For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction, between the time interval30 and 60 seconds. (given: log 2= 0.3010, log 4 = 0.6021)
OR
(a) For a reaction
, the rate is given by Rate ![]()
(i) How is the rate of reaction effected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. (log 2 =0.3010)
(i) For a first order reaction half-time of a reaction is independent of concentration of the reacting species. Half-time of a first order is given by
. So from the equation we know that reaction half-time is dependent on rate constant(k). Rate constant(k) of a first order reaction is given by
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The reaction should be of first order with respect to methyl acetate, when concentration of water is constant to be a pseudo first order reaction. So, let us check the rate constant at different time intervals, if it is constant irrespective of the concentration then is a pseudo first order reaction. Substituting the time and concentration values given in the table in the rate constant equation:

Since the rate constant is independent of the concentration of the ester, hydrolysis of esters in excess of water is indeed a pseudo first order reaction.
(ii)
M/s
OR
(a)
(i)
rate is directly proportional to concentration of B and the reaction is of second order with respect to B. So when concentration of B is doubled rate becomes four times.![]()
(ii) When A is present in large excess, the reaction becomes independent of the concentration of A, so the overall order of the reaction will be 2.
(b) ![]()

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