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

⇒

⇒

⇒ t₉₉ = 2t₉₀

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.