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]