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Show that time required to complete 99.9...

Show that time required to complete `99.9%` completion of a first order reaction is `1.5` times to `99%` completion.

Text Solution

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For `99.9%` completion, `[A]_(0)=100, [A]=0.1`
`t_(99.9%)=2.303/K "log"100/0.1=2.303/Kxx3`
Similarly `t_(99%)=2.303/K "log" 100/1=2.303/Kxx2`
`t_(99.9%)=1.5xxt_(99.9%)`
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