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Show that for a first order reaction, t_(99.9%)=10t_(1//2).

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Solution :For first order reaction
`t=(2.303)/(K)log""([R]_(0))/([R])`
The TIME required for `99.9%` completion is
`t_(99.9%)=(2.303)/(K)log""(100)/(100-99.9)`
`=(2.303)/(K)log""(100)/(0.1)`
`=(2.303)/(K)log 1000`
`=(2.303)/(K)LOG10^(3)""[logm^(n)=n log m]`
`t_(99.9%)=(3xx2.303)/(K)rArr (6.909)/(K)"….(1)"`
For the first order reaction line required line required to complete `50%` is
`t_(1//2)=(0.693)/(K)"…(2)"`
`(eq(1))/(eq(2))`
`(t_(99.9%))/(t_(1//2))=(((9.909)K))/((0.909)/(K))`
`rArr (6.909)/(K)xx(K)/(0.693)`
`(t_(99.9%))/(t_(1//2))=10""rArr""t_(199.9%=10t_(1//2)`


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