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.
1.02 Comparing Rate of Change for Reactants and Products
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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.
Impartido por:
Dr. Allison Soult
Lecturer
Dr. Kim Woodrum
Sr. Lecturer
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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