For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

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### Standard Enthalpies of Formation

- Standard Enthalpies of Formation, ΔHf°, are ALWAYS per mol of product formed.
- Standard Enthalpies of Formation are the values in Thermodynamic Tables.
- Standard Enthalpy of a Reaction is:
- SUM [ enthalpies of formation of products (times their respective stoichiometric coefficients) ] – SUM [ enthalpies of formation of reactants (times their respective stoichiometric coefficients)]

### Standard Enthalpies of Formation

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Thermochemistry 1:04
- Standard Enthalpy of Formation: Definition & Equation
- ∆H of Formation
- Example 1
- Example 2

### AP Chemistry Online Prep Course

### Transcription: 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

*H _{2} 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 O _{2} 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

*H _{2} + 1/2 O_{2} goes to H_{2}O.*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 NH _{3}; it is burned in air, which means it is going to take oxygen; and it forms nitrogen dioxide, NO_{2}, plus H_{2}O.*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 H _{2}O, multiplied by 6; add it to the ΔH of formation of NO_{2}, multiplied by 4.*0843

*From that, we are going to subtract the ΔH of O _{2}, multiplied by 7.*0851

*And then, subtract the ΔH of formation of NH _{3}, 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 NO _{2} 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 NO _{2} 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 NH _{3}, 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 NH _{3} and O_{2} gas, you produce three things: you produce nitrogen dioxide; you produce H_{2}O, 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 N _{2}, plus 3/2 H_{2}, goes to NH_{3}.*1054

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

*H _{2} + 1/2 O_{2} goes to H_{2}O; the ΔH of formation is -286 kilojoules per mole.*1071

*1/2 N _{2} + O_{2} goes to NO_{2}; 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 O _{2}--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 Fe _{2}O_{3} is equal to -826 kilojoules per mole.*1218

*The ΔH of formation of Al _{2}O_{3} 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 Fe _{2}O_{3}, 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

2 answers

Last reply by: Jason Smith

Sat Oct 10, 2015 11:47 PM

Post by Jason Smith on September 30, 2015

Hi professor, I have one more question: is "Enthalpies of Reaction" the same as "Standard Enthalpies of Formation"? Thank you.

1 answer

Last reply by: Professor Hovasapian

Fri Oct 2, 2015 2:54 AM

Post by Jason Smith on September 30, 2015

Hi professor. In a nutshell, is this what standard of enthalpies of formation mean: the enthalpy change that occurs when elements form 1 mole of a substance. Is this correct? Thank you.

1 answer

Last reply by: Angela Patrick

Sat Jan 4, 2014 3:12 PM

Post by Ferdinand Klein on October 4, 2012

For example 1, why doesn't the equation have to be per mole? I thought that the equation has to be per mole of product?