For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained : (a) Show that it follows pseudo first order reaction, as the concentration of water remains constant. (b) Calculate the average rate of reaction between the time interval 10 to 20 seconds. (Given : log 2 = 0.3010, log 4 = 0.6021)

For the hydrolysis of methyl acetate in aqueous solution, the followin

1.	For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained : (a) Show that it follows pseudo first order reaction, as the concentration of water remains constant. (b) Calculate the average rate of reaction between the time interval 10 to 20 seconds. (Given : log 2 = 0.3010, log 4 = 0.6021)
Answer» SOLUTION :(a) `K'(2.303)/(t)log""([A_(0)])/([A])"in which "[A_(0)]=0.10` `t=10 S,` `K'=(2.303)/(10)log ""([0.10])/([0.05)=(2.303)/(10)log (2)` `=(2.303)/(10)xx0.3010=0.06932=6.932xx10^(-2)S^(-1)` `t=20 S, K'=(2.303)/(20)log""([0.10])/([0.025])` `=(2.303)/(20)log(2)^(2)=(2.303)/(20)xx0.3010xx2=6.93xx10^(-2)S^(-1)` (b) Average rate during the interval 10-20 seconds `=(C_(2)-C_(1))/(t_(2)-t_(1))=((0.025-0.05))/(20-10)` `=(0.025)/(10)=0.0025=2.5xx10^(-3) MOL L^(-1) S^(-1)`