Ask Question
5 October, 23:38

A certain first-order reaction (A→productsA→products) has a rate constant of 9.90*10-3 s-1s-1 at 45 ∘C∘C. How many minutes does it take for the concentration of the reactant, [A][A], to drop to 6.25%% of the original concentration? Express your answer with the appropriate units

+4
Answers (1)
  1. 5 October, 23:50
    0
    4.66667 minutes

    Explanation:

    Rate constant, k = 9.90*10-3 s-1

    Time = ?

    Initial concentration, [A]o = 100

    Final concentration, [A] = 6.25

    The integral rate law for first order reactions is given as;

    ln[A] = ln[A]o - kt

    kt = ln[A]o - ln[A]

    t = (ln[A]o - ln[A]) / k

    t = [ln (100) - ln (6.25) ] / 9.90*10-3

    t = 2.77 / 9.90*10-3

    t = 0.28006 * 103

    t = 280 seconds

    t = 4.66667 minutes (Upon conversion by dividing by 60)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A certain first-order reaction (A→productsA→products) has a rate constant of 9.90*10-3 s-1s-1 at 45 ∘C∘C. How many minutes does it take for ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers