VIII, A, 1 



Gibbs, Williams, Galajikian: Methyl Salicylate IV 9 



a, x, and i were determined as before, and e taken as 0.005, 

 the maximum solubility. 



While neither method is capable of great accuracy, still the 

 four values of r computed according to equation (6) confirm 

 Goldschmidt's result of 0.001 as substantially correct. 



III. THE EFFECT OF NONHOMOGENEITY ON THE KATE OF 

 SAPONIFICATION 



THE SAPONIFICATION OF METHYL BENZOATE 



For this purpose the reaction of methyl benzoate with sodium 

 hydroxide was first studied. This reaction is a bimolecular one 

 when it takes place in a homogeneous medium. When, however, 

 the hydrolysis takes place in aqueous solution in which the ester 

 is but slightly soluble, the reaction proceeds sensibly as a mono- 

 molecular one. This has already been pointed out by Kremann 14 

 in the case of ethyl benzoate and amyl acetate which he took as 

 typical representatives of the class of difficultly soluble esters. 

 He failed to point out the mathematical basis for the observa- 

 tions, contenting himself with a statement of the uselessness of 

 such observations as a means of learning the mechanism of the 

 reaction. 



The mathematical basis is very evident. A bimolecular re- 

 action is one in which the velocity is proportional to the concen- 

 trations of two variables. When one of the reacting substances 

 is sparingly soluble its concentration is limited by its solubility ; 

 that is, the amount transposed is immediately replaced if an ex- 

 cess is always present, unless the velocity of solution be less than 

 that of transformation. Then the equation for a bimolecular 



dx 

 reaction becomes -^- = K(a — x) (b — x -f x), in which a and 



b are the original concentrations, x is the quantity of each 

 transposed at the time t, and (b— x-(-x) is a constant limited 

 by the solubility of the ester. That is, one variable may be 



dx 

 regarded as disappearing, and the equation becomes Vr-= K^ 



QX 



(a — x) in which K 1 = K times the molecular solubility of the 

 free ester. Kremann's results would have shown this more 

 clearly if he had used an excess of the ester so that the concen- 

 tration would remain the same to the end of the reaction. If 

 this is done, the value for K r remains constant, as is shown in 

 the case of the methyl ester in the following experiment. 



"Monatsh. f. Chem. (1905), 26, 315. 



