kinetic exercises

[andreitagm1994]

Someone knows how to do these exercises?

1. Evidence suggests that the reaction between nitrogen dioxide and fluorine in the gaseous state is a two-step process:

2NO₂ (g) + F₂ (g) --> 2NO₂F (g)

STEP 1: NO₂ (g) + F₂ (g) --> NO₂F (g) + F(g) (SLOW)

STEP 2: F (g) + NO₂ (g) --> NO₂F (g) (FAST)

s) State wich is the rate-determining step, and explain why.

b) State wich of the two steps is expected to have the higher activation energy, and explain why.

c) Give the rate expression of the reaction based on your answer.

d) What feature of the fluorine molecula makes it reasonable to propose that a fluorine atom may be formed as an intermediate in this reaction?

2. The experimental data obtained for the acid-catalysed iodination of propanone showed that the rate equation is :

Rate= K [H₃CCOCH₃ (aq)] [H⁺ (aq)]

Propose a mechanism for the reaction between iodine and propanone in the presence of an acid catalyst to form iodopropanone that is consisten with this rate equation.

[Guest Outré]

1.

a) the slowest is always the rate determining step

b) the slowest because it requires more time to result in products. therefore, what you can conclude is that the molecules require more energy to collide against each other

c) K = [NO2F] [F]

d) its high electronegativity

2.

I don't really have time to answer to it all lol but if you have the Chemistry 3rd Edition book, it has the answer. Or you might want to check your own course companion as it might refer that.

[Keel]

Someone knows how to do these exercises?

You should.

2. The experimental data obtained for the acid-catalysed iodination of propanone showed that the rate equation is :

Rate= K [H₃CCOCH₃ (aq)] [H⁺ (aq)]

Propose a mechanism for the reaction between iodine and propanone in the presence of an acid catalyst to form iodopropanone that is consisten with this rate equation.

CH₃COCH₃ + H⁺ <-> [CH₃C(OH)CH₃]⁺ [FAST]

[CH₃C(OH)CH₃]⁺ -> CH₃C(OH)CH₂ + H⁺ [sLOW]

CH₃C(OH)CH₂ + I₂ -> CH₃COCH₂I + H⁺ + I^- [FAST]