General Chemistry Principles of Chemical Equilibrium

Hi, welcome back to Educator.com.0000

Today's lecture from general chemistry is on the principles of chemical equilibrium.0003

Let's go ahead and take a look at the lesson overview.0010

We will first start off with a brief introduction and then get in right to0012

the core of everything which is basically what we call the equilibrium constant.0017

We are going to first define this equilibrium constant and then go into different0021

types of equilibria including what we call homogeneous equilibria and heterogeneous equilibria.0026

We will then go and see how we can change the value of K_eq.0033

We will also introduce something called the reaction quotient.0039

We then jump into a very fundamental principle in all of general chemistry.0042

That is called Le Chatelier's principle.0046

After that we will get into the quantitative part of the chapter which involves what we call ICE tables.0050

We will wrap everything up with a very brief summary followed by a pair of sample problems.0056

Chemical equilibrium, exactly what this is.0065

This refers to the simultaneous occurrence of a forward and a reverse reaction at the same rate.0068

We see that equilibrium is a dynamic process, not a static process.0075

For example, let's say you had two beakers here; these beakers are closed.0079

Let's say that this one beaker had H₂O in it.0086

Let's just say this is time zero.0091

At time zero, I have nothing but pure liquid water.0094

But we know this from everyday experience inside a water bottle.0097

If you let some time progress, we notice that the water level is going to drop a little.0101

That is because some of the liquid water has entered the gas phase.0108

But there is also going to be a point in time where this water level is not going to drop forever.0116

It is going to reach a minimum.0125

It reaches a minimum because as soon as the vaporization process occurs, the condensation process also occurs.0129

We say for this case that the rate of evaporation is equal to the rate of condensation.0142

If we were to write this out in a chemical reaction, this would be H₂O liquid.0158

Now we introduce a new type of reaction arrow which is this.0164

That goes to H₂O gas; this is our equilibrium arrow.0169

It basically shows that the forward direction is happening simultaneously with the reverse direction.0180

Let's now get into how we can represent equilibrium numerically.0190

Consider the following reaction.0198

Small a big A plus small b big B, equilibrium sign and then small c big C small d big D.0199

In this reaction, let the lowercase letters represent the stoichiometric coefficients.0207

You know the moles after we balance the chemical equation.0222

It turns out that after a chemical reaction has reached equilibrium, it is experimentally determined that the ratio of0226

product to reactant concentration raised to the stoichiometric coefficients is actually constant at any given temperature.0235

Basically the concentration of C raised to the c power times the0243

concentration of D raised to the d power over the concentration of A0249

raised to the a power times the concentration of B raised to the b power.0256

This whole ratio of products to reactants raised to the coefficients is equal to some constant that we call K.0262

Sometimes you are going to see this called K_eq.0272

This is formally what we call the equilibrium constant.0275

The only thing that can change the value of K_eq is temperature.0285

The equilibrium constant is temperature dependent.0290

We can typically represent K in two different ways.0302

K_c is when molarities are used; K_p is when partial pressures are used.0307

Every textbook is a little different.0326

But for the partial pressures, the typical units are going to be atmosphere or bar.0328

Once again K_c and K_p are just the equilibrium expressions when0338

molarities and partial pressures are used for K_c and K_p respectively.0343

For solutions, the equilibrium constant can be expressed in units of molarity just like we discussed.0353

However it turns out that the equilibrium constant is unitless.0361

K_c, how do we go ahead and do that?0364

Mole over liter raised to some power divided by mole over liter raised to some power.0368

It turns out that the equilibrium expression is always referenced to 1 molarity.0378

This is really moles over liter per 1 molar.0385

This is moles over liter per one molar raised to the y power and raised to the x power.0394

As you can see that after this is done, we see that all units cancel.0402

K_eq is actually one of the few unitless values in all of your general chemistry studies.0411

That is K_c.0420

It turns out that if we choose to do our problem with K_p, maybe this is going to be in atmospheres.0422

This is also relative to 1 atm, raised to some power divided by atm raised to some power.0429

We see it again that the units cancel; K_eq again is unitless.0438

What is the relationship between K_c and K_p?0451

K_c and K_p are directly proportional to each other.0455

Depending on the reaction, sometimes they are nearly identical.0461

But the exact relationship between the two is the following where0465

K_p is equal to K_c times RT over Δn where Δn is0469

equal to the moles of gaseous product minus the moles of gaseous reactant.0479

Once again K_p and K_c can be very similar.0489

However they are not quite the same thing.0495

You should really check with your instructor to see what he or she prefers.0499

For chemical equilibria, we are usually dealing with very small concentrations.0507

Because of this, we assume that pure solids and pure liquids are going to remain relatively unchanged.0512

If you have the following reaction, A solid plus B aqueous goes on to form C aqueous plus D solid,0520

it is only the aqueous species that are going to affect the equilibrium, affect K_eq.0532

Hence pure solids and pure liquids are going to be assigned an arbitrary value of 1 when incorporated into the expression for K.0546

That is they have no effect.0555

K_c for this reaction here would be simply the concentration of C times 1 divided by the0557

concentration of B times 1 which is just equal to the concentration of C over concentration of B.0566

Once again pure liquids and pure solids do not appear in the expression for K, in the K_eq expression.0573

Again this is going to be specifically for heterogeneous equilibria.0593

What are some ways where we can manipulate K?0599

The first way of manipulating K is by multiplying an entire chemical equation by a factor.0602

For example, let's take A aqueous plus 2B aqueous goes on to form 3C aqueous.0608

In this case, K as you see is equal to the concentration of C cubed0621

divided by the concentration of A times the concentration of B squared.0627

Let's go ahead and take this chemical reaction and multiply everything through by 2.0633

2A aqueous plus 4B aqueous goes on to form 6C aqueous.0638

It turns out therefore that this K_new is going to be equal to concentration of C to the sixth0647

divided by the concentration of A squared times the concentration of B raised to the fourth power.0654

We see very nicely that K_new is simply equal to the original K_c squared.0661

The rule of thumb is the following.0670

That when multiplying a reaction by a factor, K is going to be raised to that factor.0672

K is raised to this factor.0695

That is the first way of algebraically manipulating K.0699

Once again this is by multiplying through a chemical reaction by a factor.0702

The second way is by taking the reverse reaction.0709

For example, let's take not A plus 2B going to 3C0712

but 3C aqueous goes on to form A aqueous plus 2B aqueous.0717

In this case, K is equal to concentration of A times the concentration of B squared divided by the concentration of C cubed.0726

We see that this is going to be 1 over k_forward.0739

k_reverse is simply the reciprocal of k_forward.0745

The third way of changing K is by adding a series of individual chemical reactions together to form a net balanced chemical equation.0753

Let's go ahead and see.0760

For example, A aqueous plus B aqueous goes on to form C aqueous.0765

C aqueous plus D aqueous goes on to form E aqueous.0777

Let's go ahead and add these two together.0788

When we add these two together, we are going to get A aqueous plus B aqueous plus D aqueous goes on to form E aqueous.0790

You see that C is going to be cancelled out because it is going to be formed and consumed simultaneously.0803

What we want to look at now are the expressions for k for each of these.0815

Here k, I will call this k₁ for reaction one, is equal to the0819

concentration of C divided by the concentration of A times the concentration of B.0823

Here k₂ is equal to the concentration of E divided by the concentration of C times the concentration of D.0830

Here I will call this third one k_net.0841

That is equal to the concentration of E divided by the concentration of A times the concentration of B times the concentration of D.0845

We see that when we add individual reactions together to get a net reaction, the equilibrium constant0855

of the net reaction is simply equal to the product of each individual equilibrium expression constant.0864

That is the third way of algebraically manipulating K_eq.0874

Let's now go on to another aspect of the equilibrium constant.0884

It is what we call the reaction quotient.0887

The equilibrium constant is good when we actually have the values at equilibrium.0890

But what happens if we use values not at equilibrium?0894

When concentrations or pressures are inserted into the expression for K that are not at equilibrium,0899

the ratio of products to reactants is now what we call the reaction quotient symbolized Q.0906

Q is going to be equal to products raised to some power divided by reactants raised to0911

some power except that these molarities and partial pressures are not at equilibrium.0916

The significance is the following.0942

Q can be used to predict which way a reaction will shift to reobtain a state of equilibrium.0944

Basically we are going to have the following general rules.0976

If Q is greater than K, that means we have too much product, not enough reactant.0980

We are going to shift left.0987

If Q is less than K, we have too much reactant relative to product.0992

We are going to shift right.1001

Finally if Q is identical to K, we are at a equilibrium state.1005

Shifting left is the same as making more reactant.1015

Shifting right is the same as making more product.1028

At equilibrium, there is no net change; neither direction is favored over the other.1036

Once again Q can be very useful for determining what direction a reaction will shift if any.1044

Now we have come into one of the most fundamental principles from all of general chemistry.1055

This is called Le Chatelier's principle.1060

Le Chatelier's principle tells us that when a system at equilibrium is disturbed,1063

it will react to counteract the disturbance in an attempt to restore equilibrium.1068

One of the best examples we can think of is from us.1078

When we bleed, when we lose blood, what is the natural thing that our body does in order to counteract this?1082

Your body is going to try to make more blood.1091

When we lose blood, our bodies try to make more in order to counteract the lost.1092

This is a nice example of Le Chatelier's principle.1112

But now we are going to apply this to chemical reactions.1114

In Le Chatelier's principle, you notice that there is the word disturb.1119

There are several ways of disturbing a chemical system that is at equilibrium already.1124

Number one is a change in reactant or product concentration.1127

Number two if we are dealing with gases, it is going to be a change in reactant or product partial pressure.1132

The third one is the change in reaction volume.1137

The fourth one is going to be a change in temperature.1140

Let's now study each of these.1144

Basically if the concentration of reactant goes up, then we are going to shift away from it.1150

We are going to shift right.1163

If the concentration of product goes down, we don't have enough of it.1165

We are going to shift right; that is one situation.1170

If the partial pressure of a reactant goes up, partial pressure is just the same as concentration.1178

We know from ideal gas law that pressure is proportional to amount.1189

If partial pressure of the reactant goes up, we are going to shift right.1194

If the partial pressure of the product goes down, we are going to shift right.1204

For temperature, I am going to do this on the last slide because I am running out of room right now.1215

I will come back to changes in temperature and also to change in vessel volume.1228

But now let's get into some problem solving with some ICE tables.1235

We will now approach chemical equilibrium from a quantitative view.1238

An ICE table allows for one to study a chemical system at three points in time.1241

Basically what are all reactant and product initial amounts?1246

That is what the I stands for.1250

During the reaction on the way to equilibrium, what are their changes in concentration/pressure?1252

That is what the C stands for.1256

Finally what are their final volumes once equilibrium has been achieved?1258

That is what the E stands for.1262

We are going to look at this right now.1263

Consider 2 water gas goes to 2H₂ gas plus O₂ gas.1266

K is equal to 2.4 times 10^-5 at some temperature.1271

At equilibrium, the concentration of H₂O gas is 0.11 molar.1275

The concentration of H₂ gas is 0.019 molar; calculate O₂ at equilibrium.1279

The very first thing we want to do is set up our ICE table.1285

2H₂O gas goes on to form 2H₂ gas plus O₂ gas.1289

What I like to do, I just like to set up the letters I-C-E right underneath it.1296

We just fill in this table right now.1300

You are told that at equilibrium, the water is 0.11 molar and that the H₂ is 0.019 molar.1303

We don't know what O₂ is.1313

We can just call that the concentration of O₂ at equilibrium.1315

We know the expression for K is equal to 2.4 times 10^-5.1322

That is going to be equal to the concentration of H₂ squared times1328

the concentration of O₂ divided by the concentration of H₂O squared.1332

That is going to be equal to 0.019 squared times the concentration of O₂ at equilibrium divided by 0.11 squared.1338

When all is said and done, the concentration of O₂ at equilibrium is going to be 8.0 times 10^-4 molar.1349

Don't forget the units.1359

You notice that we didn't have to fill in the rest of the table1363

because we were already at halfway there to the equilibrium values.1366

This is a nice usage of the ICE table.1371

Let's go ahead and do another example though.1374

A 1 liter flask was filled with so much of SO₂ and so much NO₂ at some temperature.1377

After equilibrium was reached, 1.3 moles of NO was present.1382

The reaction is the following; calculate the value of K_c at this temperature.1385

The problem wants us to calculate K_c at this temperature.1390

We are going to go ahead and set up the ICE table--I, C, and E.1396

SO₂ here is going to be 2.00 molar initially.1406

O₂ is going to be 2.00 molar also.1411

After equilibrium was reached, 1.30 moles of NO gas was present.1415

That is what goes right here, 1.30 molar.1419

Let's fill in the rest of the table.1426

If no initial values are mentioned of SO₃ and NO, it is safe to assume that they are zero.1427

Let's go ahead and do the change.1434

The SO₂ is going to go down by some amount x.1438

O₂ is going to go down by some amount x.1441

SO₃ goes up by some amount x.1443

O is going to go up by some amount x.1452

We notice because these are all 1:1 mole ratio.1457

At equilibrium, it is just going to be the addition of the two rows added together; sum.1465

That is going to be 2.00 minus x, 2.00 minus x, x, and x.1472

But guess what? x is 1.30 molar because we were told that in the beginning; that is very nice.1479

Therefore this is 1.30 molar; this is 0.70 molar; this is 0.70 molar.1485

We know enough to calculate the value of K_c.1494

K_c is going to be 1.30 squared divided by 0.70 squared.1498

That is going to give us an answer of 3.5 for our answer.1504

This is how we use the ICE tables to help us come up with the equilibrium constant.1517

I want to quickly summarize this and then jump into sample problems.1526

Chemical equilibrium is a dynamic process where the rate of forward and reverse reactions is equal.1531

Equilibrium constant K quantifies how reactant or product favored a reaction is once it has reached equilibrium.1536

If K is between 0 and 1, we say that it is reactant favored.1543

If K is greater than 1, we say that it is product favored.1552

Le Chatelier's principle states that a system at equilibrium when disturbed will react to counteract the disturbance.1559

We already saw the concentration and partial pressures.1565

But let's go ahead and see what I left out before; this is now temperature.1573

For an exothermic reaction, if the temperature increases, it turns out we are going to shift left.1579

If the temperature decreases, we are going to shift right.1593

Basically you treat the word heat as a molecule.1599

Again if you treat the word heat as a molecule on the product side, it becomes1609

a lot more intuitive as you will see what happens when we result in the change.1616

Another thing we can disturb an equilibrium with is with the volume.1622

If the volume of the reaction vessel goes up, that means the pressure is lower.1629

That means we are going to shift to make more gases.1636

Shift toward side of reaction with more gas molecules.1640

Once again if the volume goes up, we are going to shift toward the side of the reaction with more gas molecules.1654

Those are the ways we can disrupt a system that is at equilibrium and see how it is going to counteract the stress.1662

Finally ICE tables allow for determination of the reactant and product concentrations/pressures given the equilibrium constant K.1668

That is our summary.1676

Now let's go ahead and jump into a pair of sample problems.1678

Consider 2SO₃ gas goes to 2SO₂ plus O₂.1681

Initially 12 moles of SO₃ is placed into a 3 liter flask at some temperature.1685

At equilibrium, 3 moles of SO₂ is present.1689

Calculate the value of K_c at this temperature.1692

Let's go ahead and rewrite this.1696

2SO₃ gas goes on to form 2SO₂ gas plus O₂ gas.1698

Let's go ahead and see what we can fill in, I-C-E.1707

Initially 12 moles is placed in a 3 liter flask of SO₃.1712

That is going to be 4.0 molar.1716

At equilibrium, SO₂ is present, 3 moles; that is going to be 1.0 molar.1721

Usually you can fill in two-thirds of the table with just one or two sentences.1728

SO₂ initial is not mentioned; that is zero.1733

O₂ initial is not mentioned; that is zero also.1735

We have to pay attention to the coefficents.1740

The coefficients tells us the relative amount of reactant and product that is going to be involved.1742

Really this is going to be -2x; SO₂ is going to be +2x.1747

O₂ is just going to be +x.1752

At equilibrium, we get 4.0 minus 2x, 1 molar, and x.1756

The nice thing about this is that because we know the 2x and the 1 molar, we can conclude that x is 0.5 molar.1763

When we do that, we are going to get here 0.5 molar.1776

Here for SO₃, we are going to get 3.0 molar.1780

Great, we now have all of our molarities at equilibrium.1786

K_c is equal to the concentration of SO₂ squared times the concentration1791

of O₂ divided by concentration of SO₃ squared, all equilibrium values.1797

This is going to be 0.5 squared times 0.5 divided by 3.0.1805

When all is said and done, this is going to give us a K_c of 0.056 for this reaction.1815

That is another usage of the ICE table.1827

Finally the last sample problem is sample problem number two.1831

At a particular temperature, K is equal to 3.75.1834

If all four gases had initial concentrations of 0.800, calculate their equilibrium concentrations.1837

Let's go ahead and set it up--I, C, and E.1843

0.800 molar initial, 0.800 molar, 0.800 molar, and 0.800 molar.1848

Because we have nonzero amounts of all reactants and products, we don't know which way we are going to shift.1858

This is a twist on things; now we have to solve for Q.1868

Q is going to tell us which way we are going to shift.1872

This is going to be equal to concentration of SO₃ initial times the concentration of1875

NO initial divided by concentration of SO₂ initial times the concentration of NO₂ initial.1882

That is just going to be equal to 1, 1.00 which is going to be less than K.1891

Because Q is less than K, that means we are going to shift to the right.1900

Therefore the sign that goes here is going to be ?x.1907

This is ?x; this is +x; this is +x.1911

We are going to get 0.800 minus x; 0.800 minus x.1916

This is going to be 0.800 plus x.1923

This is going to be 0.800 plus x; K is equal to 3.75.1926

That is equal to 0.800 plus x squared divided by 0.800 minus x squared.1934

We can use what we call a perfect square to solve for this.1946

The square root of both sides is what we are going to do next.1951

When we go ahead and do that, we are going to get 1.93 is equal to 0.800 plus x divided by 0.800 minus x.1956

When all is said and done, x is going to be equal to 0.25 molar.1966

All we do next is simply plug it back into all of the expressions at equilibrium.1972

The concentration of SO₃ at equilibrium is equal to the concentration of NO at equilibrium.1978

That is going to be 0.800 plus 0.25.1986

I will let you guys do the sum; that is going to be 1.050 molar.1992

The concentration of SO₂ at equilibrium is equal to the concentration of NO₂ at equilibrium.2004

That is going to be equal to 0.800 minus 0.25.2012

We are going to get 0.55 molar for these guys.2020

That is use of the ICE table when we have all four initial values present.2026

Once again you have to find which way it is going to shift.2034

You do that by evaluating Q.2036

That is our lecture from general chemistry on the principles of chemical equilibria.2039

I want to thank you for your time.2045

I will see you next time on Educator.com.2046