AP Chemistry Standard Enthalpies of Formation

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

In the last couple of lessons, we have been talking about enthalpy and energy and heat--things like that.0004

Today, we're going to continue our discussion of enthalpy, and we are going to be talking about standard enthalpies of formation; and we are going to be able to use these standard enthalpies of formation to help us bypass Hess's Law.0011

When we are given a reaction that we need to find the enthalpy for, the heat of reaction for (whether it is given off or absorbed), we can just do it by reading off of a table of enthalpies that have been prepared for a whole number of compounds.0024

It's very, very convenient, because Hess's Law, although it is very convenient in the sense that we can manipulate equations and just add them--still, it requires a little bit of work.0039

With standard enthalpies of formation, we have a standard, and then we have some things tabulated, and we can just do some really simple arithmetic and get the answer that we want--very, very powerful.0049

So, with that, let's go ahead and get started.0060

As often with these, we are going to start off with a definition, because I think it always makes it a little bit more worthwhile.0064

So, definition: the standard enthalpy of formation (and, if you have not figured it out by now, when we say "enthalpy," we just mean heat--you can use heat of formation, enthalpy...I actually prefer to use the word "heat"; I don't really care for the word "enthalpy" myself; but there it is) of a compound is the change in enthalpy that accompanies (let me spell this a little bit more appropriately here) the formation of 1 mole of the compound from its elements, with all substances in their standard states.0071

OK, so the standard enthalpy of formation of a compound is the change in enthalpy that accompanies the formation of 1 mole of the compound from its elements, with all substances in their standard states.0184

That means, if I wanted to talk about the enthalpy of formation of liquid water, it would look like this.0195

H₂ gas (because hydrogen is a gas in its standard state); and standard state means 25 degrees Celsius, 1 atmosphere pressure--normal; the way you would find it on an average day.0205

Plus O₂ gas (these are its constituent elements); they would form liquid, and the ΔH standard of formation--this is the symbol for it, ΔH_f for ΔH of formation--this little degree sign at the top means that it is standard conditions, 25 degrees Celsius, 1 atmosphere pressure; equals...some number.0218

This is what it would look like: we calculate the ΔH for the formation of a compound, 1 mole of that compound.0244

So notice, this is not balanced; now it is balanced.0251

1 mole of the product...it is always like this; this is always going to be 1, so we don't really balance this equation in terms of whole-number coefficients.0256

Because this is 1, you will end up with fractions over here.0265

So again, it is a standard; we pick a standard, and in chemistry, the standard is the mole; so that is what it is.0269

Now, let's actually list the standard states.0277

We have listed a couple of them.0279

What we mean when we say "standard states": for a gas, it is 1 atmosphere; that is standard for a gas.0281

For a solution--if we are talking about a solution in standard state--we mean 1 Molar.0291

And again, this is an older notation, an m with a line over it; you are more accustomed to seeing the capital M; Molar, mole per liter.0297

OK, an element...the state of an element is the state the element takes at 1 atmosphere and 25 degrees Celsius, which is roughly room temperature.0305

An element...the standard state of aluminum is metal; the standard state of bromine is liquid; the standard state of mercury--liquid; the standard state of oxygen is a gas.0328

The standard state of sodium is the metal; that is it--an element.0341

OK, now you should notice: standard states--this 1 atmosphere, 25 degrees Celsius, 1 Molar solution--this is not the same as standard temperature and pressure that we talked about with gases.0346

That is 0 degrees Celsius and 1 atmosphere pressure; so don't mix the two--when we are talking about a standard state, we are talking about a particular state that that compound, that element, is taking.0360

We have listed that standard as: a gas under 1 atmosphere pressure; a solution, 1 Molar; an element, the state that it assumes under roughly room temperature (25 degrees Celsius) and 1 atmosphere pressure--in other words, your average day at sea level.0373

OK, now, once again (I can't reiterate this enough): enthalpies of formation are always given, always given per mole of product, because the product is the thing that we are finding the enthalpy of formation for.0389

The enthalpy of formation is the amount of heat either generated (either given off) or absorbed when you form that product from its constituent elements.0418

We just did this one up here.0431

H₂ + 1/2 O₂ goes to H₂O.0435

Now, here is the nice part: the ΔH of the final reaction, the standard ΔH, the enthalpy of reaction (or the heat of the reaction--standard heat of reaction) is equal to the ΔH of formation of the products, minus the ΔH of formation of the reactants, when you add them together.0440

So, I didn't want to use a summation symbol; and this also includes...well, actually, you know what--let me go ahead and write out...instead of using the summation symbol, I'm going to write it out.0481

You are going to notice that I am not a really big fan of symbolism; I think it's nice, but I often think it gets a little bit in the way.0497

I will put "Sum of the ΔH formations of the products" on the right-hand side of the arrow, minus the sum of the ΔHs of formation for the reactants.0504

That is it; this is the standard definition of ΔH.0529

We use the tabulated...at the end of your book, in the appendix, you are going to see a list that is called "Standard Thermodynamic Data."0532

It is going to list a whole bunch of compounds, and you are going to have three columns in there.0540

The first or second--it is usually the first or second column--one of the columns, anyway--is usually the enthalpy, the ΔH.0546

It has a ΔH; it has a little degree--this zero degree--on it; and it has a little f.0553

It is given in kilojoules per mole.0559

Well, another column is going to be ΔG, which is free energy, which we will talk about later.0562

The other one is going to be S, entropy, which is in Joules per mole per Kelvin.0567

But again, those are other thermodynamic quantities that we will talk about a little bit later in the year, when we talk about equilibrium, spontaneity, and things like that.0576

Right now, we are just concerned with enthalpy.0584

That table in the back--that column that lists the ΔHs of formation--that is what you are going to use to tabulate this.0587

It will make more sense in a minute, when we do a problem.0593

OK, a couple of things to remember: let's see, the first thing to remember is that the ΔH of formation for elements is 0, because you are not forming the elements; the elements are already there.0597

It's very, very convenient that it is 0; it comes in very handy.0615

OK, #2: ΔHs of formation are per mole of product formed.0619

So, when you come across a reaction, make sure the reaction is balanced--balance the equation, so that you have the appropriate stoichiometric coefficients.0643

Balance the equation, because again, we are using the ΔH values of formation, but they are per mole of product; but when you are using it in an actual reaction, you may have 3 or 4 or 5 moles, which means you have to multiply the ΔH of formation by 3 or 4 or 5.0659

And again, it will make more sense in a minute, when we do an example, which we are actually going to do right now.0678

Example 1: OK, ammonia is burned in air to form nitrogen dioxide and water; what is the ΔH of this reaction?0686

OK, so let's see what this says; write out the equation.0721

Always start with an equation; this is chemistry; there is always an equation somewhere.0726

OK, so ammonia is NH₃; it is burned in air, which means it is going to take oxygen; and it forms nitrogen dioxide, NO₂, plus H₂O.0732

When we balance this, we end up with 4; with 7; with 4; and with 6.0746

Now, this is our reaction; we want to find the heat of this reaction; we want to find the enthalpy of this reaction--ΔH_rxn, the standard ΔH.0754

We are going to use the ΔHs of formation that are tabulated in the back for this, for this, for this, and for this.0767

Don't mistake the two: the final heat of reaction is for the entire reaction; what we are doing is we are using the heats of formation that are tabulated for the individual pieces of this reaction--products and reactants.0774

So, we said that the ΔH of the reaction is equal to the sum of the ΔHs of formation for the products, including coefficients, minus the sum of the ΔHs of formation for the reactants, including coefficients.0787

All right, that equals--well, let me put a circle around the (oops, I actually wanted to use red; there we go--so that we have it here) equation; ΔH.0824

We are going to take the ΔH of formation of H₂O, multiplied by 6; add it to the ΔH of formation of NO₂, multiplied by 4.0843

From that, we are going to subtract the ΔH of O₂, multiplied by 7.0851

And then, subtract the ΔH of formation of NH₃, multiplied by 4.0856

Products minus the reactants--the sum of the ΔHs of formation of the products, minus the sum of the ΔHs of the reactants.0864

All right, so now, when we look up 34, 4...so which one shall we do first?...let's do...it doesn't really matter, so let's do the NO₂ first.0875

I always like to do it this way; so we have 4 moles (that is a 4) times the ΔH of formation of NO₂ from its constituent elements.0888

We look in the back at our thermodynamic tables; it is 34 kilojoules per mole; again, it is per mole of compound.0900

That is what a ΔH of formation is; it is a standard.0907

Plus 6 moles--that is where this comes from--that is where the 6 comes from; this, times the ΔH of formation of water, which is -286 kilojoules per mole.0910

From that, we are going to subtract 4 moles, times the ΔH of formation of NH₃, which is -46 kilojoules per mole; plus 7 moles--that is where that comes from--times 0 kilojoules per mole, because again, this is an element--oxygen gas.0926

We said that oxygen gas's ΔH of formation (for elements) is 0.0953

So, it's nice; these are not elements--these are compounds--so they have actual, finite values.0958

Well, when we do this, we end up getting -1396 kilojoules; is this exothermic, or is it endothermic?0965

ΔH is negative; that means it is giving off heat; that means, when you mix NH₃ and O₂ gas, you produce three things: you produce nitrogen dioxide; you produce H₂O, and you produce heat.0976

That is what this negative means; this is exothermic--it gives off heat.0988

Now, notice: this is in kilojoules; moles here have canceled; ΔH of formation is given in kilojoules per mole--we are multiplying by the stoichiometric coefficient in the balanced equation to give us a total for the reaction, as written in the balanced equation.0994

Moles cancel; the ΔH of reaction--standard ΔH of reaction--is equal to -1396 kilojoules.1012

ΔH of reaction is in kilojoules.1024

The ΔH of formation for the individual compounds--that is in kilojoules per mole of compound formed; that is why it is standard--we have standardized it to one mole.1027

OK, now let's see: let me show you what happened in terms of Hess's Law.1040

1/2 of N₂, plus 3/2 H₂, goes to NH₃.1054

The ΔH of formation is equal to -46 kilojoules per mole.1063

H₂ + 1/2 O₂ goes to H₂O; the ΔH of formation is -286 kilojoules per mole.1071

1/2 N₂ + O₂ goes to NO₂; the ΔH of formation is equal to 34 kilojoules per mole.1083

These are just the numbers that I used, that I looked up; and O₂--that is just 0; the ΔH of formation equals 0.1095

If I used Hess's Law, I would take these equations; I would manipulate them, flip them, turn them, multiply by things, and add straight down to get the final equation that I wanted.1104

Then, I would add the appropriate changes in the ΔH.1114

Well, a lot of that has already been done for you; again, the process--the steps don't matter.1117

All that matters is where you start, where you begin; because these values have been tabulated, I can just take the products, minus the reactants; that gives me my ΔH of reaction from the ΔHs of formation that are tabulated in the thermodynamic data in the back.1122

That is it; OK, let's do another example, and I think this will suffice to explain the use of ΔH of formation.1140

Calculate the ΔH of reaction, standard, for the thermite reaction (thermite--for those of you who don't know about thermite, it is the very, very, very powerful, very...well, you'll find out in a minute, as far as the ΔH, whether it is positive or negative).1151

The reaction is: 2 moles of aluminum metal, plus 1 mole of iron (3) oxide, goes to aluminum oxide, plus 2 moles of iron.1176

So, when you mix aluminum metal with iron oxide, you end up producing aluminum oxide; you end up releasing iron metal.1201

It's kind of amazing, actually.1212

So now, let's see what we have.1214

Let's go ahead and list some ΔH of formation values: the ΔH of formation of aluminum equals 0 (aluminum is an element); the ΔH of formation of Fe₂O₃ is equal to -826 kilojoules per mole.1218

The ΔH of formation of Al₂O₃ is equal to -1676 kilojoules per mole: wow, incredibly exothermic!1245

The ΔH of formation for iron--well, iron is an element: 0.1260

Good--so now, let us take 1 mole times -1676 kilojoules per mole, plus 0.1268

That is the aluminum plus the (yes, that is right...the products) iron, minus the reactants, which is 1 mole times -826 kilojoules per mole, which is the Fe₂O₃, plus 0.1286

When we do this, we get -850 kilojoules; that is 850,000 Joules.1311

Exothermic is an understatement.1324

Exothermic releases a lot of heat; in fact, it releases so much heat that the iron that comes out is not solid iron; it is actually liquid iron; the iron is melted.1327

OK, so let's see what we have here.1341

Do we need to do anything more?--no, that is it.1347

So, we found our ΔH of the reaction; it is equal to -850 kilojoules; notice, ΔH of reaction is in kilojoules, not kilojoules per mole.1349

Kilojoules per mole is ΔH of formation; that is what we used to get the ΔH of the actual reaction.1361

That is it; that is standard enthalpies of formation--nothing too difficult; I hope that made sense.1369

With that, I will go ahead and stop there.1374

Next time, we are going to talk about calorimetry, the transfer of heat and transfer of energy directly from one object to another.1376

Until then, thank you for joining us here at Educator.com; we'll see you next time; take care.1385