Question

prove that for a first order reaction the time required for 99% completion of the reaction is twice the time required for the completion of 90% Of the reaction ​

Answer 1

Given that

Order of reaction = first

Let Initial amount = 100 g

Amount in % after 90 % completion = 100 - 90 = 10%

Amount in g [R]= 10% of 100g = 10g

Amount in % after 99 % completion = 100 - 99 = 1%

Amount in g [R]= 1% of 100g = 1g

Use the formula of first order reaction

We observe that t2 = 2t1

Hence, for a first order reaction, time required for 99% completion is twice the time required for the completion of 90% of reaction