Question

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

Answer 1

Answer:

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]