velocities of the two opposite reactions are equal to each other›. For the condition of equilibrium, then, ‹v›_{1} must be equal to ‹v›_{2} and therefore ‹k›_{1} × [‹A›] × [‹B›] = ‹k›_{2} × [‹C›] × [‹D›] or [‹A›] × [‹B›] / ([‹C›] × [‹D›]) = ‹k›_{2} / ‹k›_{1} = ‹k›_{equilibrium}.