Learning objective number seven is going to show us three other ways to

determine the delta H of a reaction.

We are going to learn how to take data from tables and

other reactions, to calculate the delta H for other reactions.

This is kind of a lengthy unit,

because we are learning three new ways of determining the delta H of a reaction.

The three ways that we had mentioned earlier, are using what's

called Enthalpy of Formation, and then Hess's Law, and then Bond Energy.

Let's begin with the Standard Enthalpy of Formation.

We need to begin by defining it,

it's a very precise definition, of what Standard Enthalpy of Formation is.

Well, the enthalpy part, that's a delta H, the change in enthalpy of a reaction.

But the little f that we see right here, means that it's a delta H for

a very specific reaction, and the little symbol we see up here says,

it's under a very specific set of conditions.

So let's look at that reaction and the conditions.

The reaction is the heat change that results when you,

form 1 mole of a compound.

Okay, we're forming one mole.

So a specific amount.

The reaction was for forming it.

But what are forming it from?

We're forming them from their elements.

These elements must be in their most stable form.

So some elements, you've got carbon for

example, that can be diamond and go, diamond and graphite.

You have to use the most stable form.

Oxygen, you have O2 and you have ozone, which is O3 in its most stable form.

It's, oxygen is a gas, under these conditions, so we use that.

So, that is the reaction.

We're going to make one mole of a compound from its elements.

Now, the condition is if they're gases, we're going to

have those gases at one atmosphere, so it's going to have an external pressure of

one atmosphere to maintain a one atmospheric pressure of those gases.

And that is what standard state conditions mean, so when we see that little

circle up there, above the delta H.

And we'll see this, in many cases in the future as well.

That is standard state conditions.

And that is one atmosphere, if they're gases.

Now we'll learn also that if it's aqueous, it's going to be one molar.

And generally those tables, are at 25 degrees Celsius.

When we see tables provided for standard enthalpy of formation.

Delta H doesn't change very drastically with temperature, but

it does change with temperature.

Standard state does not mean necessarily, it's 25 degree Celsius,

but that is the temperature that the tables we will refer to.

Are obtained in.

So, I want you to make sure you understand,

what delta H formation means before we use it, okay?

Because so many students, sure, they can use it all day long.

But they have absolutely no idea what they're using.

So here we've got a delta H of formation.

We go to the table.

We find out that the value associated with this is

a negative 1112.9 kilo-joules per mole.

So what does that mean?

Well, it is enthalpy change for a very specific reaction.

What is that reaction?

We take the definition.

The definition says, it is to form one mole of magnesium carbonate, so

that's going to be on the right hand side, we're making that, from its elements.

Well, what are the elements in there?

We've got magnesium, carbon, and oxygen.

Now these have to be in their most, stable form.

Magnesium, is a metal, it's a solid, okay?

Carbon's most stable form is graphite.

You may not have known that, but you do now.

It's graphite.

And oxygen's most stable form is O2, diatomic oxygen, so here's the reaction.

We have to balance it, keeping a one, on this side.

So we want, need one magnesium, we need one carbon, and we need three oxygens.

So how do we have three oxygens when it's O2?

Well, we can use co-efficient of three halves.

So three halves of O2 would be it.

We have to include the states.

And when we have carbon, I will sometimes write it as GR, but

it's graphite, is it stable allotrope.

And of course, oxygen is a gas, and this is an ionic compound, and

as such it is a solid.

So when we look up that value, we are looking up the delta H for this reaction.

So when this reaction occurs, it will give off 1,112.9 kilo-joules of heat.

And for every mole that reacts or is formed.

So you try it.

What is the reaction equation if you looked up a value of the heat of

formation of methane, CH4?

Well if you picked three, you'd be correct.

It is not both three and four.

It looks like well those are the same reactions.

They're just balanced differently and that is true.

But you have to choose the one that's balanced with a one here.

So one mole of the compound, from their elements and

we could write this as graphite more correctly.

Graphite.

In hydrogen is a diatomic element.

All right. If we could look at the definition of

the standard enthalpy of formation,

you should be able to predict what the value of delta H of formation of O2.

And we should have a little g here, O2 gases.

Okay, what is the value for that?

Well, it would have to be zero.

Now, if that is alluding you as to why it would have to be zero,

let's write the reaction, for the heat of formation.

It is the energy change, or

the enthalpy change, when you form one mole of that thing.

From its elements in its most stable form.

Well, what is the most stable form?

Of oxygen it's O2 gas.

So if you're going to make O2 gas from O2 gas, is it going to require any energy?

No.

There's no reaction there.

So, the delta H of formation of any elements in

their stable form is always going to be zero.

So sometimes on tests or homework problems, they're not going to be giving

you the delta h of formations of these elements, because we would expect you to

know that the value is zero by virtue of the way it's defined.

>> [LAUGH] >> Okay, so now.

We are going to determine the delta H of a reaction.

And we're going to be, when we use the heats of formation,

we're going to be obtaining standard enthalpies of reactions.

So we will only be able to calculate it under standard state conditions.

That means they're going to be carried out at one atmosphere.

Now the way this is done is using the following equation.

So our symbols are important here.

We're going to take the number of moles,

okay, moles from the balanced equation of the products.

So that's kind of your final.

We know that, delta is final minus initial.

So our final is all the heats of formations of all the products, added up.

All the heats of formations of all the reactives.

Added up.

This is our summation symbol.

And then subtract the reactant's, its products minus reactants, okay?

So that is how you use,

the delta h as a formation to determine the delta h standard of a reaction.

This method is called the direct method.

So if you see it referred to as, calculate it using the direct method.

You're being expected to go obtain values of delta H formation to get that.

So utilizing that equation, let's see if you can do it for this reaction.

Now, I'm gone to a, a table and provided for you these values.

See if you can determine the delta h, of this reaction.

Well if you collect, obtained or selected number one, you are correct.

If you did, just move on ahead and skip over my explanation.

If you did not, let's make sure we have it all straight in our minds.

The delta H of this reaction, standard, is going to be

the four moles of CO2 times the delta H of formation

of CO2 plus two moles of H2O.

Times the delta H of formation of H2O.

Now this is H2O liquid.

When you see tables, I only gave it to you for

H2O liquid, there's going to be H2O solid, H2O gas in there.

You have to make sure you look up the correct one.

So that's the sum of all of the products, minus the sum of all the reactants.

So it's be two times.

Two moles times a delta H formation.

C2H2, that's not a C.

C2H2, plus five times the delta H formation of O2.

So that's the equation taking the summation equation.

So then we can plug our values in.

The delta H of this reaction, and it'll be in standard state conditions,

would be four moles times the carbon dioxide,

which is a negative 393.5 kilo-joules per mole plus two moles.

Times the value for the H20 which is

a negative 285.8 kilo-joules per mol.

Minus two moles of C2H2 times 226.6 kiljoules per mole.

And also, let's see.

We have to add five moles times zero,

because that's the heat of formation of oxygen.

So, all of that, added up.

Is a negative 2598.8 kilojoules.

Since we're doing adding and subtracting we keep the fewest number of places to

the right of the decimal point and that's where the eight comes from.

Now this problem has got a little different twist on it.

In this problem, we know the delta h of the reaction, it is provided for us, here.

To the right of the equation.

And they're giving us heats of formation, of various substances.

We're given it for carbon dioxide, we're given it for water, we know it for oxygen.

They're asking us what is the heat of formation of the C2H6.

So see if you can solve for that.

If you selected number one, you're correct and

you'll want to just advance past my explanation.

If you did not obtain that or

you're kind of lost at how you would obtain it, watch how I work this problem.

We know that, the delta H of the reaction,

would be equal to the products minus the reactants, okay.

And, we'll have four moles of CO2.

So we'll take four mols times the delta H of formation CO2.

Plus six moles, times the delta H of formation of the H2O.

Minus two times two mols times the heat of formation of C2H6.

And then the seven moles times the O2 but that's zero so

I'm not going to even write it down.

So then we can plug in the information we know.

We know that the heat of the reaction is a negative 3119.4 kilo-joules.

Then we have the four moles, times the value for the CO2,

which is a negative 393.5 kilo-joules per mole, 'kay?

And we have six moles, times the heat of formation of the H2O,

which is a negative 285.8 kilo-joules, sorry for squeezing that in there.

Minus two times, the heat of formation of C2H6 that we don't know.

So, this is the unknown, in this equation, and

we can do the mathematics now and solve for the C2H4.

Plug those in, try it, and verify that you get the -84.7.

And it would be in units of kilo-joules per mole.

Especially if I leave that mole in there.

This one is similar to the previous one, so if you were able to do that one,

should have no trouble with this one, if you struggled with it and

you saw my explanation now maybe you can work through this problem.

Give it a try.

Well if you've got a negative 1206.9,

you'd be correct and you can move on to the next question.

If you'd like to see my explanation, I will explain it here.

The delta H of the reaction,

standard is going to be the delta H of formation of calcium oxide.

Plus the delta H of formation of CO2.

They were both at one molar.

Minus the delta H of formation of CO3.

Okay.

We have products minus the reactants.

We have been given the calcium oxide value.

Right here, so we plug that in for here.

We were given the carbon dioxide value, we plug it in for here.

We know the delta H of the reaction, we plug it in right here, and

we would then solve for, making sure you're watching the sign,

you would solve for the delta H of formation of the calcium.

It looks like I left out the calcium part.

The calcium carbonate.

Plug those numbers in and see if you don't get negative 1206.9.

More problem, here.

This one is making sure you have a good understanding of the term delta

H of formation.

I want to know the delta H of this reaction.

There's a couple of ways you can think about this, but

I'm forcing you into thinking about the connection of heat of formation of H2O,

and this reaction.

So let's think about the heat of formation of H2O, okay?

Let's just think about what we see here.

Write out that reaction, reaction to form one mole of water from its elements, and

compare it to that overall reaction, and

maybe that'll help you come up with which one is the correct answer.

Okay, if you selected number two then you are correct.

The definition of heated formation, of water would be to make one mole of water,

from its elements in their most stable form.

So that'd be hydrogen gas.

Oxygen gas, balancing it with one mole in the left hand side gives us this.

How does this reaction, compare to this reaction?

Well, it's twice as big, so it would be two times that.

That would be one way to do it,

the other way to do it would be to say okay, it'd be.

The delta H of the reaction, is two moles of H2O

minus two moles of H2, plus one mole of O2.

Okay, these are both delta H of formations equal to zero.

You will two times the delta H formation of this guy, and

you have the delta H of the reaction.

Two different ways of thinking about that to give you that answer.

Okay so we've seen the direct method, now we're looking for

the indirect method, this is called Hess's Law.

And Hess's Law states that when you do a reaction and

you're converting reactants to products, the change in enthalpy is

the same whether reaction takes place in one step or a series of steps.

This is because, enthalpy is a state function and

we've talked about how state functions are path independent.

Doesn't matter how you get from the initial to the final,

the change is always simply the final minus initial.

So it's going to utilize that, information to obtain Hess's, to work the problem.

So you're going to be given several thermochemical equations in order to use

Hess's law and the best way to learn how to utilize Hess's law,

is by way of an example.

Okay, so let's look a this example here.

So we're given three reactions, here.

We're giving one, which is graphite burning.

That's easy to obtain in a calorimeter, combustion reaction.

Another reaction is to burn hydrogen.

Same thing.

It's easy to do in a calorimeter.

One is to combust C2H6.

So you have these three exothermic reactions.

What we want to do is determine the cat, the delta H for

this reaction, which would be more difficult to do in a bomb calorimeter.

But we can utilize those reactions to obtain this.

It's kind of like building a puzzle.

We're putting the puzzle pieces together.

We want to form this reaction.

Now, we're going to go back and forth between two slides as I build it.

On the next slide.

What we need, is two carbons in the form of graphite on the left hand side.

So we look at this equation, it has a graphite we need, but

there's only one mole.

So I'm going to double that reaction.

So if we needed two, doubling the graphite would be two, it's not writing.

Two, there it is, two times the carbon graphite

plus two times the oxygen, giving me times the CO2,

and that would mean that delta h would be, for this reaction,.

Would be twice that number.

Okay? So it would be 2 times a negative 393.5.

So what did that do for us?

That got us the graphite in place.

We needed two graphites on the left hand side.

The next thing I need in place is the hydrogen.

I need three moles of hydrogen, this equation has it on the left hand side but

it's only one mole.

So let's triple that reaction equation, that would be 3H2 gas plus

three halves O2 gas producing three H2O liquid.

Now the delta H of this reaction as written would be

three times the delta H from the previous.

Reaction equation, so we flip back here, we see it's a negative 285.8,

so it's 3 times -280, that doesn't look like an eight, -285.8 kJ.

And now we have the hydrogen in place.

Now a lot of other things are coming along with it.

Now what do I need?

I need C2H6 on the right hand side.

And I need one mol of it.

Here's C2H6 it's on the left hand side and there's two mols of it.

So what I need to do is I need to take and I need to flip this equation,

put the products as a reactants and the reactants as products.

That'll change the sign of my delta H.

And I need to cut it in half.

Those two things.

So let's cut it in half and flip it.

We would have only 2 moles of CO2.

We would have only 3 moles of H2O.

We would form C2H6.

And seven halves, that's cutting that in half, of the O2.

The delta H, H for this would be, we'll take the negative 3119.6 and

cut it in half, 3119.6, cut it in half and change the sign.

It was negative, now it's positive.

'Kay?

If we take all those values, or actually, let's take the equations first.

We need to verify that if you add these equations up,

you get the equation we're looking for.

So, we can cancel out some things.

There's two carbon dioxides on the left and on the right, so they'll go away.

There's three waters on the left and three waters on the right, they can go away.

We have oxygen on the left, we have oxygen on the right.

Two plus three-halves is three and

a half, seven-halves is three and a half, so all those ac,

oxygens will cancel and that will leave me with two carbon in the form of graphite.

Plus three hydrogen gas producing C2H6 gas,

that's what we were wanting to obtain.

So we know that this value up here plus this value here plus this value here

if you add the reactions that you add up the delta h's.

That will give me a delta h.

Of this reaction equal to a negative 84.6 kilojoules.

So that's how you use Hess's law.

And now, sometimes it's not quite as straightforward.

In some of my extra videos, we'll show you some more complicated ones.

If, in a reaction,

you see that a substance is in more than one reactions skip over it.

And work on all the other ones.

And it generally will fall out in the right place when it's all said and done.

This is just an application of Hess's Law that I'm giving you a glimpse of.

There's many different ways that Hess's Law is used in chemistry.

But here we are seeing a process of dissolving an ionic compound into water.

If this is done in a one-step process,

we are talking the, ionic compound that we see here, and

we're going dissolving it in water, so it's being surrounded by water molecules.

Plopping that into water stirring is an easy heat to measure with a,

with a, thermometer to see how the temperature rises in the solution.

And doing it in one step here, we see is four kilojoules per mole.

But this could be thought of as a two step process.

Step one would be to take the ionic compound that we see here, and

break it apart into its ions, separate them from each other, so

we're basically turning it into the gas state, 'kay?

So we have a gas.

It's completely separated from each other.

The energy required to separate those ions is lattice energy.

And then we can think about, well we've got them separated from each other now we

can form attractions between those ions and the water molecules.

So this would be considered the heat of hydration.

that's doing it in a two step process of the triangle there.

And what Hess's law states is that it doesn't matter if you do it in one step,

that would be the heated solution, or you do it in two steps, that would be

the heat of hydration surrounding it by water molecules plus the lattice energy,

energy to pull them apart.

That, that would be, those energies would be the same.

So if you knew any two of those you could determine the third.

And very often, since it's impossible.

To actually measure, or, I guess maybe, I won't say impossible,

but be very difficult to measure that, value you

could determine it by knowing the other two things.

Now, I've got the definition of lattice energy, that was posted here for you.

I describe, described it, but lattice energy would be the energy required to

completely separate one mole of solid ionic compounds, into the gaseous state.

So that ends our discussion of Hess's law.

Now let's go into the third means by which we can determine the delta H

of a reaction.

This can only give us an estimate so

it gives us a close estimate of what the value is.

And we're doing it using what's called bond energies.

Bond energy, which is sometimes called bond dissociation energy,

is the energy change required, okay so I'm going to underline that word, we have it

italicized, required to break a particular bond in one mole of gaseous molecules.

So we have a mole of gaseous molecules and

we want to pull them apart, break that bond.

Energy to break that bond is a bond energy.

Or a bond association energy.

So let's look at an example.

Hydrogen is a diatomic molecule.

We started in the gas phase, which is natural for hydrogen, and break it up

apart into ions, two hydrogen, I mean not ions, two atoms, two hydrogen atoms.

What we've done here.

Has taken that one simple bond there and broken it.

So that is a depiction of that.

The, the energy to do that reaction is 436.5 kilojoules, so

that's a measurable value of how much energy it takes to break the bond.

Now we're going to see a table on the next slide that gives bond energies of

some common bonds.

Now many of these are called average bond energies, so it's on average,

this is how much it takes.

For example, up here this isn't an average.

There is actually a molecule that consists of this one bond.

If we break it apart into atoms we're going to know exactly to,

to our measurement How much energy it takes to do that.

So what would be, how would we do a average bond energy?

Well, let's say we wanted to know the bond energy of the bond

between carbon and hydrogen.

Well, there's no molecule made up of carbon and hydrogen.

We do have a molecule that's made up of methane, which is four of these,.

Now the energy to break each one of those bonds is different from one bond to

the next.

When we break the first bond, you can determine that energy and

it's a little hard to break the second one and the third one and the fourth one.

But what you do is you break each one of those bonds and divide by four and

that would be the average bond energy for that.

So in any molecule where you don't have just a diatomic molecule of that bond,.

We have to do an average bond energy.

Now since they are averages, then the best we can do is an estimate.

We're going to be able to use these values to estimate the delta H of the reaction.

So let's look at our table.

So look at the table, we have several things that are in red.

That's because those are not.

Averages.

Those are actually obtainable values because each one of those things in

red is a dia, mol, tie, is a diatomic molecule that exists.

So, there is a molecule of HF.

We all ready talked about H2.

There's N2O2F2.

These are diatomic molecules made up of those bonds that we see here.

Some of them are single, some of them are double, and some of them are triple.

Lets look at one of these here, lets look at these two.

I want you to notice that the energy to break a double bond is

not twice the energy of a single bond.

Okay so you can't just say well I'm breaking two of em so

lets double that, they're different values.

So if you're considering the bond energy of a double bond.

Make sure you're looking up the double bond, bond energy.

Look over here, look at how different these are.

An oxygen, oxygen single bond is 142 kilojuoles for

every bond that you take, but an O two double bond is way more than that.

It's way more than double.

So the next thing I want to mention to you is that this

area right here with the oxygen, carbon double bond, ,.

If it's, in general, a oxygen carbon double bond, you want to use 745.

We see this asterisk and, and

down at the bottom of the table, it tells you what that's for.

If you're dealing with CO2, and you want a value for the energy to break the CO.

CO double bond in that molecule.

So, C02 consists of a carbon, double bonded to 2 oxygens.

So, to break each one of those, on average, is 799.

So, carbon dioxide, is, a common enough reactant or

product that we need to consider.

And you would want to choose this value when you're breaking the double bond in

carbon dioxide.

If it's a double bond between CO and, C and

O in any other situation, you use the 745.

Okay, so there's tables.

Now let's look at the background.

This is kind of, a, sort of a use of Hess' Law.

We know that if we want to know the energy to.

Go from reactants to products.

So we want to know the difference in energy right here.

And since the products is higher in energy then the reactants,

this guy right here would be a positive delta H.

Okay? We had to put it a little extra energy.

This would be a, an example of an endothermic reaction, kay?

This one.

Here, we're going from here down to here.

This would be our delta H.

It is a negative delta H.

This is an exothermic reaction.

But that would be to do it in one step.

What we're going to be considering is, let's do this.

Let's take our reactant molecules and

let's pull them apart into all their little atoms.

So you'll take whatever your reactant molecules.

And you will pull them all apart.

So you just got the individual atoms now.

Okay? So there were molecules.

Now there is a bunch of individual atoms.

You're breaking a bunch of bonds.

It requires energy to break those bonds and

we call that energy, the energy required to break those bonds, the bond energy.

Okay? So.

We have that bond energy for all the bonds that we're breaking.

And that's a positive value, because it requires energy to break those bonds.

Now, once we break all those bonds,

now we want to put them back together again to form our new molecule.

So, we form our products.

Now, if it requires energy to break bonds,

it is going to release energy to form bonds.

So this is positive, releasing energy is negative.

Okay, so it's going to give off the energy to do that.

So you add up all the energy it requires to break all the bonds.

That's a positive number.

You add up all the energy released.

As you put them back together, that's a negative number.

And then you combine those two and

you'll know the delta H, you'll know this for the reaction.

So, you sum up all the bonds that are broken.

This requires energy, as I stated, and it's a positive value.

Then you sum up all the energy of the bonds formed.

This is a negative number because it's releasing energy.

That's where that negative sign is coming from.

This is not products minus reactants.

Okay?

This is add up all the energies of all the bonds broken.

That's positive because it requires energy to break bonds.

You add up, that's the summation symbol,

all the bond energies of all the bonds formed because it releases energy,

it's a negative number, and that's why we have that minus sign right there.

So we're going to use this method, and you can't use it

unless you know what bonds are broken and what bonds are formed, so

you have to write the Lewis structures for all those substances in the equation.

So you're going to utilize that method that you have learned earlier.

And from that you'll be able to determine what bonds are broken,

what bonds are formed and used that equation.

So I want you to take this one here and I want you to kind of pause the video.

And start working on doing this problem.

So start working on determining the Lewis structures of each substance.

And think about how many bonds of each type of bond is going to be broken, and

how many are going to be formed.

After you have got the Lewis structures.

Established.

Then you'll want to proceed, because I've got some questions to guide you

through determining the delta H, or estimating the delta H of this reaction.

Okay, well hopefully, you've drawn all the Lewis structures.

And to see if you've got the Lewis structures correct,

let's see if you get these right.

How many, I have a missing y there.

How many C-H single bonds will have to be broken

well if you answered twelve then you are correct if you did not answer twelve and

you don't see where those come from I would say probably the most common choice

would be as a wrong answer would be number three would be six of them but

remember there are two.

Of these molecules in the balanced reaction and

that's where we're trying to determine the delta H of.

So a Lewis structure of CH2 looks CH, sorry, C2H6 looks like this.

[SOUND] And there are two of them, so there are a total of 12.

Okay, following suit here, how many o o single bonds must be broken?

A bit of a trick question, there aren't any single bonds,

the O2 molecule is a double bond, okay?

So loose structure.

Of it is this, and we are breaking double bonds, and

how many double bonds are we breaking?

Well, we're breaking seven double bonds.

Okay, so we've got our Lewis structures all laid out.

I'm going to go ahead and draw out all the Lewis structures.

Because in our next question you're going to be using it to predict.

So I will draw the Lewis structures of everything.

We have two of these, seven of those, we have our carbon dioxide,

remember there's a different bond energy for the OO double bond in carbon dioxide.

There are four of those.

And there's water.

Which has a Lewis structure like this, and there are six of those.

So now you're ready to estimate, determine the delta h for this reaction.

Well if you picked number two you're almost correct.

The actual answer is number one.

Why were you almost there but not quite?

You did not use the correct energy for the CO double bond.

You need to make sure you're looking that up for carbon dioxide.

Now if you're nowhere close, make sure you watch and see what I have done with mine.

I'm going to back up where I've got my.

Our Lewis structure's drawn out.

And I will take you through the whole calculation.

If you've got this, then go on to the next question and skip ahead.

So we've come back to this slide so that I can have the Lewis structures and

lay out the equation for the delta H of this reaction.

So the delta H of the reaction is estimates as,

so it's close to because we're using average bond energies.

All the bonds that we break.

So we're breaking some carbon hydrogen bonds and

we figured out that we are breaking.

Six for one of these times two,

12 moles, okay, I have 12 moles of those bonds that I'm breaking.

So it's going to be 12 moles times the bond energy of the CH single bond.

I am breaking a carbon carbon single bond here and how many am I breaking?

I'm breaking two so

it'll be plus two moles times the bond energy of the C C single bond.

Okay? And then I'm breaking

some oxygen-oxygen double bonds, so I'm going to add to that 7 times

the bond energy of the O double bond O.

So that's all the bonds that we have to break.

We will subtract from that all the bonds that we're going to form.

So we come over to this side and

we see that we have some carbon oxygen double bonds.

There's two in there, times four, so we have eight moles.

Of the bond energy of an O, of a CO double bond.

Make sure you're using the one for carbon dioxide.

It's the sum of all these then we subtract, so

make sure you add them all up before you subtract from there.

So I'm also going to add to that the.

Two hydrogen oxygen single bonds times six.

So we have 12 mols of the bond energy of the oxygen and hydrogen single bond.

Okay? So there's the equation laid out for

us of all the things that we need to look up.

And we will then put the numbers in.

I'm going to do this in red.

So that we can keep track of it.

It's like kind of a busy slide.

We have the delta h as a reaction and I'm not going to put my units in.

The bond energies are in kilojoules per, per mole and they

will cancel with the moles that we have here and we'll be left with kilojoules.

But for, for area constraints, I'm just going to write my numbers.

So, I have 12 times the bond energy of the CH single bond, 'kay?

I have two times the bond energy of the CC single bond, which is 347.

I'm getting this off the table.

I have seven times the bond energy of an OO.

Double bond, it's 498.7, we can know that to more significant figures, simply

because oxygen is a diatomic molecule, we don't have to use an estimate.

We will subtract the bond energies of the former bond, so

it will be 8 times the 799, plus 12

times the oxygen hydrogen single bond, which is 460.

So all of that combined together will give us the answer of a negative 2759,

and that will be kilojoules for that balanced reaction.

So now, we're ready to do just one final question, here.

And that is, again,

kind of doing a comparison between some definitions that we've learned.

If the bond energy for hydrogen, H2, is 436.5 kilojoules per mole.

I can get that from the table.

The question here is what would be the delta H of formation of H gra-,

H, single atom H in the gas state.

Okay, so we've got the atoms not the diatomic molecule.

Okay, so we have a hint there.

It says write the reaction for both bond formation and heat of formation.

There is very specific definitions associated with that.

So lets do bond energy first.

Bond energy says if you were to take the diatomic molecule and

break it up into it's atoms, okay so

this is in a gas state and now we have them separate from each other.

What would that really look like it'd be H2 gas.

Go into 2H gas.

That is the bond energy definition.

And we know that the value for

this, the delta H of this reaction is the bond energy.

It is 436.

6.5 kilojoules, okay, for that reaction.

Now let's write the reaction for the delta h of formation.

Delta h of formation, by definition,

is to make one mole of the thing we're doing the heat of formation for.

So that's h, h, the at, atom,

not the diatomic molecule, coming from its element in its most stable form.

Well hydrogen, it's most stable form is H2.

And it's a gas.

We have to balance it keeping a one on the right hand side.

So we would do a one-half.

So if you write those reactions, this by definition is the heat of formation.

So then you can answer the question.

Well if you pick number three, then you would be correct.

Okay, that's the end of our.

Learning objective dealing with the ways to calculate delta H for a reaction.

We have three different ways to calculate it.

One is using heats of formation definition, one is you to use Hess's Law,

and the third, that we just finished up, is bond energy.

These, along with calorimetry,

are your methods of determining the delta H of reactions.

This also finishes our thermal chemistry unit and

you're ready to take your test and move on to the next unit.

7.07 Calculating Enthalpy Change

Chemistry

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