12 



If we assume that the velocity of reaction is proportional to the 

 quantity of silver oxide and of silver, we may write the equation, 



f=Kx(:-..) (1) 



Where -r- measures the velocity, x is the fraction of oxide 



decomposed and is therefore proportional to the amount of silver. 

 1— X is the fraction remaining unchanged and is therefore pro- 

 portional to the amount of silver oxide. K is a constant. 



An inspection of the equation shows that it would be represented 

 by a rising and falling curve, with a maximum of velocity at the 

 point where just one-half the silver oxide is decomposed. In order 

 to compare the assumptions we are making with our experimental 

 results we must obtain the velocity as a function of t rather than 



X 



of X. From equation 1, by integration, />i _ = K^ + C, where In 



stands for a natural logarithm and C is the integration constant. 

 If we count t from the point of maximum velocity, then at that 



X 



point K? = 0. Also, since x=(l—x), /n— ^— ^ = 0, and therefore 



C = 0. Therefore our equation stands 



lnj^ = Kt, ovj^=e^', ovx=^^ (2) 



differentiating this equation 



dx Ke^' 



(3) 



dt (1+^'^')=' 



This is the equation sought, and in order to plot its curve on a 

 suitable scale for comparison with those of fig. 3 we must choose 

 in this case also such a unit of time as to make the maximum 



velocity equal to unity. At this maximum, then, -77— 1> Ij^^t, as 

 previously shown, 



a; =- and !-.'• = -, 

 whence by equation 1, 



1 = K y-^-, and K = 4, 



whence equation 3 Ijecomes 



dx Ae" 



dt (l+e*'y 



