Determine the rate law and the value of k for the following reaction using the data provided. no2(g) + o3(g) → no3(g) + o2(g)[no2]i (m) [o3]i (m) initial rate0.10 0.33 1.420.10 0.66 2.840.25 0.66 7.10rate = 430 m-2s-1[no2]2[o3]rate = 130 m-2s-1[no2][o3]2rate = 1360 m-2.5s-1[no2]2.5[o3]rate = 43 m-1s-1[no2][o3]rate = 227 m-2.5s-1[no2][o3]2.5

Rate = 43 M⁻¹s⁻¹[NO₂][O₃]

Explanation:

We need to find the reaction order in

rate = (NO₂ )ᵃ (O₃ )ᵇ

given:

( NO₂ ) M                ( O₃ )   M         Rate  M/s

  0.10                        0.33                1.420                (1)

  0.10                        0.66               2.840                (2)

  0.25                       0.66                7.10                   (3)

When keeping the NO₂ concentration constant in the first two while  doubling the concentration of O₃ , the rate doubles. Therefore it is first order with respect to O₃

Comparing (2) and (3) increasing   the concentration of NO₂ by a factor of 2.5  and  keeping  O₃ constant  , increased the rate by  a factor of 2.5. Therefore the rate is first order with respect to NO₂

Then rate law is

= k (NO₂) (O₃ )

To find k take any of the three and substitute the values to find k:

1.420 M/s = k (0.10)M x (0.33)M ⇒ k = 43 /Ms

Then the answer is Rate = 43 M⁻¹s⁻¹[NO₂][O₃]

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year