The rate law for the decomposition of ozone to molecular oxygen is rate = k[O3]2 [O2] The mechanism proposed for this process is O3 0001 O + O2 O + O3 2O2 What is the rate law in terms of k1, k2, and k−1? Be sure to simplify the rate law before inputting your answer.

The rate law is k =

Explanation:

From the mechanism is necessary to derive the rate law from the elementary steps and explain the effects of [O2] on the rate

The first step is a reversible reaction. Assuming dynamic equilibrium is achieved, the rate of the forward reaction is equal to the rate of the backward reaction

rate(forward) = rate(backward)

k1 [O3] = k-1 [O] [O2]

[O] is not part of the rate law, so we need to express [O] in terms of other species

[O] =

from the second step

rate = k2[O] [O3]

substituting [O] from the first step

rate =

k =

The final rate law is then

k =

So, as the concentration os O2 increase the rate decrease. Also from the first step of the mechanism we can se that O2 can react to O to form back the reactant O3 resulting in decreased reaction rate.

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o