contestada

Suppose that a 1.0-L reaction vessel initially contains 1.0 mol of O3 and 1.0 mol of O2. What fraction of the O3 will have reacted when the rate falls to one-half of its initial value?

Respuesta :

Answer:

Explanation:

given reaction

                        2O₃        ⇄      3O₂

 initial                1                      1

    final           1-x                   1+ 1.5x

rate of reaction = k  [ O₃ ]² / [ O₂ ]

initial rate = k . 1² / 1 = k

final rate = k (1-x )² / (1+ 1.5x) = k/2

(1-x )² / (1+ 1.5x) = 1/2

2x² - 5.5 x + 1 = 0

x = 0.196

So fraction of ozone reacted = .196

mickymike92

Based on the rate law and rate constant of the reaction, the fraction of ozone reacted

is 0.196.

What is the rate law of the reaction?

The rate law of the reaction is determined from the equation of the reaction.

The equation of the reaction is given as follows:

  • 2 O₃ ⇄ 3O₂

The rate law for this reaction is as follows:

  • Rate = K [O₃ ]² / [O₂]

 Taking the initial and final concentration of the reactants and products:

2O₃        ⇄      3O₂

 initial      1                      1

final        1-x                   1+ 1.5x

where:

  • x is the change in concentration

rate of reaction = K [ O₃ ]² / [ O₂ ]

Substituting the values of concentration to determine the initial and final rate constant K

initial rate of reaction = K . 1² / 1 = K

final rate = k (1-x )² / (1+ 1.5x) = K/2

Solving for x ?:

(1-x )² / (1+ 1.5x) = 1/2

2x² - 5.5 x + 1 = 0

x = 0.196

Therefore, the fraction of ozone reacted

is 0.196.

Learn more about rate constant at: https://brainly.com/question/24749252

Otras preguntas