Question

The rate constant for this second‑order reaction is 0.380 M − 1 ⋅ s − 1 0.380 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 0.230 M?

Answer 1

Answer: 8.38 seconds

Explanation:

Integrated rate law for second order kinetics is given by:

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

$a_0$ = initial concentartion = 0.860 M

a= concentration left after time t = 0.230 M

k = rate constant =$0.380M^{-1}s^{-1}$

$\frac{1}{0.860}=0.380\times t+\frac{1}{0.230 }$

$t=8.38s$

Thus it will take 8.38 seconds for the concentration of  A to decrease from 0.860 M to 0.230 M .