30 CHEMICAL TRANSFORMATIONS 



solvent for the disaccharide molecule, and hence has no osmotic 

 pressure before, during, or after the reaction. Accordingly, it 

 does not come into the equilibrium equation, for no change of 

 osmotic energy takes place in connection with it, and hence such 

 reaction is practically one in which one molecule forms two or 

 vice versd. 



Nearly all the reactions of digestion and metabolism also 

 belong to this type of reaction, with the slight modification that 

 the number of simpler molecules formed is often three or more 

 instead of two, and all the conclusions deduced for the simpler 

 splitting into two molecules, or vice versd, can be applied to such 

 reactions with but slight modification. For example, the hydro- 

 lyses of neutral fats or triglycerides into fatty acids and of starches 

 into sugars belong to this category. 1 



When a disaccharide is hydrolysed (for example, maltose into 

 two molecules of dextrose), as in the equation 



C 12 H 22 O n + H 2 ^ C 6 H 12 6 + C 6 H 12 6 , 



P 2 

 the equation of equilibrium is, as deduced above, ~ = P O . e RT , 



"A 



where P B is the osmotic pressure of the hexose at equili- 

 brium, and P A that of the disaccharide, this may be written 



p2 



= K, or P| = K P A . Expressed in words, there is a constant 



"A 



ratio between the osmotic pressure of the disaccharide and the 



square of the osmotic pressure of the hexose. As a result, the 

 point of equilibrium is not fixed and independent of the concentra- 

 tion of the reacting substances as in the types of reaction pre- 

 viously discussed, but varies with the initial concentration of the 

 solution, in such a way, that in dilute solutions, the equilibrium 

 lies where the whole of the disaccharide is dissociated into its 

 constituent hexoses, while as the solution becomes more con- 

 centrated, the equilibrium point lies more and more towards com- 

 plete conversion into disaccharide. This is obvious, for if the 

 concentration of disaccharide in the solution be doubled so that P A 

 becomes 2P A , then P B only becomes ^/ 2 . P B , that is, with increas- 



1 Vide infra. 



