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For a first order reaction, show that ti...

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

Text Solution

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Use direct relation,
`(t_(99%))/(t_(90%))=(log((100)/(100-90)))/(log((100)/(100-90)))=(log100)/(log10)=2`
`:.t_(99%)=2t_(90%)`
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