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1.10a RDS Fast Equilibrium

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Chimie avancée

Université du Kentucky

4.7 (420 notes) | 50K étudiants inscrits

A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.

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4.7 (420 notes)

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KO

Jul 17, 2017

Excellent content and delivery!\n\nThis course has many well-explained examples and practice questions to reinforce the understanding of the taught concepts. Thank you!

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Sep 17, 2015

I used this course as a significant portion of my preparation for the CLEP Chemistry exam, and passed today with an A! Thank you!

À partir de la leçon

Kinetics

The study of chemical kinetics is the study of change over time. It answers questions like: How fast are reactants consumed? How fast are products formed? This unit is dedicated to the exploration of how these questions are answered. We will look at the experimental evidence of how concentration affects these rates. We will also examine what occurs on the molecular level, especially with respect to the motion of molecules, that affects rates of reactions.

1.01 The Rate of Chemical Reactions10:12

1.02 Comparing Rate of Change for Reactants and Products13:52

1.02a Obtaining a Rate Law from Experimental Data Equations4:23

1.03 The Rate Law9:01

1.04 Obtaining a Rate Law from Experimental Data16:38

1.04a Rate Law Calculations8:20

1.05 First-Order Kinetics and the Integrated Rate Law0:00

1.05a Graphic 1st Order7:10

1.06 First-Order Kinetics and the Half-Life9:32

1.07 Second-Order Reactions8:18

1.07a Graphics 2nd Order6:02

1.08 Collision Theory13:13

1.08a Activation Energy3:48

1.09 The Arrhenius Equation15:32

1.09a Graphic Activation Energy2:57

1.10 Reaction Mechanisms12:45

1.10a RDS Fast Equilibrium4:38

1.10b Rate Determining Step2:41

1.11 Catalysis4:23

Enseigné par

Dr. Allison Soult
Lecturer
Dr. Kim Woodrum
Sr. Lecturer

Transcription

We are given a mechanism in this problem and

we're going to derive a rate law for this reaction.

Is that a slow step which is a second step and a fast equilibrium before it.

It's got symbols that we may not be familiar with, so

let's get familiar with it.

E stands for enzyme.

So this is a substance that is a catalyst, a biological catalyst.

S stands for substrate.

That is your reactant in a sense that is going to be converted to P product.

And then it has ES and the question there is what is ES?

Well a substance that appears in an early step and then later gets consumed and

therefore is cancelled out in the overall reaction is an intermediate.

So ES is an intermediate of this reaction.

If we look at the overall reaction here,

we would be cancelling the intermediate, but we also cancel the enzyme,

and that would give me the substrate being converted into product.

So this reaction will happen in a biological system

with the enzyme catalyzing the reaction.

Now we were taught that the write a rate law for

the overall reaction we can look at the slow step,

which is the second step of the mechanism and use the coefficients of the reactant.

Which shows a one coefficient, that will become the power.

Goes to the first power here, for the reaction.

But we never want to leave a rate law in terms of an intermediate,

something that is getting produced and later consumed in the reaction.

Whenever the second step which is the slow step has an intermediate in that rate law,

we can use a fast equilibrium that precedes it

to replace that intermediate with actual reactants.

So let's see how it's done.

Because the first step is an equilibrium we can write a rate law for

the forward as well as the reverse reaction.

So the rate law for the forward reaction would be the rate constant of that forward

reaction times a reactants, which is e to the first power and s to the first power.

As we look at that reaction and we know it's reversible,

we can also write the rate law for the reverse reaction,

thinking about it going in this direction instead.

So the rate of the reverse reaction would be the rate constant

on that reverse reaction times e times s.

Now since this is an equilibrium we know that the reaction

will have the forward reaction occurring at the same rate as the reverse reaction.

So we can set this portion and

this portion equal to each other because their rates are equal.

Now the rate constants aren't equal, but the rates are equal.

Now why are we doing this?

We are doing this because we want to come up with something where I am replacing

this ES with actual substances that are reactants.

If I take this equation and divide both sides by K sub r,

we get that e sub s is equal to everything we see on the left side.

Well a Kf and a Kr are two constants.

When you divide constants you're going to get a constant.

So I'm just going to replace this with, I'll call it K1.

For the rate constant of that first reaction would be [E][S].

Now, what I can do is replace ES, this, in this reaction.

And that's going to give me rate is equal to K, and then let's put that K1.

[E][S]..

Well we can part combine our k's into another k.

All right, so we have a k is equal to [E][S].

So now we have got a rate law in terms of enzyme and substrate.

And it's okay to enzymes in rate laws, but it's not okay to have intermediates.

So this would be the rate law, what we see written right here.

Would be the right law of the overall reaction where a slow step

has an intermediate in it and is preceded by a fast equilibrium.

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