Reaction before Complete Equilibrium. 475 



it inicontact with water, and you will never get a saturated 

 solution in the time of a very few minutes. A further 

 question is, Why should C of the solid not change during the 

 reaction with temperature, while C T of the solution does change 

 with the same? But let us pursue the same conception 

 further. When a solid transforms into a liquid, or liquid 

 into a solid, or solid into a solid and liquid into a liquid, e. g., 

 in a system of ice and water, we might (as in case of solution 

 of solid salt) according to this conception have for both 

 opposite reactions separately the same equations 



(JW.Co(2 T + K) and (J)'=c'a(2 7 + K), 



and sincej according to the above conception, above, below, 

 and at equilibrium both opposite reactions take place simul- 

 taneously, the total velocity 



dt _ /dT\ _/^Y 

 dr \dr) \dr/ 



ought to be 



=6'a(2 T +K)- c / a(2 T +K), 



i. e.. not only at the point of equilibrium, but also above 

 or below the point of equilibrium, no separation or solution 



of the solid (in case of the system water ^ ice, no separa- 

 tion of ice below zero, and no melting of ice above zero) 

 would be possible. Let us now assume that the velocity- 

 constants of the two opposite reactions (of ice melting and of 

 ice separation) are different, which, however, does not follow 

 from the decomposition of equation (a'), still we ought to get 



in this case with the above conception in case water ^ ice for 

 the total velocity above, below, and at equilibrium, 



2- = Mo (tr + K) - c"C ($ T + K) = (K' - K") {$r + K) = K'"(2 T + K) 



instead of the actually found equation 



Xow let us assume for a while that since the velocity-constants 

 change with temperature, we get for ice ^ water the equation 



( lL=: c{ t -t)(t-t ov + K), or ^=ofo-0(2 T + K), 



because K'" is to be put =c(f — t). Should we assume that 



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