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

To find out how long it would take for the concentration of A to decrease from 0.860 M to 0.230 M in a second-order reaction, we can use the integrated rate law.

Step-by-step explanation:

The reaction in question is second order and the rate constant is 0.380 M⁻¹⋅s⁻¹ at 300 ∘C. To find out how long it would take for the concentration of A to decrease from 0.860 M to 0.230 M, we can use the integrated rate law for a second-order reaction:

t = 1 / (k * [A])

Substituting the given values:

t = 1 / (0.380 M⁻¹⋅s⁻¹ * 0.860 M)

t ≈ 2.80 s

Answer 2

Answer: 8.38 seconds

Step-by-step explanation:

Integrated rate law for second order kinetics is given by:

`(1)/(a)=kt+(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)`

`(1)/(0.860)=0.380* t+(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 .