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1.02 Comparing Rate of Change for Reactants and Products

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Advanced Chemistry

Universidad de Kentucky

4.7 (422 calificaciones) | 51K estudiantes inscritos

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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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!

CW

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!

De la lección

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

Impartido por:

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

Transcripción

We are ready to look at our second learning objective

in our chemical kinetics unit.

We are going to see that not all substances

appear and disappear at the same rate.

In our previous learning objective

we didn't have any coefficients other then 1.

But coefficients will start we didn't have any coefficients other then 1.

But coefficients will start

playing a role, and we will see how they play a role.

We are going to learn how to write, what is called

a rate expression. We are going to use it to compare the rate

of appearance or disappearance of one substance.

in a reaction to the rate of appearance

or disappearance of another substance.

To demonstrate how coefficients

play a role, we are going to do a demonstration.

We are going to end the power point and you will see me

making little booklets

and then we will resume the power point and see

how the numbers play out

and how the coefficients come into play.

We are going to simulate a chemical reaction.

The reaction we are going to simulate is 2 A

going to B.

This is my A, this is my reactants.

I have 18 sheets of paper here.

I am going to take this, and take

two of them and produce a booklet. It will be

a little 4 page booklet, and that will be B.

We are going to see that we can either monitor

the reactants, to determine the rate of the reaction.

Or we can monitor the product

and we are going to see how the coefficients come

into play as we do that.

So what we are going to look at now

is how we can monitor

the reactant, which was our sheets of paper

and our product

which is the booklet, and determine the rate of reaction.

What we need to look at is how much we have now.

We had started with 18 sheets What we need to look at is how much we have now.

We had started with 18 sheets

then as we count through these I now have

10 remaining.

If I look at the booklets I

started with no booklets, and I now have

four that are there.

So we are going to see how you can either

monitor the reactants or monitor

products and determine the rate of the reaction.

So you saw me do the demonstration

in which I took two pieces of paper

and made one booklet.

So this reaction might and made one booklet.

So this reaction might

be used to represent that process.

Two pieces of paper [we will call that A]

turning into a product of B [the booklet].

Now we recorded this process

for 30 seconds.

Initially, I had 18

piece of paper.

I counted when it was all said and done

and there was still 10 piece of paper left over

for the booklets, I didn't have any booklets to begin with

but I produced 4 booklets.

So if we take those numbers

and we ponder the change So if we take those numbers

and we ponder the change

so we could talk about the change in A

and we could talk about the change in B

we can put a little delta in here for the change in B

as time goes by.

If you think about those numbers

and I will let you do that before we switch slides

you will be able to make a comparison

of the rate of the disappearance

of A, verse the rate of the appearance of B.

Now for the demonstration

did B appear at twice the rate that A is disappearing?

Or did A disappear at twice

the rate that B is appearing?

Or did they disappear and appear at the same rate.

So take those numbers and I will go back

to that slide

and think about the value of the chance in concentration.

And the rate in which is appears and disappears.

Did you say that

A disappears at twice the rate that B is appearing?

If you did, then that would be correct.

Now we can use this

right here to

compare A to B.

The minus sign is there because

the change in A with respect to time was

negative. It was the change in A with respect to time was

negative. It was

10 - 18 here.

That would be a negative 8 [-8].

We put a minus sign in there to make it a positive 8. That would be a negative 8 [-8].

We put a minus sign in there to make it a positive 8.

We had over here

4 minus 0, it is already

positive so that is 4.

So we do not need a minus sign.

Since this was 8 and this was

four we are going to have to divide by 2

and that was right here

in order to make these two portions

here equal to each other.

That is how we define the rate of the reaction.

We will include that coefficient

of 2 down here in the denominator

to make define the rate of the reaction. So when I use of 2 down here in the denominator

to make define the rate of the reaction. So when I use

the word rate here, I am talking about about the rate

of the reaction.

We can monitor in terms of A

we can monitor in terms of B.

For this reaction

as we look at it, lets put the full numbers in here.

We have a negative 1/2

change in concentration in A, that was. We have a negative 1/2

change in concentration in A, that was.

The value of 10-18

or negative 8.

and that was divided by the 30 seconds.

Over here we had

4 - 0, again that is

divided by 30 seconds

and whether you monitor it in terms of A

or monitor it in terms of B

you are going to get the same number.

That is the rate of the reaction.

Now I have a numerator here, that I have written down as one because

it is not molarity, it is the number of pieces

that I have, but

the values are going to be exactly the same

because I incorporated the coefficient of 2 into

the expression.

Now we can look as any generic reaction.

No matter what the coefficients are

and we can define the rate of the reaction

with what we call a rate expression.

This is the rate expression.

So it a change in concentration of the reactants.

You put the minus sign in for reactants.

We include the coefficients You put the minus sign in for reactants.

We include the coefficients

all along the way.

So we could define the rate of the reaction all along the way.

So we could define the rate of the reaction

in term of the reactants.

or in terms of the products, it does not matter.

It will always give us the same rate

for the reaction, and we call this a rate expression. It will always give us the same rate

for the reaction, and we call this a rate expression.

If a problem says, write the rate expression

for a reaction, this is what you are going to be writing.

Now we need to look and see how we use this.

Here I have a reaction

in which there is all sorts of different coefficients.

Ones and Fives and sixes and threes. in which there is all sorts of different coefficients.

Ones and Fives and sixes and threes.

Somebody goes in, and they monitor

the rate at which BrO3 minus

is disappearing and that is what this is

saying, the change in concentration

which respect to time of BrO3

how fast is it disappearing?

It is disappearing at a rate of 1.5 x 10^-2

molarity per second.

And this is at some instance in time, because remember the rate

change is not the same through out the course of the reaction.

And this problem, wants us to say

it is disappearing at this rate, at what rate is

the Br- disappearing.

Well we can come up with a rate expression the Br- disappearing.

Well we can come up with a rate expression

and work this problem.

Now the rate, in terms of

the BrO3 is the change of

concentration of BrO3-

with respect to time

put a minus sign in there, and that would define the rate of the reaction.

Or we could do it in terms of the

Br-. It would be the change in concentration of Br-

with respect to time.

Put a minus sign in there so it is positive

and also put a 5 in there

so they can be equal.

Now they gave me this value, I am going to

put a box around it.

It is 1.5 x 10 ^ -2 molarity per second.

There asking me for this portion.

And I am going to put a circle around it.

They are asking me for that portion.

So how do I solve for that portion?

I will have to take, and I am going to write the whole thing again.

and multiply both sides by 5.

If I multiply both side by 5.

The five cancel and I will have solved for

portion I have put a circle around.

I will have solved for a negative

Br-

over delta t

and that would be 7.5 x 10 ^ 2

Lets use some language. Instead of saying

the value of and they give you this.

They might use the words

the rate of disappearance of BrO3- is

1.5 x 10 ^ -2.

So let me say that in words and write it down.

The rate of disappearance

of BrO3- is

that is a way of writing in expressions with

symbols, but the words would be the rate of disappearance.

If they said instead of these symbols here

they might say, 'What is the rate of

disappearance

of Br-?

That would be how you use words for that expression.

OK we are going to work this problem.

It is based on the same reaction as the previous problem.

Wording is a little different, we are solving for a

component of the reaction.

Lets look at how it is written here.

There are actually using the words now

Br- disappears at a certain rate. There are actually using the words now

Br- disappears at a certain rate.

At what rate is Br2 appearing?

Let turn those words into symbols.

The rate of disappearance Let turn those words into symbols.

The rate of disappearance

of Br- would be the

change on concentration of Br-

with respect to time.

Now that would be a negative value, because it is disappearing.

It is telling me, by using the word disappearing,

wouldn't give me a positive number, they have include that minus sign.

it is 7.5 times 10 ^ -2.

They want to know as what rate

Br2 is appearing.

So symbols for that would be

change in concentration So symbols for that would be

change in concentration

of BR2

with respect to time

it is a product, it is appearing it is already

positive. I do not need that negative sign to turn it positive.

That is what I am trying to find.

I will being with a rate expression.

Rate is equal to.

I am only going to do it

for the substances I am interested in, there is

no point for everything in there.

But to compare those substances

to the rate of the reaction and to each to other

we have to include the coefficients.

So, the negative change in the concentration of

Br- divided 5 delta t

would equal

the change of concentration of Br2 would equal

the change of concentration of Br2

and lets include the 3 delta t.

If we look carefully at the symbols that are represented right here.

Where is that found here in this expression?

It is right here. Everything except for the five.

So I am going to put that value in

7.5 x 10 ^ -2

and I will include that 5

so that the rate of the reaction is equal to that.

I am going to go ahead and put

these symbols back in with the 3

and think about where is

this portion found, it is right here.

I want to solve for that, that i just put a box around.

So I am going to multiply both sides by 3.

3 times 7.5 x 10 ^ -2 divided by 5

would give me what I am looking for.

The value for that, is 4.5 x 10 ^ -2

molarity per second.

That is equal to the change in concentration

of Bromine with respect to time

so this is how fast it is appearing.

So that is how you use a rate expression to compare

if you know the rate of change for one substance

and compare it to the rate of change for the other substance

you can also obtain the rate of the reaction.

The rate of the reaction would simply be

plugging this portion right here The rate of the reaction would simply be

plugging this portion right here

divide by 5, and that would give me the rate of the reaction.

So that is the end of our learning objective 2.

We are using a rate expression

which we have learned how to write to compare the rate

of the appearance or disappearance of any one substance

to the rate of appearance or disappearance of another substance.

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