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PROFESSOR: OK, let's get started.
Everyone can go ahead and take 10 more seconds on today's
clicker question.
This is about hybridization, which we will go over more to
start today, but just to see where we are
as a starting point.
All right, great.
So most of you got this.
The question is what was the hybridization of an atom if we
know that it has two hybrid orbitals.
This is really, really important, so I'm just going
to write this out for that 20% of you that didn't see this
right away.
So we know that when we have hybrid orbitals that come from
s and p orbitals, these are our atomic orbitals we have to
choose from to hybridize.
So if we know, for example, that we need two hybrid
orbitals, that means we needed to have started with two
atomic orbitals.
So that means what we would form are two s p hybrid
orbitals and we'd be left with our other two p
orbitals left over.
So any time you figure out that you're going to need two
hybrid orbitals, and we'll go over when you would find this
out in just a minute, but any time that happens you know it
has to come from an s and a p, so that's why you'll end up
with s p hybrid orbitals there.
And exam 2 is happening in just one week from today, so
next Wednesday.
The semester moves really fast. So I can't believe we're
already up to exam 2 talk, but we are.
So, you should have all received two handouts when you
walked in today, your notes and then an exam 2 info sheet.
Did everybody get that?
If you didn't get it, just raise your hand and a TA will
come to you with that.
I see there's a few around, if one of the TA's can jump up
and fill those in.
All right, so basically, as with exam 1, all the
information you need to know about what's going to be on
exam 2 is on this handout.
Specifically exam 2 is going to cover
lectures 10 through 17.
Today is lecture 16, so that means it's going to cover
today's lecture and then a bit into Friday's lecture as well,
so I'll be really clear on Friday where exam 2 material
ends, where you can kind of switch off the exam 2 part of
your studying.
And it will also be on problem-sets 4 and 5.
So it's going to be on two problem-sets instead of three
problem-sets, but as you know, there are a lot of different
concepts covered in these two problem-sets.
So it's still a lot to be thinking about and you need to
make sure that you really understand all of these
different concepts.
So this is explicitly listed in this sheet here, so make
sure you look it over.
But I just want to say is well, the concepts that we've
covered since exam 1 have been talking about covalent bonds,
going along with that are all the Lewis structures that
we've done.
Then we talked about ionic bonds, so you need to make
sure you can solve those ionic bond problems. And then we
started talking about different types of bonding in
terms of thinking about MO theory, and also vesper theory
and hybridization.
And then today, and a little bit on Friday, you'll be
responsible for some of this new material, which is going
to be a little bit of thermochemistry.
Not too much on exam 2, most of that will be saved for exam
3 material, but an intro to thermochemistry, that will
also be on the exam, just as far as these two lectures.
So, basically this is all written out here, and I also
want to mention in terms of your MO diagrams and MO
theory, I sent out an email about this, but I also wrote
it out here so you have it written down in terms of what
you are responsible for in terms of ordering the energy
levels of molecular orbitals.
So that's all written down explicitly right here, this is
what you need to know for the exam.
Also it's helpful for the problem-set as well.
And also, in terms of MO diagrams, I think that you
will already gone over this pretty explicitly either last
week in recitation or yesterday, in terms of what
you need to do to get full credit when you're actually
drawing an MO diagram.
Just in case you don't have that information or you didn't
write it down, I wrote this down for you here as well in
terms of things like making sure to draw your energy
arrows, or remembering to label all
of your atomic orbitals.
We don't want anyone getting points off on the exam for
doing something silly where they do understand what's
going on, but they're just not writing it
in the correct form.
So all this information is on your sheet here.
And also on this sheet mentions that you will be
getting optional extra problems on Friday.
This is just like for exam 1.
The way that you should approach these extra problems,
you don't need to turn them in, you're not required to do
them, but if you want to do well on the exam we really,
really encourage that you do these problems. And in terms
of having it be worth your time when you're solving these
problems, what you want to do is make sure you're at the
point where you can put your notes away, where you can sit
alone without your friends, and really approach these
extra problems as if they're exam problems. This way if
there's some conceptual leap that you haven't quite made
yet, but you don't realize it because you've been using your
notes or using the conversations with your
friends, you want to make that leap and run into those
problems alone in your room when you have two hours to
figure it out.
You don't want to do this leap on an exam.
That's not the ideal situation.
So especially with Lewis structures where sometimes you
can run into confusions that you need to work through and
make sure it all makes sense in terms of your head how to
do this, make sure you take the time to do that before the
exam and don't have to end up struggling with that on the
exam in a time situation.
All right, so the extra problem solutions will be
posted on Sunday.
And in terms of office hours for next week, my office hours
are going to be on Monday from 2 to 4, I'm moving them up,
and your TAs will move up their office hours as well, so
check with them in recitation or on the website.
So that's pretty much it for exam 2.
Are there any questions about what you're responsible for or
anything regarding exam 2?
OK, and in terms of the room it's in Walker again, so at
least this part should be familiar.
You'll be used to going to your row, it'll be the same
row, it's all written here again, but you can just do
exactly what you did for exam 1.
All right, so one last class announcement, and this is that
if you've been looking at our first sheet of the whole year,
sort of the course overview sheet, you know that our class
in terms of grading is out of 750 points, and 50 of those
points are for attendance and something called occasional
in-class quizzes.
So what we're going to do is start filling in some of those
points because we only have 36 classes worth of attendance.
So we have a couple of points to make up in terms of
quizzes, and these are going to be clicker quizzes, and let
me explain how this is going to work.
Basically, in at least for my part of the class, what we're
going to do, because what I really care about is that
people are answering the clicker questions, it's really
valuable for me.
I don't necessarily care if you get them right or wrong
during class, because this is the point at which you're
still learning them.
So these quiz points, which could be up to 2 points per
class, will be in addition to attendance points, and you'll
get full credit if you answer the quiz question.
So you need to be answering the questions because the
quizzes will not be announced.
So this is another reward for people that are always
participating.
These clicker questions are really valuable to us in terms
of the feedback, it really gives me an idea of where the
class is as a whole.
So really, it's important that you are answering these
questions and this is a way to get your quiz points up, so in
addition to attendance, you'll get 2 quiz points.
We'll have our first unannounced quiz today, so you
won't know what problem it is beforehand, so I suggest you
make sure you're answering all the clicker questions, which
most of you do anyway.
And I will just say I recognize that some of you are
just scurrying across campus to get here on time.
So I'll never make that first question a quiz question,
because I know how hard it is to get from one place to
another on this campus, and I know most of you do a great
job of sneaking in right at the last minute when you have
that long run.
So we won't make the first question a quiz question, but
be on the look out for a quiz question today.
All right.
So let's get to today's topic.
We're going to finish talking about valence bond theory and
hybridization.
So what we've already done is really cover all the theory
that we're going to cover behind how hybridization
works, and behind how valence bond theory works.
But what we haven't covered is how do we actually solve
problems doing this in the really quick way.
So we're going to do that.
This is going to be a more practical lesson in
hybridization.
And once we do that, we'll move on to now talking about
energies and enthalpies of chemical reactions.
So this is a real shift.
We spent a while talking about single atoms, and then we've
spent most of the material, since exam 1,
talking about bonding.
So now we're taking it one step further and we're going
to actually get to talk about some chemical reactions.
So we'll do that once we finish up with our
hybridization unit here.
All right, so how do we go about determining
hybridization in complex molecules?
It's actually incredibly simple, and all that you need
to do is think about for the given atom that you're looking
at, what you want to do is add up the number of atoms that
are bonded to your central atom, or the atom you're
considering, and add to that the number of lone pairs.
And what you end up with is the number of hybrid orbitals
that you need.
All right, so the theory behind hybridization and
picturing everything, that can be pretty complicated, and I
do want you to understand that.
But if, for example, on an exam situation, you don't
necessarily have to think through everything, you can
just use this very quick way to figure out the number of
hybrid orbitals that you need.
And what we just said is that any time we need two hybrid
orbitals, we know that the hybridization of that atom is
going to be s p, because it's made up of
one s and one p orbital.
So let's say we need three hybrid orbitals.
What is our hybridization of our atom in this case?
Yeah, it's s p 2.
We want to take three atomic orbitals to make three hybrid
orbitals, so it's going to be s p 2.
So what if we need four hybrid orbitals, what's the
hybridization going to be in this case?
STUDENT: S p 3.
PROFESSOR: S p 3.
All right, great.
So, really that's it.
Hopefully hybridization just got a lot simpler for any of
you that have been struggling a little bit with it.
But let's go ahead and do an example to make sure that we
can all do this.
And let's take a look -- first let me mention
an exception here.
The one exception to thinking about how many hybrid orbitals
that you have are in the case where you have single bonded
terminal atoms. So in the case of a terminal atom that is a
single bond, you're not going to hybridize it.
So in that case, just don't even change anything it's just
going to be one of the p orbitals, or unless you're
talking about hydrogen in which case will be an s
orbital that overlaps.
All right, so we can illustrate this with an
example of formal chloride.
So this is a good example because we're just dealing
with our carbon as our central atom here, and we have three
terminal atoms, two of which are single bonded, so we're
not going to hybridize those, and one which is double
bonded, the oxygen, so we will hybridize that one.
So in terms of thinking about the carbon, what is a
hybridization around the carbon atom?
STUDENT: [INAUDIBLE]
PROFESSOR: All right, I'm hearing some mixed answers.
So let's think.
So the carbon atom is bonded to three different atoms, and
it has no lone pairs.
So what's the hybridization?
STUDENT: S p 2.
PROFESSOR: Good s p 2.
So, if we're talking about the c h bond, we're going to say
that it's carbon, it's a sigma bond because it's a single
bond, and then it's carbon 2 s p 2 bonded to a
hydrogen 1 s orbital.
All right, so let's take a look at the carbon
chlorine bond here.
Again, carbon is still 2 s p 2, it's the same carbon atom,
and it's going to be a sigma bond because it's a single
bond, but what about the chlorine, what atomic orbital
is going to be bonding in the chlorine?
STUDENT: [INAUDIBLE]
PROFESSOR: All right.
I'm hearing a little bit of mixes, they're all p's, which
is a good start.
It turns out that it's going to be the chlorine 3 p z.
So the reason that it's p is because we're not hybridizing
it, and the reason that I specifically say that it's the
p z instead of the p x or the p y, is remember that the
z-axis, that's our bonding axis, that's our
internuclear axis.
So any time we have a p orbital that's involved in a
sigma bond, it has to be the z orbital because that's the
only one that has the right orientation to overlap along
that z-axis.
So we're going to say it's carbon s p 2, and then
chlorine 3 p z.
All right, so let's look at the last bond here, which is a
carbon oxygen double bond.
We know that any time we have a double bond it's made up of
one sigma bond plus one pi bond.
So, when we look at the carbon again, that's 2 s p 2.
But what about the oxygen, what's the hybridization of
this oxygen atom?
S p 2.
So we're going to have carbon 2 s p 2, and then
oxygen 2 s p 2.
It's s p 2 because the oxygen is bonded to one atom plus two
lone pairs, so we're going to have a total of
three hybrid orbitals.
All right, so this is not our only bond, we have a double
bond, so we also need to talk about the pi bond.
So if we talk about the pi bond, we can say that's carbon
2 p y, then the oxygen 2 p y.
We could alternatively say, and be correct, that it was
the carbon 2 p x and the oxygen 2 p x.
Either one of those is fine.
All right, so that's an example of how we can assign
very quickly and very easily what the hybridization is of a
given atom.
And then also fully describe the symmetry and the atomic or
hybrid orbitals that make up bonds.
So this is a lot of your problems on the p-set, so if
you haven't finished that section, hopefully you'll be
able to get through that section pretty quickly when
you go back to it.
And let's take a look at a little bit of a more complex
molecule here, which is ascorbic acid or vitamin C.
And this is a good example because it's starting to look
a little more complicated, but not too much more complicated.
But the reality is if you know how to do assigning the
hybridization this way, even if I gave you a 1,000 atom
protein, you should still be able to get
it completely correct.
Vitamin C is a really important molecule to actually
think about in terms of thinking about its shape and
its hybridization.
Vitamin C is an antoxidant.
So we might remember that from when we were talking about
free radicals.
Free radicals can cause oxidative damage.
These antioxidants, some of our vitamins, including
vitamin C are antioxidants.
So that's something that a lot of people in the general
population know if they're into their health.
But another really important thing about ascorbic acid is
that it's an enzyme cofactor.
And a cofactor just means that it's some type of molecule or
atom, in this case it's a molecule, that's required by
an enzyme in order for the enzyme to
carry out its chemistry.
And the chemistry that the enzymes that vitamin C are a
cofactor for are responsible for putting oxygen, o h
groups, onto collagen molecules in order to form
what is the collagen triple helix.
And collagen is the main constituent of bones, of
joints, of connective tissue, of many important structural
parts of our bodies.
So you can imagine that vitamin C is very important
because we need to keep our collagen intact.
Collogen is actually one of the most prevalent proteins in
our entire body, and it makes up a large part of our cells.
So you can imagine if we don't have collagen that is in this
triple helix, we're going to run into some problems. Does
anyone, or I'm sure many of you do know, but can someone
tell me what the name is of the disease if you have a
deficiency in vitamin C?
STUDENT: Scurvy.
PROFESSOR: Yeah, scurvy.
So scurvy, most often associated with ships because
of the 1500's and 1600's and 1700's when boating technology
was far above biochemistry technology, so people could
make ships and go on these long journeys, but people did
not yet understand they needed to keep certain vitamins in
their bodies, such as vitamin C. Maybe also, they didn't
know about -- well they didn't even have the structure, so
they couldn't have realized that vitamin C is polar and
won't stay in their bodies for very long, it will get
excreted out through their urine.
So they were not thinking about how quickly they could
run into trouble in the long sea voyages.
And I don't know how many of you had a unit on explorers in
elementary school --
I did in fifth grade, and one that always sticks in my head
is Magellan's trip around the world, who led the first fleet
of ships around the globe.
What they did not mention to us in fifth grade was that 90%
of his men died on that journey.
And I understand we were young kids, I didn't want to know
that at the time, but it's interesting and it's really
important, and the reason for many of
these deaths was scurvy.
And it's totally preventable.
And, in fact, the cure for scurvy, which is as simple as
taking vitamin C in any form, was known for -- since the
fifth century people started figuring this out, but the
problem was there was so much false information as well that
it was not a general practice on ships to be treating men
that were suffering with vitamin C or using it as any
kind of preventative measure.
The other somewhat interesting thing I want to say about
vitamin C is that it was part of the first ever published
clinical study.
And this was a controlled clinical study that was done
in the 1700's, the first one that's ever been reported, and
not surprisingly it was done on a ship and it was done by
James Lind who was a Scottish naval surgeon.
And basically, he took 12 of his men, so this was not a
very large clinical study like they do today, but he took 12
men that were suffering from scurvy, and he gave them a
diet mostly of just carbohydrate mush, because
when you have scurvy, one of the first signs is bleeding
gums and teeth falling out, they couldn't
really eat too much.
But he supplemented their diet.
He put them in pairs of two and
supplemented with 1 of 6 things.
So two of them got a pint of cider a day
to see if that helped.
Two of them got half a pint of seawater -- maybe not the best
group to be in.
Even worse group to be in is two of them got diluted
sulfuric acid.
Tuned out that didn't help so well.
Other things where some got spices, not so bad.
And then two of them actually got citrus fruit.
So no big surprise ending to this clinical study, the
people with citrus fruit were completely cured and back to
duty within a week, and I'm not sure exactly what happened
to everyone else, but let's hope they got
switched over as well.
But it's just neat to think about even back then, people
were doing these controlled scientific studies, this was
one of the first cases.
Unfortunately, it was another 40 years between this study
being published and the British Navy requiring
supplements of citric acid.
So, really, we've come a long way in terms of disseminating
scientific knowledge to the community, and that's a really
important part of being scientists -- not just making
these discoveries, but also passing along, for example, to
these ships full of people that need to know it.
So, in terms of scurvy, we don't see too much of it
today, but who should be concerned about
suffering from scurvy?
Do we have to worry about our cats or dogs at home, are they
getting enough of their vegetables and vitamin C?
It turns out we don't have to worry.
Primates are who should be concerned.
So we need to be concerned, other
primates need to be concerned.
It turns out that most other mammals actually biosynthesize
vitamin C, so we don't have to worry about most
of our pets at home.
Unless you have a guinea pig and they don't, in fact,
biosynthesize vitamin C, so I think they supplement most
guinea pig pellets with vitamin C, but maybe keep an
eye on your guinea pig.
We primates and guinea pigs are the only types of mammals
that don't actually biosynthesize their vitamin C.
So again, vitamin C is a cofactor, that means it needs
to bind to an enzyme, it means that the shape of vitamin C
and the hybridization are going to be really important.
So let's take a look and talk about some of these carbon
atoms here.
And you can look at your structure of vitamin C in your
class notes while you're doing this, and first I want you to
tell me what the hybridization is of that carbon a in the
vitamin C molecule.
And let's take 10 more seconds on that.
OK, 78%.
We'll do another question like this in a minute.
So if we can switch back to the notes we'll
take a look at that.
I think we can get better, I think we can get to the 90's
with these hybridization, we just need to follow the rules.
So if we're thinking about vitamin C and we're talking
about carbon a, how many things is carbon a bonded to?
STUDENT: [INAUDIBLE]
PROFESSOR: four things.
We have two hydrogen atoms, an oxygen, and
then another carbon.
So that means we need, if it's bonded to four things, we need
four different hybrid orbitals, so it needs to be s
p 3 hybridized.
All right, let's try this again but just shouting out,
what is the hybridization of carbon b?
Good.
OK, I'm going to say that was about 90% of you I heard,
which is excellent.
S p 3.
What about carbon c here?
Yup, s p 3 again.
Bonded to four things it's s p 3.
What about carbon d?
S p 2, great.
Carbon e?
STUDENT: S p 2.
PROFESSOR: All right, and this last one, the last 10% of you
can join in.
Carbon f?
Great, s p 2.
All right, so we can also think about shape once we know
hybridization and how many atoms a carbon
atom is bonded to.
What would we say that the shape is of these s p 3
orbitals -- or the shape of the carbon atom.
Yeah, it's tetrahedral.
So this is going to be tetrahedral.
And what about carbon d, e, and f?
What is the geometry of carbon d, e, and f?
Good, trigonal planar.
All right.
So let's actually take this one step further from just
talking about the hybridization of the
individual atom.
Let's talk about a couple of these bonds.
I won't go through all of them.
I think you can get the idea with just a few, but you can
go back and try all of them later.
So let's start with thinking about carbon b hydrogen bond.
Is this going to be a sigma or a pi bond?
Sigma.
We know it's sigma, because any time we have a single bond
it's a sigma bond.
So carbon b, we just said, is carbon, it's
going to be 2 s p 3.
And then hydrogen, what's the atomic orbital we're talking
about here?
1 s.
All right, great.
So let's take a look at another one, let's look again
at carbon b, but now let's talk about it bonding to its
oxygen there.
So again, are we going to be talking about a
sigma or a pi bond?
A sigma bond, because again, it's a single bond here.
So what we'll say again is carbon 2 s p 3, and let's take
a look at this oxygen that it's bound to.
Is this oxygen -- what is the hybridization of this oxygen
right here?
What is it?
OK, good, it's s p 3.
It's bound to two atoms, plus it has two lone pairs, that's
a total of four things.
So it needs to have three different hybrid orbitals, so
we'll say it's oxygen 2 s p 3.
So let's skip to a different carbon now, let's talk about
carbon d, specifically the carbon d oxygen bond.
So let's do a clicker question for this one.
Let's get into the 90's, if you don't mind, this would be
very good to do.
It'll make me feel better about you finishing your
problem-sets tonight.
So if you talk about the symmetry and the hybrid or
atomic orbitals that contribute, we're talking
about the c d oxygen bond.
So go ahead and look at your notes and see
what that bond is.
We already identified the hybridization of carbon d, so
we really just need to think about the oxygen here.
All right, let's take 10 more seconds.
OK, I'll take that, 84%.
That's not bad.
So again, we're looking at a sigma bond here.
It looks like the one people got it confused with was
talking about oxygen s p 2 versus s p 3.
Remember, you just need to look at what it is actually
bound to, so for carbon d, the oxygen's bound to two atoms
plus two lone pairs, so it's going to be s p 3 hybridized.
So we'll call this a sigma bond where we have carbon 2 s
p, now it's s p 2, and oxygen 2 s p 3.
All right, let's take a look at just one more here.
So, let's look at the carbon d bond with carbon e.
So if we can switch back to the class notes just to see
that, though you guys can see it on your notes as well.
So in terms of that, we're talking about a double bond
now, so we know that we have to have one sigma bond and one
pi bond to completely describe our double bond.
So for the sigma bond, again, it's going to be
carbon 2 s p 2.
And now we're talking about the other carbon.
What is the hybridization of the other carbon?
Yup, s p 2.
So carbon 2 s p 2.
All right, but we're not done there.
We also need to talk about the pi bond.
So for the pi bond, we'll talk about carbon 2 p y, and then
the second carbon 2 p y.
All right, so I filled in a few more in your notes, so you
can maybe cover those up and make sure that you get all of
those correct is a good self test before you finish off any
in your problem-set.
Does anyone have any questions about figuring out
hybridization?
Yes?
STUDENT: [INAUDIBLE]
PROFESSOR: I'm sorry, what is that?
In this bond here?
Oh, OK, that's a good question.
So the question was why is there not hybridization in
these p orbitals here.
So in order to form a double bond, we need to have a pi
bond forms, and a pi bond forms from two
unhybridized p orbitals.
And if you can kind of picture those p orbitals, if our
molecules are this way and they're coming up here, we
need to have electron density above the bond
and below the bond.
So if they're hybrid, then they're going to be spread out
into that tetrahedral geometry.
We want to have them parallel to each other and they need to
be the p bonds, they need to be the p orbitals.
Does that makes sense?
So any time we have a double bond, the pi part of the bond
is going to be p orbitals.
Yes?
STUDENT: [INAUDIBLE]
PROFESSOR: No, absolutely not.
It would be absolutely correct to put 2 p x and 2 p x here,
you just can't put z, because that's the one that's going to
be involved in the sigma bond.
And the other important thing is if you do put x for one,
you have to make sure you put x in the other, because they
do need to be able to interact with each other.
You can't have one x and one y, but sure, you can have both
x or both y.
OK, so let's shift gears a little bit and start talking
about bonding in terms of chemical reactions.
So we're going to start today with talking about bond
energies, and also something called bond enthalpies.
And the first point is just to bring us back to something
we're very familiar with.
When we talked about covalent bonding, a concept that we
have been discussing is bond dissociation energy.
That's just the energy that's required to put into a
molecule in order to break that bond.
This is something that we saw when we were talking just at
the very beginning of our unit on covalent bonds.
That energy difference is the energy difference between when
we have the c h 4, if we're talking about methane, versus
one of those hydrogen bonds breaking, one of those c h
bonds breaking.
So we've talked in the past about bond energy, but what I
want to introduce to you today is a very related concept,
which is called bond enthalpy.
So that's delta h here is what we call bond enthalpy.
So, this is talking about instead of the change of
energy accompanied with a bond breaking, we're talking about
a change in heat accompanied with a bond breaking.
So whether when that bond breaks it requires heat, or
whether it gives off heat when you break that bond or talking
about a reaction.
That's what we're talking about when we're
talking about enthalpy.
It turns out that enthalpy is, in fact, very related to
energy, and we can relate it with this equation here that
bond enthalpy is equal to bond energy plus the change in
pressure times volume.
So we could, in fact, go back and forth between bond
energies and bond enthalpies, but the reality is if we're
talking about gases, which we are in many cases, then what
we find is the difference between bond enthalpy and bond
energy is only 1% to 2%.
So it's not very significant.
And, in fact, if we're talking about solids or liquids, now
we can say that the difference between bond energy and bond
enthalpy is going to be negligible.
So, for the rest of this class today, we're going to stop
talking about bond energy, which I did just for a moment,
to orient you back to a discussion that was familiar,
we're going to switch our discussion to bond enthalpies.
One reason that we like to talk about enthalpies is
unlike energy, which can be a little bit more tricky, bond
enthalpies are easy to measure.
It's easy to measure how much heat a reaction gives off or
how much heat a reaction takes in.
So we can also talk about something, which is called the
standard bond enthalpy, so any time you see this symbol here,
which looks like a knot, so it looks like a delta h knot,
that refers to the standard bond enthalpy.
Any time we we're talking about a standard, whether it's
enthalpy or, as we'll see in the next lecture, standard
entropy or a standard free energy, we're just saying that
the molecules that we're talking about are in their
standard states, which means they're in their pure form.
If they're a gas it means they're at one bar.
Usually this is referring to room
temperature, so 298 kelvin.
It's often that when you look up tables of different bond
enthalpies, you'll see them in their standard state and it
will be at room temperature.
So if we talk about standard bond enthalpy, let's talk
about what this is for these c h bonds, which
we just saw in methane.
So if we're going from c h 4 to breaking one of these c h
bonds, what we see is that the bond enthalpy is 438
kilojoules per mole.
The fact that it's positive tells us that we need to put
in that much heat into the system in order
to break that bond.
So we now know the enthalpy of a c h bond, but, of course,
methane is not the only kind of c h bond
that you can envision.
We could talk about the c h bond, for example, in ethane
or trifluoromethane or trichloro or tribromomethane,
and what you'll find is that the amount of enthalpy that
that reaction is associated with depends on the exact type
of c h bond that you have. So, for example, we can see that
we have slightly different, not vastly different, but
slightly different changes in enthalpy depending on which
kind of c h bond that were breaking.
And what I want to point out also is that all of these
delta h's, all of these standard bond
enthalpies are positive.
Any time we have a positive delta h, we call this an
endothermic reaction.
And again, endothermic just means that the
reaction takes in energy.
It's very similar to what when we're talking about energy
whether something's taken in or released, whether it's
positive or negative.
If you have positive delta h, it's endothermic, it takes in
heat in order for the reaction to go.
So we can think about instead of talking about all of these
individual c h bonds, instead we can talk about what the
average value would be for all c h bonds.
So we can say that the average of any c h bond is 412
kilojoules per mole.
So you can imagine if you're looking these things up in
terms of a reference table or maybe in the appendix of your
textbook, it would be an impossibly huge number of
pages to write every single different kinds of c h bond.
So that's why instead they talk about the average bond
enthalpies.
So for c h, again, that's 412, and all of the enthalpies we
just saw fall within about 8% of that.
And in your book you can look up any type of bond that's
listed here.
So we saw c h is 412.
You can look up a c c bond, you can look up a c c double
bond, a c c triple bond, and so on.
And you'll note that what these are are mean bond
enthalpies, so they're the average of all different types
of c h bonds or c c bonds that you can imagine.
So, you might be asking what's so important about being able
to look up and think about these different bond
enthalpies.
And the reason that it's important is because if you're
looking at a reaction, no matter how complicated that
reaction is, you can actually figure out what the enthalpy
of the entire reaction is by adding up all the individual
mean bond enthalpies of the products, and all the
individual mean bond enthalpies of their reactants
and thinking about the difference between those two.
So we'll do that in just a second, and the reaction that
we'll do it with is the oxidation of glucose here.
So, c 6 h 12 o 6, one mole of glucose plus six moles of
oxygen, gives us six moles of carbon dioxide and
six moles of water.
So what we can find out, and hopefully what we will match
up when we look at using the different bond enthalpies, is
that the enthalpy of this entire reaction is negative
2816 kilojoules per mole.
So in this case we're saying that delta h is negative.
Does anyone know what it's called
when delta h is negative?
Everyone knows, great.
So it's exothermic.
This is an exothermic reaction, the reaction
releases heat.
So I just want to mention before we go on, if you look
at almost any freshman chemistry textbook, what
you'll find is this oxidation of glucose reaction is used a
lot in talking about thermochemistry.
And one reason it's talked about is because it's very
convenient to talk about something where we start with
one mole of glucose and end up with 12 moles of products.
That's going to be helpful what we're doing a practice
problem to see exactly how you deal with it when there's
different numbers of moles.
But the other reason you always see this reaction is
because it's an incredibly important reaction.
The oxidation of glucose is going on all the time in our
body, this is our main source of energy for all animals.
So let's think a little bit about why this reaction is so
important and so prevalent.
It turns out that if we're talking about plants, plants
do the reverse reaction.
So plants take carbon dioxide and water and they turn it
into glucose or energy for us, energy stored in the bonds of
glucose plus oxygen.
So if we do the reverse reaction, this is actually
going to require energy.
Where do plants get this energy?
Yeah, this is just photosynthesis here.
This is the photosynthesis where plants are turning
carbon dioxide and water into sugar and into oxygen.
So what happens when we eat the plants or when we eat
animals that have eaten the plants is that we perform the
reverse reaction now, which is what I just showed you, the
oxidation of glucose.
And even though it's not the products of the reaction that
are particularly valuable to us, we just breathe out the c
o 2, and actually we'd rather have less of that than more in
our environment, but what's important here is instead the
energy or the enthalpy that's given often in this reaction.
So as I said, this reaction has a negative enthalpy of
2816 kilojoules per mole.
That's a lot of enthalpy and a lot of energy.
So we actually end up using that energy to fuel most of
what is going on in our bodies.
So instead of storing it as sugar, once we oxidize the
sugar, now we just store it as ATP, and as you know, ATP is
the currency of energy in the cells.
So this is why you see this reaction again and again and
again in just about any chemistry textbook that you
open up as sort of a general reaction.
The reason it's used is it's just so important and so
prevalent in terms of thinking about our bodies and how we're
staying alive and using energy.
All right, so let's go ahead and use this as an example of
what we just said, which is using the bond enthalpies of
the products and the reactants to figure out the enthalpy of
the entire reaction.
So the way that we do this is we add up all of the
individual mean bond enthalpies of the reactants,
and we subtract from that all of the individual bond
enthalpies of the products.
So we can think about what this will tell us.
If you think about the fact if the bonds are stronger in the
products than they were in the reactants, you can go ahead
and click in and tell me if you think we'll have a
negative or a positive delta h here.
All right, let's take 10 more seconds on this.
OK, great, so negative is correct and some people got
totally mixed up.
Didn't just get one thing -- so let's just focus on the
right answer to start with.
The correct answer is that it's negative, it's an
exothermic reaction.
And actually, lets switch to our class
notes to explain why.
And also, this was the quiz question for today, so whether
you got the answer correct or incorrect, you get full quiz
points if you did, in fact, answer.
But let's still focus on the right answer here and see why
it's correct.
So if we're talking about the bonds being stronger in the
products, that basically means that we ended up releasing a
lot of heat when we made those bonds and the products, and we
didn't have to use up too much of that heat to break all the
bonds and the reactants.
So that's why we say that delta h
is going to be negative.
You could also just do it not conceptually, just plugging it
into the equation, but it's better to kind of understand
exactly why that is.
It's because it takes more energy -- you gain more energy
forming the products than you take
breaking up the reactants.
So if we have the opposite case here, if the bonds are
stronger in the reactants, now what we're going to find is
that the delta h of the reaction is positive and what
you're dealing with is an endothermic reaction.
So let's go ahead and do this with our example of the
oxidation of glucose.
So, I've just written out the glucose molecule so you can
see all of its individual bonds here since we're going
to be using that information.
So let's start by talking about the bonds that are
broken in terms of thinking about all of
the reactants here.
So if we're talking about the sugar molecule itself, what we
have is we have seven c h bonds.
And if we add up all the o h bonds, we have five of those.
How many c o single bonds do we have?
What do people think?
Yeah. five c o single bonds.
What about c c single bonds? five of those as well.
C o double bonds?
Just one c o double bond.
And then we have our oxygen molecule to worry about and we
have six o o double bonds.
So if we add up all of the bonds are broken, and we
subtract from that the bonds that are formed, those
strengths, the c double bond o, we have 12 of those.
And how many o h bonds do we have?
Right, 12 as well.
So if we add up all the bonds broken, what we end up with is
12,452 kilojoules per mole, and it's talking about per
mole of glucose that's oxidized.
And in terms of bonds formed, what we see is 15,192
kilojoules per mole.
So all we need to do to figure out the change in enthalpy,
and when it has this subscript r here, I meant to mention,
that means the enthalpy for the reaction.
We just subtract these bonds here from the ones that we
ended up forming.
So basically, what we're saying -- excuse me, these are
the ones that we broke.
It's 12,452 minus 15,192, which we formed.
So we end up with an enthalpy of reaction of negative 2,740
kilojoules per mole of glucose that's oxidized.
All right, so if you remember the number that I told you
before, it doesn't exactly match up with what we had said
is the exact number.
The exact number is 2,816.
It is within 3% though, that's pretty good.
Because remember, we're not using exact bond enthalpies
here, what we were using is average bond enthalpies.
So it makes sense that we're going to be a little bit off.
But if all you have in front of you is the information on
bond enthalpies, mean bond enthalpies, this is a great
way to figure out the enthalpy of an overall reaction.
We can think of a different way, however, to think about
the enthalpy of an overall reaction, and this is talking
about the heat of formation.
And the heat of formation or delta h formation here is
equal to the heat of -- or the enthalpy of reaction, if we're
talking about forming one mole of compound from its pure
elements, which are in their standard states.
So basically, there's a table that you can look up which
tells you the enthalpy of formation for any compound
that you're interested in, and this is actually an appendix
to of your textbooks, and you will need to use this to solve
your problems on p-set.
But let me say that, for example, if we're talking
about water here, that's formed from hydrogen and
oxygen, if we're talking about the elements in their pure
forms at one bar, standard states, and room temperature.
And if we look this up in our textbook, we would find that
the delta h of formation for water is negative 286
kilojoules per mole.
Similarly we can look up the same thing, for example, for
carbon dioxide, and what we'll find here is that our delta h
of formation that we look up is negative 393 .
5 kilojoules per mole.
We can also look up or think about what our delta h of
formation would be for oxygen.
Does anyone have a guess here?
Yup, it's actually going to be zero.
Oxygen already is in its most stable state, so any time
we're talking about an element in their most stable state,
it's going to be zero.
That's going to be the delta h of formation, there
is none, it's zero.
We can also look up what it was for sugar, for glucose, c
6 h 12 o 6.
So that coming from its most stable forms, it was the most
pure form of the elements, is going to be negative 1260
kilojoules per mole.
All of this information has been tabulated and is in
tables, you can refer to them, and they're also in the back
of your textbook, so you can refer to those in terms of
your problem-set.
So let's go ahead and solve a problem actually using the
delta h's of formation.
Let's see if we can do any better to coming close to the
reality for the oxidation of glucose.
So if we're going to calculate the delta h for the reaction
for the oxidation of glucose, or actually for any reaction
at all, what we want to do is take the delta h of formation
of the products and subtract from that the delta h of
formation of the reactants.
So let's go ahead and do that.
And in terms of talking about the oxidation of glucose, if
we talk about delta h of this reaction, what we need to take
is 6 times delta h of formation of c o 2, since
we're forming, that's in our products, we're forming six
moles of c o 2.
Plus 6 times delta h formation of h 2 o, since we're forming
six moles of h 2 o.
And we subtract from that the heat of formation of all of
our reactants.
So we only have one mole, so we just say delta h formation
of c 6 h 12 o 6, and then in addition to that we have six
moles delta h formation of oxygen, of o 2.
All right, so this is our equation here and at this
point what we would do is we would look up what all of the
delta h formation values are for c o 2, h 2
o, sugar, and oxygen.
So what we would find is that we end up having 6 times
negative 393 .
5 -- that's what we had just looked up and
told you for c o 2.
Plus 6 times negative 285 .
8 for the water.
Minus 1 times 1260, so it's negative 1260
for our sugar here.
And then what we're going to end up with is having minus,
and what is it for oxygen again?
So minus 0.
So what we end up with in terms of the delta h for the
entire reaction here, is we end up with negative 2816, and
it's going to be kilojoules per mole of glucose.
All right, so let's see how this matches up, and hopefully
you can actually remember that this matches up actually
perfectly here.
So what we are going to see is that, in fact, what we
calculated versus what is experimental
is dead-on the same.
So there's actually one more way to figure out the
enthalpies of a reaction, I'm going to go over
it just very briefly.
And that's based on the fact that enthalpy is a state
function, and by state function what I mean is that
it doesn't matter how you got from point a to point b, all
that matters is the difference between the two.
So another example of a state function is altitude, for
example, on a mountain.
So if you're talking about altitude, you go from point a
to point b on the mountain, and it doesn't matter how you
got there -- you could have climbed all the way up the top
of the mountain, then went back down and eventually
landed on point b.
Or you could have gone straight from
point a to point b.
The change in altitude between point a and point b do not
depend on the path, they're independent of path.
So this difference right here does not matter on how you got
there, it does not make a difference how you got there,
altitude is a state function.
So similarly, we can say that enthalpy is a state function
as well, if we're talking about part reaction a here,
which has our reactants going to our products in 2 b here.
It doesn't matter how we got there.
We can actually calculate the change in enthalpy at all
different points here.
All that we actually have to worry about at the end of the
day is the difference between a and b.
So that's what we mean by state function.
Actually, we're not going to have a chance to get to fully
explaining the consequence of this, which is Hess's law,
which allows us to add and subtract different reactions.
So what I'm going to say is the last two problems on your
problem-set you won't have to do because we're not going to
get a chance to cover in class on Friday.
So I'll send out an email about this as well, so that'll
be pushed back.
So, it will be on the exam, so you want to do them at some
point, but you don't need to actually turn those in.
So those are the last two problems, I believe.