Ask Question
9 August, 21:11

The reaction ch3nc - > ch3cn is first order. it takes 156 seconds for the concentration of reactant to fall from 0.100 m to 0.0500 m. how much time would it take for the concentration of reactant to fall from 0.0500 m to 0.0400 m?

+3
Answers (1)
  1. 9 August, 21:25
    0
    Reaction of interest is: CH3NC → CH3CN

    Given: Reaction obeys 1st order kinetics

    Also, it takes 156 s for the concentration of reactant to fall from 0.100 m to 0.0500 m. Hence, half life of reaction (t1/2) is 156 s.

    We know that, for 1st order reaction:

    k = 0.693 / t (1/2)

    = 0.693 / 156

    = 0.0044 s-1

    We also know that, for 1st order reaction

    t = 2.303/k log (initial conc. / final conc.)

    ∴ Time required for the conc. of reactant to fall from 0.05 m to 0.04 m = 2.303/k log (initial conc. / final conc.)

    = 2.303/0.0044 log (0.05/0.04)

    = 50.72 s.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “The reaction ch3nc - > ch3cn is first order. it takes 156 seconds for the concentration of reactant to fall from 0.100 m to 0.0500 m. how ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers