Question

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show that in case of a first order reaction time required for 99% completion a reaction is twice the time required for complexion of 90% of reaction​

Answer 1

Answer:

For a first order reaction, the time required for 99% completionis

t1 = 2.303/k Log 100/100-99

t1 = 2.303/k Log 100/100-99= 2.303/k Log 100

t1 = 2.303/k Log 100/100-99= 2.303/k Log 100= 2x 2.303/k

For a first order reaction, the time required for 90% completion is

t2 = 2.303/k Log 100/100-90

t2 = 2.303/k Log 100/100-90= 2.303/k Log 10

t2 = 2.303/k Log 100/100-90= 2.303/k Log 10= 2.303/k

Therefore, t1 = 2t2

Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.

Hope it helps!!