Question

The rate constant for this second‑order reaction is 0.990 M − 1 ⋅ s − 1 0.990 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 0.650 M 0.650 M to 0.380 M?

Answer 1

Answer : The time taken would be, 1.10 s

Explanation :

The integrated rate law equation for second order reaction follows:

`k=(1)/(t)\left ((1)/([A])-(1)/([A]_o)\right)`

where,

k = rate constant =
`0.990M^(-1)s^(-1)`

t = time taken = ?

[A] = concentration of substance after time 't' = 0.380 M

`[A]_o` = Initial concentration = 0.650 M

Now put all the given values in above equation, we get:

`0.990=(1)/(t)\left ((1)/((0.380))-(1)/((0.650))\right)`

`t=1.10s`

Hence, the time taken would be, 1.10 s