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Show that in a first order reaction, tim...

Show that in a first order reaction, time pequired for fompletion of `99.9%` is 10 times of half-life `(t_(1//2))` of the reaction.

Text Solution

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When reaction is completed `99.9%, [R]_(n)=[R]_(0)-0.999[R]_(0)`
`k=(2.303)/(t)log ""([R]_(0))/([R])`
`=(2.303)/(t )log ""([R]_(0))/([R]_(0)-0.999[R]_(0))=(2.303)/(t)log 10^(3)`
`t=6.909//K` ltbr For half-life of the reaction
`t_(1//2) =0.693//K`
`(t)/(t_(1//2))=(6.909)/(k)xx(k)/(0.693)=10`
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