1.

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

Answer»

Solution :For first order REACTION, `t=(2.303)/(k)"log"(a)/(a-x)`
At 99% COMPLETION`x=99%` of a i.e., 0.99a
`therefore t(99%)=(2.303)/(k)"log"(a)/(a-0.99a)=(2.303)/(k) log 10^(2)= 2XX(2.303)/(k)`
At 90% completion MEANS that `x=90%` of a = 0.90 a
`therefore t(90%) =(2.303)/(k)"log"(1)/(a-0.90a)=(2.303)/(k)log 10=(2.303)/(k)`
`therefore (t(99%))/(t(90%))=((2xx2.303)/(k))//((2.303)/(k))=2.`


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