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

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

### Enthalpy & Hess's Law

- ΔH = q for constant pressure processes. ΔH is called the change in Enthalpy of a system. It is also referred to as the Heat of reaction.
- A State Function is a thermodynamic property that does not depend on the path used to achieve a given state.
- Enthalpy is a State Function.
- Hess’ Law allows you to manipulate given chemical equations, so that, when you add the resulting equations, you achieve the equation desired. The Enthalpy of the desired equation is the sum of the Enthalpies of the individual equations.

### Enthalpy & Hess's Law

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 1:43
- Defining Enthalpy & Hess's Law
- Example 1
- State Function
- Example 2
- Example 3

### AP Chemistry Online Prep Course

### Transcription: Enthalpy & Hess's Law

*Welcome back to Educator.com; welcome back to AP Chemistry.*0000

*Today, we're going to continue our discussion of thermochemistry.*0003

*We're going to talk about enthalpy and Hess's Law.*0007

*I have to, before I begin...I wanted to discuss--just take a couple of minutes to talk about--thermodynamics and thermochemistry, and a lot of the terms that are sort of being thrown around, and some of the equations.*0010

*Back in the early...well, not early days; in the turn of the century, there is a saying about thermodynamics; a very famous thermodynamicist says this; he said, "None of us really understands thermodynamics; we just get used to it."*0025

*Now, of course, that is not completely true; we understand it; but to a large extent, a lot of it really is true.*0037

*Thermodynamics is a very, very strange thing; heat behaves in very difficult ways--very unusual ways.*0043

*These terms, like enthalpy and heat and energy and work, and pressure-volume...I know that a lot of that is very, very odd; it's difficult to wrap your mind around; it's difficult to sort of get a good, intuitive feeling for what is going on.*0049

*My recommendation for dealing with that is: don't really worry about it too much.*0065

*A lot of the comfort that comes from dealing with thermochemistry and thermodynamics, aside from some of the stuff like today's (it's actually not that bad)...a lot of it just comes from familiarity.*0069

*You will be seeing it over and over and over again, and you will be doing the problems over and over again, so that you will get more of a sense.*0079

*So, if you don't understand it the way you do other things, I wouldn't worry about it too much; that is just the nature of thermodynamics.*0085

*Again, just to throw something out there to make you put your mind at ease: it's like that for everybody.*0092

*With that, let's go ahead and get started.*0099

*OK, so let's start with a definition.*0103

*We're going to define enthalpy: enthalpy (which we use the symbol H for, and you will understand in a minute, when we actually equate it to heat) is equal to the energy of a system, plus the pressure of the system, times the volume of the system.*0108

*So, let's say I had some container of gas; it's going to have a certain energy that is associated with it; it's going to have a certain pressure that is associated with it and a certain volume that is associated with it.*0132

*By definition, that is the enthalpy; if I take the energy, plus the pressure, times the volume, I get the enthalpy.*0142

*Now, like most thermodynamic properties, we don't really know what absolute enthalpies are; the only thing we can actually measure (which is science: in science, we measure things) are changes in enthalpy.*0149

*So, ultimately, we are going to be concerned with ΔH; and, in fact, all of thermodynamics is going to be concerned with the Δ of some properties.*0162

*Later on in the course, we will talk about ΔS; we will talk about ΔG, which is free energy entropy; and H is enthalpy.*0169

*We are concerned more with Δ, and, a little bit later in this lesson, we will show why the Δ is important...in any case, just so you know...*0177

*The definition of enthalpy is the energy plus the pressure times the volume of the system.*0189

*Now, let's recall what we did last time: we said that the change in energy of a system is equal to q, plus the work, which was -PΔV.*0194

*Now, let me rearrange this; let me bring this -ΔV over to this side; q equals the change in energy, plus PΔV--now, let me just set that aside for a minute.*0209

*Now, let me come over here, and let me take my definition of H as equal to E, plus PV, and now I'll do ΔH.*0222

*Well, ΔH, which is final minus initial, equals Δ of this, which is the ΔE plus Δ of PV.*0236

*Well, in this particular case, if we keep the pressure constant, that means we can pull this out of the Δ; we end up having ΔH=ΔE+PΔV.*0247

*Notice what we have: we have that q, which is the heat that is transferred, is equal to the change in energy plus the pressure times the change in volume (constant pressure).*0266

*And we have that ΔH, just based on the definition, equals ΔE, plus pressure, times Δvolume.*0276

*These are the same; q equals ΔH at constant pressure.*0282

*So, this is a constant pressure situation here, OK?--at constant pressure (which is pretty much what we are doing--what we do most of our chemistry in, at a constant atmospheric pressure), q equals ΔH.*0293

*In other words, the enthalpy is nothing more than the heat that is transferred--the energy that flows as heat.*0307

*In other words, if a certain reaction takes place, and it gives off a certain amount of heat or it takes in a certain amount of heat, that heat is equal to the enthalpy.*0314

*So, I can talk about the enthalpy by just referring to the heat; all I have to do is deal with the heat--it happens to be the same as the enthalpy, under constant pressure conditions.*0325

*That is why constant pressure is really, really nice, because under those conditions, I don't have to deal with enthalpy; I just deal with the heat; they are equivalent.*0335

*Heat is a very easy thing to measure; you just run a reaction, and you measure how hot something gets or how cold something gets; that is your enthalpy.*0343

*It is a thermodynamic quantity that is related to heat; under constant pressure conditions, it is the heat--it is equivalent to the heat; so that is really nice.*0351

*This is the important thing to know.*0362

*Heat and enthalpy are equivalent in magnitude under constant pressure conditions.*0366

*OK, so when we talk about the enthalpy of a reaction, ΔH _{rxn}, well, that is equivalent to the heat of the reaction.*0370

*We will often talk about the heat of the reaction.*0380

*OK, now, ΔH of the reaction is equal to...actually, you know what, no; I'm not going to...I'll save this for the next time; there is no need to confuse us with any information that we don't need right away.*0392

*OK, so let's just do a quick example to get a feel for what is going on--a little bit of stoichiometry.*0408

*Example 1: When 1 mole of C _{6}H_{12}O_{6}, which is glucose, is fermented to ethanol at a constant pressure, 67 kilojoules of heat is released.*0415

*The system releases heat; heat is flowing out of the system; it is negative.*0458

*How much heat is released when 7.6 grams of glucose is fermented?*0467

*1 mole of glucose is fermented to ethanol at constant pressure; 67 kilojoules of heat is released--exothermic.*0490

*The enthalpy is negative; now heat and enthalpy is the same--constant pressure.*0499

*How much heat is released when 7.6 grams of glucose is fermented?*0505

*OK, so let's write out what this looks like to get a sense of how we sort of start these problems, and what it looks like, notationally, for a chemist.*0509

*So, C _{6}H_{12}O_{6} is fermented to 2 C_{2}H_{5}OH (ethanol--this is regular drinking alcohol), plus carbon dioxide gas.*0518

*And we often write the ΔH over here as -67 kilojoules; so we see that heat is released; ΔH is negative; it is an exothermic process.*0534

*That is what it means: exothermic--ΔH is negative; ΔH is negative--it is exothermic; it is giving off heat, which means, in addition to this product and this product, one of the other products is that much heat.*0543

*That is why we write it; imagine heat as that third product that also comes out of the reaction.*0556

*This heat comes from the bonds in the carbon, hydrogen, and oxygen; that is where it is coming from.*0563

*OK, let's draw a little energy diagram, so you see what is going on here.*0569

*C _{6}H_{12}O_{6}: this is H, enthalpy (heat at constant pressure), and this is just the reaction coordinate.*0576

*The reaction coordinate just means that the reaction is proceeding in that direction.*0588

*Well, there are some...here we have the C _{2}H_{5}OH, ethanol, plus our CO_{2}.*0594

*Now, what this means--this -67 kilojoules--as it turns out, thermodynamically, the reactants--there is more heat in these bonds; when going from glucose to ethanol as CO _{2}, the amount of energy in these bonds is actually 67 kilojoules less.*0605

*So, this difference right here, from this point to this point, is the 67 kilojoules.*0627

*Because, again, energy cannot be created or destroyed, the energy in these bonds is returned into the energy of these bonds.*0634

*But now, I have an excess amount of energy; what am I going to do with it?--well, the reaction just releases it as heat.*0642

*That is what this says; it is going to a lower heat--this excess heat is what is given off.*0647

*That is why it is -67; this is what it actually looks like.*0655

*The products are thermodynamically at a lower energy.*0658

*OK, so let's go ahead and calculate: well, C _{6}H_{12}O_{6} is 180 grams per mole; we have 7.6 grams of it, times 1 mole--180 grams; that gives us 0.0422 mol, and I hope that I did my arithmetic correctly.*0663

*Well, I have 0.0422 mol, and it is telling me that it releases 67 kilojoules per mole (that is what the problem says).*0694

*It is a simple arithmetic problem.*0705

*-2.8 kilojoules of heat is released.*0708

*2,800 Joules of heat is released with 7.6 grams of glucose.*0723

*7.6 grams of glucose is not very much; it's a handful--not even a handful; it is 2,800 Joules; that is a lot of heat.*0729

*There is a lot of heat in those bonds; that is why the human body metabolizes glucose--it breaks it down, not into alcohol and CO _{2}--it breaks it down completely into carbon dioxide and water, and all of the energy that is released--the body uses that energy to produce a molecule called adenosine triphosphate, and it is the adenosine triphosphate that runs the body.*0743

*That is our energy currency, and it all comes from the energy that is stored in the bonds of C _{6}H_{12}O_{6}.*0767

*Well, just 7.6 grams produces 2,800 kilojoules of energy!*0773

*You can imagine the amount of glucose we actually take in, in the form of carbohydrates and other things; the body requires a lot of energy to run.*0778

*All right, now let's talk about something called a state function.*0790

*Let me...all right, let's define a state function.*0798

*A state function is a property (you could call it a state property; I don't know why they call it a state function, but it is a property) that does not depend on the path taken to achieve that state.*0803

*OK, so let's say I have something here and something there; these are two states; I can get from this state to this state--I can either go this way, directly, or I can go this way and come back; I can go this way, this way, this way, this way, this way, this way.*0846

*Now, as it turns out, there are certain properties that are not state functions, like heat and work.*0862

*So, for example, if I went from here to here, I would have to do a certain amount of work.*0872

*Clearly, if I went from here to here to here to here to here to here, I am doing more work.*0876

*But, as it turns out, energy (which, as we said, is equal to heat plus work, neither of which is a state function)--as it turns out, energy is a state function.*0880

*As it turns out, it doesn't matter how I get to the final state--all that matters is where I started and where I ended up.*0894

*That is why we are concerned with ΔS: ΔH, ΔG--those are state functions; enthalpy is another state function.*0900

*So, we said that enthalpy is equal to the change in energy, plus the pressure, times the change in volume, right?*0907

*Well, this is a state function; pressure is a state function--it doesn't matter how I get there; at a certain point, and at the pressure that I start and end up with, it is just a certain pressure; it doesn't matter how I get there.*0916

*Volume: it doesn't matter how I go from 3 liters to 5 liters; I can go up to 18 liters, drop down to .1 liter, and then go up to 5 liters; I have still just gone from 3 to 5--the net effect is the same.*0929

*So, ΔH is also a state function.*0942

*It is a state function because it is the sum of two state functions.*0944

*Energy is a state function despite the fact that neither of these is a state function.*0948

*This is also quite extraordinary, that that is the case.*0953

*OK, so as far as chemistry is concerned, now: chemistry--if we start with certain reactants, and we want to end up with certain products, well, as far as the enthalpy is concerned, it doesn't matter how I get there.*0957

*I can get there in 2 steps; 15 steps; 147 steps.*0974

*Now, yes, there are areas of chemistry where we are concerned about the steps, but as far as a thermodynamicist is concerned, all he cares about is the enthalpy at the beginning and the enthalpy at the end.*0978

*It is a state function; it doesn't matter how you get there; all that matters is that you get there.*0988

*It is the two states that matter; that is all that matters.*0993

*Because of that, we can actually take a reaction that we are interested in, and perform it, and if we want to find the enthalpy of that reaction, well, if we have enthalpies of other reactions that we can use to get to our final reaction, we get our final enthalpy.*0996

*OK, so now, this is the idea of Hess's Law.*1014

*And, rather than talking about it or writing about it, the best thing to do is just do a problem, and of course, it will make sense.*1021

*So, let me write out Hess's Law here.*1026

*Well, I won't write it out; we'll just do an example, and it will make sense.*1032

*OK, we want to find the ΔH for the reaction of sulfur, plus oxygen gas, going to sulfur dioxide gas.*1036

*In other words, there is a certain heat of reaction associated with this; either it absorbs heat to create SO _{2}, or in the process of creating SO_{2}, it releases heat.*1059

*I want to find the enthalpy--the heat of the reaction.*1069

*Well, how do I do that?*1072

*OK, so we want to find ΔH for the reaction; well, as it turns out, we just happen to know that we have a couple of reactions at our disposal.*1074

*We know that, if I take sulfur plus three-halves oxygen gas, in the process of creating sulfur trioxide gas, we happen to know that the ΔH of that is -395.2 kilojoules.*1088

*We also happen to know that sulfur dioxide gas, 2 moles of that, plus oxygen gas, goes to 2 sulfur trioxide gas, and we happen to know that the ΔH of that equals -198.2 kilojoules.*1110

*So, we have this reaction that we know the ΔH for; we have this reaction that we know the ΔH for.*1134

*Hess's Law says it doesn't matter how we get to our final reaction, if we can come up (oops, no, we don't want these stray lines here) with a way of manipulating these equations (switching them, multiplying them by coefficients, a lot like you do for linear equations in linear algebra or algebra courses that you have taken).*1140

*If we can fiddle around with them and add the equations to come up with a final equation--this one that we want--well, we will just add the ΔHs, and we will get the final enthalpy of the reaction, because again, ΔH is a state function; it doesn't matter how you get there, as long as you get there.*1166

*So, let's see what we are going to do.*1184

*So, how can I fiddle with these equations in order that, when I add them vertically, I end up with this equation?*1186

*OK, well, let's see; let's reverse...let's see, what can I do?*1194

*I am trying to create SO _{2} gas; so notice that SO_{2} here is on the right-hand side, but in these equations, SO_{3} is on the right-hand side.*1201

*The only equation that has SO _{2} in it is over here; I want to reverse it, and there is one SO_{2} here, but there are 2 SO_{2}, so I am going to flip equation 2, and I'm going to divide it by 2.*1212

*OK, so let me...actually, let me rewrite everything again, so that we have it on one page.*1228

*We have S + O _{2} going to SO_{2}; that is the reaction that we want; and we are given S + 3/2 O_{2} goes to SO_{3}; ΔH equals -395.2 kilojoules.*1241

*Our second equation (we'll call that #1) is 2 sulfur dioxide gas, plus O _{2}, goes to 2 SO_{3}, and the ΔH for that is -198.2 kilojoules.*1262

*OK, so we said we want this equation from these equations.*1279

*We are going to flip this equation, #2; so, we are going to reverse #2 and divide it by 2.*1284

*When we reverse it and divide it by 2, this 2 SO _{3} comes to the left, and it ends up becoming just SO_{3}; this 2 SO_{2} ends up on the right, but becomes SO_{2}; and this O_{2} also ends up on the right, but it becomes 1/2 O_{2}.*1299

*Now, what happens to the ΔH?--well, exactly what you think.*1315

*If you flip a reaction--if you flip an equation--you change the sign of ΔH; if you divide an equation by 2, you divide the ΔH by 2.*1318

*So now, the ΔH is no longer -198; it is going to equal +99.1 kilojoules.*1327

*And now, we leave the...here, our other equation is S + 3/2 O _{2} goes to SO_{3}, so S is on the left--there is one S on the left--so let's leave that one alone.*1337

*We have S + 3/2 O _{2} going to SO_{3}; that ΔH stays the same: -395.2 kilojoules.*1351

*Now, we just add straight down.*1367

*Everything that is on both sides--if there is something on the left and something on the right, they cancel.*1369

*SO _{3} on the left; SO_{3} on the right; it cancels.*1375

*S comes down; that is taken care of.*1379

*I have 3/2 O _{2} on the left; I have 1/2 O_{2} on the right; 3/2 minus 1/2 is equal to...well, there you go: O_{2} (two halves)...+ O_{2}.*1383

*O _{2} is taken care of; now, the only thing left is the SO_{2}.*1400

*There you go; I have the final equation that I wanted by messing around with the equations for which I did have information.*1405

*Now, all I do is: I just add the ΔHs.*1413

*When I add the ΔHs: +99.1, minus 395.2; we get -296.1 kilojoules.*1416

*I used reactions that I knew, manipulated the equations, made the appropriate changes to the enthalpy, and I added straight down, added straight down; now I have the enthalpy for this reaction.*1426

*This is Hess's Law; I can use reactions that I do know to find a reaction that I want.*1439

*OK, let's do another example here.*1447

*Let's do this on a new page.*1454

*Calculate the ΔH for the synthesis (which means formation) of di-nitrogen pentoxide gas, N _{2}O_{5}, from its elements.*1461

*OK, elements--so, the reaction that we want is nitrogen gas, plus oxygen gas, goes to N _{2}O_{5} gas.*1489

*Now, we have to balance this: so, we have 5 and 2, so we'll put a 5 here; we'll put a 2 here.*1505

*We'll put a 2 here; now it's balanced.*1512

*This is the equation that we want; OK, now here are the equations that we have at our disposal.*1515

*Equation #1: we have H _{2} + 1/2 O_{2} goes to H_{2}O; the ΔH of that is equal to -285.8 kilojoules.*1521

*Our second equation is: N _{2}O_{5} + H_{2}O goes to 2 HNO_{3}, which is nitric acid; the ΔH of that is -76.6; exothermic, exothermic.*1540

*3: we have 1/2 N _{2} + 3/2 O_{2} + 1/2 H_{2} goes to HNO_{3}; the ΔH of this is equal to -174.1 kilojoules.*1565

*So, our task is to take these three equations, manipulate them by multiplying by coefficients, reversing them, and then adding them straight down and arranging them in such a way so that, when they add, they add to the final equation.*1585

*OK, so let's see what we have; what can we do?*1604

*I notice I have N _{2}O_{5} on the right, and here I have N_{2}O_{5} on the left, so let me just leave that one alone for now.*1608

*Let me see: I have 2 N _{2} on the left, 5 O_{2}; I have 2 N_{2}, and the only equation here that has N_{2} in it is this one, so I'm going to multiply equation #3 by 4.*1623

*So, multiply #3 by 4, and what I end up with, when I multiply the third equation by 4--I get 2 N _{2} + 6 O_{2} + 2 H_{2} goes to 4 HNO_{3}, and the ΔH also gets multiplied by 4.*1638

*Negative...so this is multiplied by 4 minus 696.4 kilojoules; so whatever I do to the equation, I do the same thing to the enthalpy.*1673

*OK, now let's see what I have; HNO _{3}...I have 4 HNO_{3}, but I don't have any HNO_{3} here, so I need 4 HNO_{3}s on the left.*1683

*I have 2 HNO _{3}s here, so I'm going to flip this equation and multiply it by w.*1695

*So, we'll flip 2 and multiply by 2.*1699

*When I flip this and multiply by 2, I get 4 HNO _{3} goes to 2 N_{2}O_{5} + 2 H_{2}O, and now the ΔH for this...I have flipped it, and I have multiplied it by 2, so now this is a positive 153.2 kilojoules.*1711

*Now, #1--let me see: #1 equation: H _{2}; I have...what do I have?...I have 2 H_{2}s on the left; I have an H_{2} here; I need to cancel the H_{2}s.*1741

*I have 2 H _{2}Os on the right; I have an H_{2}O on the right here; so I have used equation 3 and used equation 2; I need to get this on the left, and I need to multiply it by 2, so I'll do the same thing.*1755

*I'll flip #1 and multiply by 2, and I end up with the following equation.*1765

*I get 2 H _{2}O goes to 2 H_{2} + O_{2}; and again, the ΔH is going to be +571.6 kilojoules.*1776

*And now, I should end up with what I have; so let me see here; let me...we are concerned with this equation, this equation, and this equation; so let's see what cancels.*1797

*H _{2}O; 2 H_{2}O, 2 H_{2}O; 2 H_{2}, 2 H_{2}, right?--this is on the left of the arrow; this is on the right of the arrow.*1815

*They are on top of each other, but it is where they are on as far as the arrows are concerned.*1824

*I have 4 HNO _{3}, 4 HNO_{3}; 2 N_{2}, so I'll bring that down; that takes care of the 2 N_{2}.*1830

*I have 6 oxygens on the left; I have 1 oxygen on the right; so, 6 minus the 1 leaves 5 oxygens on the left.*1841

*So, that is taken care of and that is taken care of; now, the only thing left is the 2 N _{2}O_{5}, which is on the right, so I am not adding it.*1853

*That is on the right-hand side of the arrow; so I get 2 N _{2}O_{5}, which is exactly what we want: 2 N_{2} + 5 O_{2} goes to 2 N_{2}O_{5}, exactly what we found.*1866

*Now, let's just add the ΔHs straight down, and when we add them, we end up with 28.4; again, you might want to check my arithmetic; I'm notorious for bad arithmetic.*1880

*So, this reaction: nitrogen gas plus oxygen gas to form di-nitrogen pentoxide: it is positive 28.4 kilojoules, so this is endothermic.*1893

*That means, in order for this reaction to go forward, I actually have to add heat to it.*1906

*Or, if I leave it alone, and there are other circumstances where there is something happening--there is enough heat that it can pull from the surroundings--it will pull that heat from the surroundings in order to make this go.*1911

*So, there you have it--Hess's Law.*1924

*We want to find the enthalpy, the heat of a given reaction; well, if we have other reactions at our disposal that we know the heat for, and we can rearrange those equations in a certain way--we can fiddle with the enthalpies appropriately--add the equations; we will get the equation that we are looking for, and we will get the enthalpy of that equation.*1926

*OK, thank you for joining us today at Educator.com for the discussion of enthalpy.*1947

*We will see you next time; goodbye.*1953

2 answers

Last reply by: Professor Hovasapian

Thu Jul 21, 2016 6:28 PM

Post by Adel Althaqafy on July 20, 2016

Hi Prof

when you multiply the equation by 2 why change - minus to positive in the amount of the enthalpy change

2 answers

Last reply by: Professor Hovasapian

Sun Jul 3, 2016 7:21 PM

Post by Jeffrey McNeary on June 29, 2016

at 15:40, you stated that delta H is a state function, and 15 seconds later you stated that h is NOT a state function. If q is equivalent to delta, then how are they not the same kind of function?

4 answers

Last reply by: Kaye Lim

Mon Jul 4, 2016 5:56 PM

Post by Kaye Lim on June 8, 2016

For endothermic process, if product bonds store less amount of Energy compared to reactant's bonds (product has lower Energy than reactant), then why it takes more Energy heat to break the product bond into free elements and also into free atoms?

It confused me how the bonds in product have lower Energy in endothermic process because it is further down the graph, yet require more Energy to be broken into free atoms. Please explain what was wrong in my reasoning process.

3 answers

Last reply by: Professor Hovasapian

Fri Jun 3, 2016 6:23 PM

Post by Tram T on May 17, 2016

Dear prof. Hovasapian,

-For endothermic reaction, I see that the bonding energy or the energy to break all the products of the reaction into free atoms is higher than the bonding energy of reactants because the Energy gap going from products to free atoms is larger than the Energy gap going from reactants to free atoms.

But I can't wrap my mind around how going from system (reactants) with lower Energy (Energy stored in the reactant bonds) to system with higher Energy releasing Energy heat. Please explain! Thank you!

2 answers

Last reply by: Jason Smith

Wed Dec 30, 2015 7:59 PM

Post by Jason Smith on December 23, 2015

Hi professor. How did you arrange the [delta E = q - P delta V] equation to get [q = delta E + P delta V]?

Like, how did you get the q isolated on one side of the equation without having to divide?

Sorry for silly question, math really isn't my strong suit :P

1 answer

Last reply by: Professor Hovasapian

Sun Sep 27, 2015 1:39 AM

Post by Gaurav Kumar on September 26, 2015

Hi Professor Hovasapian,

For examples 2 and 3 are the equations that you wrote after the equation that we were trying to reach given in the problem, or did make up those equations? For example, was N2O5 +H2O ----> 2HNO3 given in the problem for example 3 or did you make it up?

Thank you

1 answer

Last reply by: Professor Hovasapian

Tue Dec 2, 2014 2:45 AM

Post by Muhammad Ziad on November 29, 2014

Also, are the three equations in problem 3 given or do you have to come up with them on your own?

1 answer

Last reply by: Professor Hovasapian

Tue Dec 2, 2014 2:50 AM

Post by Muhammad Ziad on November 29, 2014

Thank you for this lecture. I have a question about the delta H values in example 2. Where do you get these values from? Are they supposed to be memorized or is there a table we can find the values? I'm starting to understand everything much better except getting the delta H values.

1 answer

Last reply by: Professor Hovasapian

Fri Nov 28, 2014 12:33 AM

Post by Minjae Kim on November 27, 2014

If Î”E = q + w and Î”H = q, then why is Î”H = Î”E + PÎ”V? Why not Î”H= = Î”E - PÎ”V?

1 answer

Last reply by: Professor Hovasapian

Thu May 1, 2014 9:16 PM

Post by Mark Andrews on April 20, 2014

In example 2

2S02 + O2 -----> 2SO3

To be balanced this shouldn't this be 2SO4?

1 answer

Last reply by: Andrea Baric

Sat Apr 26, 2014 9:18 PM

Post by Hyun Cho on December 22, 2013

i have a question concerning 18:31 when you are giving examples on Hess's Law. When you tell us that S+3/2O2> SO3+change in enthalpy of -395.2KJ, is that stated in the problem or is it just something that we should know as basics of chemistry? if we should already know it, could you tell me where I can learn those kind of stuffs?

1 answer

Last reply by: Professor Hovasapian

Mon Dec 23, 2013 8:46 PM

Post by Hyun Cho on December 22, 2013

i have a question concerning 18:31 when you are giving examples on Hess's Law. When you tell us that S+3/2O2> SO3+change in enthalpy of -395.2KJ, is that stated in the problem or is it just something that we should know as basics of chemistry? if we should already know it, could you tell me where I can learn those kind of stuffs?

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Last reply by: Professor Hovasapian

Mon Oct 21, 2013 3:42 AM

Post by Cynthia Alvarez on October 21, 2013

Given: K+1/2Cl2 -> KCl DeltaH=-437kJ

How much heat is absorbed/released when o.5moles Cl2 is formed from KCl?

Would this be the answer?: 437kJ absorbed

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Last reply by: Professor Hovasapian

Wed May 15, 2013 2:07 AM

Post by Nawaphan Jedjomnongkit on May 14, 2013

Thank you for the lecture. From what you explain about state function and did mention that q is not state function but when it comes to enthalpy, it is a state function. So am I right if I think that q normally it is not a state function but it will be only when P is constant ?

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Last reply by: Professor Hovasapian

Mon Mar 18, 2013 10:23 PM

Post by Joseph Grosse on March 18, 2013

Example 1: When 1 mole of glucose is fermented to ethanol, 67KJ of energy is released. When 7.6 g of glucose (aka ~.0422 moles) goes through fermentation the energy released is -2.8KJ.

Why would the sign of energy released switch between the two situations, when the only variable, is that of the the mass of the compound that is being fermented?

Thank you.

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Last reply by: Professor Hovasapian

Mon Jul 23, 2012 7:15 PM

Post by Marlon Kalicharan on July 22, 2012

How come you have for the 1.H2+HALF02 YIELDS H20.I am lost.Where does that come from?

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Post by NGAWANG TSERING on March 8, 2012

hi it stuck automatically at example 2 of state function