INJUEY AND RECOVERY 103 



On replacing the tissue in sea water, therefore, we 

 start with M = 70.69 and A =2,027.96, but this value of 

 A is at once augmented by the conversion of S into A. 

 In order to find the amount of this augmentation we 

 must know the value of S. 



During exposure to NaCl the reaction R — >8 — >T 

 occurs. The value of 8 may be easily calculated by 

 employing formula (1) and substituting the appropriate 

 constants. We thus obtain 



The value of R at the start in sea water is taken as 1,041.77 

 and that of 8 as 2.7. In the solution of NaCl the values of 

 Kj^ (thevelocity constant of the reaction E — >8) and jST^ 

 (the velocity constant of the reaction 8 — >■ T) are taken 

 as 0.04998 and 0.02856 respectively (see Table V). Hence 

 the value^' of 8 at the end of 15.9 minutes is 447.26. 

 When the tissue is replaced in sea water 8 is rapidly 

 converted into A so that the total value of the latter 

 becomes 447.26 + 2,027.96 = 2,475.22. 



On replacing the tissue in sea water A — 2,475.22 and 

 M = 70.69. The resistance due to A and M after any 

 given time T in sea water is obtained by modifying 

 formula (1) which becomes 



Resistance = 2,475.22! _^ j | e~^^ Tr_—KmTr J 



+70.69 (^e~^^^*j +10 (4) 



in which T ^ denotes the time which has elapsed after 



" In general the greater the rise in recovery the greater the value of S, 

 while the greater the fall the leas the value of S. 



