Question

The decomposition of NOBr follows second order kinetics. The rate constant is found to be 0.556 M-1 s-1. If the initial concentration of NOBr in the container is 0.25 M, how long will it take for the concentration to decrease to 0.025 M?

Answer 1

It will take 64.75 seconds for the concentration of NOBr to decrease to 0.025 M

Step-by-step explanation:

The integrated rate law for decomposition of NOBr is -

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

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

Here [NOBr] is 0.025 M, k is 0.556
`M^(-1)S^(-1)` and
`[NOBr]_(0)` is 0.25 M

Plug in all the values in above equation-

`(1)/(0.025)=(1)/(0.25)+(0.556* t)`

or, t = 64.75

So it will take 64.75 seconds for the concentration to decrease to 0.025 M