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Lecture Comments (4)

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?

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 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

Transcription: Thermochemistry Example Problems

Hello and welcome back to 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 coefficients0148

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

We will see you next time, bye. 3545