80 MASS ACTION 



As has been shown by Guldberg and Waage in 1867, the velocity 

 with which two substances react depends upon 



(1) The product of their active masses, i.e., the number of equiva- 

 lents of each of the substances present in the unit volume. 



(%) Another factor, depending upon the nature of the two sub- 

 stances, the temperature and other physical conditions of the reaction. 



Thus, if m and m be the number of gramme equivalents of the 

 two substances A and B present in the unit volume of the solution, 

 the speed of the reaction would be measured by k x m x m where k 

 is a constant, It is obvious that, other things being uniform, the 

 rapidity of a reaction will depend upon the frequency of collisions be- 

 tween substances which are to react. If the number of equivalents 

 m of one substance be doubled, the number of collisions per unit time 

 between the molecules of A with those of B will be doubled. So, too, 

 if m' be also doubled, the number of collisions per unit time will be 

 again doubled, so that the speed of the reaction will now be repre- 

 sented by A; x I 2m x 2m', i.e., four times as great as before. 



If one of the substances is insoluble in water (or so little soluble 

 that there is always some of it present in the* solid state) its active 

 mass is constant and the speed of the reaction then varies only as the 

 active mass of the soluble substance varies. 



Now when two substances react they, as a rule, form two other 

 substances, which may react upon each other, re-forming the original 

 two. This occurs in all so-called reversible reactions. In such cases 

 equilibrium is attained when the speeds of the reaction in the two di- 

 rections are equal. Consider the case to which allusion has already 

 been made and let a represent the number of equivalents of K^COg per 

 unit volume, b the active mass of BaSO 4 (constant because insoluble), 

 c the number of equivalents of K 2 S0 4 , and d the active mass of 

 BaC0 3 (again constant). Then the condition for equilibrium would 

 be 



k (a - x) (b - x) = k' (c + x) (d + x) 



but since b and d are constant they are not altered by the subtraction 

 or addition of x, which represents the number of equivalents of K 2 CO a 

 (or BaS0 4 ), which undergoes the change. 

 Hence the equation may be written 



k (a - x) (b) = k' (c + x) (d) 

 k (c + x) d 

 k' ~ (a - x) b 



or if single equivalents were taken of K 2 C0 3 and BaS0 4 



k(l-x) (b) = k' x d 

 k x d 



(1 -x)b 

 and since r i g constant, we see that equilibrium is reached when 



x 



attains a certain value. 

 1 - x 



