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

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

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## Table of Contents

## Transcription

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### Thermochemistry Example Problems

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
- Example I: Find ∆H° for the Following Reaction
- Example II: Calculate the ∆U° for the Reaction in Example I
- Example III: Calculate the Heat of Formation of NH₃ at 298 K
- Example IV
- Part A: Calculate the Heat of Vaporization of Water at 25°C
- Part B: Calculate the Work Done in Vaporizing 2 Mols of Water at 25°C Under a Constant Pressure of 1 atm
- Part C: Find ∆U for the Vaporization of Water at 25°C
- Part D: Find the Enthalpy of Vaporization of Water at 100°C
- Example V

- Intro 0:00
- Example I: Find ∆H° for the Following Reaction 0:42
- Example II: Calculate the ∆U° for the Reaction in Example I 5:33
- Example III: Calculate the Heat of Formation of NH₃ at 298 K 14:23
- Example IV 32:15
- Part A: Calculate the Heat of Vaporization of Water at 25°C
- Part B: Calculate the Work Done in Vaporizing 2 Mols of Water at 25°C Under a Constant Pressure of 1 atm
- Part C: Find ∆U for the Vaporization of Water at 25°C
- Part D: Find the Enthalpy of Vaporization of Water at 100°C
- Example V 49:24
- Part A: Constant Temperature & Increasing Pressure
- Part B: Increasing temperature & Constant Pressure

### Physical Chemistry Online Course

### Transcription: Thermochemistry Example Problems

*Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.*0000

*Today, we are going to talk about Thermal Chemistry.*0005

*We just finished our discussion with several examples of Thermodynamics.*0007

*Discussing energy and heat and work and things like that.*0011

*I thought it would be best instead of discussing Thermal Chemistry theoretically and then doing problems, I think it is best just do the problems.*0015

*And then within the problems itself, discuss anything that needs to be discussed theoretically which is actually very little.*0021

*Much of this is stuff that you already know from General Chemistry.*0031

*Just a couple of the problems are a little bit more sophisticated than what you are you are used to, except for the first which is going to be very basic.*0034

*Let us jump right on in.*0041

*This one was nice and easy, it says find δ H for the following reaction.*0044

*We have titanium dioxide + chlorine gas goes to titanium chloride and oxygen gas.*0049

*Notice we have solid gas, solid gas.*0059

*Finding the δ H for the reaction is really simple.*0063

*Again we know this from general chemistry.*0066

*Let us see what color I’m going to use, I think I’m going to do blue.*0068

*We know that the δ H is equal to what would you say the δ H of the reaction and*0074

*we remember that this little degree sign up above there represents standard conditions.*0086

*Standard conditions happen to be 25°C or 298 K and 1 atm pressure.*0090

*It is just we need a particular standard so we choose that standard.*0097

*It is equal to the sum, I have forgotten how to write this term here.*0102

*The sum of the δ H is of formation for the products - the sum of the δ H is of formation for the reactants including coefficients.*0113

*Again, this is something that we already know from general chemistry, it is products – reactants.*0130

*Whenever you see a standard table of thermo chemical data or thermodynamic data, you are going to have the enthalpy a formation listed.*0136

*You are going to have the standard free energies and you are going to have the entropy listed.*0144

*And you can use those tables, you just take the enthalpy of the products, subtract the enthalpy of the reactants making sure to include the coefficients*0148

*and that gives you the actual enthalpy change for the heat of reaction under constant pressure conditions,*0160

*because under constant pressure conditions the enthalpy is equal to the heat.*0169

*Let us just go ahead and calculate this.*0176

*All we need is a table of thermodynamic data so we have δ H for elements and their standard state is 0.*0177

*This is chlorine gas CO₂ so that is going to be 0 and O₂ gas that is going to be 0.*0193

*We only have to worry about that one and that one right there.*0198

*When we look up the enthalpy for the Ti Cl4, we have -803 kJ/ mol × 1 mol.*0203

*That was canceled and we add 2 at the enthalpy for this which is 0 and we subtract the sum of the enthalpy of formation for the reactants.*0220

*And we get, for the titanium dioxide, we look it up and -945 kJ/ mol.*0231

*Again, the coefficient on that is 1 so it is × 1 mol + 0.*0240

*It is going to be -803 --945 so we have a δ H for this reaction is going to equal 142 kJ or 142,000 J.*0248

*Make sure you watch out for the units, it is really important.*0265

*When you are looking at the table of thermodynamic values, the enthalpy of formation and the standard free energies of formation those are in kJ/ mol.*0270

*The entropy is going to be in J/ mol/ K.*0282

*Later on, when we actually get to the equation you are familiar with, δ G = δ H - T δ S, when you are mixing does up and*0287

*when you are working with entropy, as well as enthalpy and free energy, we want to make sure the units match.*0295

*We either need to convert to J or we need to convert the entropy to kJ.*0301

*142 there you go, that is it.*0310

*This is positive enthalpy which means this is an endothermic reaction.*0314

*In other words, in order for this reaction to both go forward, it actually has to pull 142 kJ of heat from the surroundings, it has to go into the system.*0318

*Let us go ahead, that was nice and easy.*0333

*Let us go ahead and this one says calculate δ U for the reaction in problem 1, assuming the gases behave ideally.*0336

*This is different, now we are not calculating enthalpy, we are calculating the change in energy.*0344

*Let us go ahead and write our equation again just so we had on this page.*0351

*We had a titanium dioxide which was solid + we had 2 Cl2 which was gaseous and it went 2 Ti Cl4 the titanium 4 chloride which was solid + 02 gas.*0355

*Here is what is interesting.*0379

*Notice that we have solid gas and a solid gas, when we run these reactions, when calculating heats of reaction,*0380

*If any of the reactants or products are gaseous as they are in this case, the reaction has to be one of the bomb colorimeter.*0399

*A bomb colorimeter, if you remember from general chemistry or if you do not I will explain it right now.*0436

*It is basically is a colorimeter where the volume is held constant.*0441

*We put everything in sight of this thing called a bomb and it is immersed in water and we run the reaction inside.*0445

*We generally tend to keep the temperature and we tend to keep the volume.*0452

*The volume is absolutely fixed because it is a bomb, it is a single volume, there is no expansion.*0458

*This makes sense because if you are going to produce, for example this reaction produce oxygen gas, you have to contain that gas.*0463

*You cannot just let it escape into the atm which is why we have to constrict and make sure it stays in one place.*0468

*There are some things that we can control.*0478

*In this case, we are controlling volume and we also make sure the temperature stays constant so the only pressure that ends up changing.*0480

*Let us go ahead and see what is happening here.*0489

*The transformation is this.*0492

*Let us go ahead and see, the transformation is we have our reactants and it is going two products.*0496

*It is going to be at a given pressure, the reactant is going to be at a given pressure P1, a certain temperature, and a certain volume.*0511

*We are going to end up keeping temperature and volume the same experimentally but we are going to end up with a new pressure.*0520

*This is what happens in terms of the state variables P1 TV to P2 TV.*0526

*The relation between δ H and δ U, we are looking for δ U.*0533

*We already calculated δ H from the previous problem.*0538

*The relation is remember δ H = δ U+ δ PV.*0542

*Remember the relation H = U + PV, in this case δ H = δ U+ δ PV that is the actual relation.*0560

*But notice we are holding volume constant so it comes out.*0569

*What you end up with is δ H = δ U+ δ P × V.*0573

*Now δ H = δ U, δ P is just the final pressure - the initial pressure inside that bomb × the volume.*0585

*Again, this says assuming the gases behave ideally.*0600

*The ideal gas law is PV = nrt therefore P = nrt/ V.*0604

*Therefore, pressure 2 = n2 RT/ T.*0615

*Remember, T is constant and V is constant, R is a constant, it is the gas constant.*0622

*The only thing that changes is the number of moles and P1 = n2 × RT/ V.*0625

*Therefore, we will put these expressions into here and we end up with δ H = δ U + n2 RT/ V – n1 RT/ V × V.*0635

*The V's cancel out and we are left with δ H = δ U + n2 – n1.*0658

*I will pull out the RT δ n × RT.*0665

*δ n is just the difference in the number of moles of the gaseous components of products and reactants.*0673

*δ n is just the number of moles and product - the number of moles in the reactant of the gaseous products.*0680

*The solids they do not count, they do not contribute to anything in terms of actual pressure of things like that.*0688

*It is the gas that contributes to the pressure.*0694

*This is our basic equation that we are going to need.*0696

*What we are going to do is we are going to move this over the other side to get the δ U because δ U is what we want.*0698

*Let us go ahead and calculate δ n.*0707

*δ n = the number of moles of gaseous products - the number of moles gaseous reactants.*0711

*Again that is just the coefficients δ n = 1 -2 = -1.*0722

*δ H = δ U+ δ N × RT move this over and we are left with δ U = δ H – δ n × RT.*0735

*We already calculated δ H, I’m going to do this in terms of, that was 142 kJ.*0754

*We have 142,000 J.*0762

*I’m just going to go ahead and convert them to J because R is here, I’m going to use the 8.314 J/ K mol - δ n is -1, R is 8.314 J/ mol K, and then RT temperature at 298, 25°.*0767

*When I do this, I end up with δ U= 144000 J or 144 kJ.*0794

*What is important was what we derive, this relation right here.*0817

*It all comes from the definition of enthalpy H = U + PV.*0821

*δ H = δ U + δ PV.*0828

*V is held constant so I can pull that out so I just get δ H = δ U+ δ P × V.*0832

*Pv = nrt so I put that in and I'm left with this final relationship here.*0845

*Enthalpy energy is the relationship that we used.*0852

*Let us go ahead and see what else we can do here.*0856

*This is one of the problems that you probably did not see when you are in general chemistry and it is a very important problem*0866

*because it is going to allow you to calculate enthalpy is a different temperatures.*0874

*All your enthalpy is that we calculated in general chemistry where from the table of thermodynamic data and all that is done at 25°C 1 atm pressure.*0880

*They told you this at the time but of course you did not do anything with it, that enthalpy reaction depends on the temperature which we run the reaction.*0891

*The reaction was at 25°C vs. Reaction that is 225°C, the enthalpy is going to be different.*0900

*We need to find a way to calculate the enthalpy at a new temperature.*0907

*That is what I do with this problem.*0911

*The following data holds at 1000 K.*0915

*We have nitrogen gas + 3 mol of hydrogen gas goes to 2 mol of ammonia gas.*0919

*Sorry about that, it should be right there, the δ H for this reaction at 1000° = -123.8 kJ/ mol.*0926

*Here is some more data here, the substance nitrogen gas its molar heat capacity is 3.50 × R.*0939

*Hydrogen has a molar heat capacity of 3.47 R and ammonia has a molar capacity of 4.22 R.*0948

*We want to calculate the heat of formation of NH3 at 298 K.*0955

*We want to find the heat formation NH3.*0962

*In other words, we want to find this.*0965

*We want to find N2 + 3H2 goes to 2 NH3.*0967

*We want a find δ H of formation at 298.*0976

*The δ H of formation is for formation of 1 mol.*0981

*We actually have a something like this.*0987

*This is actually what we want, the δ H of formation is the heat of a reaction for the formation of 1 mol of whatever it is that we are looking for.*0998

*The reactants and their elements in their most basic state, nitrogen gas, hydrogen gas, goes to 1 mol of ammonia gas.*1010

*This is what we are looking for.*1020

*Let us get started.*1022

*We will write this again so δ H values are generally temperature dependent.*1025

*In other words, the enthalpy change for a reaction depends on the temperature at which a reaction is run.*1043

*Let us begin with this one, let us begin with δ H = H of the products - the H of the reactants.*1074

*That is what δ is, just products – reactants that is the definition of the δ.*1087

*I will go ahead and differentiate this with respect to T.*1090

*We have differential with respect to temperature not time.*1094

*Differential with respect to temperature so we have DDT of this δ H = DH DT – DH sub R DT.*1101

*Products and reactants, DH DT we know what that is.*1120

*Anytime you have an enthalpy divided by a temperature that is the definition of heat capacity.*1132

*DH DT that is the definition of constant pressure heat capacity.*1138

*These are standard, let us just go ahead and put the standard, in other words, the heat capacity at 25°C and 1 atm pressure.*1146

*Since that is the case, this D of the δ H with respect to temperature, this thing right here that = δ of the heat capacity,*1155

*in other words, the heat capacity of all of the products - the heat capacity of the reactants.*1180

*That is what we are saying here, that is what this is.*1186

*Because DH DT = heat capacity is δ H is actually δ CP.*1190

*We are just taking the heat constant pressure heat capacity of the products - the constant pressure heat capacity of the reactants.*1197

*This is this and this is this.*1208

*Since we have that, let us go ahead and move this over there and we have the following.*1217

*We have the D of δ H standard = δ CP × DT.*1222

*Let us go ahead integrate this.*1235

*If we integrate this from temperature 1 to temperature 2, we integrate this from temperature 1 to temperature 2.*1238

*Here, what this becomes is just δ H at temperature 2 - δ H at temperature 1.*1246

*Write the integral of the D just go away and you are left with δ H and you do the T2 - T1 = the integral from T1 to T2, the difference in the heat capacity DT.*1264

*Therefore, I just take this and move it over here and I get the change in enthalpy of the given reaction*1282

*at a particular temperature T2 = the change in enthalpy at a given temperature.*1291

*The way I keep this standard is actually the standard does not refer to temperature, the standards refers to pressure.*1302

*When you see this thing right here, it actually means 1 atm pressure.*1309

*It just happens that most of the work that we do in general chemistry happens at 25°C.*1313

*We also include that 25°C and its degree sign but you often see the degree sign for different temperatures*1319

*because that really just refers to the 1 atm pressure.*1326

*It is when we change the pressure that this degree sign tends to disappear.*1328

*I will move this over + the integral from T1 to T2 of δ CP DT.*1333

*This is the equation that we wanted.*1343

*This is the equation that we are going to use.*1344

*Here is what is going on, it is telling me that if I want to calculate the change in enthalpy of the reaction at a given temperature like 1000°,*1348

*I start by calculating the δ H at the temperature that I do know which is 25°.*1359

*Then, I integrate the difference in the heat capacities of the products and the reactants from one temperature to the other.*1365

*That is what this says.*1375

*I have a way of calculating the enthalpy change for any reaction if I already know the change at 25°C*1376

*and if I know the constant pressure heat capacities for the reactants and the products.*1385

*Constant pressure heat capacities are tabulated for everything.*1390

*These are just because I have to look them up in a table, the way you would in other bit of information in chemistry and physics.*1395

*Let us go ahead and do this problem.*1401

*This is the import the equation.*1404

*Again δ CP, let me write that down, where this δ CP = the sum of the heat capacities of the products - the sum of the heat capacities of the reactants.*1406

*That is all this is.*1430

*Something very important to remember that is why I put several asterisk by it.*1433

*When we calculate enthalpy for elements are going to be 0 like the O₂ and the Cl₂ it was 0.*1441

*Heat capacity is never 0.*1451

*Therefore, when you are dealing with elements you cannot ignore the heat capacity.*1453

*The heat capacity for every product in every reacted must appear in this equation.*1458

*The heat capacity for elements is not 0 like it is for the δ H of formation.*1468

*The δ H of formation of elements is 0.*1484

*All heat capacities must be accounted for and again multiplied by the respective coefficients.*1491

*Here is what we have, we have that δ H at a given temperature = the δ H at some initial temperature + the integral from T0 to T of the change in the δ CP DT.*1530

*We can set up this integral up in two ways.*1549

*I will go ahead and show you the two different ways to set up and choose one to actually work with.*1558

*We can set up this way.*1562

*We are looking for the δ H in this particular problem, we are looking for the δ H of formation at 298°K,*1564

*because the information they give us is 4000 K so we can set up this way.*1570

*We can go δ H at 298 = δ H at 1000 + the integral from 1000 to 298.*1577

*It is going to be from this temperature to that temperature.*1608

*Generally, we tend to increase temperature.*1611

*In this case, they gave us the δ H at a higher temperature.*1613

*They want to know what it is for the normal or lower temperature.*1616

*It is what you are given to where you are going, that is the integral.*1619

*This is going to end up being negative of the δ CP DT or I can set up this way.*1624

*I can set up as δ H 1000 = δ H of 298 + the integral of 298 to 1000 of δ CP DT.*1634

*In this case, some people like to go, they like the lower limit integration to be a lower number, it is a personal choice.*1651

*It does not really matter.*1657

*In this case because this is the number you are looking for, you are given this one, you are going to calculate this one.*1658

*When you do this arithmetic, you are just going to move this over to actually get that value.*1663

*Let us go ahead and do it this way.*1670

*The δ H at 298 is going to equal δ H at 1000 - the integral 298 to 1000 of δ CP DT.*1677

*Let us go ahead and calculate δ CP.*1695

*δ CP is equal to the capacities of the products - the capacities of the reactants multiplied by the coefficients.*1698

*We will do this in red.*1708

*Our equation was, I’m just going to write the equation right here.*1709

*I had N2 + 3H2 = 2NH3.*1715

*For NH3= 4.22 × R.*1725

*Remember it was 4.22 R.*1732

*So × 8.314 and then there are 2 mol × 2, that is our products.*1734

*We will do the N2.*1743

*The N2 was 3.50 R, 8.314, and there is 1 mol of it + for H2 it was 3.47 R 8.314.*1746

*And of course, there are 3 mol of it, that is that.*1763

*Our δ CP δ of the heat capacities of products and reactants = -45.48 J K.*1767

*We can actually run the reaction.*1784

*Our δ H of 298 = the δ H.*1787

*That is what we are doing, we decided to use this version of it.*1800

*Not this version but this version, δ H of 298 it was the δ H to 1000.*1804

*That was given to us, the δ H 1,000 was -123.8 kJ.*1808

*So -123,800 J - the integral from 298 to 1000 of the δ CP which we just calculated as - 45.48 DT.*1814

*You can do this by hand or put a mathematical software, you end up with the following.*1840

*When you actually do this calculation, what you are going to end up with is - 91,873 J but notice this is for the formation of 2 mol of the NH3.*1851

*We were looking for the δ H of formation which by definition is the formation of 1 mol.*1870

*We just divide that by 2.*1874

*Our δ H of formation, sorry about that not the reaction.*1877

*The δ H of formation is going to be at 298 going to be - 45.9 kJ/ mol, that is our final answer, that is what we wanted.*1884

*And it is all based on this right there.*1905

*This the δ H at 1 temperature is the δ H at any given temperature is that 298 - 25°C that we normally get from thermodynamic cables +*1912

*the integral up the difference of the heat capacities of products and reactants.*1925

*It is very important equation.*1930

*Let us move on to the next one.*1936

*This look like it is going to be long.*1937

*Hopefully, it is not too bad.*1939

*Part A, using a table standard thermodynamic data, calculate the heat of a precision of water at 25°C.*1942

*This one is very easy, not a problem.*1948

*Calculate the work done in vaporizing 2 mol of water at 25°C under constant pressure of 1 atm.*1951

*It should not be too bad because we know what the definition of work is.*1958

*We know the work is external pressure × change in volume.*1961

*I will go ahead and write that down here so work = external pressure × change in volume, that is just the definition of work.*1968

*Find a δ U for the vaporization of water.*1978

*We are probably end up using that relationship δ H = δ U+ δ PV and we are going to probably work with that and manipulate that little bit.*1984

*The CP for water vapor is this, the CP in other words the constant pressure heat capacity for liquid water is this.*1997

*Find the enthalpy of vaporization of water at 100°C.*2003

*We are going to do what we just did.*2007

*We are going to find enthalpy in a given temperature and we are going to use it to find enthalpy at a higher temperature.*2009

*Let us jump right on in and see what we can do.*2016

*Part A, use a table standard thermodynamic data, calculate the heat of a precision of water at 25°C.*2020

*Part A, our reaction is this, we are going H to a liquid to H2O gas and we want to find δ H of this reaction.*2030

* This is the δ H of vaporization, in this reaction we are taking the liquid and we are converting it into a gas.*2043

*Δ H of vaporization happens to equal the δ H of the reaction which is this reaction which is equal to.*2050

*Let us look this one up, it is this – this, what you end up with is -241.814 - -285.830 so you end up with δ H of vaporization = 44.016 kJ/ mol.*2062

*That is our answer.*2097

*A 25°C 44.016 kJ of heat needs to be put into the liquid water in order to convert all of it to gaseous water, that is all the says.*2099

*The δ H of vaporization is, how much heat is required to affect this transformation.*2112

*If I'm at 25°C, in order to convert all that water I need to put in this much heat.*2118

*Part B, what is the work required to affect this transformation?*2125

*We said that the work = the external pressure × the change in volume.*2133

*We know what the external pressures is 1 atm, this is happening under atmospheric conditions.*2139

*You are just boiling water, it is what you are doing.*2143

*Our δ V = the volume of the gas - the volume of liquid.*2147

*In other words liquid has a certain volume, you are converting that all to water vapor gas.*2155

*The change in volume is the difference between the volume that the gas occupies and the volume of liquid occupies.*2160

*We have a little bit of calculation to do to find out what the volume of the liquid is.*2166

*We only know what the volume of the gas is, we can just go ahead and use what we know from general chemistry which is 22.4 L/ mol.*2172

*Let us see what we can do as far as the volume of liquid is concerned.*2179

*The volume of liquid, volume of the liquid water, here is how you calculate it.*2183

*25°C, the density of water is 0.9970 g/ ml.*2190

*There are 1000 ml in 1 L and there are in 1 mol of water 18 g.*2200

*G cancels g, ml cancels ml, I'm left with mol and L.*2214

* The number is going to be 55.39 mol/ L.*2219

*I need the molar volume so I need L/ ml.*2228

*I’m going to this, when I reciprocated it I end up getting 1/ 55.39 mol/ L gives me 0.01805 L/ mol.*2232

*1 mol liquid water at 25°C occupies 0.01805 L.*2255

*This is just a simple conversion.*2262

*I have 2 mol of water so I just multiply by 2 and I get 0.03611 L.*2267

*2 mol of water at 25°C liquid water occupies 0.03611 L.*2282

*That is the volume of liquid water.*2288

*Let us go ahead and find the volume of the gas.*2293

*The volume of the gas is very simple.*2296

*We have 22.414 L/ mol.*2299

*I have 2 mol of water so I have 44.828.*2306

*Please confirm my arithmetic, notorious for arithmetic mistakes.*2315

*The volume of our gas is that, δ V our change in volume is the volume of the gas - the volume of liquid.*2318

*It is equal this - that so 44.828 -0.03611.*2328

*I'm left with 44.792 L.*2341

*Our δ V is that, now we can go to our problem.*2347

*How our work = our external pressure × change in volume.*2352

*Our external pressure is 1 atm, in this particular case our change in volume was 44.792 L.*2360

*Therefore, our work = 44.792 L atm that is a unit of energy but we need to convert that J.*2370

*Therefore, 44.792 L atm is × 8.314 J =.08206 L atm.*2383

*Those are just the two different versions of R, the gas constant in J/ mol K L atm/ mol K.*2400

*That is your conversion factor.*2407

*When you do this, you end up with the work = 4.5 kJ/ 2 mol.*2410

*Per mol = 2.25 kJ/ mol.*2422

*There we go.*2430

*What this says that if I have 1 mol water at 25°C, the amount of work that I do in converting that liquid water to gas,*2432

*When I'm converting it to gas I’m pushing away the atm, the gas, the liquid, the water vapor is expanding.*2444

*It is doing work, it is pushing against the atm.*2451

*The amount of work it is doing per mol of water is 2.25 kJ.*2453

*That is a lot of work.*2457

*Part C, it wants you to find the change in energy for this transformation.*2462

*We know that δ H = δ U.*2468

*I will just start with the definition of enthalpy H = U + PV.*2473

*Therefore, δ H = δ U+ δ PV.*2482

*Pressure is constant, it is the atmospheric pressure so that P comes out.*2493

*Therefore, what you have is now δ H = δ U+ P δ V.*2498

*δ H = δ U + δ V that is just the work.*2510

*We just did that, that is just work.*2516

*Therefore, our δ U = δ H – W.*2519

*We already calculated δ H that was 44.016 kJ/ mol.*2526

*We calculate the work which was 2.25 kJ/ mol.*2534

*Therefore, the change in energy is 41.766 kJ/ mol.*2540

*In going from liquid to gas at 25°C the energy of the system increased by 41.76 kJ, that is a huge amount of energy, that is all this is saying.*2555

*This is all based on just simple basic mathematical relationship that we already know,*2568

*the handful of equations that we already know from our work with work and heat and energy in the first law of thermodynamics.*2572

*We are just applying it to chemical situations.*2580

*Let us see what we have for D.*2585

*Let us see if I have another page actually available.*2587

*I do, great.*2590

*Part D, they want to know what the δ H is at 100°C instead of 25°C.*2593

*The δ H that we actually calculated was this thing right here was the first that we get part A.*2605

*This is a 25°C, that is the δ H in 100 ℃.*2612

*First of all, let us go ahead and work in K so 25°C = is the same as 298 K and 100°C =373 K.*2619

*Therefore, from the equation that we have from the previous problem our δ H at 373 K, 1 atm pressure its equal to the δ H at 298 K +*2637

*the integral from 298 to 373 of the difference in the heat capacities of products and reactants, gas or water vapor, liquid water, DT.*2660

*That is it, nice and simple.*2676

*Let us go ahead and do this.*2678

*Let us go ahead and write up the whole thing δ H 373.*2683

*Let us go ahead and do that is going to equal δ H at 373 is going to equal δ H of,*2693

*Let me write this again to keep it on one page + the integral from T0 to δ CP DT.*2705

*Therefore, δ H 373 = δ H at 298 which we calculated which was part A 44.016 + the integral from 298 to 373.*2718

*And now the difference in the heat capacities, they gave us those in the beginning part of a problem in part D.*2735

*Gas, water, that was 33.577 and that is the molar heat capacity for water vapor - 75.291 molar heat capacity for the liquid water.*2741

*We end up with 44.016 + -41.714 × 75.*2763

*Just doing this by hand, this is this number and this is a constant so it pulls out.*2780

*This integral of the DT is just δ T or δ T, between 290 - 373 is 75.*2785

*What you end up with is δ H at 373 = 40.9 kJ/ mol.*2791

*This tells me now at 25°C if I'm converting liquid water to water vapor, I have to put it that much heat 44.016 kJ/ mol of water.*2806

*At 100°C, I only have to put in 40.9 kJ/ mol.*2818

*Notice, this is less and that makes sense.*2824

*I'm at a higher temperature so I do not have to work as hard.*2826

*In order boil water, I have to get it to 100.*2829

*A part of this energy here, about 4 kJ is used just to get my water from 25°C to 100°C and I can start boiling it off.*2834

*If I’m already at 100℃, all that heat that I put in automatically goes towards just boiling it off.*2843

*I do not have to raise it because it is boiling point temperature.*2848

*This makes sense, empirically it makes sense.*2851

*We know from experience.*2854

*Let me write that down.*2856

*Notice, it takes less energy to vaporize water at 100°C than it does at 25°C.*2859

*And this makes sense.*2887

*If you ended up with a higher number than 44 then something happened, something went wrong.*2893

*A lot of times we get so wrapped up in the mathematics and I'm guilty of this just as much as anybody else,*2898

*it is probably more so because to the mathematics that we do not pull back and stop and take a look if this actually make sense physically.*2902

*It is very important that we want you to be able to develop mathematical skills that are important to solve these problems*2910

*but we want you to understand what is happening conceptually.*2917

*We want you to realize, wait a minute I have a 25°C of vaporizing water, a certain amount of heat is going to required for that, that is the δ H.*2920

*If I’m at 100°C that is the boiling point of water already.*2930

*I should stand back and say I should require less energy because I’m already at the boiling point.*2936

*Therefore, the number that I get should be lower, sure enough it is lower 40.9 vs. 44.016.*2942

*It makes sense physically.*2952

*Let us round of this discussion of thermal chemistry with one more problem here and let us see what we can do.*2957

*Liquid water has a molar volume of 18.0 cc/ mol, if the temperature is held constant.*2966

*Temperature is constant and the pressure is increased by 15 atm.*2975

*We take it from 1 atm to 15 atm, calculate the enthalpy change then calculate the enthalpy change by 15° increase in temperature when the pressure is held constant.*2980

*Here what we are doing is we have a certain volume of water, we are asking if I hold the temperature constant in the first case and increase the pressure,*2993

*what is the enthalpy change for that particular process?*3004

*In the second process, if I hold the pressure constant the change in temperature what is the enthalpy change?*3011

*That is what is happening here and the molar heat capacity 75.3 J/ mol K.*3018

*Enthalpy change, remember one of the basic equations is the following.*3029

*Let us go back to blue on this one , DH = the constant pressure heat capacity DT + DH DP.*3032

*DP this is the general equation, remember that the change in enthalpy of the system, I can change the temperature,*3049

*I can change the pressure, that is how I can change the enthalpy.*3056

*The change is equal to the constant pressure heat capacity × the differential change in temperature + this pressure dependence in enthalpy.*3063

*That is the rate at which the enthalpy changes per unit change in pressure × DP.*3076

*If I hold the temperature constant this just goes to 0 only this matters.*3082

*If I hold pressure constant this goes to 0 only this matters.*3086

*We are going to do both.*3089

*This equation is not valid for gas, it is valid for liquids and for solids.*3093

*All of the equations that we developed so far they are valid for every single state of aggregation solid, liquid, gas, for every single phase.*3097

*It does not really matter, we just need to have the particular values that are important to be able to solve the problem.*3106

*The only thing that we need to recall is that for a liquid this DH DP at constant temperature it is actually equal to the volume of a particular sample.*3113

*Let us go ahead and do the first part of temperatures held constant and the pressure is increased at 15 atm.*3137

*In this particular case, the temperature is being held constant so that is going to go to 0.*3143

*What I had is DH = DH DP T × DP.*3162

*The DH DP = V so I have is DH = V DP.*3174

*If I integrate this I end up with the following.*3183

*I end up with δ H = V × the change in pressure so that is what I'm going to look at.*3187

*I have the volume and I'm going to find the change in pressure.*3199

*I already know what the change in pressure is, it is just 1 atm to 15 atm.*3203

*I just need to know what the volume is.*3208

*Let us go ahead and make some conversions here.*3212

*I have 18.0 cm³ × 10⁻⁶ m³/ cm³ I need to convert this cm³ to m³ because when I multiply by a pressure*3216

*which is going to be in a Pascal, the Pascal is defined in terms of the cubic meter not the cm³ .*3240

*We have to watch for the units.*3247

*When I go ahead and do that, I would actually change the 15 atm to Pascal, 15 atm × 1.01 × 10⁵ Pascal/ 1 atm.*3250

*I’m just making some conversions here, = 15. 15 × 10⁵ Pascal and this is going to be 18 × 10⁻⁶ m³.*3273

*Therefore, my δ H = V × δ P.*3291

*My V is this, my δ P is this, I will just multiply them.*3294

*So δ H = 18.0 × 10⁻⁶ m³/ mol and this is going to be 15.15 × 10⁵ Pascal, when I do this, my δ H going to be 27.27 J.*3300

*I use my basic equation in this particular case temperature was held constant so this term is unimportant.*3337

*This for a liquid happens to equal V so I get this and I’m left with δ H = V δ P.*3344

*And what the volume is, I just have to make sure to be in appropriate unit.*3352

*I know what the change in pressure is, I just need to make sure it is in the appropriate unit.*3355

*I will go ahead and calculate the J.*3359

*Notice, this is not very much 27.27 J.*3361

*For constant pressure, let us do the same thing.*3366

*We go ahead and move this back up here.*3370

*For constant pressure, let us rewrite the equation.*3380

*The basic equation it never hurts to actually write the equations that you are supposed to know and memorize over and over again.*3383

*DH = CP DT + DH DP constant temperature DP.*3389

*And this particular case, it is the pressure that is held constant so this goes to 0 that means there is no change in pressure.*3399

*DP is 0 that means this is 0.*3405

*What I have is DH=CP DT.*3409

*When I integrate this, I end up with δ H = the constant pressure heat capacity × δ T the change in temperature.*3414

*Therefore, the δ H I should go ahead and put P to let me know that this is happening under constant pressure = CP they gave it to us already.*3427

*This is 75.3 J/ mol K and the difference in temperature was 50°.*3440

*50°C the degree increment, the δ of Celsius is the same as the δ of K.*3455

*Therefore, this is 50 K.*3462

*What we end up with is 1130 J/ mol.*3466

*That is a very big difference between 27 J/ mol and 1130 J/ mol.*3474

*Basically, what this says is that if I have a liquid and if I put some pressure on it, it does not really change the enthalpy of the system.*3481

*In other words, we are not really investing a lot.*3489

*If I just change the temperature by 50° all of a sudden invested a lot of energy into this.*3493

*The enthalpy has gone up by 1130 J.*3502

*Clearly, for liquid and for a solid, temperature has a much greater affect on the enthalpy of the system than pressure.*3507

*Obviously, if I put like 10,000 atm pressure on it, that is going to be different but normally with the levels of pressure*3515

*that we work with the laboratory which is maybe 1 to 20, 30, 40, something like that.*3524

*It really does not matter.*3529

*The effect of pressure on enthalpy is for the most part negligible, that is really all that this is saying.*3531

*That took care of the thermal chemistry via examples.*3541

*Thank you so much for joining us here at www.educator.com.*3543

*We will see you next time, bye.*3545

1 answer

Last reply by: Professor Hovasapian

Fri Jan 16, 2015 9:55 PM

Post by tiffany yang on January 16, 2015

In example four, part B asked about the work, so we need to find change in volume. However, can delta V really be V of gas - V of liquid?

I thought that delta V would just be V of the gas....because no gas presents initially...something I learned by looking at the solution of my hw problem

Thanks Raffi!

1 answer

Last reply by: Professor Hovasapian

Sun Dec 21, 2014 8:45 PM

Post by David Llewellyn on December 18, 2014

In example II the delta(H)o 1000 of the reaction is given as -123.8 kJ/mol but at the end of the problem you divided the delta(H)o 298 obtained by delta(H)o 1000 - Int[delta(Cp)dt]1000 298 by 2 to give the delta(H)o form. The answer you obtained is close to the tabulated value so shouldn't delta(H)o 1000 just be the heat of reaction in Joules rather than Joules/mol?