Question

What is the concentration of A after 50.5 minutes for the second order reaction A → Products when the initial concentration of A is 0.250 M? (k = 0.117 Mi⁻¹min⁻¹)

Answer 1

Step-by-step explanation:

This is a second order reaction, so we will use the integrated rate law for a second order reaction, then rearrange it to solve for concentration after time t:

`(1)/([A]t)` -
`(1)/([A]init)` = kt

`(1)/([A]t)` = kt +
`(1)/([A]init)`

[A]t refers to the concentration after time t, and [A]init refers to the initial concentration.

We know k, t, and [A]init, so plug those into the equation:

`(1)/([A]t)` = (0.117M⁻¹*min⁻¹)(50.5 min) +
`(1)/(0.250M)`

= 5.9085M⁻¹ + 4M⁻¹

= 9.9085M⁻¹

= 1 / 9.9085M⁻¹

[A]t = 0.101M

I hope this helps! :)