60% of a first order reaction was completed in 60 min, then 50% of the same reaction can be completed in: [log 4 = 0.60, log 5=0.69]

Step-by-step solution

For a first order reaction, the rate constant k is given by: k = 2.303 t ( a a-x ) Given that 60 % of the reaction is completed in 60 minutes: k = 2.303 60 ( 100 100-60 ) k = 2.303 60 ( 100 40 ) = 2.303 60 (2.5) Using the given values, (2.5) = ( 10 4 ) = 10 - 4 = 1 - 0.60 = 0.40 k = 2.303 60 0.40 The time required for 50 % completion (half-life) is: t_ 50 % = 2.303 k ( 100 100-50 ) = 2.303 k 2 Since 4 = 0.60 , we have 2 2 = 0.60 2 = 0.30 Substituting the value of k : t_ 50 % = 2.303 2.303 60 0.40 0.30 t_ 50 % = 60