The time needed for completion of 80 % is y times the half-life period of a first order reaction. What is the value of y ?

Step-by-step solution

For a first order reaction, the rate constant k is given by k = 2.303 t ( [A]₀ [A]_t ) . The half-life period t_ 1/2 is given by t_ 1/2 = 0.693 k . For 80 % completion, the amount remaining [A]_t = [A]₀ - 0.80[A]₀ = 0.20[A]₀ . The time taken for 80 % completion is t_ 80 % = 2.303 k ( [A]₀ 0.20[A]₀ ) = 2.303 k (5) . Using (5) 0.699 , we have t_ 80 % = 2.303 0.699 k 1.609 k . Given t_ 80 % = y t_ 1/2 , we substitute the expressions: 1.609 k = y 0.693 k . y = 1.609 0.693 2.32 . Answer: 2.32