﻿508 Dr. T. Ewan on the Rate of Oxidation 



calculated on the assumption that the rate of the reaction is 

 proportional to the partial pressure of the oxygen. Or 



-t= K 'p « 



On integration this becomes 



— log p = K'£ + const. 



We have p=po, when £ = 0, p being the pressure of the 

 oxygen at the beginning of the experiment. 

 We have, therefore, 



const. =— logp , 

 and 



K 



Po 



t s p 



(la) 



Table I. — Phosphorus and Moist Air. 



Temperature = 20°*2 to 20°*4. 



Pressure of aqueous vapour . =17*8 millim. 

 ,, phosphorus-vapour =0*113 „ 



Time, in 

 minutes, 



Total 



Partial 



pressure of 



Oxygen 







from the 



beginning. 



t. 



pressure. 

 P. 



K'. 



K. 







773-1 



1578 







25 



750-6 



135-3 



•00267 



42-0 



50 



729-7 



1140 



•00282 



43-1 



75 



7143 



99-0 



•00271 



40-1 



100 



697-4 



82-1 



•00284 



42-3 



130 



682-2 



669 



•00286 



42-1 



Temperature =20°*5 to 20°'68. 



Pressure of aqueous vapour . =17*9 millim. 

 phosphorus-vapour =0*113 „ 







7646 



156-0 







25 



749-5 



1409 



•00177 



27-2 



50 



732-9 



124-3 



•00197 



28-2 



70 



723-2 



1146 



•00191 



29-0 



100 



7071 



98-5 



•00200 



29-8 



130 



693-5 



84-9 



•00203 



300 



171 



677-4 



68-8 



•00208 



30-3 



201 



6674 



58-8 



•00211 



30-4 



246 



653-2 



44-6 



•00221 



31-4 



The vapour-pressure of phosphorus was determined by 

 Joubert, loc. cit. 



