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.

CDS 13 Reaction Energy and Bond Energy II

General Chemistry: Concept Development and Application

Meet the Instructors