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

<p><strong>4.27&nbsp;</strong>Let, initial concentration be [R]_°</p><p>Concentration at 90% completion be (100-90)/100)* [R]_°</p><p>? Concentration at 90% be 0.1 [R]_°</p><p>Concentration at 99% completion be (100-99)/100)* [R]_°? Concentration at 99% be 0.01 [R]_°</p><p>_{we know time, t= 2.303/K log R0 / R}</p><p>Time taken for 90% completion is&nbsp;</p><p>T₉₀ = 2.303 / K log R₀/ 0.1 R₀&nbsp;</p><p>T₉₀ = 2.303 / K log 1_&nbsp;/ 0.1</p><p>T₉₀ = 2.303 / K log 10_&nbsp;/ 1</p><p>T₉₀ = 2.303 / K</p><p>Time taken for 99% completion is</p><p>T₉₉&nbsp;= 2.303 / K log R₀/ 0.01 R₀&nbsp;</p><p>T₉₉&nbsp;= 2.303 / K log 1_&nbsp;/ 0.01</p><p>T₉₉ = 2.303 / K log 100_&nbsp;/ 1</p><p>T₉₉&nbsp;= 2 X 2.303 / K</p><p>T₉₉&nbsp;= 2 T₉₀</p><p>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.</p>

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