NITRIC ACID AND BARIUM DIOXIDE. 179 



sensitive to ^ of a degree, gives that of the acid, the latter being 

 kept as near as possible to that of the calorimeter. The cor- 

 rections for cooling are made according to the ordinary processes. 1 

 It is with the aid of all these data that the quantity, Q, is 

 calculated. Further, a weight of anhydrous barium dioxide 

 equivalent to the above weight of nitric oxide absorbed (3Ba0 2 

 for 2NO) is dissolved in the excess of dilute hydrochloric 

 acid, which liberates Q L . The concentrated nitric acid, which 

 has dissolved the nitric oxide, is then mixed in the calorimeter 

 with the hydrochloric solution of barium dioxide. The whole is 

 thus brought to the final state of dilute nitric acid and dilute 

 barium chloride, liberating a measured quantity of heat equal 

 to Q 2 . Lastly is dissolved the same weight of the same pure 

 nitric acid as used at the outset, in the same volume of dilute 

 hydrochloric acid, which liberates a quantity of heat, P. 



As a check, a weighed quantity of barium dioxide is added to 

 the solution, which should, and in fact does, produce the same 

 quantity of heat as if it were dissolved in a solution containing 

 only hydrochloric acid ; which proves that the nitric acid 

 employed is very free from nitrogen trioxide. 



This being established we have all the data for the calcula- 

 tion. Let the initial system be 3 ; 2NO; 3BaO ; m(2HN0 3 

 + fiAq) ; 6 HC1 dilute ; these bodies being taken separately ; 

 and let the final system be 3(BaCl 2 + H 2 dissolved + (m -f 1) 

 2HN0 3 dilute). We can pass from one to the other according 

 to the two following cycles : 



FIRST CYCLE. 



3BaO + 30 = 3Ba0 2 anhydrous liberates + 18-0 Cal. 



6HC1 dilute + 3Ba0 2 .. > ... ... Q l 



Reaction of NO on the concentrated nitric acid ... Q 



Reaction between the two mixtures Q 2 



Sum ... + 18-0 Cal. + Q + Qi + Q 2 



SECOND CYCLE. 



3BaO anhydrous + 6HC1 dilute liberates + 27'8 x 3 



(according to the author's experiments) or ... + 83' 4 Cal. 



2NO + 3 + water = 2HN0 3 dilute x 



m(HN0 3 + wAq) + water 



Sum + 83-4 Cal. + x + P 



whence we deduce the value of x : 



+ 18-0 Cal. + Q + Q! + Q 2 = 4- 834 Cal. + x + P. 



The experiments gave x = 344 Cal., a value which is 

 slightly too small. 

 4. The third method is based on the use of nitric peroxide. 



1 " Essai de Me*canique Chimique," torn. i. p. 208. 



N 2 



