Question

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.

Answer 1

For the second-order reaction 2 NOCl → 2 NO + Cl2, it will take 3.32 seconds for the concentration of NOCl to decrease from 0.076 M to 0.042 M.

Step-by-step explanation:

The reaction 2 NOCl → 2 NO + Cl2 is second order with respect to [NOCl], meaning the rate law can be written as rate = k[NOCl]2. Given that k = 3.2 M−1s−1, we can use the integrated rate law for second-order reactions to find the time it takes for the concentration of NOCl to change from 0.076 M to 0.042 M. The integrated rate law for second-order reactions is ¼¾ ϑ¾ = k(t-t0), with t as the time elapsed, t0 as the initial time, and ϑ and ϑ0 as the final and initial concentrations, respectively. Solving for t gives us

t = ¼¾ [ϑ0^{-1} - ϑ^{-1}] / k

Substituting the given values we get:

t = ( ¼¾ [0.076−1 - 0.042−1] ) / 3.2

t = ( ¼¾ [13.16 - 23.81] ) / 3.2

t = ( ¼¾ [-10.65] ) / 3.2

t = 3.32 seconds

Answer 2

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

Step-by-step explanation:

Integrated rate law for this second order reaction-

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

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

Here,
`[NOCl]_(0)` = 0.076 M, k = 3.2
`M^(-1)s^(-1)` and [NOCl] = 0.042 M

So,
`(1)/(0.042M)=[3.2M^(-1)s^(-1)* t]+(1)/(0.076M)`

or, t = 3.3 s

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