Question

The decomposition of ozone into oxygen is believed to occur by the following mechanism:

O₃   <-->   O₂ + O       fast, equilibrium

O +   O₃ --> 2 O₂       slow

The rate constant for the overall rate law is 30.6 at 87^oC. If you start with a mixture of 1.5 M O₃ and 3.0 M O₂, what is the initial rate of the reaction?

Answer 1

The first step is a reversible equilibrium which can be written as below:

Where are the rate constant of the forward and backward reactions.

Now, the second step is written as

k2 is the rate constant of the second step.

Now, net rate of the reaction is basically net rate of decomposition of O3 in all the three steps.

Note, that O3 is being decomposed in step a and step c while it is forming in step b.

Hence, for the rate of reaction we can write

Now, O is an intermediate in our mechanism. Hence, applying steady state approximation to the concentration of O, we can write

Now, we can substitute the value of [O] given in equation (e) in equation (d) to get

Note that the first reversible steps are fast and the second step is slow.

Hence,

Hence, the denominator in the rate expression can be approximated as

Hence, the rate expression becomes

Where the overall rate constant k is given by

It is given that

Hence, the initial rate of the reaction can be calculated as

Hence, the initial rate of the reaction is 23 M per unit time.

Note: since the unit of time is not mentioned in the rate constant, we can not write a unit of time for the rate of our reaction.