Question

NO2 (g) reacts with O2 4 NO2 (g) + O2(g) to produce N2 O5 (g) as shown by the equation below: (g) = 2 N2 O5 (g).

Experimentally the rate orders were determined and rate law written as shown below: Rate = k [NO2]2[O2 ].
Calculate the value of k if the initial concentration of NO2 was 0.250 M and initial
concentration of O2 (g) was 0.150 M. The initial rate was 1.25 x 10-2 M.s-1 .

Answer 1

`k=1.33M^(-2)s^(-1)`

Step-by-step explanation:

Hello there!

In this case, according to the given information for the definition of the rate law, which is second-order with respect to NO2 and first-order to O2, we can solve for k as both concentrations are given as well as the initial rate of reaction:

`k=(r)/([NO_2]^2[O_2])`

In such a way, we can just plug in the given values to obtain the correct rate constant with the appropriate units:

`k=(1.25x10^(-2)Ms^(-1))/((0.250M)^2(0.150M))\\\\k=1.33M^(-2)s^(-1)`

Regards!