The time required for 80 % of a first order reaction is " y " times the half-life period of the same 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, [A]_t = [A]₀ - 0.80[A]₀ = 0.20[A]₀ . The time t_ 80 % is t_ 80 % = 2.303 k ( [A]₀ 0.20[A]₀ ) = 2.303 k (5) . Using (5) 0.699 , t_ 80 % = 2.303 0.699 k = 1.61 k . We are given t_ 80 % = y t_ 1/2 , so y = t_ 80 % t_ 1/2 = 1.61/k 0.693/k = 1.61 0.693 2.32 . Answer: 2.32