Half life of a first order reaction is 900 second. If initial concentration of reactant is 0.08 , mol dm ⁻³ find concentration that remains after 35 minute ?

Step-by-step solution

Given t_ 1/2 = 900 s and t = 35 min = 35 60 = 2100 s. Number of half-lives n = t t_ 1/2 = 2100 900 = 7 3 . For a first order reaction, the concentration remaining after n half-lives is given by: [A] = [A]₀ 2^n Substituting the given values: [A] = 0.08 2^ 7/3 [A] = 0.08 4 2^ 1/3 Since 2^ 1/3 1.26 : [A] 0.08 4 1.26 = 0.08 5.04 0.0159 mol dm ⁻³ Answer: 0.0159 , mol dm ⁻³