General Chemistry Entropy & Free Energy

Hi, welcome back to Educator.com.0000

Today's lesson from general chemistry is on entropy and free energy.0003

This is basically going to be our second and final lecture on thermodynamics.0009

As usual we are going to start off with a brief introduction.0013

We are going to get into something very important in thermodynamics which is what we call entropy.0017

It is going to be related to spontaneous processes.0023

After we get into entropy, we are going to go through the second and third laws of thermodynamics followed by problem solving.0027

After entropy, we are going to then get into what is called Gibbs free energy followed by definition and problem solving.0038

Then we are going to go ahead and conclude with a summary and a pair of sample problems.0047

We have consistently said, you have heard me say over and over again that nature favors states of low energy.0055

In other words, reactions occur to obtain a lower energy state in the end.0063

This is also known as a downhill reaction.0069

In other words, nature tends to favor reactions that start in a high energy state and go to a low energy state.0072

That is going to be the natural tendency.0079

In this lecture, we are going to complete our discussion from thermodynamics and quantify the above statement.0081

In the end, we are going to see that nature not only favors0089

states of low energy but also states of what we call high entropy.0091

Consider the following closed reaction vessels that contain gas molecules.0099

Let me go ahead and draw a closed reaction vessel here.0104

We can make this one gas particle; the red one can be another gas particle.0109

The black one can be another gas particle; we are now going to...0115

Because these are gases, gases naturally expand; what can happen is the following.0120

What are some possible different configurations for these three gas particles if they naturally expand?0128

Situation number one is when all three gas particles are on the same side.0135

We can have a second situation when only the blue and the black are together and the red is different.0140

We can have another situation where the blue is by itself and the red and black are together.0148

You notice that these two situations here are actually the same0162

where we have two gas particles on one side versus one on the other.0169

It turns out that configuration number two actually has more what we call microstates.0177

Once again configuration number two is going to have what we call more microstates.0196

We can even do one more to show you how this is possible.0204

Now red and blue can be together with black by itself.0209

This is also considered a configuration number two.0215

Configuration number one, there is only one way to put three particles on the same side.0220

Configuration number one has one microstate.0227

Because gases expand, we are going to see that configuration number two is going to be the more probable configuration.0237

Why?--because there are a higher number of ways to do it.0246

There are a higher number of microstates to achieve, configuration number two, where one gas particle is on one side.0250

Two gas particles are on the other side.0257

We symbolize the term microstate with capital W.0260

For configuration number one, W is 1; for configuration number two, W is simply 3.0264

We can quantify that this is going to be more probable.0273

In other words, this is more likely to occur.0281

The equation we use to quantify this is following.0286

S is equal to k_B times the natural log of W where S is what we call entropy.0289

I want you to think of entropy as basically disorder or chaos.0297

We see that S is directly proportional to W.0308

In other words, as W goes up, so does S.0313

What does that mean though?0318

We already said that a high value of W means it is going to be more likely to occur0321

which says that very high entropy values are going to relate to more likely processes.0326

If W large, S is large; process more likely to occur.0337

In this equation, we see that there is a proportionality constant called k_B.0351

k_B is what we call the Boltzmann constant.0356

That is equal to 1.38 times 10^-23 joules per kelvin.0363

Because natural log of W is going to be unitless, that means the units of entropy are also joules per kelvin.0369

This entire equation is known as the Boltzmann equation.0377

S is equal to k_B natural log of W.0383

The units of entropy are going to be joules per kelvin.0386

Let's go ahead and relate to entropy to what we have already discussed0393

in our first lecture from thermodynamics which involved heat flow and energy.0397

If you recall from our first thermodynamic lecture, if heat leaves the system, Q is going to be a negative value.0402

If heat enters the system, Q is going to be a positive value.0410

Entropy is going to be a state function and can be related to Q where ΔS is equal to Q over T.0415

Temperature, because you see capital T, has to be in kelvin here.0426

Again because this is a state function, what we are interested in is0433

a change in entropy just like we talked about for ΔE or ΔU.0437

If heat enters the system, Q is going to be positive which means ΔS is going to be positive.0448

Let's see if that makes sense.0460

Remember if heat enters the system, we are supplying thermal energy for molecules to absorb.0462

If they are going to absorb this thermal energy, that gives them more molecular motion such as vibrations.0470

That increases the number of states an electron can occupy.0477

Basically we are increasing the amount of disorder.0483

We are increasing the amount of chaos.0485

Yes, it does make sense that ΔS is a positive value.0488

ΔS is also related to heat flow during phase changes.0494

It is found that Q is equal to n times ΔH of the phase change.0500

This is from our first thermodynamics; from thermo I.0506

Therefore it follows that ΔS is equal to Q over T which equals to nΔH of the phase change over the temperature.0514

What does that mean then?0525

That means for endothermic processes such as melting and vaporization, endothermic means Q is going to be a positive value.0527

ΔS is positive.0540

This agrees with the previous slide where when we put energy in, everything gets a little more chaotic and a little more disorderly.0542

For example, as you go for melting, you go from a very orderly state, a solid, to a less orderly state, to a liquid.0553

When go from a liquid to a gas, a liquid is less orderly.0563

A gas is going to be the most disordered state.0567

Everything is a lot more chaotic there.0572

ΔS is going to be positive when heat enters the system.0576

Here we have the phase change backing up this point.0580

In other words then, we can summarize this very nicely.0586

If a process is likely to occur, then ΔS is going to be a positive value.0590

We say that the process is expected to be spontaneous if ΔS is positive.0596

This brings us into our second law of thermodynamics.0605

The second law of thermodynamics states that a process will be spontaneous if ΔS total is greater0609

than 0 where ΔS total is equal to ΔS of the system plus ΔS of the surroundings.0616

In other words, nature is going to favor processes that contribute to the overall entropy of the entire universe.0623

In other words, nature favors processes where the amount of entropy or disorder increases as a result of the process.0631

We see this occurring in nature quite a bit.0640

If you think of a virus, what does a virus naturally do?0643

A virus is going to naturally expand and go from cell to cell consuming the resources there and moving on and on.0648

If viruses expand, they get more disorderly, more chaotic.0658

We see this in herd migration.0662

Herd migration, flocks of animals tend to naturally expand from area to area.0667

They tend not to stay just in a single location.0678

Of course the universe itself, the universe theoretically started off as a single point.0682

It is just constantly expanding and expanding.0689

Again we see that entropy is seen in several naturally occurring examples.0694

Then you may be wondering, if ΔS positive is going to be for a spontaneous process, what does it mean when entropy is zero?0704

That brings us into the third law of thermodynamics.0715

Zero entropy really does not exist because the conditions to obtain zero entropy do not exist experimentally.0718

The third law of thermodynamics states that entropy is zero only for a perfect flawless crystal at absolute zero.0728

But absolute zero has not been obtained experimentally.0736

Only in the most extreme conditions where we have zero thermo motion, that is molecules cease to move, entropy is completely zero.0741

In other words, entropy is always going to be a nonzero quantity in practice.0750

Let's now focus on problem solving involving entropy.0758

The following are typical types of questions involving entropy.0762

Sometimes a question can ask you simply to predict the sign of ΔS given a chemical reaction.0765

This is going to be an example of a qualitative process, of a qualitative problem.0770

Basically when the process is a physical change, ΔS is going to be positive for0777

any endothermic process just like we talked about, including melting, vaporization, and sublimation.0781

When you go from a solid to a liquid, from a liquid to a gas, or from a0788

solid to a gas, ΔS of these processes are going to be expected to be positive.0793

They are opposite processes, are all exothermic which means ΔS is expected to be negative.0801

Liquid to solid, gas to liquid, and gas to solid.0807

Freezing, condensation, and deposition; again ΔS is expected to be negative.0812

Let's go on to another type of problem.0822

Another type of problem involves a quantitative problem.0828

What you want to do here is pay close attention to the physical states.0834

From a solid to a liquid to a gas, entropy increases.0838

We have already seen that.0841

What you want to keep track of is the change in physical states from reactant to product side.0843

Let's go ahead and look at an example.0850

Fe solid plus O₂ gas goes on to form Fe₂O₃ solid.0856

We are going to need two of these, three of these, and two of these.0864

Excuse me... we will need four irons.0869

You notice that we start off with a solid and a gas.0873

But on the product side, we have no gases here; no gases on product side.0878

If we are eliminating the highest disorderly state which is a gas, we are starting off with it.0887

We are not winding with it; we are actually getting less disorderly.0894

ΔS for this process expected to be a negative value.0898

Let's go on to another type of problem.0903

If the reaction involves all gases, what you want to keep track of0906

is the net change in gas molecules from reactant to product side.0909

For example, N₂ gas plus 3H₂ gas goes on to form 2NH₃ gas.0913

On the left side, we start off with four gas molecules.0923

On the right side, we wind up with only two gas molecules.0930

For all intents and purposes, gas molecule A is going to be the same as gas molecule B for entropy counting purposes.0935

Because we start off with four gas molecules and we only wind up with two,0945

we are actually decreasing the amount of disorder in this process.0949

ΔS is expected to be negative for this example.0953

Sometimes the problem is going to ask you to calculate ΔS0958

for what is called the system, surroundings, and total where0960

ΔS total is equal to ΔS system plus ΔS of the surroundings.0963

We have talked about system and surroundings in thermodynamics I.0967

If you recall, Q of the system equals to ?Q of the surroundings.0972

This is going to come into play.0978

If a refrigerator coolant absorbs heat from stored food items, the heat then vaporizes the coolant which boils at -27 degrees Celsius.0979

ΔH of vaporization is equal to -22.0 kilojoules per mole.0988

Calculate ΔS total when 1.471 moles of the coolant vaporizes exchanging heat with the food items that are stored at 4 degrees Celsius.0993

What we have to do is the following.1002

We know we are using this equation for sure.1005

ΔS total is equal to ΔS of the system plus ΔS of the surroundings.1007

Let's go ahead and define the system; it doesn't matter which one as the coolant.1018

Let's go ahead and define the surroundings as the food items inside the refrigerator.1024

This is then going to be equal to...1031

We know that ΔS is equal to Q of the system over T.1035

The ΔS of the surroundings is going to be equal to Q of the surroundings over T.1043

But we also know that this relationship here where Q of surroundings is equal to ?Q of the system.1050

We can then go ahead and plug everything in.1060

That is going to be equal to nΔH of vaporization over the temperature of the coolant.1062

That is going to be equal to ?nΔH of vaporization but this time over the temperature of the food.1072

Really the only difference is the temperature that the system and surroundings are stored at1080

if we assume complete transfer of heat from the system to the surroundings.1085

When all is said and done, we are going to get -0.248... don't forget the units.1092

Here this is going to be kilojoules per kelvin or -248 joules per kelvin.1098

Again you must be careful on how you define your system and surroundings.1108

Be consistent throughout the entire equation.1113

Finally another type of problem involves standard molar entropies of formations which is the1118

change in entropy when a compound is formed from its constituent elements under standard conditions.1123

Recall that we covered standard molar enthalpies of formation and used the summation equation.1131

Similarly ΔS of the reaction is going to be equal to summation nS of the products minus summation nS of the reactants.1137

These values here again just like for the enthalpy, you are going to look up in the appendix.1148

Don't forget that n is going to be the stoichiometric coefficient.1157

Unlike standard molar enthalpies of formation, known compound is going to have a standard molar enthalpy of formation.1169

Why?--because the zero is only obtained under the conditions of the third law of thermodynamics.1175

Let's go ahead and examine our... excuse me... this is very important.1182

Again you never can assume that the molar entropy of formation is a zero value ever.1191

It is only true for ΔH and, we are going to learn, for one more--what we call ΔG later.1200

Let's examine our final state function.1207

This is an introduction to Gibbs free energy; free energy is ΔG.1210

It is defined as the amount of energy available to do work.1215

We said that mother nature favors high entropy already and low energy.1220

It is that energy that we are really talking about.1225

We will see that the spontaneous processes will all have ΔG less than zero.1227

To go ahead and do this, we can do ΔG is equal to ΔH minus TΔS.1241

This is the equation that relates all three thermodynamic properties together.1258

We already said that a positive ΔS is going to be a spontaneous process.1266

We want ΔG to be negative for a spontaneous process.1279

What conditions give you ΔG negative?1287

ΔG negative is going to be spontaneous under the following conditions--for very large ΔS values and at high temperatures.1290

As you can see, if ΔS is large and temperature is large,1312

this term is going to be larger than this term giving us1318

an overall negative ΔG value; very large ΔS at high temperatures.1322

Basically ΔH is going to be a relatively negative value; ΔH, small as possible.1329

Again this is what we call Gibbs free energy.1345

Just like our previous state functions, we have standard molar free energies of formation1350

where ΔG of the reaction is equal to summation nΔG of the products minus summation nΔG of the reactants.1355

Like enthalpy, the free energies of formation are zero for all compounds in the standard state.1363

Something like O₂ gas, H₂ gas, N₂ gas, Cl₂ gas, Br₂ liquid, and I₂ solid just to name a couple of examples.1369

These are all going to have a ΔG of formation which is equal to ΔH of formation equal to zero.1385

But again S, the molar entropy of formation, is not going to be zero whatsoever.1397

Let's now study the temperature dependence of ΔG.1409

It is found that we already said that ΔG of the system is equal to ΔH of the system minus TΔS of the system.1414

The reason why this is important is because of the following--this term right here.1424

ΔS of the system is in this equation.1429

But the second law tells us that ΔS of the universe has to be a positive value.1433

If we go strictly by entropy, we need to know the ΔS of the total, not just of the system or the surroundings.1446

However this equation here tells us that all I have to know is ΔG of the system; that is all.1454

Let's go ahead and see what conditions are going to be right for this.1466

We can go ahead and make a chart here; ΔH, T, and ΔS.1474

Then this is going to be sign of ΔG.1481

Let's go ahead and delete temperature here--sign of ΔG, and then ΔH, ΔS.1488

Let's say ΔH is a negative value--if this is less than zero and ΔS is a negative value.1495

If ΔH is negative and ΔS is negative, it will be very difficult to get ΔG to be negative.1505

The only time ΔG can be negative is going to be at a low temperature here.1516

What if ΔH is negative and ΔS is positive?1523

If ΔH is negative and ΔS is positive, then this sign of ΔG is going to be negative at high temperature.1531

What if ΔH is positive and ΔS is positive?1547

If ΔH is positive and ΔS is positive, ΔG is going to be negative at high temperatures.1551

Finally what if ΔH is positive and ΔS is negative?1560

If ΔH is positive and ΔS is negative, ΔG will never be negative according to the equation at any temperature.1565

This is going to be greater than zero at all temperatures.1574

Again as you can see, ΔG of the system is greatly influenced by1579

not only the signs of ΔH and ΔS but also the kelvin temperature.1585

Finally let's go ahead and examine the effect of reactant and product concentration on the sign of free energy.1595

It is found that ΔG of the reaction is equal to ΔG⁰ plus RT natural log of Q.1602

Remember what Q is?--Q is what we call the reaction quotient.1609

This equation relates reactant and product concentration to the free energy not at1618

standard conditions which means the concentrations are not going to be 1 molar.1623

Or the pressures are not going to be 1 atm; or partial pressures not equal to 1 atm.1630

Finally it can also be shown at equilibrium that ΔG not of the reaction is equal to ?RT natural log of K1639

where this is our equilibrium constant which we have discussed so much in the previous sections.1647

This is any equilibrium constant.1654

This can be K_a, K_b, K_p, K_sp, K_f, etc; any of them.1656

Basically this equation tells us that if K is large, the reaction is product favored.1667

ΔG of the reaction is a negative value; that makes sense.1672

If K is very large, that means the reaction is highly product favored.1677

In other words, the reaction as written is going to be likely to occur.1682

If the reaction is likely to occur, ΔG of that reaction should be negative.1687

Let's go ahead and summarize our thermodynamics lecture before getting into our problems.1694

We see that nature not only favors states of low energy but also now high entropy.1701

Entropy is basically disorder or chaos.1707

It is the focus of the second and third laws of thermodynamics.1710

Basically what we have seen in this lecture is that there is a series of equations that govern entropy and free energy.1713

They allow us to calculate both of them under a given set of conditions.1722

Let's go ahead and do sample problem number one.1729

Calculate ΔS of the reaction under standard conditions for the following.1732

We have aluminum oxide reacting with 3H₂ going on to form two aluminums plus three waters.1738

Anytime you see a balanced chemical equation and it asks you simply to calculate1746

ΔS not of the reaction, you are going to use the summation equation.1751

Use ΔS of the reaction is equal to summation n S molality of all products minus summation n S molality of all reactants.1755

I am just going to setup the problem for you.1772

Again you are going to look up these values in the appendix of your textbook.1775

Let's go ahead and do the products first.1779

This is going to be 2 moles of aluminum times the standard molar entropy of formation of aluminum solid.1781

Because it is summation, I am going to add... plus.1799

Let me put that in parentheses.1803

Plus 3 moles of the water vapor times its standard molar entropy.1804

This is all in brackets; that is going to be subtracted from the reactant part.1813

Here this is going to be 1 mole of Al₂O₃ solid times1819

its molar entropy plus 3 moles of H₂ gas times its molar entropy.1830

Again you must pay attention to the physical states because the molar entropy of formation for say water vapor1848

is going to be different than from liquid water which is going to be different than from solid water.1856

Once again pay attention to the physical states.1862

When this is all said and done, you are going to get your answer1865

in of course units of joules per kelvin or kilojoules per kelvin.1868

It depends what the question specifies of course.1874

Let's go ahead and do sample problem number two.1880

The reaction N₂O₄ going to 2NO₂ is not spontaneous under standard conditions.1883

Calculate the temperature at which the reaction becomes spontaneous.1889

What that translates to is, in other words, at what temperature does ΔG become negative?1893

Anytime you are asked to calculate a temperature at which something becomes spontaneous,1902

pretty much you always use this equation: ΔG is equal to ΔH minus TΔS.1908

They are asking you when does this change sign?1916

In other words, when does ΔG go from positive to negative?1920

What you simply is you set this whole thing equal to zero.1926

Zero is equal to ΔH minus TΔS; you solve for temperature.1931

Temperature is going to be equal to ΔH over ΔS.1936

This is all going to be standard conditions of course.1941

To get ΔH, you are going to use the summation equation.1946

To get ΔS, you are going to use the summation equation1950

which means you have to look up the formation values in the appendix.1954

Look up formation values in appendix; appendices.1959

When you do this, you are going to get a temperature value.1969

At temperatures lower than T, ΔG is going to be positive.1976

At temperatures higher than what you calculate for T, ΔG is going to be a negative value.1987

Once again anytime you are asked to calculate the temperature at which something becomes spontaneous,2006

almost always you are going to use this equation: ΔG is equal to ΔH minus TΔS.2015

Set ΔG equal to zero and solve for the temperature.2021

Finally our third sample problem, calculate ΔG of the reaction at 25 degrees Celsius2029

for the formation of 1 mole of C₂H₅OH gas from C₂H₄ gas and H₂O gas.2036

The first thing we probably want to do is translate this into a balanced chemical equation.2047

The reactants are C₂H₄ gas; this is going to react with H₂O gas.2054

That is going to form 1 mole of C₂H₅OH gas.2063

Let's see if everything is balanced.2069

Two carbons, six hydrogens, one oxygen; yes, this is nicely balanced.2071

You are asked to calculate ΔG of the reaction at 25 degrees Celsius.2078

Again ΔG of the reaction can simply be calculated from ΔH of the reaction minus TΔS of the reaction.2083

Again to get the ΔH, you are going to use the summation equation.2094

To get the ΔS, you are going to use the summation equation.2099

Again the kelvin temperature is already specified for you right there.2102

That gives you ΔG of the reaction; they then ask you to calculate K_p.2108

The only equation in this lecture that involves the equilibrium constant is ΔG is equal to ?RT natural log of K.2114

We can then insert this ΔG of a reaction in there.2126

K therefore is equal to 10 raised to the -ΔG⁰ over RT2131

where of course R is equal to 8.314 joules per k mole.2140

Always double check your work.2151

Make sure that K is a positive value because again2152

we saw in previous lectures that K can never ever be negative.2158

That is our lecture from general chemistry on thermodynamics.2165

I want to thank you for your time and attention.2169

I will see you next time on Educator.com.2171