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

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 -

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

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-

                            $\frac{1}{0.025}=\frac{1}{0.25}+(0.556\times t)$

                            or, t = 64.75

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