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Question

(a) For a reaction A+BP,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 23.1 minutes for 50% completion. Calculate the time required for 75% completion of this reaction. (log 2 =0.301, log 3 =0.4771, log 4 =0.6021)

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Solution

(a) (i) Rate will become four times because rate [B]2.
(i) 2

(b) Time taken for completion of 50% of a reaction is half life period.
t1/2=23.1 minutes

Now, t1/2=0.693k

or k=0.693t12=0.69323.1=0.03min1

For 75% completion of reaction
x =0.75 a

t=2.303klogaax

=2.3030.03logaa0.75a

=2.3030.03loga0.25a

=2.3030.03log10025

=2.3030.03log 4

=2.3030.03×0.6021=46.2 minutes

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