For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained, a) Shat that it follows pseudo first order reaction, as the concentration of water remains constant. b) Calcualte the average rate of reaction between the time interval 30 to 60 seconds.

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) Shat that it follows pseudo first order reaction, as the concentration of water remains constant. b) Calcualte the average rate of reaction between the time interval 30 to 60 seconds.
Answer» Solution :a) The chemical equation for the hydrolysis reaction is: `CH_(3)COOCH_(3) + H_(2)O OVERSET(H^(+))tounderset("Excess")(CH_(3)COOH) + CH_(3)OH` Let us substitute the values in the rate equation for first order reaction and calculate the value of rate constant. Case I. `k=2.303/tlog(a/(a-x))=(2.303)/(30s)LOG([0.60M])/([0.30M])=2.303/(30S)log2=2.303/(30s) xx 0.3010 = 0.0231s^(-1)` Case II. `k= 2.303/tlog(a/(a-x)) = (2.303)/(60S)log([0.60]M)/([0.15M])=2.303/(60 MIN) log4=(2.303)/(60s) xx 0.6021 = 0.0231s^(-1)` As the value of k in both the case in the same, this show that the reaction is pseudo first order reaction. b) Average reaction between time interval 30-60 sec is: Average rate = `(0.15-0.30M)/(60-30)s = (0.15M)/(30s) = 0.15 mol L^(-1)s^(-1)`