HYDROGEN INTO ATOMS 171 



We assume that tlie coefficients a and /? are constant; that is, that they 

 are independent of ;;/ and n, although they may vary with the temperature. 



The energy (watts) carried away from the wire (per cm.) hecause of 

 the dissociation, we have called Wn- If ^i be the heat (calories) necessary 

 to dissociate i g. of hydrogen, then we have 



W/) = 4.19 Jt (/ f/i (a2iii2 — (^2^2) ■ (16) 



Here d is the diameter of the wire. The quantity in parenthesis is the 

 difference between the rates at which hydrogen molecules are absorbed and 

 are given up by the wire and therefore is equal to the rate at which hy- 

 drogen is dissociated. 



If we place 



then (16) becomes 



oi = \W / {4.19 n d qi) (17) 



0) = a2"/2 — /52W2 (18) 



Combining this with (15), we obtain 



0) = /?i«i - ai7;/i (19) 



whence 



i/5iWi = ai;;;i + w ^^20) 



Now we have assumed that equilibrium exists between the hydrogen 

 atoms and molecules absorbed by the wire. According to the law of mass 

 action, the concentration of the hydrogen molecules in the metal is pro- 

 portional to the square of the concentration of the atoms. But no and wi 

 must be proportional to these concentrations. From Equation 20 we thus 

 obtain 



(ttiWi +<«)^ = A(a2m2 — (o). (21) 



Here A is a constant which is proportional to the dissociation constant 

 of hydrogen dissolved in the metal. We may now obtain a relation between 

 this quantity A and the true dissociation constant K of gaseous hydrogen 

 (outside the wire). 



We have 



K = (p',r-/P'2 (22) 



where />'i and p^2 are, respectively, the partial pressures of hydrogen atoms 

 and molecules corresponding to equilibrium in the gaseous phase. 



By Equation 5 (Part I) we have from (22) by placing Mi = i and 

 Mo = 2 



K = V^^i^r; (my/m, (23) 



where T2 is the temperature of the filament. 



