The rate constant is found to be 3.9 x 10-5 m-1 s-1. calculate the rate of formation of ozone when the concentration of oxygen atoms is 3.0 x 10-14 molar and the concentration of diatomic oxygen is 1.3 x 10-14 molar.

The answer is 1.5 X 10 ^32 M/s.

Determine the Chemical reaction:

O + O₂ → O₃

Note:

The Ozone is formed when an oxygen atom collides with a molecule of oxygen.

Given: 

k = 3.9 x 10-5 M-1 s-1

Solution:

c(O) = 3.0 x 10⁻¹⁴ M

c(O₂) = 1.3 x 10⁻¹⁴ M

Required:

Find the rate of formation

rate of formation = k·c(O)·c(O₂)

rate of formation = (3.9 x 10⁻⁵ M⁻¹·s⁻¹) (1.3 x 10⁻¹⁴ M) (3.0 x 10⁻¹⁴ M)

rate of formation of ozone = 1.5 x 10⁻³² M/s

1.5x10⁻³² M/s

Explanation:


1) The ozone is formed when oxygen atoms (O) react with oxygen diatomic molecules (O₂)


2) The chemical equation for the formation of ozone is:


O + O₂ → O₃


3) The rate of the formation is:


rate = k[O][O2]


4) You are given the three values k ,[O] and [O2], so just plug in and calculate:

rate = 3.9·10⁻⁵ M⁻¹·s⁻¹ · 1.3·10⁻¹⁴ M · 3.0·10⁻¹⁴ M. = 1.5x10⁻³² M/s

Where is the following?