FERMENTS, OR ENZYMES 75 



4. The velocity constant in the case of inorganic catalysts remains un- 

 changed throughout the reaction, whereas in the case of enzymes it 'be- 

 comes either less or greater as the process proceeds. When a substance is 

 changed by catalytic action, it is, of course, constantly being diminished 

 in concentration so that less and less of it remains to be acted on. This 

 implies that there are fewer molecules present for the same amount of 

 catalyst to act on and consequently that the amount changed in a unit 

 of time is progressively less. At any moment, therefore, the rate of 

 catalysis will be proportional to the amount of substance (substrate) 

 left. To understand this we must refer back to what we have learned about 

 mass action. If we suppose that two substances A and B react to form 

 two other substances C and D, then, by the law of mass action, the reac- 

 tion will not go on to completion but will stop when a certain equilibrium 

 is reached. The reaction can be represented by the equation 

 A + B + C + D, which means that it proceeds at a rate proportional to 

 the reacting molecules. In some cases this reaction goes on until either 

 A or B has practically disappeared (that is, the equilibrium point is very 

 near the right of the equation), as is the case in the inversion of cane 

 sugar: 



C 12 H 22 1X + H 2 = C 6 H 12 6 + C 6 H 12 6 



Taking place as it does in an excess of water, and there being very 

 little tendency for the reaction to go in the opposite direction (cf. re- 

 versible action page 25), the only thing which will influence its velocity 

 is the concentration of cane sugar; in other words, the velocity of the 

 reaction at any moment will depend on the concentration of the cane 

 sugar still left undecomposed. This can be determined by means of 

 an equation.* 



The value of such an equation is that it gives us a figure K, represent- 

 ing the amount of inversion that would occur in each unit of time if the 

 cane sugar were kept in constant concentration. When, for example, 

 it is stated that K for a particular strength of acid acting on cane sugar 



*If x be the amount of sugar inverted in time t, C, the concentration of the sugar not yet in- 

 verted, and if we use a figure called a constant (K) to express the fundamental rate of the reaction 



(which will therefore be different for different reactions), then = KC. But C can not be the same 

 at any two consecutive periods of time, because the reaction is going on continuously. This renders 



it necessary to use the notation of the differential calculus, and we have ~ = KC. The sign 5 



ot 



indicates that the reaction is a constantly changing one so that 5x and 8t represent such infinitely 

 small amounts that they can not be measured. By methods of integration, however, it can be shown 

 that the above equation may be written: 



K= * log. nat. -i-, 



I 2 i v-2 



thus permitting us to find the value of K (d C 2 being the concentrations of cane sugar at the 



Any two determinations during the course of the reaction can be used for calculating K. These 

 equations apply only to cases in which but one substance is changing (monomolecular reaction). 

 When two substances are involved, the equation is more complicated. 



