Question

Question #a42c0

Answer 1

After `22.45"s"`

Explanation:

This is a 1st order reaction of the form:

`Ararr"products"`

As the reaction is 1st order the following expression applies:

`[A]_t=[A]_0e^(-kt)`

`[A]_0` is the initial concentration of `A`.

`[A]_t` is the concentration of `A` after time `t`.

`k` is the rate constant.

Taking natural logs of both sides gives:

`ln[A]_t=ln[A]_0-kt`

`[A]_0=100%`

`[A]_t=11%`

`:.ln11=ln100-kt`

`kt=ln100-ln11`

`t=(4.6-2.4)/(0.098)`

`t=22.45"s"`