[SQUEAKING]
[RUSTLING]
[CLICKING]
SARAH HEWETT: All right.
We can probably get started.
So good afternoon.
And today, we are going to start talking about the final lab
that we are going to talk about in the lecture
and the final lab that you haven't
heard about yet for the course, which is the catalase lab.
And some of you guys started that yesterday.
And some of you will start it today.
And then the rest of you will have to wait a little bit.
But the catalase lab is our biochemistry type lab that we
were doing in 5.310.
And catalase is an enzyme if you couldn't tell by the name.
A lot of enzyme names--
you guys are familiar from your biology classes-- end in ase.
So catalase is an enzyme.
And enzymes are proteins-- just a review of your general bio--
that carry out specific chemical reactions.
And they're produced by different living organisms.
They work in the cells to do a very wide range of chemistry.
And there are many different graphical representations,
weird pictures that you can find on the internet that
try to depict what an enzyme actually does in cartoon form.
And so here are a couple of examples of that.
So you have your enzyme and then you have
what's called your substrate.
And so that is the thing that the enzyme is doing chemistry
on.
The substrate comes in.
It interacts with your enzyme to form an enzyme substrate
complex.
Then the chemistry happens.
And then the enzyme releases the product.
So sometimes the products will be rearranging the substrate.
Sometimes it will be taking two substrates
and joining them together.
And sometimes it will be taking one substrate
and breaking it apart.
So there's a bunch of different reactions
that can happen in different enzymes.
Similar thing-- if you want to look at your enzyme
as a little Pacman kind of guy.
You form the enzyme substrate complex.
You form your products.
And then they break apart.
So that's the general idea of how enzymes
works in a very quick nutshell.
So our specific enzyme is catalase.
And here are a couple of pictures of catalase or artist
renditions of catalase.
So this is a space filling model of what
it would look like if all of the atoms were kind of 3D balls.
So you can see it has four different subdomains here.
And then these darker parts in the middle
with the little green dot, those are your heme center,
so these are iron molecules and that's
where the chemistry happens.
So these are the active sites of the catalase.
And you can also-- you may have seen in other courses proteins
represented like this, which has a little bit more
of the features of their secondary and tertiary
structure.
So you can see the alpha helices and beta sheets.
And we're going to talk more about the structure
of the active site and how the chemistry happens
and what all of those squiggles mean next week
when we talk about catalase.
AUDIENCE: [INAUDIBLE]
SARAH HEWETT: Yes.
Or wait.
Each sub-domain has one active site.
So it has one heme in each of the four parts.
And we'll talk more about that next week
when we go a little bit deeper into the structure of catalase
and how it works.
And then the other important thing
you need to know about catalase is it's a large molecule.
Proteins generally are.
There are polypeptides-- lots and lots of amino acids.
And so its weight is 240 kilodaltons.
And a kilodalton is 1,000 mass units.
So it's 240,000 grams per mole.
And you'll need that for some calculations
that you'll do in the second half of the lab.
And we'll go over that again in the next lecture.
But to give you an idea of the size
of that thing and the molar mass, it is very large.
Why do we need catalase?
So there are many redox reactions
that are happening in your body where you do electron transfer
reactions and you oxidize or reduce different things
in your metabolism.
And so electrons are transferred around cells.
You may have heard of the molecule NADPH or NADH.
Those transfer electrons between different molecules
in your cells and help with metabolism.
And while these electrons are being transferred around,
you also know that we breathe oxygen
and we need oxygen to do-- to perform
many of our life processes.
So you have a lot of oxygen in your body as well.
And during a lot of these metabolic processes
you can get electrons that are transferred to oxygen.
So if we have our oxygen and it gains an electron,
we have what is called superoxide.
So now this has a negative charge
and an unpaired electron.
And so it is a superoxide radical.
And what do we know about radicals?
AUDIENCE: [INAUDIBLE]
SARAH HEWETT: They would like to be stable.
So they are not stable because they
have this unpaired electrons.
So they will either give up this electron
or try to take an electron from something else
in order to have their electron be--
or not be unpaired anymore.
So these are very reactive in the body.
And this is actually a problem.
And we don't want this in our cells
because it is so reactive.
So there is another enzyme called superoxide dismutase
that breaks this apart into elemental oxygen
and oddly enough hydrogen peroxide.
But it turns out that hydrogen peroxide
in high enough concentrations is also toxic to your cells.
And hydrogen peroxide can form hydroxy radicals,
which are also very damaging to your cells.
There's a couple of reactions that hydrogen peroxide
can do in the body.
So the first is if you have hydrogen peroxide
and then you have some of this superoxide around--
so it's a radical anion there.
You can make oxygen, hydroxide ions, which are OK,
and then hydroxy radicals.
So then you form more radicals and those can go and damage
your DNA, your proteins, the different fatty acids
in your body, and cause all kinds of problems.
Another thing that can happen is if there's just hydrogen
peroxide around and it picks up an electron, it can form water
and another hydroxyl radical.
So these are the reactions that we do not
want to happen in our body because it forms these--
it just keeps propagating.
And then when radicals react with other things that have all
paired electrons, it generates more radicals
and you propagate the radical formation
throughout your cell it causes a lot of damage.
So we need a way to get rid of the hydrogen
peroxide in our body and that is what catalase is for.
And all organisms have catalase.
You can extract it out of humans, all types of mammals,
even plants have it.
Cells do not want hydrogen peroxide in them.
So there are different forms of catalase
that you can get from any organism-- bacteria, all
the things.
The catalase that we're going to be using in the lab
was extracted from cows.
So just a fun fact.
And when we're talking about enzymes,
the reason that enzymes are so good and so efficient
at their jobs is because they catalyze chemical reactions.
And a lot of you are chemical engineers,
so you probably know about catalysis.
So what does it mean for something to be a catalyst?
Anyone?
Please give me a quick catalyst definition.
AUDIENCE: [INAUDIBLE] reaction.
SARAH HEWETT: Yeah.
It speeds up a chemical reaction.
And it's not used up during the process.
So if we look at our energy reaction diagram,
if you have energy on this side and then reaction
progress on this axis, you can say
that you have reactants that are starting at a certain energy
level.
And then you have some energy that you
have to put into the reaction to get it to go.
And then in a lot of cases, if your reaction is exothermic,
your products maybe at a lower energy level.
So it'll release energy overall.
But in order to get the reaction to go
you have to put energy in.
So this part of the curve is our delta H of our reaction.
So it would be the energy released
if it's exothermic, which in this case this reaction is.
But no matter what you need to put in some energy
to get the reaction to go and this
is called the activation energy.
Occasionally abbreviated E sub a.
So this is what we care about when
we're talking about reaction kinetics
and what a catalyst does.
So a catalyst functions and it makes the reaction faster
by lowering this energy barrier.
So if the yellow line is the reaction without a catalyst,
then with a catalyst it will be smaller.
And the products and the reactants
will have the same total energy, but the energy
you need to put in to get the reaction to go
is going to be less.
And so the reaction can go faster
and it can go in milder conditions.
And we can talk about how enzymes
work to lower this activation energy.
So does anybody know what methods enzymes
use to lower the activation energy of a reaction?
Yes?
AUDIENCE: Proximity?
Just getting the reactants together?
SARAH HEWETT: Yeah.
Proximity.
So it gets the reactants close together.
If you have your reactants in a beaker
and they're all floating around, they
have to find each other before they can react.
And so if you have an enzyme that
has a specifically designed active site
to hold these two molecules close together,
then they will react.
What else?
AUDIENCE: It can stabilize a transition state?
SARAH HEWETT: Yeah.
It can stabilize a transition state.
And how does it do that?
AUDIENCE: [INAUDIBLE]
SARAH HEWETT: Yeah.
So it bonds to the substrate and it
can have either electrostatic or even sometimes covalent
interactions.
And it stabilizes the transition state.
And part of the way that it does that is it
holds the molecules in the right orientation for them to react.
So even if you have two molecules,
if they're supposed to interact like this to react
and they're in a beaker and they hit on this side
or upside down, the reaction will not happen.
So we can stabilize transition states.
Your transition state and put molecules
in the correct orientation.
Anything else?
There is one more thing that they can do.
OK.
What's your idea?
AUDIENCE: If they raise the temperature, it will increase--
collisions could increase.
SARAH HEWETT: So increasing the temperature of a reaction
will get it to go faster.
Right?
But that's just because then you will
have more molecules that have the necessary activation
energy.
So you're not necessarily lowering the activation energy
that the reaction needs.
But the last thing that they can do
is they can provide alternate reaction pathways.
So sometimes you may think that this will just
be a one-step reaction.
If you had an enzyme, occasionally it'll
look like this.
It might provide an extra intermediate in there
that helps to lower the overall energy.
And so that's another way that it
can lower this overall activation energy
is by providing an alternate mechanism
other than maybe just two molecules reacting together.
There might be a third intermediate
that helps actually lower the overall energy.
So those are possible ways that the enzymes can work
to lower the activation energy.
And so our goals for the catalase lab are--
well, there are a few goals that we have.
But our goals for days one and two
are to determine the activation energy of hydrogen--
the decomposition of hydrogen peroxide as catalyzed
by the catalase enzyme.
So we're going to try and figure out
what the activation energy is after it's already
been catalyzed.
And in order to do this, we are going
to need to first determine the order of the reaction
with respect to hydrogen peroxide
and be able to write a rate law.
That'll help us characterize the kinetics.
So in day one of this lab, you're
going to do that first part.
And you're going to figure out what the order of the reaction
is.
And in order to have that make any sense
we need to talk about kinetics and how we discuss kinetics
in a chemical sense.
So kinetics deals with the speed of chemical reactions.
And if we use everybody's favorite generic reaction
a plus b goes to c plus d.
You can write the rate of this reaction-- you can express it
in a number of ways.
So the first way is to think of the rate
of the reaction as the change in the reactants over time--
so a and b.
And it has a negative sign because you
are losing the reactants.
Hopefully they're going away as the reaction progresses.
And it is equal to the rate of the appearance
of c and the appearance of d.
So these do not have negative signs
because you are making your product.
So that is one way to express the rate of the reaction.
And when you do it this way, you need to account for the fact
that there are different molar ratios.
So the reactants and the products
will disappear in form at different rates
depending on how many of them you need in the reaction.
So to account for that, we multiply by the reciprocal
of this coefficient.
We can also write a generic rate law
like this where your rate equals your rate constant, which
you can calculate and determine in lab
and is dependent on the temperature
times the concentration of your reactants raised
to a power and your concentration
of your other reactants raised to a different power.
And you can also write an integrated rate law,
which you may have seen before.
So integrated rate laws deal with one reactant at a time.
And so if we look at a, we can say-- we can write--
if there's only a we can write the rate of a equals the rate
constant times a.
Or we can say that the rate is equal to the disappearance
of a overtime.
And you can take the derivative of this.
So if we move everything over, rearrange it--
So if we integrate both sides of this--
if we integrate this side from your initial concentration
to your concentration at some temperature
and this side from t equals 0 to some time t,
then if you do that math out you will get the natural log of--
this is your integrated rate law if your exponent here equals 1.
So obviously, that will change if you have different exponents
here.
And then when you do your integration
you will get different values.
But this is a way to relate the concentration
at your initial concentration to a concentration
at a final temperature or a final time.
So that's one thing that you've probably
seen before in some of your other chemistry classes.
And we can also talk about--
well, we'll talk about that in a second.
You may also have seen in some of your other classes enzyme
kinetics or heard of Michaelis-Menten kinetics.
And the Michaelis-Menten model of enzyme kinetics
says that if you have an enzyme and you
have a bunch of substrate, the reaction can only
go as fast as there is when there's only one
substrate in each enzyme.
So if all of the active sites are full of substrate,
then that is as fast as the reaction can go.
And it will proceed at a steady state.
So you can make graphs of the reaction
rate versus how much concentration of substrate
you have.
And you can get this Michaelis-Menten constant here,
which is a measure of how much the enzyme--
what its affinity is for the substrate essentially.
So does the enzyme need a whole bunch of substrate
before it'll start going quickly?
Or can it go quickly and efficiently
with just a tiny concentration of substrate?
So we're not going to worry so much
about those parameters for our lab right now.
But this is something you may have seen.
And one of the assumptions of the Michaelis-Menten kinetics,
which is actually kind of important to us
is that you have your enzyme and your substrate.
They make the enzyme substrate complex.
And then they go to enzyme and product.
And they don't go in the reverse reaction.
So that's kind of important to us
that if we are trying to measure the rates of our reactions,
that we don't have to contend with the reactions being
catalyzed in reverse.
All right.
So now we can take a little bit of a closer look
at the rate law for just a normal chemical reaction.
And pretty much the most important thing
you want to know about this is that the units of the rate
are always moles per liter per second.
So whenever you are trying to calculate a rate constant,
you want your units to line up.
So if your rate--
if we want this in molarity per second--
if your rate equals--
let's say these are both equal to 1 just for now.
Then what are our units over here?
M and M. So what are the units of k have to be?
So that's how you can figure out what the units of your rate
constant are because it always needs
to multiply with your concentrations and your k
to equal moles per liter per second.
And that is going to be the units of our rate
that we care about in our kinetic equations.
The rate constant is different at different temperatures.
So if you change the temperature of the reaction,
then your rate constant will change
and you'll have to remeasure that.
And x and y are values that need to be determined.
They may be the same as little a and little b but not always.
So the kinetics of a reaction are
determined if it's a multi-step reaction
if their reaction mechanism has more than one part in it.
They may not all be reflected in this overall chemical equation.
You may not see them, but the rate
is going to be dependent on your slowest step.
And that slowest step is what is going
to determine your rate law.
And it may not always match your overall reaction equation.
So be careful when you are determining these things.
You can't always just assume that it is
from the little a and little b.
So how are we going to determine the order for our reaction?
And this is what are going to do on day one.
We will use our rate law expression
to determine the order of the reaction.
So you're going to--
according to our rate law here.
It's the only reactant that we have in this case
is hydrogen peroxide, and we need
to determine what its order is.
So if we change the concentration of the peroxide,
we should change the rate, yes?
So if we do a bunch of reactions with a bunch
of different concentrations of peroxide,
we can measure the rates, and then
use the relationship between those two things
to figure out our concentration of--
or our order of the reaction.
So the way that you're going to do this in lab
is you will have a series of solutions,
and these are your solutions.
You'll have a stock hydrogen peroxide solution
that you're going to make up.
It's going to be about 4% hydrogen peroxide.
You'll use a phosphate buffer.
Why do we need to use a buffer for this reaction?
What's a buffer, I guess is the first question?
AUDIENCE: It helps the solution from changing PH2.
SARAH HEWETT: Perfect.
So a buffer helps the solution, it prevents a solution
from changing PH.
So you can make a buffer out of certain PH,
and then if acid or base gets added to it,
it resists the change in PH.
It keeps the constant.
Why is that important for our reaction?
AUDIENCE: Maybe [INAUDIBLE].
SARAH HEWETT: Yeah.
So enzymes are most functional at a set PH.
So your PH of your body is around 7,
so we want to keep the PH of our reaction
somewhere around our physiological PH,
so our buffer is going to be set to PH 6.8, which
is going to help make sure that the enzyme is
in its active state.
It is not denatured.
If you change the PH too much, you'll
change the protonation state of the different parts
of the protein, and it can fall apart.
So we want to make sure that this
is happening at the correct PH.
We'll add some enzyme, and then we
will measure the rate of the reaction.
And we're going to do that using this apparatus right here.
So you'll have one of these.
You'll have a water bath, so that we can
keep the temperature constant.
So you'll just fill this with room temperature water,
and then you can monitor the temperature
throughout your reaction to make sure that it does not change.
And then this is your reaction vessel
here, your pressure tube, because you will be putting
all of your solutions in here.
You will add your peroxide and your buffer,
and then you will put this cap on.
And this Teflon cap here, it's important to note,
has an O-ring on the bottom.
And so you want to make sure when you're
doing this reaction that you're O-ring
is in there, because that'll make a good seal,
so you won't have your products escaping
because our product is a gas.
So you'll screw this on.
And then there's a hole in the top here.
You can inject your enzyme really fast.
You're going be doing this in partners.
Somebody can inject the enzyme really fast.
And then you want to connect your pressure tube.
And this tube is connected to a pressure sensor.
And this pressure sensor will be connected
to your computer, which is why you need to bring
a laptop for the catalase lab.
And we will give you the Logger Pro software.
And it will send the data directly to your laptop.
And it'll plot it for you.
And you'll get a curve that, hopefully,
looks something like this.
And we will measure the pressure of oxygen formed in kilopascals
versus time in seconds.
And it'll graph this thing for you.
And the slope of the graph is going
to be your rate of kilopascals per second.
And we want to take the rate at this first part of the graph
where it is linear, and why do we
want to take the rate at the beginning of our reaction?
Yes.
AUDIENCE: Reaction rate is dependent on amount
of reactants?
SARAH HEWETT: Yeah.
So the reaction rate is dependent on the amount
of reactants, and the--
so the rate of this reaction obviously changes over time,
and we set up our reactions so that we know
the concentration of peroxide.
And the only time in this reaction
that we're theoretically going to know
the concentration of peroxide is at the beginning,
before some of it has reacted and the rate changes.
So we want to measure the initial rate so
that we know our concentration, and that we get a--
we have a consistent place to take our data
from on each measurement.
So once we have done all of these different reactions--
you'll do all four of them-- all of the same procedure
in the same tube-- you'll make five different graphs.
You'll have this.
Then how do we analyze this data and get information
about the order out of it?
So here we are going to have to do a little bit of math.
So here are some of the steps of the data analysis.
The first thing that you're going to have to do
is calculate the concentration of the peroxide that you used.
And we are going to give you 30% hydrogen peroxide, which is--
its volume per volume-- so there's 30 milliliters of H2O2,
and 70 millimeters of water.
So if you remember that your molarity is
your moles of your solute over the total volume
of the solution, you can turn this milliliters, assuming
a density of about 1, into grams, grams to moles,
and you can calculate the molarity of this solution.
You're then going to take this 30% hydrogen peroxide that we
give you, and dilute it 13.3 milliliters to 100 milliliters.
So then you can do another dilution equation
to figure out your final concentration of your hydrogen
peroxide that you're going to be using in the lab.
So that's a first step.
And then I'm going to kind of rearrange
a couple of these other steps, just
because it'll make this a little bit easier to go through.
But the first thing that we're going to do
is the easy part, which is get your rate of oxygen formation
in kilopascals per second.
And that we're just going to read straight
from the slope of the graph.
So that's pretty easy.
So we'll have our rate of oxygen formation
in kilopascals per second.
Great.
So now we want to take a look at our rate law.
And we have our rate of oxygen formation,
but what's in our rate law?
Hydrogen peroxide.
So we need to find a way to go from oxygen and kilopascals
per second to moles per liter per second of hydrogen
peroxide.
So the first way that we're going to do that
is to go from kilopascals per second here to molarity
of oxygen. And we can do that using our gas constant here.
So if you divide by the gas constant,
multiply by 1 over 8.314 moles, liters times kilopascals,
we can cancel out our kilopascals.
And we'll have moles per liter, which is good,
but then we still have this K, which
is our temperature in Kelvin, so we also
have to divide by our temperature in Kelvin.
And this will get us our rate of oxygen
in moles per liter per second, which is pretty good.
We're close.
But we still need to get from oxygen to hydrogen peroxide.
So how can we get from oxygen to hydrogen peroxide
using this information?
AUDIENCE: [INAUDIBLE]
SARAH HEWETT: It would be--
so the ratio is 2:1, right.
So there's two hydrogen peroxides
in this reaction for every one oxygen.
But these coefficients are in moles.
And this is in moles per liter.
So we need to figure out how many moles of oxygen we have
before we can use our stoichiometry ratio here.
So how are we going to figure out how many moles of oxygen
we have?
We need to know what our volume of oxygen was.
And for that we can go to our pressure tube.
So you'll have some of this is going to be your solution down
here, so that'll be liquid.
And then we need to calculate the volume of the gas
above our reaction, which is where
oxygen is going to be formed.
So we'll need to know the volume of this,
and the volume of the pressure tube.
And the volume of the pressure tube has already been measured,
and we'll give that to you.
And you guys are going to measure
in the lab what the volume of your reaction vessel is.
You're going to fill this thing with water,
weigh it, then weigh it empty, and then you
can calculate by difference in the density of water
what your volume of your whole tube is.
And then you know your reaction volume is 25 milliliters,
so you can subtract those and get the volume of the gas.
So if we know these two together,
you'll get the volume of the gas above your reaction.
So if we have our rate of oxygen formation in moles per liter
per second, we can multiply that by our volume
that we calculate, so liters of oxygen.
And then we have our rate in moles of oxygen per second.
And now we can use our stoichiometry,
because we have it in moles, so we can take our rate in moles
per second, and then we know that it is 1 mole of oxygen
for 2 moles H2O2.
Then we get our rate H2O2 in moles per second.
And what are the units of rate that we want?
Moles per liter per second.
So we have it in miles per second.
We're so close.
Then we just need to divide by the liters of our reaction,
which if you go back a couple of slides, is 25 milliliters,
so all of our reaction volumes are 25 milliliters.
So if you convert that to liters, divide by 0.25
liters, and then we will have our rate
in moles per liter per second, which is what we want.
Now this still doesn't get us to a,
but we can rearrange this equation
by taking the natural log of both sides.
So if you have the natural log of the rate
equals the natural log of K plus,
then the A comes down, a times the natural log
of your hydrogen peroxide concentration, then--
we've just figured this out from all of that math.
We know this concentration.
And this is going to be our y-intercept.
So if we have y equals b plus mx,
if we make a graph of the hydrogen--
or the natural log of the peroxide concentration
and the natural log of the rate, then our slope
is going to be our order.
Hopefully, we haven't lost you.
So the way that you're going to treat all this data is you can,
instead of having to do this calculation out
for every single trial, you can make a giant spreadsheet.
And in the lab manual, it tells you what data
we want you to collect.
And this is all of it here.
And then you can make a spreadsheet that
does all of the calculations for you, that gets you down
to the natural log of rate, and the natural log
of hydrogen peroxide.
So this is an example of a pretty good way
to present your data.
And I know it's kind of hard to see on the printed out slides,
but these will be on Stellar if you want to look at them.
And then your graph will look something like this.
And you'll have a straight line.
And so from that straight line, you
can figure out what your order of your reaction
is based on your slope.
So that was a lot.
So once we have that done on day one,
then, if you remember our second goal for the lab,
it was to figure out the activation
energy of the reaction.
And so the way that we're going to do
that is to use the Arrhenius equation,
and that says that the rate constant is
equal to A, which is this collision frequency factor,
times e to the negative activation energy over RT.
And if we remember the things that can change the rate,
we can change the concentration and that'll
affect the rate, and also the temperature.
So if we do our reaction at different temperatures,
then we can determine our activation energy.
So what you're going to do is you're
going to pick one of the trials from day one.
Usually people pick the middle one because this is easy.
It's just 1 milliliter, no decimals.
Just easy to measure and it's right in the middle.
You'll pick one and you'll hold the concentration of peroxide
constant, and then we will vary the temperature.
So you'll do seven runs at varying temperatures.
So you'll pick a temperature in each of these ranges
and you will measure the rate.
Again, you'll get the same graph, the measure
the rate of oxygen per second.
And then you can do all of this same math
to get your information in concentration of peroxide
to go from oxygen to peroxide, because that's
what we care about.
The keys to success for this reaction
are to wait for the peroxide and the buffer
to reach the correct temperature before you add the enzyme.
So when you have your reaction here in this tube,
you'll put it in your water bath and then
you want to give it a few minutes
to stir and come to equilibrium at the right temperature,
because it'll all be at room temperature.
And then if you're trying to do your reaction at 0 degrees or 5
degrees, your reaction part won't actually be at 5 degrees
even if your water bath is.
So give it a few minutes to equilibrate
before you take your measurements
and before you inject your enzyme.
Another key to success is to start
at the hottest temperature and then add ice to cool it.
It's easier to control the cooling
than it is to control the heating with these hot plates.
The hot plates tend to get really excited
and then they'll heat your solution up
way above what it needs to, and then it won't be able to cool.
So if you start at your hottest temperature and then add ice,
you'll be able to slowly lower the temperature
and do it in a more controlled fashion.
The other important thing about this reaction
for day one and two, and I'm sure your TAs will emphasize
this when they tell you in lab and you guys have a chance
to have this in front of you, is to do everything really
quickly.
So the reaction starts as soon as you add the enzyme.
So you'll have your micropipette,
you'll stick it right in that hole,
add the enzyme really fast, and then
you want to attach your pressure tube
and then hit go on your computer to start collecting the data.
And that all has to happen in like, a second so
that you don't miss that beginning of your data
collection.
So those are some important things.
If we talk about the data analysis for this reaction,
it's kind of similar to the other one.
So you'll calculate the concentration of peroxide,
you'll calculate the volume of air above your reaction.
You'll have the initial rate of oxygen formation
in kilopascals per second.
Then you can get the initial rate of oxygen formation
in molarity per second and moles per second.
Then you can change your rate of oxygen formation
to the rate of peroxide decomposition, same math.
And then using the rate law that you determined on day one,
you're going to calculate the rate constant.
So if we have our rate, our general rate law,
we'll measure this in the lab or we'll calculate this, I guess.
We will know this from day one.
So you'll plug in the value that you
get from your first set of calculations right there.
And then we measure this in the lab
or we can calculate/measure this.
So we can calculate our rate constant,
which is, if you just rearrange this, it'll be k--
so that's not so bad.
And then in order to get our activation energy out of this,
we are again going to take the natural log of both sides.
I think that I did that out correctly.
No plus sign here.
All right, so this is your linear form of this equation.
So we have, again, y equals b m and x.
So if we do our reaction at different temperatures,
then we will know the temperature of our reaction.
And then we will get k from our rate constant from our rate
law, and you'll calculate that for each of your trials,
as well, based on the rate.
And then our slope is going to be the activation energy
over the ideal gas constant.
So we can get the activation energy
from the slope of a graph of 1 over t
times the natural log of k versus the natural log of k.
And then you can also get your collision frequency factor,
this A term, from your y-intercept,
so you'll also be able to determine that.
Then you'll use your Linus program
to get the errors in all of these things,
and that is essentially day one and day two of the lab.
Do you guys have any questions about that?
There's a lot of math.
So again, there's another data chart
that you can make for day two of the lab that
has all of the information that you're going
to need for these calculations.
And then this is hopefully what your graph will look like.
It should be linear with a negative slope,
and then the negative will get taken
care of with that negative sign in the equation.
AUDIENCE: This will be posted on Stellar?
SARAH HEWETT: Yes.
Yeah, all of the slides will be posted on Stellar
so you can see it in a larger form,
because I know they didn't print out super well.
OK.
So now we can take a look at the reaction
that you're going to do in a slightly different form.
And you may have seen this before,
but we're going to do it anyway.
So hydrogen peroxide, you may have some at your house.
They sell it in a drugstore.
It's like, 3% so it's not as concentrated as the stuff
that we're going to be using in the lab.
But nonetheless, hydrogen peroxide
is a thing that you can buy.
And it naturally decomposes over time,
so this reaction is always happening.
If you have a bottle of hydrogen peroxide,
usually the caps are vented because if you seal it
for too long, you'll build up pressure of oxygen
and that's a problem.
The stuff that we have in the lab is 30%,
so that's definitely vented, and we store it in the refrigerator
so that we slow down the reaction
and keep our product for longer.
So it is always happening and you can speed it up
in a number of ways.
So one of the ways is obviously using enzymes, catalase.
That'll speed up the reaction.
But there are other catalysts that you can use, as well.
So have you guys seen the elephant toothpaste reaction
before?
Yes.
Do you guys know what the catalyst
is that we use for this?
There's a few you can use, but does anyone remember?
So we're going to react hydrogen peroxide with potassium
iodide in this case.
And you can use a couple different ones,
like manganese dioxide, I believe, works.
You can use different iodide salts,
but potassium iodide is the one that I have today.
And we're going to do a little bit of an experiment.
So we said that concentration effects rate, right?
So we will try our first reaction
with some hydrogen peroxide.
And I'm going to pour out about 10 milliliters of this
and then dilute it so that our new concentration is about 7%.
Yeah, that's right.
So pour that in there.
And then we want to use a potassium iodide solution.
So we can pour some of this in here.
And then we have the oxygen bubbles,
and who did the catalase reaction yesterday in lab?
Anybody who's here?
So what did you guys see when you added the enzyme
to your peroxide in the tube?
Bubbles, good.
But it didn't foam up, right?
Because they just popped and formed oxygen.
So the way that we can get this to foam is by adding some soap.
This is just powdered soap.
And we can add a couple of scoops of that.
And now hopefully, if all goes well,
when I add these together, we will see the decomposition
of hydrogen peroxide.
I can probably hold it up so you can see what's happening.
So what's happening in there?
It's bubbling, good.
Excellent.
So what are the bubbles?
Oxygen. All right, so it's taking its time.
So while that is going, we can set up another one.
And this time, we won't dilute the hydrogen peroxide,
so we'll just use 40 milliliters of hydrogen peroxide.
Need our soap.
Yeah.
Yeah, so it looks like toothpaste maybe.
Toothpaste that an elephant would use, yeah.
What?
Elephant toothpaste.
Yeah.
I also, in the course of doing some research for this,
found that some people call it old foamy, like a guyser.
So you may have heard it by that name.
All right, and then we add the same amount
of potassium iodide.
Hopefully I measured this out properly.
So if that was 7% hydrogen peroxide,
what do we think is going to happen if we use 30%?
Faster.
Let's find out.
So can you see the steam coming off of it?
Yeah, so what does that mean about the reaction?
It's exothermic.
Yeah, it's very hot.
If you were to come up here and touch this, it is quite warm.
But yeah, so that went faster, right?
So we can do a little bit of math
about how much faster the potassium iodide makes
the reaction go than just what would happen in your bottle
at home.
And so our activation energy of hydrogen peroxide decomposition
that is uncatalyzed, just happens in nature,
is 75 kilojoules per mole.
So if we use our Arrhenius equation,
then we can plug in 75 kilojoules or 75,000 joules
to match our or gas constant units.
And then the activation energy of the decomposition
as catalyzed by potassium iodide is
about 56 kilojoules per mole, so it lowers it,
makes the reaction go faster.
If you do this math out--
I don't want to get my numbers wrong here--
it is about 2,500 times faster or 2,400 times faster
than what would happen in nature.
Now, if we talk about catalase, and you guys
will calculate the actual activation
energy of the catalase-catalyzed reaction,
I will start and tell you that it is less than 10 kilojoules
per mole.
So that is a pretty significant change, right?
Going from 75 kilojoules per mole down to 10 or less.
That's a lot.
So if you do this math out, where you go from 75,000 joules
to 10,000 joules, does anyone have any guesses
as to how much faster the catalase is
than just normal decomposition?
A lot.
So that's a lot faster.
So you can be thinking about that
when you are thinking about what happens
in the cells of your body because you are forming
these superoxide radicals and hydrogen peroxide
all the time in your cells.
And this reaction is happening on this scale, or the reaction
that you just saw is happening this many times faster
in your body, and you will see it happen also in your flask.
So any questions, thoughts?
No?
All right, well, thank you guys.
You can head to lab.