A first order reaction proceeds to 90 % completion. What will be the approximate time taken for 90 % completion in relation to t_ 1 / 2 of the reaction?

Step-by-step solution

For a first order reaction, the rate constant k is given by k = 2.303 t ( [A]₀ [A]_t ) . For 90 % completion, [A]_t = [A]₀ - 0.90[A]₀ = 0.10[A]₀ . Substituting this into the rate equation, k = 2.303 t_ 90 % ( [A]₀ 0.10[A]₀ ) = 2.303 t_ 90 % (10) = 2.303 t_ 90 % . The half-life t_ 1/2 is given by t_ 1/2 = 0.693 k , which implies k = 0.693 t_ 1/2 . Equating the two expressions for k : 2.303 t_ 90 % = 0.693 t_ 1/2 t_ 90 % = 2.303 0.693 t_ 1/2 3.323 t_ 1/2 . Answer: 3.32 times of t _ 1 / 2