DISSOCIATION. 9 



heat developed by the reaction is such that it increases the 

 gaseous volume if the reaction take place under constant 

 pressure, it consequently increases the pressure if kept at 

 constant volume. The effects are such that the heat liberated 

 increases the gaseous volume in a proportion greater than the 

 condensation, the latter being calculated upon the hypothesis of 

 a total combination effected at the initial temperature of the 

 system. In other words, the pressure of a gaseous system 

 cannot diminish, generally speaking, through the fact of an 

 exothermal reaction, when it takes place at constant volume, and 

 gives rise only to gaseous products. 



But dissociation being an endothermal reaction the increase of 

 gaseous volume due to that action is more than counterbalanced 

 by the diminution of volume due to the absorption of heat, and 

 consequently the pressure can never be increased by dissociation. 



3. Let us calculate these changes. 



The pressure depends on the temperature developed, and on 

 the state of condensation of the products. Let t be the tempe- 

 rature developed by the real reaction, taking place at a constant 

 volume, and supposing the whole of the heat liberated to have 

 been employed in heating the products. Let V be the sum of 

 the volumes of the gaseous bodies which form part of the initial 

 system, supposing them reduced to and 760 mm. 



At the temperature t the final system contains a certain 

 number of gaseous bodies. Further, let V x be the reduced 

 volume which these bodies would occupy if they could be 

 brought without change of state to and 760 mm. 



V 1 



The ratio of the reduced volumes ^ = expresses the con- 



V rC 



densation produced by the reaction. It is applicable to every 

 pressure and temperature according to the ordinary laws. 



An arithmetical value can easily be found for this ratio for 

 every chemical reaction of which the formulae are related to the 

 molecular volumes. For example 



1 2 

 H 2 4- = H 2 (gaseous) gives -^ = - 



i 2 

 CO + = C0 2 (gaseous) gives L = - 



K o 



Now let us calculate the pressure developed during the 

 reaction, occurring at constant volume and at the temperature 

 t ; the initial temperature being zero, and the initial pressure h. 



Admitting the laws of Mariotte and Gay-Lussac, the pressure 

 will become 



h X \(l + at) 

 a being equal to ^} s , as is known. 



