Show that for a first order reaction, time required for `99.99%` of the reaction to take place is 10 times the time required for the completion of hal
Show that for a first order reaction, time required for `99.99%` of the reaction to take place is 10 times the time required for the completion of half of the reaction.
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For first order reaction:
`t=2.303/k log a/(a-x)`
Ist case, `a=100%, x=50%, (a-x)=100-50 = 50%`.
`t_(1//2)=2.303/k log2(100%)/(50%)`
`=(2.303)/k log 2= (2.303 xx 0.3010)/k=0.693/k` ...........(i)
IInd case, `a=100%, x=99.9%, (a-x) = (100-99.9)=0.1%`
`t_(99.9%) = 2.303/k log (100%)/(0.1%) = 2.303/k log(10^(3)) = 2.303/k log(10^(3))= 2.303/k xx 3`................(ii)
Dividing eqn. (ii) by eqn. (i),
`(t_(99.9%))/t_(1/2)= (2.303 xx 3)/(0.693) = 9.97 = 10`
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