This course will cover the topics of a full year, two semester General Chemistry course. We will use a free on-line textbook, Concept Development Studies in Chemistry, available via Rice’s Connexions project. The fundamental concepts in the course will be introduced via the Concept Development Approach developed at Rice University. In this approach, we will develop the concepts you need to know from experimental observations and scientific reasoning rather than simply telling you the concepts and then asking you to simply memorize or apply them. So why use this approach? One reason is that most of us are inductive learners, meaning that we like to make specific observations and then generalize from there. Many of the most significant concepts in Chemistry are counter-intuitive. When we see where those concepts come from, we can more readily accept them, explain them, and apply them. A second reason is that scientific reasoning in general and Chemistry reasoning in particular are inductive processes. This Concept Development approach illustrates those reasoning processes. A third reason is that this is simply more interesting! The structure and reactions of matter are fascinating puzzles to be solved by observation and reasoning. It is more fun intellectually when we can solve those puzzles together, rather than simply have the answers to the riddles revealed at the outset. Recommended Background: The class can be taken by someone with no prior experience in chemistry. However, some prior familiarity with the basics of chemistry is desirable as we will cover some elements only briefly. For example, a prior high school chemistry class would be helpful. Suggested Readings: Readings will be assigned from the on-line textbook “Concept Development Studies in Chemistry”, available via Rice’s Connexions project. In addition, we will suggest readings from any of the standard textbooks in General Chemistry. A particularly good free on-line resource is Dickerson, Gray, and Haight, "Chemical Principles, 3rd Edition". Links to these two texts will be available in the Introduction module.
CDS 13 Reaction Energy and Bond Energy II
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General Chemistry: Concept Development and Application
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Energy Changes and Reaction Energies
Chemical reactions involve energy changes, most commonly with the transfer of heat into or out of the reaction. Many chemical reactions are performed specifically because of the release of heat or other forms of energy. In this module, we develop a means to measure these energy transfers and we use these measurements to develop laws which govern energy transfers. These laws permit us to calculate and predict energy changes during reactions and to understand the energy of a reaction in terms of the energies of the bonds between atoms breaking and forming during a reaction.
Meet the Instructors
John Steven HutchinsonDean of Undergraduates and Professor of Chemistry
Now that we have a means to calculate the reaction energy associated with any
reaction, we're going to develop a means by which we can interpret what those
energies actually mean. We're going to interpret them in terms of
bond energies. To understand this, we're going to go
back to the way that we began in the previous lecture, by considering what is
it that all reactants and products have in common.
We drew a diagram before that looked something like this, where we had the
energy or the enthalpy along a particular axis, and we said we could start with the
reactants, and we could wind up for example with the products.
And we needed to pass through some kind of an intermediate which was common
between the reactants and the products, and the last time, we chose the elements
to be that intermediate. But another option could have been to
choose the atoms, because of course during the course of the chemical
reaction, the atoms are conserved. We have exactly the same atoms of the
reactants that we have of the products. So let's imagine in this case then that
the intermediates are going to be the atoms, in which case as we go from the
reactant to the atoms, what we must be doing is in fact breaking all of the
bonds. We have to be taking all of the bonds
that exist in the reactants and breaking all of them to break them up into atoms.
Correspondingly as well, to get from the products to the atoms requires us to
break all the bonds in the products. So the energy here is the bond energies
of the reactants, and the energy here is the bond energies of the products.
Let's consider that in a little bit more detail here.
The bond energy is defined to be the energy associated with taking the atoms
together, and separating them into A and B.
In this particular case, notice that this is, when those molecules or atoms are in
the gas phase, because we are ignoring the interactions between the molecules.
So this is only useful to us when we talk about things in the gases.
Notice, this is always an endothermic process.
It always costs us energy to break a bond.
In fact, we defined a bond as being the lowering of energies of the electrons
being shared when the electrons are shared between the atoms.
Therefore, in order to separate the atoms, we have to add energy back in.
Let's apply this in a particular context here by considering the reaction in which
we have a hydrogen molecule reacting with a fluorine atom to form a hydrogen
fluoride molecule plus a hydrogen atom. Turn to this diagram that we've drawn
back over here. We will consider the energy change
associated with this by starting off with the reactants, which are a hydrogen
molecule plus a fluorine atom, and the products will wind up as a hydrogen
fluoride molecule plus a hydrogen atom. The atoms in between, according to the
previous diagram, are a hydrogen atom plus another hydrogen atom plus a
fluorine atom. The energy we are interested in is this
energy, here, going from reactants to products, that's the delta H of the
reaction. To go from H2 plus F to H plus H plus F,
is in fact the bond energy of the H2 molecule.
To go from the HF to the plus H, to H plus H plus F, requires us to break the
HF bond. So, this is the bond energy of the HF.
Using our usual arguments, now, we can see that the delta H, going from here to
here, is the same as going along any pathway.
That pathway would take us first to break the H2 bond and then to form the HF bond.
Overall then the delta H of the reaction is the energy to break the bond in the
reactants, which is the bond of the energy of H2, minus the energy to break
the bond in the products, because we are now forming the bond rather than breaking
it. So this is minus the bond energy of the
HF bond, and that's the equation I've written here on the slide.
If we actually plug the numbers in by looking up the bond energy of H2 and
looking up the bond energy of HF, the bond energy of H2 is 432 kilojoules.
The bond energy of HF is 565 kilojoules, and if we take the difference between
those two numbers, we discover that the heat of the reaction is minus 133
kilojoules. That means that the reaction is
exothermic. It releases energy.
The diagram I've drawn above is correct. We went down in energy when we went to
the H2 plus F to the HF plus H, but an interesting question to ask now is, why
is this reaction exothermic? Because now we have some extra
information. In fact, we can see that the reason why
this reaction is exothermic is because of the previous line.
And the previous line tells us in fact that the bond energy of the H2 is smaller
than the bond energy of the HF. Well what does that mean?
It means we have broken a weaker bond and formed a stronger bond.
The reaction is exothermic because we formed a strong bond by breaking up a
weaker bond, and this is always going to be true.
Exothermic reactions result when we form stronger bonds, perhaps more of them, and
break weaker bonds, perhaps fewer of them.
Let's apply this to another particular reaction here.
Let's consider the reaction in which we burn hydrogen and make water out of it,
again everything in the gas phase. I've given you some particular data here,
the data required to break the hydrogen bond, to break the oxygen bond, and also
to break the OH bonds in water. Let's draw a diagram again of what this
looks like. So here we have the energy of the that we
have the reactants, two hydrogens plus an oxygen.
The products are lower in energy as we shall see to make two waters.
We know that's true because this is an enormously exothermic reaction, explosive
even. The atoms in between of course are, let's
see, four hydrogen atoms plus two oxygen atoms.
So to go from the reactants to the atoms requires us to break two hydrogen bonds
and one oxygen-oxygen double bond. To go from the products to the
intermediates would require us to break four HO bonds.
In fact, what we can compare then is the energy to go from H2O to 2H plus O, is
given here as 490 426.9. The energy to go from an H2 to an H is
given here as 436. And the energy to go from the O2 to 2O is
498. That means we can actually add all these
numbers together and say that delta H is equal to 2 times the first energy that we
have given here, 436 kilojoules, plus 498.2 kilojoules to break the OO bond,
and then minus now, 2 times 926.9 kilojoules, keeping in mind that what
I've given you here is the energy of breaking up one water molecule.
But we need to break up two water molecules, to go from the products to the
intermediate here. And when we do this calculation, we
actually get minus 483.5 kilojoules per mole of the reactants burned.
Notice again that this is a negative number.
This is a strongly exothermic reaction. And we can tell why this is such a
strongly exothermic reaction, in large measure because in comparing these
numbers, the principal difference between the reactants and the products has to do
with the fact that there are three bonds that we need to break in the reactants,
whereas there are four bonds which are formed in the products.
Consequently, we actually get energy out because we are forming more bonds than
the bonds that we broke. This suggests that it would be very
helpful for us to have tabulations of bond energies so we could do these
calculations all the time and interpret where the energy comes from in an
exothermic reaction, or where it goes in an endothermic reaction.
There exists such tables, and we're going to now redo this example, by using
tabulated bond energies, to figure out the energy of exactly the same reaction
again. This time we're going to go and look up
in the tables, and find out, what is the energy associated with breaking the H2
bond, that's 436. What's the energy required to break the
O2 bond, that's 498. And in the tables we find that the energy
associated with breaking an OH bond is 467 kilojoules per mole.
So now if we try to calculate the delta H of the same reaction, lets see, what do
we need to do? We need to take double the bond energy of
the H2, because we're going to break 2H two bonds.
We need to take the bond energy of the O2, and then we need to subtract from
that the bond energy associated with forming an OH bond, and keep in mind,
there are four OH bonds which are to be formed when we create the water molecules
in the diagram up above. If we take the numbers which I've given
to you here, plug them into the formula that we've just determined, what we in
fact discover is that we get minus 498 kilojoules per mole as the exothermicity
of this reaction. But that's not the number we got above.
Up above we got minus 483. Below we got minus 498.
The question then becomes, why wasn't this exact?
Why didn't we get the exact answer? The reason for this turns out to have to
do with the nature of the tabulated data. We have here that the bond energy, I'll
go back to the previous page, the bond energy associated with an OH bond is
given as 467. That's an interesting number.
We need to figure out where that number comes from, why that number exists.
On the previous slide, I had given you that to break both of the bonds in a, in
a water molecule costs us 926 kilojoules per mole.
But it turns out that the two H, OH bonds don't cost the same amount.
Either one of them would be the same, but once I've broken the first bond, the bond
energy of the second bond is not the same anymore.
We illustrate that with the data which is in front of you here.
To break the first bond, here I've taken one of the hydrogens off the water
molecule. It cost me 492 kilojoules.
Now I'll take the second hydrogen away from the OH which remains, and that cost
us only 434 kilojoules. That's very interesting.
What that means actually, is removing the first hydrogen actually weakens the
remaining bond between the O and the H in the remaining molecule.
In addition, the number we gave for the OH bond energy on the previous slide was
467 kilojoules, which is in fact neither this number nor this number, nor exactly
the average of those two. It's a different number altogether.
What that number actually comes from is taking the OH bond energy in many
different molecules and averaging them over.
Recognizing that the strength of an OH bond differs from one molecule to the
next, because it depends upon whether that oxygen is attached to another
hydrogen or a carbon or any other type of atom out there, or to nothing at all, as
we've seen in this particular case. Let's examine that in a little bit more
detail by considering average bond energies as given here.
For example, it turns out that if I just want to break one of the bonds in
methane, it costs me 435 kilojoules per mole.
I might think then that to break all 4 of the bonds in methane should take 4 times
435. That number is a little bit over 1700,
it's actually 1740. But the energy actually required to break
all four of the bounds is 1663, not equal to, in fact, substantially less than, 4
times 435. That immediately tells me that the energy
required to, to break the second bond, once I've broken the first one, is not
the same as the energy required to break the first one.
And to break the third one, after having broken the first and the second, is
different yet, and breaking the fourth one is different yet.
It's also true that if I look at CH bonds in very similar kinds of molecules, here
are some interesting comparison. Let's say that we have a methane where we
have a single chlorine atom in it, and we try to pull a CH, a hydrogen, off of that
molecule. So we're now going to take chloromethane,
a single chlorine on a methane, and pull off a hydrogen.
It takes 422 kilojoules. Let's take dichloromethane, a methane
with two chlorines on it, and pull off one of the hydrogens.
It takes a different number, 414. Let's take trichloromethane and pull off
the single hydrogen on there. That requires 400 kilojoules per mole.
So, the strength of a CH bond clearly depends upon what the carbon is bonded to
on the other side. Notice, though, that the numbers don't
differ by all that much. We might be able to average the CH bond
over those three molecules, and come up with a CH bond which is on average useful
for approximations. That means when we calculate the energies
of reactions using bond energies, we're only using averages.
We're using average bond energies, which means that we're only estimating the
energy associated with the reaction by using bond energies.
Here's a good example for us to try. This is the reaction for burning
acetylene, C2H2 in the gas phase. It reacts with oxygen to make two carbon
dioxides and a water, and here we actually have the bond energies from
tabulated bond energies associated here. Lets think about what we have to do in
terms of our usual diagram over here for the energies.
Here are the reactants here. We have an acetylene molecule, and we
have five halves of a mole of oxygen molecules.
I know I'm going to release energy because burning acetylene releases a lot
of energy. I got two carbon dioxides and a water
molecule. The intermediate in our diagram here are
going to be the separated atoms. That's going to be two gaseous carbon
atoms plus two gaseous hydrogen atoms plus five gaseous oxygen atoms.
To make it from acetylene and oxygen to these atoms, clearly, the first step in
the process in going to involve breaking the triple bond in the acetylene, as well
as breaking two CH bonds in the acetylene.
Remember that acetylene is the molecule HC triple bond CH.
Plus, we need to break the oxygen-oxygen bond in five halves of a mole of
oxygen-oxygen doubly bonded. Going from carbon dioxide to these atoms
will cost us an amount energy, which is let's see, four times the bond energy of
the C double bond in carbon dioxide. So I need a 4 there.
And 2 times the bond energy in an OH bond, but notice I'm subtracting that
amount out, because when I go from here to here I am tearing up the bond energies
of the reactants from the products to the intermediates, it is the bond energies of
the products. So I subtract them out.
That gives rise to the equation I've written here on the screen.
I think it's exactly the same equation. I can plug all of those numbers in and
come to a determination of what the final value is, and when I plug all those
numbers in, I get that the delta H of this reaction is minus 1,248 kilojoules
per mole of acetylene burned. This is a number very much less than
zero, that's a very strongly exothermic reaction.
Where did the big exothermicity come from?
Staring at the data it's clear that it has to come form forming products,
forming strong bonds in the products. Where are those strong bonds, they're
clearly right here in the carbon dioxide which has been formed.
We formed four very strong carbon oxygen double bonds in the two carbon dioxide
molecules form from burning a single molecule of acetylene.
Of course acetylene has its own strong bond in it, but keep in mind, we need
only break a single carbon-carbon triple bond to set up the possibility of forming
four carbon dioxide very strong bonds. That is why, in fact, burning acetylene
produces an enormous amount of energy. In fact, this the reason why burning
hydrocarbons of any type produces an enormous amount of energy.
It's not so much that we want to tear up the original fuel, it's that by tearing
up the original fuel and reacting with oxygen, we can produce an enormous number
of very strong carbon dioxide bonds. The exothermicity associated with burning
hydrocarbon fuels is almost entirely a consequence of the formation of very
strong carbon dioxide bonds, which releases an enormous amount of energy.
We can use these kinds of bond energies to estimate the chemical reaction energy
in a large number of reactions. Keep in mind two things.
One is, these are only estimates, because we're using average bond energies.
And secondly, it is important that all of the substances be gases as you'll notice
here, in each case the reactants and the products were all gases, because in doing
so we don't have to worry about the energy associated with the interactions
between molecules.
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