Question

A second order reaction is 40% completed after one hour. What is the rate constant for this reaction?

Answer 1

`1.1xx10^-2` `"M"^-1` `"min"^-1`

Explanation:

The rate law for a second-order reaction is

`1/([A])=1/([A]_0)+kt`

We can say that the current concentration `[A]` is `0.6` and the original concentration `[A]_0` is `1`. This is true because the reaction is `40%` complete, which means that `60%` still has to be reacted, and `60%` of an original concentration `1` is `0.6`.

`t=60` since `60` minutes have elapsed.

`1/(0.6)=1/1+k(60)`

`1.1xx10^-2=k`

The rate constant is `1.1xx10^-2` `"M"^-1` `"min"^-1`.