Question

The rate constant for this second‑order reaction is 0.610 M − 1 ⋅ s − 1 0.610 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶products How long, in seconds, would it take for the concentration of A A to decrease from 0.860 M 0.860 M to 0.260 M?

Answer 1

Answer: It takes 3.120 seconds for the concentration of  A to decrease from 0.860 M to 0.260 M.

Explanation:

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

k = rate constant = $0.610M^{-1}s^{-1}$

$a_0$ = initial concentration = 0.860 M

a= concentration left after time t = 0.260 M

$\frac{1}{0.260}=0.860\times t+\frac{1}{0.860}$

$t=3.120s$

Thus it takes 3.120 seconds for the concentration of  A to decrease from 0.860 M to 0.260 M.