For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:
t/s 0 10 20
[CH3COOCH3]/mol L-1 0.10 0.05 0.025
(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 intervals 10 to 20 seconds. (Given: log 2 = 0.3010, log 4 = 0.6021)
OR
(a) For a reaction A + B → P, the rate is given by
Rate = k [A][B]2
(i) How is the rate of reaction affected 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.
(a) [A]1= 0.1 mol L-1 at time = 0 seconds.
[A]2 = 0.05 mol L-1 at time = 10 seconds.
The rate constant (K) for a first order reaction is given as,
K = 
K = 
K = 0.069 sec-1
Now, at time = 20 s, [A]2 = 0.025 mol L-1
At time = 10 s, [A]1= 0.05 mol L-1
K = 
K = 
K = 0.069 sec-1
Since, the value of rate constant at any time is same i.e. 0.069sec-1. Therefore, the given reaction follows pseudo first order reaction.
(b) Average rate of reaction between the time intervals 10 to 20 seconds is
= 
= 
= 0.0025 mol L-1 sec-1
OR
(a) For a reaction A + B → P, the rate is given by
Rate = k [A][B]2
(i) If the concentration of B is doubled,
New rate of reaction, rate’ = k[A][2B]2
= 4k[A][B]
Rate’ = 4
rate
The rate of reaction will become 4 times if the concentration of B is doubled.
(ii) Overall order of reaction if A is present in large excess will be 2. As A is present in large excess therefore the reaction will be independent of the concentration of A and will depend only on the concentration of B.
(b) Given, reaction is of 1st order,
t1/2 = 30 mins
K = 
= 
= 0.0231 min-1
[R] at 90% completion = [R0] - 
[R0] = 
Time required for 90% completion of this reaction,
K = 
t = 
= 
= 99.69 min
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