contestada

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?

Respuesta :

JannetPalos

Answer:

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

Explanation:

The integrated rate law for decomposition of NOBr is -

                     [tex]\frac{1}{[NOBr]}=\frac{1}{[NOBr]_{0}}+kt[/tex]

where [NOBr] is concentration of NOBr after "t" time, [tex][NOBr]_{0}[/tex] is initial concentration of NOBr and k is rate constant

Here [NOBr] is 0.025 M, k is 0.556 [tex]M^{-1}S^{-1}[/tex] and [tex][NOBr]_{0}[/tex] is 0.25 M

Plug in all the values in above equation-

                            [tex]\frac{1}{0.025}=\frac{1}{0.25}+(0.556\times t)[/tex]

                            or, t = 64.75

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

Otras preguntas