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The decomposition reaction 2 NOCl → 2 NO + Cl_2 has a rate law that is second order with respect to ​[NOCl], where k = 3.2 M^{-1}s^{-1} at a certain temperature. If the initial concentration of NOCl is 0.076 M, how many seconds will it take for ​[NOCl] to decrease to 0.042 M at this temperature? Do not enter units with your numerical answer.

Respuesta :

Answer:

It will take 3.3 s for [NOCl] to decrease to 0.042 M.

Explanation:

Integrated rate law for this second order reaction-

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

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

Here, [tex][NOCl]_{0}[/tex] = 0.076 M, k = 3.2 [tex]M^{-1}s^{-1}[/tex] and [NOCl] = 0.042 M

So, [tex]\frac{1}{0.042M}=[3.2M^{-1}s^{-1}\times t]+\frac{1}{0.076M}[/tex]

or, t = 3.3 s

So, it will take 3.3 s for [NOCl] to decrease to 0.042 M.

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