HYDROGEN INTO ATOMS 211 



Similarly, from (2) we obtain, since Mg = 2, and Mi = i 



In order to find the value of K^ we place 00 = which is evidently 

 the condition for equilibrium. Equations 4 and 5 then reduce to 



did ' 



Whence by (9) 



JU2 = 



(13) 



-(s)- 



«2 



K,= (^).^. (14) 



It should be noted that 6 and 9x cancel out in the derivation of this 

 equation. 



Since the factors a and v are not functions of the pressure, Equation 

 14 expresses the law of mass action, which in this case follows auto- 

 matically as a consequence of the mechanism we have assumed. From 

 Equations 4 and 5 we can determine the values of the two factors in the 

 second member of (14). Thus 



(15) 

 (16) 



("■ + -e) 



2 



^^ = 7 ^- (^7) 



The value of 6 may be found by solving (4) and (6) as simultaneous 

 equations. We thus obtain 



e = "4: — = -— r — -T' (18) 



The two Equations 17 and 18 give us a complete solution of our prob- 

 lem. They enable us to calculate the dissociation constant K^ in terms of 

 ttij ct2, Vj and experimentally determined values of co. 



