IN LIVING MATTER 37 



in the change from the one phase to the other, and the absolute 

 temperature of the reaction ; for changes in either of these affect 

 the value of the constant K. 



The effects of changes in C and T upon the point of equilibrium 

 can best be understood by following out the energy changes as the 

 substances pass in the reaction from the one condition to the other. 

 The matter can best be understood by graphic illustration as in the 

 diagram on page 36, in which are represented the energy changes 

 at each instant as the substance passes from the one form to the 

 other according to the simplest type of equation, that in which 



The base line XX' represents the zero line of energy exchange, ordi- 

 nates above the base line representing quantities of energy set free by 

 the reaction at any stage, and ordinates below, energy required in order 

 that the reaction may proceed. At X the substance is all in the form 

 A and at X' in the form B, at intermediate points fractions are in the 

 two forms proportionate to the distances from X and X'. The 

 horizontal line above or below XX' represents by its 'height above or 

 below XX' the amount of chemical energy set free (positive when 

 above XX' and negative when below) when a grm. molecule passes 

 from the form A to the form B. The curved line represents the 

 osmotic energy set free at each point in the reaction when a grm. 

 molecule changes form at the particular osmotic pressures present at 



that point. The height of this line above or below XX' is given 



p 

 numerically by the expression RT log p 6 , and the sign is reversed 



in plotting it so that at each stage the difference in the heights 

 above XX' of the straight line and the curved line give the value of 



H, the energy set free at that stage in the reaction, for a grm. 



p 

 molecule changed, according to the equation H = C - RT log ^~ . 



^A 



Tracing now the value of the amount of energy set free in the 



reaction at each stage, we have at X where P B = 0, and therefore 



p 

 log p B = log = - oo . Therefore at X, and points close to it, 



p 

 - RT log p 2 has a large positive value, which is to be added to the 



constant C in order to give the energy set free. Hence no value of 

 C positive or negative can cause the equilibrium point to lie quite at 



X. As the reaction proceeds, however, and more and more of B is 



p 

 present, P B rises in value, and the positive value of - RT log p 2 



