Chem 101 Homework 9: Rate Laws Question 6

Question

The second order reaction A → Products takes 13.5 s for the concentration of A to decrease from 0.740 M to 0.289 M. What is the value of k for this reaction?

Answer=______M⁻¹s⁻¹

Answer 1

The integrated rate law for a 2nd order reaction is

1/[A] = 1/[A]_o + kt

[A] = 0.289 M

[A]_o = 0.740 M

t = 13.5 s

1/0.289 = 1/0.740 + 13.5 k

3.46 = 1.35 + 13.5 k

k = 0.156 M^-1s^-1