Question

For the second-order reaction below, the initial concentration of reactant A is 0.24 M. If the rate constant for the reaction is 1.5 x10–2 M–1s–1, what is the concentration of A after 265 seconds?2A --> B + Crate = k[A]20.12 M0.19 M0.95 M4.0 M5.2 M

Answer 1

The concentration of A after 265 seconds is 0.19 M.

Step-by-step explanation:

In order to find the concentration of A after 265 seconds, we can use the integrated rate law for a second-order reaction, which is:

1/[A]t = kt + 1/[A]0

Given that the initial concentration of A is 0.24 M, the rate constant is 1.5 x 10-2 M-1s-1,

we can substitute these values into the equation and solve for [A]t:

1/[A]t = (1.5 x 10-2 M-1s-1) (265 s) + 1/0.24 M

Simplifying the equation gives:

[A]t = 0.19 M

Therefore, the concentration of A after 265 seconds is 0.19 M.

Answer 2

Concentration of A after 265 seconds is 0.12 M

Step-by-step explanation:

Integrated rate law for the given second order reaction is-

`(1)/([A])=(1)/([A]_(0))+kt`

where [A] is concentration of A after "t" time,
`[A]_(0)` is intital concentration of A and k is rate constant.

Here
`[A]_(0)` is 0.24 M, k is 0.015
`M^(-1)S^(-1)` and t is 265 S

Plug in all the values in the above equation-

`(1)/([A])=(1)/(0.24)+(0.015* 265)`

or, [A] = 0.12

So concentration of A after 265 seconds is 0.12 M