For a first order reaction, the time taken for 87.5 % completion is t₁ and the time taken for x % completion is t₂ . If the ratio t₁ t₂ is 1.5 , the value of x is :

Step-by-step solution

For a first order reaction, the amount of reactant remaining after n half-lives is given by C_t = C₀ 2^n . For 87.5 % completion, the amount of reactant remaining is 100 % - 87.5 % = 12.5 % , which is 1 8 of the initial concentration. Since 1 8 = ( 1 2 )^3 , it takes 3 half-lives. Thus, t₁ = 3 t_ 1/2 . Given t₁ t₂ = 1.5 = 3 2 , we have: 3 t_ 1/2 t₂ = 3 2 t₂ = 2 t_ 1/2 In 2 half-lives, the amount of reactant remaining is ( 1 2 )^2 = 1 4 , which is 25 % . Therefore, the percentage of reaction completed is 100 % - 25