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

For a first order reaction, show that the time required for `99%` comp

1.	For a first order reaction, show that the time required for `99%` completion is twice the time required for the completion of `90%` of reaction.
Answer» For a first order reaction , the time required for `99%` completion is `t_(1) = (2.303)/(k) "log" (100)/(100 - 99)` = `(2.303)/(k) "log" 100` = `2 xx (2.303)/(k)` For a first order reaction , the time required for `90%` completion is `t_(2) = (2.303)/(k) "log" (100)/(100 - 90)` = `(2.303)/(k) "log" 10` = `(2.303)/(k)` Therefore , `t_(1) = 2t_(2)` Hence , the time required for `99%` completion of a first order reaction is twice the time required for the completion of `90%` of the reaction .

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

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