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Lecture Comments (25)

0 answers

Post by Abdullah Alqahtani on June 24, 2017

YOU are very bad lecturer

1 answer

Last reply by: Professor Franklin Ow
Wed Jul 20, 2016 11:02 PM

Post by Magic Fu on July 20, 2016

In 15:44. Why Styrofoam has smaller heat capacities, and still need lot of heat? Shouldn't it be the opposite instead?

0 answers

Post by Peter Ke on September 16, 2015

At 45:00, why did you multiply 1.00 g/mL x 100mL because 1 Liter = 1000 mL so I was thinking that you should multiply it by 1000 and get 1000 grams of solution?

Also, when you got -339 kJ shouldn't it be 339 J without the kilo? Because if you were to set up the equation -100g(4.18 J/gC)(24.21 C - 23.40 C), don't the grams and celsius cancel each other out and you would be left with J?

Please correct me if I'm wrong.

0 answers

Post by Peter Ke on September 15, 2015

At 23:12. I understand how you got Qreaction = - Qbomb, because you subtract Qbomb on both side but I don't understand how you got -Cbomb(DeltaT). Please explain. Thx!

0 answers

Post by Peter Ke on September 15, 2015

At 16:04, can you show me how you got 30.7, cause the answer I got was like 495.1 which totally made no sense. So, please show me how you got 30.7. Thx

0 answers

Post by Peter Ke on September 15, 2015

At 8:59, why 488 is negative? Please explain it in a easier way. Thx

1 answer

Last reply by: Professor Franklin Ow
Mon Jun 1, 2015 5:41 PM

Post by Ivan de La Grange on May 31, 2015

For Sample problem 1, shouldn't it be .0327 grams?

2 answers

Last reply by: Denise Bermudez
Sun Mar 8, 2015 10:04 AM

Post by Denise Bermudez on March 2, 2015

hi professor
can you please explain or re-do how you got the answer of 30.7 degrees celsius. I set the equation correctly but i cant figure out how you got your answer with it. The part that confused me was the *4.18 J/ g celsius.


0 answers

Post by Saadman Elman on January 31, 2015

Please look at 13:00 - 14:00 min, You did first part of the problem but you didn't do the second part of the problem. The question was asked Calculate delta U for burning of fuel if the system gives off 875 joule. When all is said and done, the answer that you will end up is -998.6  Joule which is the answer for delta you. You skipped that part. Please verify my answer.

0 answers

Post by Saadman Elman on January 9, 2015

Correct me if i am wrong. You said Insulator such as Styrofoam has small heat capacities. Since they retain the heat so i think they have higher heat capacities.

1 answer

Last reply by: Ryan Lozon
Sun Dec 21, 2014 11:34 AM

Post by Ryan Lozon on December 21, 2014

You mean insulators such as styrofoam having larger heat capacities than metals, right? Because they require more energy to raise one gram by 1 Celsius?

3 answers

Last reply by: Professor Franklin Ow
Mon Aug 4, 2014 11:31 PM

Post by brandon joyner on July 1, 2014

Why does 725 have a negative sign?

2 answers

Last reply by: Peidong, He
Sun Mar 16, 2014 11:53 AM

Post by Peidong, He on March 15, 2014

should insulators, such as Styrofoam, have a higher specific heat capacity?  

0 answers

Post by KyungYeop Kim on July 9, 2013

Regarding a specific situation in which a nonspontaneous reaction under standard conditions becomes spontaneous at lower temperature, how can I describe this phenomenon in relation to enthalpy, entropy, and free energy? and how can I explain it in terms of temperature change? I've succeeded so far in determining that ΔG>0 since it's nonspontaneous under standard conditions, but what about ΔH and ΔS?

Given the equation = ΔG = ΔH*TΔS; I think the fact that temperature(T) can go either from positive to negative or negative to negative seems to confuse me. Are we assuming, in saying lowering temperature, that the T goes from negative(-) to negative(-)?

I know it's a complex problem, and I apologize if I'm asking too much, but I would like to know what the answer is and why. Thank you always!

0 answers

Post by KyungYeop Kim on July 9, 2013

I have a question about effective nuclear charge. As you go down a group in the periodic table, why is it that the effective nuclear charge decreases? from what I know, is it true that as the attraction decreases down the group, it somehow counterbalances the increase in nuclear charge? I'm confused.

Energy & Chemical Reactions

  • State functions are path-independent.
  • The first law of thermodynamics states that energy is conserved during a chemical reaction.
  • Energy can be transferred through heat and work.
  • Calorimetry problems involve either constant-volume or constant-pressure scenarios.
  • Enthalpy is the amount of heat transferred under constant pressure.

Energy & Chemical Reactions

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
  • Lesson Overview 0:14
  • Introduction 1:22
    • Recall: Chemistry
    • Energy Can Be Expressed In Different Units
  • The First Law of Thermodynamics 2:43
    • Internal Energy
  • The First Law of Thermodynamics Cont'd 6:14
    • Ways to Transfer Internal Energy
    • Work Energy
    • Heat Energy
    • ∆U = q + w
  • Calculating ∆U, Q, and W 8:58
    • Changes In Both Volume and Temperature of a System
  • Calculating ∆U, Q, and W Cont'd 11:01
    • The Work Equation
    • Example 1: Calculate ∆U For The Burning Fuel
  • Calculating ∆U, Q, and W Cont'd 14:09
    • The Heat Equation
  • Calculating ∆U, Q, and W Cont'd 16:03
    • Example 2: Calculate The Final Temperature
  • Constant-Volume Calorimetry 18:05
    • Bomb Calorimeter
    • The Effect of Constant Volume On The Equation For Internal Energy
    • Example 3: Calculate ∆U
  • Constant-Pressure Conditions 26:05
    • Constant-Pressure Conditions
  • Calculating Enthalpy: Phase Changes 27:29
    • Melting, Vaporization, and Sublimation
    • Freezing, Condensation and Deposition
    • Enthalpy Values For Phase Changes
    • Example 4: How Much Energy In The Form of heat is Required to Melt 1.36 Grams of Ice?
  • Calculating Enthalpy: Heats of Reaction 31:22
    • Example 5: Calculate The Heat In kJ Associated With The Complete Reaction of 155 g NH₃
  • Using Standard Enthalpies of Formation 33:53
    • Standard Enthalpies of Formation
  • Using Standard Enthalpies of Formation 36:12
    • Example 6: Calculate The Standard Enthalpies of Formation For The Following Reaction
  • Enthalpy From a Series of Reactions 39:58
    • Hess's Law
  • Coffee-Cup Calorimetry 42:43
    • Coffee-Cup Calorimetry
    • Example 7: Calculate ∆H° of Reaction
  • Summary 47:12
  • Sample Problem 1 48:58
  • Sample Problem 2 51:24

Transcription: Energy & Chemical Reactions

Hi, welcome back to

Today's lesson from general chemistry is going to be our first discussion on thermodynamics.0002

We are going to be discussing energy and chemical reactions.0010

Let's go ahead and go over the lesson overview.0015

We are going to first start off with a brief introduction.0018

Because this is going to be totally different than all of the previous material we have been seeing.0021

We are then going to discuss the very important first law of thermodynamics.0027

Then the rest of the chapter is highly quantitative.0034

Once we have the introduced the different parameters that are at a focus0037

in thermodynamics, we are going to learn how to calculate each of them.0042

There is going to be different situations when we both have a volume and temperature change.0045

There is also going to be a situation when we have no volume change.0051

Then there is going to be a situation when we have no pressure change.0057

That is what pretty much this chapter is about, studying, introducing these very important physical parameters.0061

Then learning how to calculate them in each of their different situations.0069

We are going to finish up the chapter with a discussion of enthalpy.0074

Followed by our summary and then two sample problems.0078

Let's go ahead and get into the lesson now.0083

Remember we introduced the definition of chemistry way long time ago.0088

We said that chemistry is the study of matter and the changes that it undergoes.0094

We have already discussed the three states of matter.0098

We have already discussed the types of physical and chemical changes.0101

But there is always something that is associated with physical and chemical changes.0105

That is a change in energy.0111

When we talk about energy, energy can be measured and expressed in different units.0115

If you have a utility bill, especially the electricity, that is going to be usually expressed in terms of kilowatt hours.0123

If you are going to look at the side of a cereal box or anything0131

or any type of beverage, some nutritional information, nutritional information tends to use calories.0136

But in the physical sciences, we tend to use what is called the Joule.0142

The Joule is going to be abbreviated capital J.0147

The relationship between all three is the following.0150

One kilowatt hour is 3.6 times 106 joules; one calorie is 4.184 joules0153

Now that we know the units of energy, let's get into some more theory.0163

What we are interested in in general chemistry is what is called internal energy.0171

You have probably taken physics by now.0178

You probably have already learned what is called kinetic energy and potential energy.0181

Where potential energy is just related to position and kinetic energy is the energy related to motion.0187

What internal energy is, internal energy is going to be the sum basically.0197

Internal energy is the sum of a system's potential energy and its kinetic energy.0203

Depending on what textbook you use, internal energy will be either symbolized with capital U or with capital E.0209

We are going to now introduce another term, what is called the state function.0219

Internal energy is an example of a state function; a state function is path independent.0223

In thermodynamics, it is pretty easy to always identify what a state function is.0231

A state function name is going to tend to be capitalized.0235

What do I mean by path independent?0240

Let's go ahead and look at the following reaction.0243

A solid plus B solid goes on to form AB solid.0246

This reaction is the direct combination of A in the solid state reacting with0259

B in the solid state to form the compound AB in the solid state.0265

Let's say that this reaction had associated with it a certain change in energy.0269

This is the most direct route; but I can do this in a stepwise motion.0278

For example, I can first go from A solid to A gas.0283

I can then go from B solid to B gas.0290

Once I have A and B in the gaseous state, I can combine the two.0295

Once in the gaseous state, that can then go on to form AB gas.0300

AB vapor can then condense back down to the solid.0308

What a state function means, what path independent means is the following.0314

If I take the direct route or if I do the stepwise route, in the end0320

the change in energy is going to be the same regardless of the path.0330

Δenergy same regardless of path.0334

Again state functions are path independent.0345

We really don't care about the step by step process that we take.0348

We only really care about the comparison between the initial and the final state.0352

Because of that, we are then often interested in not just internal energy but the change in internal energy.0359

What I am going to be using is the symbol ΔU to represent change in internal energy.0367

When we talk about internal energy, there are essentially two ways to transfer internal energy in and out of a system.0377

When I use the word system, I mean maybe a chemical reaction for example.0386

The two ways to transfer internal energy in and out of a system are through work and through heat.0392

Work is going to be symbolized lowercase w.0401

Heat is going to be symbolized with lowercase q.0404

Because they are lowercase, they are not state functions.0407

Remember from the previous slide that state functions tend to be capitalized.0410

What you want to pay attention to are the signs.0418

If a system performs work, then the work of the system is going to be negative.0424

If a system has work performed on it, we say that the sign is going to be positive.0440

Let's now look at the signs for q.0457

If heat enters a system, we say that the sign of q is going to be positive.0460

If heat leaves a system, we say that the sign of q for the system is going be negative.0473

You have to know these signs because when you are going to be given a0485

word problem, you have to be able to derive the sign from the sentences.0489

A lot of this many of you will recognize this comes immediately from physics.0496

We define work energy as the product of applied force over a certain distance.0512

Many of you will recognize the equation, work is equal to force times distance.0513

Heat energy basically, that is more literal.0516

Heat energy basically is going to be essentially thermal energy.0519

The change in internal energy equals the sum of work and heat energies.0525

The equation for the first law of thermodynamics is ΔU is equal to q plus w.0533

A very common application of the first law of thermodynamics is when the system experiences both a change in volume and in temperature.0542

Now we are going to learn to calculate the previous three parameters; ΔU, q, and w.0553

This is a very traditional type of problem.0564

This involves a piston and a cylinder like you have in your automobile engine.0566

The cylinder and piston are warmed by a flame.0571

The gases in the cylinder expand, doing work on the surroundings by pushing the piston outward.0574

If the system absorbs 559 joules of heat and does 488 joules of work, what is ΔU?0579

Basically we have a piston here inside a cylinder.0587

Here we have a lot of a hot gas inside.0594

This hot gas is going to push against the piston.0599

It is going to apply a force against it.0604

You are told that the system absorbs 559 joules of heat and it does 488 joules of work.0608

You see immediately how important it is to pay attention to the sentences because you have to derive the signs.0615

ΔU is equal to q plus w; let's go ahead and look at q.0621

You are told that the system absorbs 559 joules.0628

That means heat is entering the system; that is going to be +559 joules.0632

In addition, w, you are told that the system is doing, it is performing 488 joules of work.0638

That is going to be a negative sign.0646

When all is said and done, we get an answer of 71 joules for ΔU.0650

It is quite straightforward to use this equation that represents the first law of thermodynamics.0657

Let's now introduce the equations for work and for heat because each of them have their own.0666

Let's first introduce the work equation.0673

Work is equal to -Pexternal times ΔV.0677

In this case, Pexternal, this is going to be a constant external pressure.0682

That is the key; the pressure has to be constant or else you cannot you this equation.0690

Of course ΔV, we immediately recognize as change in volume or Vfinal minus Vinitial.0697

When fuel was burned in the cylinder equipped with a piston, the volume expands0708

from 0.255 liters to 1.45 liters against an external pressure of 1.02 atm.0712

During this process, 875 joules of heat is given off.0719

Calculate ΔU for the burning of the fuel.0723

Work is equal to ?Pexternal times ΔV.0730

You are told in the problem that the external pressure is 1.02 atm; times ΔV.0737

That is going to be Vfinal minus Vinitial or 1.45 liters minus 0.255 liters.0749

When all is said, we are going to get an answer of -1.22.0757

Our units are going to be atmospheres times liters together.0762

We always want to try to get to joules.0768

The conversion factor between liter atmosphere and Joules is the following.0772

One liter atmosphere equals 101.3 Joules.0777

We are just going to use dimensional analysis.0782

-1.22 atmosphere times liter times 101.3 joules over 1 liter atmosphere.0785

That is going to be equal to -123 joules.0793

That is going to be our value for w.0797

Let's see if this sign makes sense.0800

Does it make sense that the sign of work here is going to be negative?0803

We have a piston.0808

The piston is exerting a downward pressure of 1.02 atm.0811

If my system is the...0818

If gas is trapped inside the piston, they have to do work to push against this downward pressure.0822

You are told that the piston is going to expand from 0.255 to 1.45.0828

In order for that increase in volume to occur, the system will be performing work to go against the downward pressure.0834

Yes, the negative sign of w does make sense.0843

The system is performing work; very good.0847

Now that we have introduced the work equation, let's now tackle the heat equation.0852

The heat equation for all you pre-meds is mCaΔT.0858

You will recognize that as MCAT; you will hear q being the MCAT equation.0863

M is just your mass, usually in grams.0869

ΔT is your change in temperature; Tfinal minus Tinitial.0872

C is something new; C is what we call the specific heat capacity.0878

The specific heat capacity is in units of joules per gram degree Celsius.0882

What specific heat capacity is, it is formally defined as the following.0887

It is the energy required to raise the temperature of 1 gram of a compound by exactly 1 degree Celsius.0891

What does the specific heat capacity physically mean?0900

If you think of a metal, metals tend to get very hot very very quickly.0905

They are excellent conductors of thermal energy.0910

It doesn't take much to heat a metal.0914

We say that metals are going to have relatively smaller heat capacities.0918

They don't retain the heat as much.0921

The opposite of metals, remember insulators.0925

Insulators such as Styrofoam are going to have smaller heat capacities.0928

It take a lot of energy to heat up Styrofoam.0933

That is what we know and have experienced.0937

One very important compound of course is going to be liquid water.0943

You should just know its specific heat capacity to be 4.18 joules per gram degree Celsius.0948

That is relatively large, especially compared to metals.0953

If you think about it, yes it does require lots of energy to heat up water.0956

So q is equal to mcΔT.0962

Now let's go ahead and look through an example.0966

You have a 0.055 kilogram sample of water and 27.5 degree Celsius.0969

You are told that it absorbs 725 joules of heat.0975

Calculate its final temperature.0978

q is equal to mcΔT.0982

Let's plug in what we know and see if we can solve for what is remaining.0988

You are told that the water absorbs 725 joules of heat.0994

That means heat is entering the system, which means it is going to be a positive sign.0998

q is equal to +725 joules; that is equal to the mass of the water.1003

Remember we want to use mass in units of grams.1011

That is going to be 55 grams.1014

Times the specific heat capacity of water which is going to be 4.18 joules per gram degree Celsius.1018

Finally times ΔT where ΔT is equal to T2 minus T1.1029

T2 is what we are trying to solve for.1034

The initial temperature is 27.5 degrees Celsius.1038

We are in good shape; it looks like we only have one unknown.1044

That is the final temperature which is what we are trying to solve for anyways.1049

T2 is going to be equal to 30.7 degrees Celsius.1053

Always double check; does it make sense that T2 is greater than T1?1058

Yes, absolutely because you are told that the sample of water is absorbing heat.1064

That heat is going into the system, into the water to warm it up.1068

Yes, our final temperature had better be greater than our initial temperature.1073

Again for thermodynamics, it is always a good habit to double check your work and ask yourself, does it physically make sense?1077

There is different ways of calculating the amount of energy that is involved in a process.1088

One way is using a constant volume process.1096

Now we introduce the word calorimetry.1101

All calorimetry is, it is the determination of the amount of energy transferred in a chemical process.1103

We can do this using a constant volume process.1111

This is what we call bond calorimetry.1115

The energy released from a combustion reaction can be determined experimentally in a bomb calorimeter.1118

Let me go ahead and provide a rough sketch of what a bomb calorimeter is.1125

Basically you have a small cup; inside the cup is going to be your reactants.1131

Here is the key; this cup is going to tend to be made of stainless steel.1142

We know stainless steel to be very very hard and very very rigid.1154

Because it is quite rigid, its volume is not going to change, at least on the macroscopic scale.1158

That is how we guarantee a zero change in volume.1167

Stainless steel maintains volume, maintains a fixed volume during the reaction.1172

This little cup of stainless steel, that contains your reactants.1183

This is essentially hooked up to a fuse.1187

This is why we call it a bomb calorimeter.1190

Basically you are going to ignite the fuse.1192

The fuse is going to provide a spark, a source of energy.1195

You are going to have basically a combustion reaction occur inside the stainless steel apparatus.1199

On the outside of this stainless steel apparatus is a vessel.1206

This is going to be some type of insulative material.1213

This vessel then is going to have a thermometer embedded in it.1221

The thermometer is going to allow us to determine what the temperature change is during the reaction.1227

Of course the temperature is going to go up because it is a combustion reaction.1234

Sometimes the bomb calorimeter is going to have water too.1239

This is going to be with or without water.1244

Basically the water is going to provide a very nice heat sink for us1247

so that we can dissipate the thermal energy coming off of the combustion reaction.1251

What we are assuming is the following.1261

That all of the heat energy that is given off by this combustion reaction is going to be essentially completely transferred to the surroundings.1263

Basically the surroundings of this experiment are going to be the bomb calorimeter itself.1274

If there is the water, it is also the water.1282

Anything that is capable of absorbing the energy.1285

All of this is equal to zero.1292

q of reaction plus q of bomb plus q of water is equal to zero.1294

There is no heat lost to the outside environment.1298

All of it is going from the reaction inside the bomb calorimeter to the bomb calorimeter itself.1301

If there is water, to the water itself.1308

The equation for the heat of the bomb calorimeter is simply equal to the heat capacity of the calorimeter times ΔT.1312

Please make a note that this is going to be independent of its mass; mass independent.1321

Something very interesting happens.1333

What we want to look at is now the effect of constant volume on our equation for internal energy.1335

We know that ΔU is equal to q plus w, which equals to q minus PΔV.1341

If the volume is not changing, ΔV and this whole term essentially goes to zero.1350

We very nicely have the relationship of ΔU is strictly equal to q.1356

That is, in a constant volume process, all of the energy that is1362

transferred is in the form of heat, not in the form of work.1368

All energy transferred as heat.1374

Again constant volume process, ΔU equals to q.1383

Let's go ahead and now look at a typical problem.1393

You have 1.550 grams of C6H14; it is undergoing combustion in a bomb calorimeter.1400

Right away, bomb calorimeter, immediately I would underline that because that tells me that ΔV is zero.1407

So ΔU is q; q of the reaction of course.1416

You are told that the temperature changes from 25.87 to 28.13.1422

Remember that is what is expected; the temperature is going to increase.1426

Remember this is a combustion reaction after all.1429

They are always going to be giving off energy in the form of heat.1431

Calculate ΔU in kilojoules per mole.1436

You are told that the specific heat capacity of the calorimeter is 5.73 kilojoules per degree Celsius.1439

Let's go ahead and solve the problem; you are asked to calculate ΔU.1446

But because this is bomb calorimetry, ΔU is equal to q of the reaction.1450

In addition we know that q of the reaction plus q of the bomb calorimeter1457

plus q of water is equal to zero from the previous slide.1464

But do you see water being mentioned in this example?1468

There is none; this term is not applicable.1472

We have the relationship that q of the reaction is equal to ?q of the bomb.1479

This is equal to ?C of the bomb times ΔT.1486

Let's go ahead and plug in everything we know.1494

That is going to be equal to -12.95 kilojoules.1496

The question asks us for ΔU in units of kilojoules per mole.1508

You see here that we have the mass of the reactant.1513

All we have to do is get that to moles; then divide our kilojoules over the moles.1517

That is going to be 1.550 grams times 1 mole over its molar mass which is approximately 86 grams.1523

That is going to give us x moles of the reactant, C6H14.1536

When all is said and done and we take the kilojoules over the moles, we are going to get -718.5 kJ per mole.1544

When you do these problems, always pay attention to the units that are requested for ΔU.1554

That is how we do a typical bomb calorimetry problem.1561

In addition to having a calorimetry performed under a constant volume process, we can also do a constant pressure process.1567

This is going to be a slight mathematical derivation here.1576

We know that ΔU is equal to q plus w, which is equal to q minus PΔV.1580

From thermodynamics, there is going to be a new state function that is going to introduced right now.1586

This state function is what we call enthalpy; enthalpy is going to be symbolized ΔH.1592

By definition from thermodynamics, ΔH is equal to q plus PΔV.1597

We will call that equation two; let's combine equations one and two right now.1602

ΔH is equal to U plus ΔV, which is equal to q minus PΔV plus PΔV, which is equal to q.1608

In other words, at constant pressure conditions, enthalpy and heat are identical.1619

They are the same thing.1626

In addition, if the volume change is negligible, you see from the equation that ΔH and U are approximately the same value.1629

This is how we define enthalpy.1640

Enthalpy is essentially heat energy under the constraint of constant pressure.1642

The remainder of this lecture deals with calculating enthalpy under various conditions.1652

There is numerous ones; there is about four to five.1662

Which just shows you how important enthalpy is to this chapter.1664

The first way of calculating enthalpy involves phase changes.1669

Every phase change which we know to be physical changes of course is going to be associated with a change of energy.1674

Melting, vaporization, and sublimation; think about it.1682

If you are going to melt something, you have to put energy into it.1686

They all require energy.1688

Those processes that require energy are what we call endothermic processes.1692

Endothermic processes tells us the following.1697

That the sign of ΔH is going to be a positive value.1700

The opposite directions, freezing, condensation, and deposition, all give off energy and are exothermic processes.1706

For exothermic processes, the sign of ΔH is going to be a negative value.1714

Enthalpy values for phase changes are commonly given in units of kilojoules per mole under standard conditions.1721

What do we mean by standard conditions?1733

Basically standard conditions are provided for us to level the playing field1735

because there is so many different parameters we can perform experiments at.1739

It is very for us to come up with a universal accepted condition.1744

That is going to make everything a lot easier.1755

Standard conditions are defined as the following.1757

If we are dealing with a gas, it is going to be 1 atmosphere of pressure.1760

If we are dealing with a solution, it is going to be a 1 molar concentration.1765

The temperature is going to be 25 degrees Celsius.1772

Again 1 atm, 1 molar, and 25 degrees Celsius.1776

Let's go ahead and look at the following question.1784

How much energy in the form of heat is required to melt 1.36 grams of ice?1787

You are given the ΔH of these phase changes to be 6.00 kilojoules per mole.1792

Anytime you see the letters f-u-s, that means a fusion.1798

That is going to be for the melting process; melting process.1802

Anytime you are dealing with a phase change, it is quite simple.1809

q is equal to n times ΔH of the change.1812

Very straightforward equation where n is going to be moles of your compound.1821

ΔH of the phase change is going to be like we said in kilojoules per mole.1827

It is pretty straightforward.1833

We have to get the 1.36 grams of the ice to moles.1834

1.36 grams times 1 mole over 18.016 grams.1837

Then we want to multiply this by kilojoules per mole.1845

That is going to give us 0.453 kJ for our final answer.1851

Does it make sense that the sign of ΔH here is positive?1857

Remember this is a melting process so absolutely this is going to be an endothermic process.1861

Basically you want to add ΔH as a new conversion factor, kilojoules per mole; very good.1867

That is the first way of calculating enthalpy we are going to talk about via a phase change.1877

The next way of calculating enthalpy is what we call the heat of a reaction.1884

Remember that we said that every chemical process is going to have a change in energy associated with it.1888

This change in energy is what we call ΔH of the reaction.1895

ΔH of the reaction is going to be symbolized ΔHr.1898

It is usually given to the right of a chemical reaction.1903

Basically this is telling me that if I am going to perform this reaction exactly one time,1908

and I got the perfect amounts, mole to mole ratios of reactants,1916

it is going to release exactly 906 kilojoules of energy in the form of heat.1919

Calculate the heat in kiloJoules associated with the complete reaction of 155 grams of ammonia.1927

Anytime you are given a ΔH of a reaction problem, all you are going to do is1934

you are going to using ΔH of reaction as a conversion factor; again new conversion factor.1938

Basically the conversion factors are the following.1947

You can have -906 kilojoules for every 4 moles of ammonia.1950

You can have -906 kilojoules for every 5 moles of O2, etc.1959

We have seen this before where we are using the stoichiometric coefficients directly into a conversion factor.1966

Again I want to remind you, this is all the same material.1973

We are just using a slight twist.1976

But it is completely within your realm of what you know how to do already.1979

Let's go ahead and solve the problem; 155 grams of ammonia.1985

Again this is stoichiometry because we are asked to relate any two items in the chemical reaction.1990

One of those items is energy.1995

Times... we have to get everything into moles.1998

Remember that is always our first step.2000

Divided by roughly 17 grams of ammonia.2002

Then I am going to use my conversion factor.2005

That is going to be -906 kilojoules divided by 4 moles of ammonia.2009

When all is said and done, I am left with -2065 kilojoules as our final answer.2016

Again this is how you deal with what is called heats of reaction.2025

They are typically given to the far right of a chemical reaction.2027

The next scenario with calculating enthalpy is what we call standard enthalpies of formation.2035

Standard enthalpies of formation is going to be symbolized as ΔHf.2042

The formal definition is the energy change associated with the formation of a compound2047

from its constituent elements in their standard states at 1 atm in 25 degrees Celsius.2053

Let's take apart this definition; let's talk about water vapor.2060

The elements that are involved are of course hydrogen and oxygen.2067

We have already discussed these.2072

Hydrogen exists as a diatomic gas under standard conditions.2074

Oxygen exists as a diatomic gas under standard conditions.2078

The reaction representing the formation of water vapor under standard conditions is H2 gas plus 1/2O2 gas goes on to form water vapor.2084

This reaction is -241.8 kilojoules for the heat of formation.2097

Every textbook always has an appendix.2104

You are going to find in the appendix a very comprehensive table of heats of formation for many many compounds.2108

You probably don't have to memorize them.2118

You will either be given the information or be able to look it up in the appendix when you do problems.2120

There are two notes that you want to be aware of now.2127

ΔH of formation and all ΔH values, ΔH reaction, ΔH formation, etc, they are all affected by the reaction, the coefficients.2132

ΔH is proportional to the coefficients.2145

If I take the same reaction and I double it, I multiply it through by 2,2148

the ΔH value is going to get increased proportionally.2154

If I take the reverse reaction though, the sign is also going to flip.2160

For the reverse reaction, it is the same magnitude but opposite sign.2166

How do we go ahead and use ΔH of formation to solve ΔH of the reaction?2174

A typical problem is represented as the following.2183

Calculate ΔH of the reaction for the following given the heats of formation.2185

Again if you are not given it, you need to be able to look it up in the appendix.2190

This is going to be a very easy equation to remember.2196

ΔH of the reaction equals the following.2202

Summation n, ΔH of formation of all products, minus summation n ΔH of formation of all reactants,2206

where n is going to be the stoichiometric coefficient from the balanced chemical equation.2212

That is why again it is so important to make sure your chemical equation is balanced.2223

Or else this problem is going to be trouble.2227

We are going to take the products first.2232

You notice that only Al2O3 and Fe2O3 are given to you.2235

What is not given is iron solid and aluminum solid.2241

The reason why is because they are the elements in their standard state--aluminum solid monatomic and iron solid monatomic.2244

What you want to remember is the following.2254

That for elements in their standard states, ΔH of formation is formally going to be equal to zero.2256

That is what you will always want to remember.2273

That requires you to know those elements in their standard state.2277

Again most of them are monatomic, except the following.2281

H2 gas, N2 gas, F2 gas, Cl2 gas, Br2 liquid, I2 solid.2286

Also of course remember mercury is going to be a liquid under standard conditions.2301

Again these are the ones that you want to remember.2309

These are going to be zero ΔH of formation.2312

The rest are typically going to be monatomic solids.2316

Let's go ahead and do this; we are going to look at the products.2324

That is going to be aluminum oxide.2327

That is going to be -1675.7 kilojoules per mole.2329

My coefficient for aluminum oxide here is just 1.2336

The other product, iron, is just 0.2342

Remember summation tells me I am going to add everything together.2347

That takes care of my products.2350

I am now going to subtract this with the same equation for the reactants.2351

This is going to be for iron oxide.2360

Minus 1 mole of iron oxide times -824.2 kilojoules per mole.2362

Plus 0 for aluminum solid; summation products minus summation reactants.2371

We are going to get -852 kilojoules for the ΔH of the reaction.2381

Again it is a very easy equation to know; it really sticks out.2387

This is a summation equation again for ΔH of formation values under standard conditions.2392

Yet another way to calculate enthalpy is from a series of reactions.2400

We introduce now a very important law from thermodynamics is what we call Hess's law.2406

Hess's law tells us basically the following.2412

That the whole is equal to the sum of the parts.2414

That is, the change in enthalpy for an overall reaction equals to the sum of the enthalpy changes for each individual step.2417

Again the whole is equal to the sum of the parts.2425

Imagine we have a reaction through a three step process.2432

If I am to combine all these reactions, all we are going to do is combine reactions.2437

We are going to add up everything.2442

We are going to sum all reactants on one side.2444

We are going to sum all products on one side.2449

We are basically adding chemical reactions together.2453

We go ahead and do that; let's see what we get.2457

2NO gas plus O2 gas plus N2 gas plus O2 gas plus 2N2O gas goes on to form all of the products.2460

2NO2 gas plus 2NO gas plus 2N2 gas plus O2 gas.2477

You want to treat reactions basically as any algebraic equation.2490

Now we can go ahead and cancel like terms on opposite sides of the equation.2497

Let's see what we can get rid of.2505

One O2 here is gone; that means one O2 here is gone.2510

This N2 is gone which means we only have one N2 remaining right here.2514

2NO on the left is going to cancel with the 2NOs on the right.2519

We have a net balanced combined reaction of 2N2O gas plus O2 gas going on to form 2NO2 gas plus N2 gas.2523

This is our overall net equation.2539

Basically what Hess's law is telling me is that when I combine these three reactions,2543

ΔH of the overall reaction is simply equal to the sum of the individual ΔHs.2549

ΔH1 plus ΔH2 plus ΔH3; again this is Hess's law; very good.2556

The final way to calculate enthalpy is what we call a coffee cup calorimeter.2566

It is called coffee cup calorimetry because the calorimeter is good old Styrofoam.2578

Styrofoam, we know to be a very inexpensive readily available insulator.2584

This type of calorimetry is going to more or less guarantee constant pressure conditions2590

because it is going to be whatever the atmospheric pressure is at the time.2596

q of the reaction is going to be equal to ?q of the solution.2603

Basically for coffee cup calorimetry, you have a coffee cup with a aqueous solution usually with your reactant in here.2608

All it is is the thermometer inserted and usually some type of lid is here used.2619

You see that this reaction is pretty much going to always be at whatever the atmospheric pressure is going to be.2627

Coffee cup calorimetry is not used for combustion reactions but instead for aqueous reactions.2632

That is a big difference right there; aqueous reactions, not combustion.2641

The system is equal to q of the reaction.2647

If the reaction is giving off energy, where is it going to go?2650

It is basically going to go to the aqueous environment, to q of the solution.2654

We are basically assuming that the Styrofoam material, because it is such a good insulator,2661

it is not involved in the transfer of energy; assuming Styrofoam not involved.2667

That is why you don't see any term for the Styrofoam material.2675

Basically all of the energy that is given off by the reaction or absorbed by the reaction2681

is going to come from or go to the aqueous solution that the reaction is placed into.2686

q of reaction is equal to ?q of solution.2693

In other words, q of the system equals ?q of the surroundings.2696

We know the equation for q; we have introduced it before.2703

It is the MCAT equation; that is equal to ?mcΔT.2706

Let's go ahead and look at the following reaction.2711

Silver nitrate is reacting with hydrochloric acid to form silver chloride and nitric acid.2714

You have 50 milliliters of 0.1 molar silver nitrate and 50 milliliters of 0.1 molar HCl.2720

You mix it in a coffee cup calorimeter.2727

The temperature rises from 23.4 to 24.21; calculate ΔH of the reaction.2730

You are told that the density of the solution is 1 gram per milliliter.2738

You are given the heat capacity of the solution.2742

We are assuming it to be that of water.2745

We are combining 50 milliliters of one reactant with 50 milliliters of another.2748

Both of them are aqueous.2753

We are assuming that the density of solution is roughly 1.2755

We can get the mass of the solution.2757

1.00 gram per milliliter times the 100 milliliters of the solution is going to give me 100 grams of the solution.2761

We know that q of the reaction is equal ΔH of the reaction2773

because this is constant pressure after all; coffee cup calorimetry.2781

That is going to be equal to ?q of the solution.2785

That is equal to -100 grams times the heat capacity which is given.2790

Times the ΔT which is going to be 24.21 minus 23.40 degrees Celsius.2796

We are going to get an answer of -339 kilojoules.2804

Once again ask yourself, does it make sense that my sign of ΔH here is negative?2809

You are told that the temperature increases, which means it is warm to the touch.2815

Which means it is going to be giving off energy; this is an exothermic process.2819

Yes, our sign of ΔH had better be negative; this answer does absolutely make sense.2823

That is how we do coffee cup calorimetry.2830

Let's now take a moment and summarize the key points from this opening introductory chapter on thermodynamics.2834

We learn the first law of thermodynamics is basically a restatement for the law of conservation of energy.2841

That energy can neither be created nor destroyed.2846

Everything you have is not going to be lost at all; it is conserved.2849

However we introduced the mathematical form of the first law, where ΔU is equal to q plus w.2857

This is going to be true for any closed isolated system.2865

Basically after our brief introduction, we spent a lot of time doing calculations, calculating these different parameters of thermodynamics.2869

To calculate these parameters, you must consider the conditions.2877

Do we have volume constant?--do we have pressure constant?2880

We saw that the equations are different for each of the situations.2885

We learned a very important experimental technique called calorimetry.2890

Calorimetry is used to determine the energy change of a chemical reaction under either constant volume or constant pressure conditions.2894

Constant volume conditions, we did this using a bomb calorimeter.2905

Constant pressure conditions, we did this using a coffee cup calorimeter.2915

Finally we spent a great deal of time on enthalpy.2920

We learned many many ways of calculating enthalpy.2924

Remember enthalpy is essentially heat transferred under the constraint of constant pressure.2927

Let's now spend some time on a pair of sample problems.2935

Here we are given a chemical reaction; you are given a ΔH of reaction.2941

Right away I know that this is going to be a type of a stoichiometry problem.2946

I am going to be incorporating the ΔH of the reaction as a conversion factor.2951

Again we are going to be incorporating this as a conversion factor with of course the coefficients.2961

What mass of C4H10 is required to produce 1500 joules of heat?2972

Produce, what does produce mean?2980

Produce means it is going to be giving off heat.2982

What is my sign?--is it positive or negative?2987

It is going to be negative; -1500 joules.2990

We are going to be using ΔH of the reaction as a conversion factor.2997

The reactant we are interested in is a C4H10; this equation is already balanced.3000

1 mole of C4H10 over -26 of 58000 joules.3006

I am then going to go ahead and multiply this by the molar mass of the compound, 58 grams roughly over 1 mole.3018

When we get this, we get 32.7 grams of C4H10 required.3029

Once again this was a nice stoichiometry problem where we used ΔH of the reaction as a conversion factor with the stoichiometric coefficients.3039

Pay attention to this.3051

If we had forgotten the negative sign, we would get then a negative mass, which physically doesn't exist.3054

Right away again always check if your signs are correct.3067

Again if we had not placed this as -1500 joules, we would have gotten a negative mass, which is not going to make sense.3072

Let's now move on to the final sample problem two.3086

Consider the following series of chemical reactions.3090

Right away you see three reactions; ΔH 1, 2, and 3.3092

You pretty much know this is a Hess's law type of problem.3097

The question asks us to calculate ΔH of the reaction for N2O gas plus NO2 gas going to 3NO gas.3100

If I try to add up these reactions right away, I do not get the balanced overall reaction.3113

This means that I must manipulate my previous three reactions.3118

Must manipulate; must change at least one reaction.3123

Remember we went over the two ways to manipulate a chemical reaction.3134

We can either multiply through by some type of factor; multiply through.3139

Or we can reverse the reaction.3148

Remember if you multiply through all of the coefficients by a factor, ΔH gets multiplied by the same factor.3154

If you reverse a reaction, remember the sign of ΔH also changes.3170

Those are the two warnings that you must remember whenever manipulating a chemical reaction.3179

I see that NO2 here is on the reactant side.3188

But the only place where NO2 appears is on the product side here.3191

For the first reaction, I am going to go ahead and I am going to reverse it.3195

2NO2 goes to 2NO plus O2.3199

In addition, I only see one NO2 in the whole problem, but I have two here.3206

Which means I am going to have to multiply through by 1-1/2 to get rid of that 2, to reduce it to 1.3214

That means I have -1-1/2 ΔH1.3221

ΔH2 is just N2 plus O2 going to 2NO.3230

That is going to remain exactly the same.3236

Finally the third reaction is 2N2O going to N2 plus O2.3238

That looks good right now because you see that N2O is on the reactant side in the overall reaction.3249

Here it is on the reactant side in this reaction.3254

But what is the difference?--it is the factor of 2 again.3257

Now we are going to multiply through by 1-1/2 again to get rid of the 2.3260

That going to be 1-1/2 ΔH3.3271

When all is said and done, when I add up all three of these equations that we have changed,3278

that we have made changes to some, we get the overall reaction, N2O plus NO2 going to 3NO.3286

Hess's law tells me that my ΔH of the reaction is simply going to be equal to the sum of the individual ΔHs.3295

-1-1/2 ΔH1 plus ΔH2 plus 1-1/2 ΔH3.3304

This problem is a great illustration of Hess's law.3314

That was our first lecture on thermodynamics.3319

Thank you for using; I will see you all next time.3324