Question

A second order reaction is 50 % complete in 19 minutes. How long will it take to be 89 % complete? Do not give units in your answer. Just give the numbers Answer

Answer 1

Given, 2nd order reaction 50% complete in 19 minutes. 89% = ? (time) The rate law for a second order reaction is, -r_A = RC_A^2 integrating from C_A0 to C_A, i.e. 1/C_A - 1/C_A0 = k t C_A rightarrow conc at given time t C_A0 rightarrow initial conc k = rate constant As we know, half life of a chemical reaction tells us how much time is needed for concentration od reactant to reach half of its initial value. So have, 50% reaction is completed in 19 minutes. Therefore C_A = 1/2 C_A) => half of initial conc remains after one half life => 2/C_A0 - 1/C_A0 = kt initially C_A0 = 1 (no reaction occurs) current conc C_A = 0.5 => 1 = k times 19 Therefore k = 0.0526 M^-1 min^-1 After 89% conversion, concentration left is 0.11 i.e (1 - 0.89) because 89% of original conc 1 is 0.89 which is reacted Now new C_A = 0.11