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

Let, initial concentration be [R]_{°}

Concentration at 90% completion be ((100-90)/100)×[R]_{°}

∴ Concentration at 90% be 0.1[R]_{°}

Concentration at 99% completion be ((100-99)/100)× [R]_{°}∴ Concentration at 99% be 0.01[R]_{°}

We know, time

Time taken for 90% completion is

⇒

⇒

Time taken for 99% completion is

⇒

⇒

⇒ t_{99} = 2t_{90}

Hence, the time taken to complete 9% of the first order reaction is twice the time required for the completion of 90% of the reaction.

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