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Lecture Comments (14)

1 answer

Last reply by: Professor Hovasapian
Thu Jan 25, 2018 4:43 AM

Post by Magic Fu on January 22 at 10:32:47 AM

I have a question:
2SO2 (g) O2 (g) <-> 2SO3 (g)
The equilibrium cannot be established when _____ is/are placed in a
1.0-L container.

a. 0.25 mol SO2(g) and 0.25 mol O2(g)
b. 0.75 mol SO2(g)
c. 0.25 mol SO2(g) and 0.25 mol SO3(g)
d. 0.50 mol O2(g) and 0.50 mol SO3(g)
e. 1.0 mol SO3(g)

1 answer

Last reply by: James Dohm
Sun Feb 8, 2015 10:01 PM

Post by Hong Zhao on May 5, 2014

Why concentration and pressure are the same thing?

1 answer

Last reply by: Professor Hovasapian
Thu Mar 13, 2014 11:24 PM

Post by Christian Fischer on February 17, 2014

Hi Raffi, I enjoy your videos so much!! I have a qeustion regarding "Bridging the gap between Le Chatelier and Waters equilibrium reaction" I hope you get time one day: Here it comes:

We know that water self-ionizes from the following equilibrium reaction
2H2O = OH(-) + H3O(+)

And the equilibrium constant is

So let’s assume we add 1mol HCl to 1Liter water:  

According to the equilibrium constant the concentration of OH(-) is
[OH]=Kw/[1mol] =10^-14  

So from this, can we conclude the following (Is this correctly understood?) :
1) In very acidic solutions (1mole of HCl) waters self-ionization still occurs and the same amount of OH(-) is still produced as in a neutral solution BUT a big amount of the added H3O(+) reacts with OH to form water.
2) According to Le chatelier, by adding 1 mole of  HCl, Waters self-ionization equilibrium should shift to the left (due to the high concentration of H+) such that only 10^-14 [OH] Is produced from the reaction between water molecule for the Kw to be 10^-14. This means that waters self-ionization happens on a SMALLER scale so only 2x10^-14 water molecules split per unit time, and not 2*10^-7 such as at pH=7,.

So the equilibrium constant always remains constant, but unlike most equilibrium expressions involving both reactants and products, the concentrations of reactions at equilibrium do not need to be  equivalent - [OH]=10^-14 and [H+]=1Molar which are not at all equivalent but they still make the equilibrium constant true.  

1 answer

Last reply by: Professor Hovasapian
Wed May 15, 2013 1:45 AM

Post by Nawaphan Jedjomnongkit on May 14, 2013

Thank you for the lecture but I still don't get why when we add inert gas , change the P but it will have no effect with the equilibrium, on the other hand when we change volume without adding anything but it will change the position of equilibrium? Even inert gas have noting to do with the reaction but it make the pressure change in the same way when we reduce the volume, right? And if we think like when the room so small they will tend to change in the way to have less people in the room equal to we decrease volume and when we have the same size room but already have so many people inside the room equal to we put inert gas in the same flask, or I miss some points there?

1 answer

Last reply by: Professor Hovasapian
Tue Apr 30, 2013 4:33 AM

Post by Antie Chen on April 30, 2013

Hello Raffi, just want to confirm the part of pressure. If the (delta)n>0, when the pressure increase, the reaction want fewer pressure and move to left(the side have smaller sum of coefficient). And if (delta)n<0,when the pressure increase, the reaction will move to left.
If pressure decrease, these directions also turn to another orientation.

1 answer

Last reply by: Professor Hovasapian
Mon Mar 25, 2013 7:38 PM

Post by Harpreet Singh on March 25, 2013

you made a mistake when you were trying to balancing the equation.

1 answer

Last reply by: Professor Hovasapian
Wed Nov 7, 2012 1:40 PM

Post by tanya bond on November 6, 2012

This isn't a question, just wanted to say I just wrote my chem midterm and I'm pretty sure I got an A. These videos cleared up the stuff I was having trouble understanding and the material seemed so easy after! I was skeptical about this site but it turned out great, so thanks!

Le Chatelier's principle & Equilibrium

  • Le Chatelier’s Principle allows you to decide how a system already at equilibrium will respond when you disturb that equilibrium.
  • Disturbances are: Adding/Subtracting species, changing the temperature, changing the Pressure/Volume.
  • Upon disturbance a system will move in the direction that offsets the disturbance.

Le Chatelier's principle & Equilibrium

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
  • Le Chatelier 1:05
    • Le Chatelier Principle
    • Concentration: Add 'x'
    • Concentration: Subtract 'x'
    • Example 1
    • Change in Pressure
    • Example 2
    • Temperature: Exothermic and Endothermic
    • Example 3
    • Example 4

Transcription: Le Chatelier's principle & Equilibrium

Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.0000

Today, we are going to close out our discussion of equilibrium with a discussion of Le Chatelier's Principle in relation to equilibrium.0004

Now, that doesn't mean that we are just never going to talk about equilibrium again; we are--equilibrium is the central concept in chemistry and most sciences.0011

This was a general discussion about equilibrium--getting a little bit of practice in the techniques and how to subtract and add reactants to decide which way a reaction is going to go.0019

All of these things will show up when we discuss acid-base, solubility product, complex ion, equilibria...so, very, very important.0031

So now, we are going to discuss a qualitative method of deciding what happens when you actually (once a system is at equilibrium) place a stress on that system.0040

Your intuition is actually absolutely correct; when you place a stress on any system, the system is going to react in a way that it wants to offset that stress.0050

In chemistry, in an equilibrium situation, we call that Le Chatelier's Principle.0059

Let's jump in and get started.0064

Once again, it is a way of predicting the effect that a change in concentration or pressure or temperature has on the system that is already at equilibrium.0067

If we have a system at equilibrium, we can do a couple of things to it to stress it out: we can change the concentration of a particular reactant or product (in other words, add or subtract reactant or product); we can change the pressure of the system, either by directly changing the pressure or by changing the volume, or by adding an inert gas; or we can actually change the temperature (drop the temperature or increase the temperature).0077

When we do these things, the equilibrium position is going to change.0102

Now mind you, when we change concentration and when we change pressure, the equilibrium position might change, meaning there might be more or less of reactant or product; but the equilibrium constant does not change.0106

In other words, the equilibrium itself doesn't change; the position changes.0120

However, that is not the same with temperature, because, remember: the equilibrium constant is actually dependent on temperature.0125

So anyway, we will see more of that when we start doing some of the problems.0131

OK, so let's just go ahead and write out the principle as our fundamental thing to get started with.0135

The principle is: If a change is imposed on a system already at equilibrium (let me go ahead and not use that symbol; let me go ahead and just write out "at equilibrium"), the position of the equilibrium will shift in the direction that offsets the change.0146

Now, let me just describe what that means in terms of a chemical reaction.0200

So, let's just take aA + bB is in equilibrium with cC + dD.0205

Now, when we say it will shift in the direction that offsets the equilibrium, once the system actually reaches equilibrium, if I stress it out somehow, it will move in the direction to offset that change--the change that I have made.0213

What we mean is that it will either shift right or it will shift left; in other words, it will produce more product, or it will produce more reactant.0224

That is all that that means--shift equilibrium.0232

So remember, when we actually start a reaction, it moves in the forward direction; but eventually, when enough product develops, it starts to also move in the reverse direction.0235

At some point, the rate of the forward and reverse reactions is the same; that is what we call the equilibrium point--when the concentrations of this, this, this, and this are constant--when there is no more change.0244

That doesn't mean the reaction has stopped; it's a dynamic equilibrium--things are still going, except there is no net change.0255

So, a shift in equilibrium means will it move forward? Will it move back? And I can do things to push it forward and back.0260

That is essentially what chemical engineers do: they try to find ways to push reactions forward--the reactions that would not otherwise want to move forward.0267

How can we stress it out to make it go forward?0277

OK, the first...so we said that we can do a couple of things to it: we can affect, we can change (well, not change--we can--yes, let's go ahead and say change) the concentration by adding or subtracting; we can change the pressure of the system (that is another method by which we can affect the system), or the temperature.0280

Temperature, pressure, and concentration are the things that we can control, that we can actually do to a system at equilibrium.0314

When we effect these changes, what happens?0322

The first thing we are going to talk about is concentration.0325

Here we go: How does a system respond to a change in concentration?0333

OK, well, like this: If we add species X (and by species X it could mean any one of these--it could be A; it could be B; it could be C; it could be D)--I have this system here; it is in a reaction vessel; I can add or subtract any one of these species; here is how the system will react.0338

Well, if I add a species, X, well, the system is going to do what is necessary to offset the addition, which means (so, system offsets the addition) that it will consume X, which means it will deplete X, which means that the reaction will shift accordingly.0361

What "accordingly" means: it means away from X.0411

In other words (we will do an example, but I'll just do it here), if I have this system at equilibrium, and if I add B, that means I have pumped more B into here.0419

Well, the system wants to offset my addition of B, so it wants to reduce the concentration of B.0429

The only way to reduce the concentration of B is by actually using it up.0435

The only way it can use it up is by shifting the reaction forward to use up B.0440

Yes, it also ends up using up A, but the idea is that it is using up B.0445

It wants to offset what I have done to it: I have increased the concentration of B--it wants to push the concentration back down.0449

In order to push the concentration back down, it has to use it up.0455

In order to use it up, it moves the reaction forward; it produces more product, less reactant.0459

That is what it means; so in this case, we say that the equilibrium shifts to the right, towards the product.0465

Now, what happens when we subtract a species?0471

Let's say I have A, B, C, D in equilibrium, and let's say I siphon off...let's say O2 gas is one of the products that is formed--what happens if I siphon off O2 as is produced?0475

Well, the subtraction of X goes like this.0487

It is always the same: Subtract species X; well, the system will offset the subtraction...the system wants to offset the subtraction; that means it will...since I am subtracting X, it wants to do the opposite, which means it will produce X; that means that the reaction will shift in a direction which produces X.0491

It is that simple; you are just following the train of logic which produces X.0538

In the example of, let's say, aA + bB is in equilibrium with cC + dD, if I end up subtracting B (meaning if I pulled B away from a system already at equilibrium), well, the system is going to respond by wanting to increase the concentration of A.0545

In order to increase the concentration of A (in other words, produce A...I'm sorry, B), the reaction will shift in the direction that does that.0564

Well, the only way to produce B is to shift this way.0572

In other words, use up this and this so that more B is produced.0575

So, the reaction is going to shift to the left.0579

That is all you are doing here.0583

Let's do an example, a real-life example.0585

It's going to be Example 1: I have the equation 2 SO2 + O2 goes to 2 SO3.0592

All of these are gaseous: sulfur dioxide, oxygen gas, sulfur trioxide.0601

Now, we want you to predict the direction the reaction will shift when its equilibrium is disturbed (so this is a system already at equilibrium).0605

One disturbance we are going to do is: we are going to add SO2.0633

The second disturbance we are going to do: we are going to remove SO2.0640

The third disturbance we are going to do is: we are going to remove O2.0646

OK, well, so we have a reaction here; now, we want (the system is at equilibrium)--we add sulfur dioxide.0652

Well, if we add sulfur dioxide, the system wants to now reduce the sulfur dioxide--it wants to do the opposite of what we are doing.0663

We are adding; it wants to subtract; that means lower the sulfur dioxide concentration again.0670

In order to lower it, it has to use it up; in order to use it up, it is going to shift it to the right.0675

So, the equilibrium is going to shift to the right.0680

You can either write "shift to the right," or you can just do a right arrow; I tend to use arrows.0682

I like symbolics, when it works--when it is clear.0686

That is it; it is that simple.0690

What happens if I remove SO2?0691

Well, if I remove SO2 from an equilibrium mixture, the system is going to want to produce more SO2 to compensate for what I have taken away.0694

In order to produce more SO2, it has to shift the equilibrium that way.0702

That means product is going to decompose, O2 and SO2 are going to form, until the SO2 level comes back to such that the equilibrium constant is brought back to where it is.0706

So, in this particular case, it is going to shift to the left.0720

What happens if I actually remove...you know what, let's change this; I don't want to remove this; I want to remove SO3.0726

Now, what happens if I remove SO3?0733

Well, I go over here; SO3 is on the product side; if I remove SO3, if I pull it out of the flask, the system is going to want to compensate for that loss by producing SO3.0735

The only way to produce SO3 is by pulling the reaction forward, depleting this, depleting that, producing SO3.0748

The reaction is going to shift to the right.0757

That is all that is going on here.0760

Find the stress that you are putting on the system, and find out what the system has to do in order to do the opposite of that stress--it's that simple.0762

OK, so now, let's go ahead and talk about stressing out the system via pressure.0773

Now, instead of changing the concentration of a given species, let's see what happens when we increase the pressure or decrease the pressure on the system.0780

So, this is going to be a change in pressure.0790

This is the second of the effects that we can run on a system at equilibrium.0795

Well, as it turns out, there are three ways to change the pressure.0800

OK, we can add or remove reactant or product--gaseous; OK, I'm going to put "gaseous" in; so again, when we talk about pressure, we are talking about gases.0804

So, if I want to change the pressure, one of the ways that I can do it is by adding or subtracting actual gaseous product--reactant or product, if one of those species happens to be a gas.0823

If it's not a gas, then I can't do anything about the pressure.0835

Add or remove--well, by adding or removing a reactant or product, I handled that the same way that I did just a moment ago; I'm just adding or removing a reactant or a product.0837

I handle it--it's the same as changing the concentration.0849

That is what I am doing; when I am adding or subtracting a gas, yes, I am changing the pressure, but what I am doing is actually changing the concentration, because remember: pressure and concentration are actually the same thing.0852

They are just variations of the same thing.0862

So that one--I handle it just by addition or subtraction of reactant, like we did a moment ago.0866

The other way to change it is by adding an inert gas.0873

If I pump in an inert gas (in other words, a gas that doesn't react with anything, that just takes up space), I have changed the pressure of the system; how does that system react when I have actually added an inert gas?0877

We will see the answer in just a minute.0890

3: I can change the volume.0893

Change the volume: so, you know PV=nRT; if the volume goes up, the pressure goes down; if the volume goes down, the pressure increases, all else being equal.0896

The relationship between volume and pressure is an inverse relationship.0908

So, these three ways are the ways that I can actually change the pressure of the system.0911

Now, let's see how each one reacts.0916

Well, 1: as we said, handle the same way as the addition or the subtraction, changing the concentration.0918

This is handled the same as a change in concentration, because that is it; that is all you are doing.0929

"As a change in concentration..." so we already know this one.0938

2: If we add an inert gas, as it turns out, adding an inert gas--it changes the total pressure, but it doesn't change the equilibrium, because it's not reacting with any of the species.0943

So, adding an inert gas changes nothing--no change.0954

Yes, total pressure will increase in the system, but because the gas is not really reacting with anything, it doesn't really shift the equilibrium one way or the other.0964

The same concentrations of reactant and product are in there, and the Keq, which is the product of the products, over the product of the reactants.0972

Those concentrations don't change, so the Keq doesn't change.0988

Well, nothing is changing; so there is no change.0992

3: OK, this one--there definitely is a change.0996

If I do a volume decrease, that is the same as a pressure increase; these double arrows--they just mean "implies"--it's following a train of logic.1002

Volume decrease is the same as a pressure increase; the system now wants to offset the pressure increase; "wants to offset" means it will seek out a way to decrease the pressure back.1020

That means it will shift in the direction of fewer gas particles.1054

I won't do an example just yet; I want to actually talk about a volume increase, and then I will do the example, and it will make sense.1069

So, let's follow the train of logic here: if I do a volume decrease for a system at equilibrium, well, if I decrease the volume, the pressure is going to increase.1074

If the pressure increases, we know that the system is going to offset that increase, so it is going to want to seek out a way to decrease the pressure back.1084

Therefore, it is going to shift in the direction that has fewer gas particles, because fewer gas particles means less pressure.1091

That is what the system is going to do.1099

Now, what happens if we have a volume increase?1101

OK, so a volume increase: well, it's going to be just the opposite.1107

A volume increase means a pressure decrease, so the system wants to offset by a pressure increase.1110

How does it do that?--well, it shifts in the direction of more gas particles.1133

More gas particles means higher pressure; so once again, if I increase the volume of the system, I decrease the pressure.1149

Well, by decreasing the pressure, the system needs to offset that decrease that I just made, so it is going to want to increase the pressure.1158

The only way that a system has to increase the pressure is by shifting the reaction in a direction that produces more gas particles.1165

So, if the products happen to have more gas particles, it will shift in the direction of products, which means it will shift to the right.1173

So now, let's do an example.1180

Oh, I should give one last note here: I'll put N.B. for nota bene; if reactants and products have equal numbers of gas particles, there is no change.1184

I can increase the pressure, decrease it...do whatever I need to; there is no change, because now, the system has nowhere to go.1211

There are equal numbers of particles on the reactant side and on the product side, so there is nowhere for the system to go.1218

The total pressure will increase or decrease accordingly; the system will stay exactly the same--there will be no shift.1224

In other words, it won't move to the right to create more product; it won't move to the left to create more reactant.1230

OK, so let's do Example 2; let's do this in red.1236

Example 2: we have: CH4 gas + H2O gas is in equilibrium with carbon monoxide gas, plus hydrogen gas.1247

We have gaseous components everywhere--in reactants and products--we have 1, 2, 3, 4; OK.1261

Let's see: oh, I'm sorry, this is 3 H2; I was going to say, "There is something wrong here"; let's double-check: C, C, H2O gas; we have H2; that is 4; that is 6; that is 6; 1 oxygen; 1 oxygen; yes, OK.1270

We have 1 mole of CH4, 1 mole of H2O, 1 mole of carbon monoxide, and 3 moles of hydrogen gas.1288

OK, the question is: What happens if we (hmm, temperature--interesting; OK) reduce the volume?1295

This system--let's say, all of a sudden, we pump all of this from a 1-liter flask into a 250-milliliter flask; we reduced the volume; what happens to the equilibrium position?1319

Again, the equilibrium constant doesn't change; that is the whole idea behind the shift; the constant wants to stay the same, so it has to shift in a direction that will keep it the same.1333

OK, what happens when we reduce the volume?1344

OK, so I'm going to do some symbolism here: a down arrow is decrease; and up arrow is increase.1347

A volume decrease implies a pressure increase; a pressure increase means that a system wants a pressure decrease.1352

In order to get a pressure decrease, I need fewer gas particles.1368

The only way to get fewer gas particles is to shift the reaction to the left.1376

The reason is: I have (let me do this in blue) 1 mole, 1 mole; I have 1 mole...I have 3 moles here.1386

The products have 4 moles of gas particles; here, I only have 2 moles of gas particles.1396

If I increase the pressure, the pressure will want to decrease to offset; the only way that the system has in order to decrease it is by shifting the reaction that way, by diminishing the product and producing more reactant.1401

4 moles, 2 moles; there are fewer moles of gas particles if the reaction goes this way.1414

In that case, it will actually reduce the pressure as far as it can to reach equilibrium.1422

That is all that is going on here.1428

OK, let's say the opposite: What happens if we just directly decrease the pressure?1431

Well, if we decrease the pressure, the system is going to want to offset it by increasing the pressure.1445

The system wants to increase pressure; the only way to increase pressure is to move the reaction in a direction that produces more gas particles.1453

So, it will shift to the right.1465

That is it; it's all you are doing--follow the train of logic; everything will fall out.1468

OK, now let's talk about the third change that we can make to a system at equilibrium, and that is going to be changing the temperature (heating it up, cooling it down); what happens to the equilibrium?1474

Now, remember: we said that the equilibrium position changes when we change the pressure or change the concentration.1487

In other words, the actual values of the concentrations change, but the equilibrium constant does not change.1495

That is the whole idea behind a constant.1501

With temperature, it is different, because the equilibrium constant is actually dependent on the temperature.1507

For most practical purposes, it is not really a problem, but you should know that the Keq is temperature-dependent.1514

That is important; so, at a different temperature, you are going to have a different Keq.1525

An equilibrium constant--yes, it is a fingerprint for that reaction, but it is a fingerprint for that reaction at a specific temperature, which is why the problems always give you what the temperature is.1529

It doesn't mean you have to use the temperature in a calculation; it just means "Know that it is at that temperature."1540

OK, so let's talk about what happens with temperature.1547

Temperature is actually really easy to deal with; let's do a quick review.1555

Exothermic means ΔH is (not equals) negative; ΔH is less than 0.1560

It also means that heat is released; heat is produced, in other words.1577

Endothermic is the opposite; that means the ΔH is positive (not equals, but is; the equals sign is used for numbers); in other words, heat is absorbed.1591

Or, a better way of saying it is "is required."1610

OK, so exothermic--now we will talk about how we deal with heat, as far as Le Chatelier's Principle is concerned.1617

Exothermic--heat is released; heat is produced; imagine that heat is a product--that is it.1627

Not only do you produce a chemical product, but you are also producing heat in the process.1634

Treat heat like a product; that is it.1639

Heat is a product, so you write it on the right side of the arrows.1643

Endothermic means heat is required; well, if heat is required, that means heat is a reactant--treat heat like a reactant--put it on the left side of the arrows.1650

Well, if I am writing heat as a product or a reactant, I can imagine it like a species.1666

If I increase the temperature of something, that means I am actually increasing the concentration of heat.1679

Now, I treat a change in temperature the same way that I treated an increase in concentration.1686

Well, let's write out what this looks like; let's do an example, actually.1693

Actually, you know what, let me just do a rough breakdown; so let me do aA + bB goes to cC + heat.1696

This is an exothermic reaction; I have written it; ΔH is negative.1706

I have written it as if heat is just another product; it is one of the things that is produced.1711

It produces a chemical, C, but it also produces heat.1716

Now, if I have this situation at equilibrium, what happens if I increase the temperature?1720

Well, if I increase the temperature, that means I am increasing the concentration of heat.1726

Well, if I increase the concentration of heat, the system wants to react by decreasing the concentration of heat; that means it wants to use up the heat--that means it is going to shift that way in order to use up the heat, to pull it away--just like the first part, when we did concentration.1732

If I do a temperature decrease on an exothermic reaction--well, with a temperature decrease, the temperature is going to want to offset by going up.1749

Well, the only way the temperature can go up is by producing more heat from the equation itself.1759

In order to produce more heat, the reaction has to shift to the right.1764

It is going to shift toward products, toward the production of heat--yes, and C, but of heat.1770

So again, when you are dealing with temperature, take a look to see which--whether your reaction is exothermic or endothermic.1776

If it's exothermic, write heat as one of the products; if it's endothermic, write it as one of the reactants.1783

And then, go back to treating it like it's just a reactant or a product that you are increasing or decreasing the concentration of.1788

So let's do an actual example.1796

We have: N2 + O2 is in equilibrium with 2 NO; so nitrogen gas and oxygen gas in equilibrium with nitrogen monoxide.1799

The ΔH for this process is 181 kilojoules.1813

I apologize; I tend to write my kilos with a capital K; I know it's probably bad practice, but...it's all right; it's not supposed to be, but...I'm going to be like that.1818

This is a positive 181 kilojoules; so this is an endothermic reaction.1826

OK, this is endothermic, so I'm going to write this reaction with heat as one of the reactants.1831

I am going to write heat, plus N2, plus O2, is in equilibrium with 2 NO.1836

So now, let's see what happens.1845

What can we do?...well, OK; if we increase (at part A) the temperature, what happens?1852

Well, if we increase the temperature, that means we are increasing the concentration of heat, because it is one of the reactants--it's an endothermic reaction.1861

It is going to want to decrease the concentration of heat to offset.1869

The only way to decrease it is to deplete it, which means it is going to push the reaction to the right.1872

So, the reaction will shift right.1878

In other words, shifting right means it will produce more nitrogen monoxide; in the process of producing more nitrogen monoxide, oxygen will deplete, N will deplete, and heat will deplete--it will use up the excess heat that I have put in there.1881

Well, if I do a temperature decrease--if I decrease the temperature--the system wants to increase the temperature back up.1895

The only way it has (the system has) to increase the temperature is by producing more heat.1903

The only way to produce more heat is by moving the reaction that way, to produce more heat, more N2, and more O2.1908

It is going to shift left.1915

That is all these shifts mean: is it going to move towards formation of more product or formation of more reactant?1919

Formation of more product means "shift right"; formation of more reactant means "shift left."1926

That is it; that is all that is going on here.1933

Let's do another example.1936

We have: 2 SO2 + O2 going to SO3 (well, it's 2 SO3; it's not balanced--and that is going to be 3 O2; there, that is a little bit better).1937

So, we have sulfur dioxide gas and oxygen gas, sulfur trioxide; I think I may have left the equation unbalanced earlier on; I apologize for that, if so.1956

So now, it says that the ΔH for this reaction is -198 kilojoules.1966

Well, a negative ΔH is exothermic; exothermic means it is releasing heat--it is producing heat--so I'm going to rewrite this as: 2 SO2 + 3 O2 is in equilibrium with 2 SO3 + heat.1973

Or, I can just write "energy"--and again, heat and energy are just the same thing.1990

So now, heat is a product; it is one of the things that is formed in this reaction.1993

Well, if I increase the temperature of this reaction at equilibrium--if I increase the temperature, the system is going to want to decrease that temperature back down.1998

The only way the system has to decrease the temperature is by shifting the reaction that way--using up heat and using up SO3 to produce more reactant; so it is going to shift left.2008

If I decrease the temperature, the system is going to want to offset by increasing the temperature.2022

The only way the system has to increase temperature is by producing more heat.2029

How does the system produce more heat?--it moves to the right to produce more sulfur trioxide and to produce more heat to compensate for the heat that I took away.2032

So, it ends up shifting right.2043

That is it.2049

So, we have a system at equilibrium, and there are a couple of things that we can do to it--well, three things that we can do to it.2052

We can add or subtract reactant or product; we can change the pressure of the system; or we can change the temperature of the system.2059

Each of those things causes the reaction to actually move in a direction to reestablish equilibrium.2069

That is the whole point: I have upset the equilibrium balance; the system needs to do what it can, if it can, to reestablish the balance.2076

That doesn't mean it can always do that; like we said, if we have 2 moles of gas on the reactant side and 2 moles of gas particles on the product side, and all of a sudden I either...let's say I change the pressure of the system by increasing it, well, there is no direction that the system can actually move in to reduce that pressure change, because both sides have 2 moles.2085

There isn't one side with a...so, basically, the system stays exactly where it is.2108

I have stressed out the system, but the system doesn't have a means of escaping, of alleviating that stress, so it stays there.2112

More often than not, it will--it will have an escape route; it will have an avenue by which to offset the stress.2120

But, it doesn't mean it will have; so you have to watch out for that.2126

OK, so let's do a final example here, and it will sort of be an all-encompassing example: What happens when we do all kinds of things to it?2130

So, let's have the equation example; let's do H2 (I really, really need to write a little bit more clearly, or at least slow down; I know that I tend to write quickly--I apologize) gas, plus I2 gas, is in equilibrium with hydrogen iodide gas.2140

Now notice: I didn't say hydroiodic acid; this is a gas--this is not aqueous.2164

When you take hydrogen iodide gas, and you bubble it in the water, that is when it dissociates into hydrogen ion and iodide ion.2169

That is when we call it hydroiodic acid.2177

This is just hydrogen iodide; it's not acidic--this won't hurt...well, I won't say it won't hurt you, but it won't hurt you as an acid.2179

OK, so now, what happens when H2 gas is added?2187

Well, if H2 gas is added, the system is going to want to move in a direction that actually depletes what I added.2195

It wants to offset it; in order to deplete H2, it has to move to the right, so it's going to shift right.2202

I'm going to use symbols here.2208

Part B: What happens when I2 is removed?2211

Well, if I remove I2, the system is going to want to offset it by producing more I2 that was there.2217

So, it is going to want to pull the reaction back this way, so it's going to shift to the left.2223

It is going to produce more reactant; yes, it will also produce more of this, but the idea is to produce more of this.2227

C: HI is removed.2235

Well, if I remove hydrogen iodide, the system is going to want to offset by creating more hydrogen iodide.2241

How do you create more iodide?--you shift the reaction to the right.2247

You don't do it; the system automatically shifts to the right (I'm sorry; I should be a little bit more precise).2252

You are putting a stress on the system; the system is reacting by moving to the right to offset the stress that you put on it.2257

Part D: Argon is added--argon gas.2266

Well, argon is a noble gas; it is inert; so we add an inert gas.2271

Well, if you are adding an inert gas, you are changing the total pressure of the system, but there is not really much that you can do; so, in this particular case, nothing is going to happen--no change.2276

OK, C, D, E: And...volume is doubled.2296

Well, if we double the volume, it is going to be sort of the same as the argon gas; even though it's inert...2308

Volume is doubled; that means the pressure is decreased, right?--volume is doubled; that means you are making the volume bigger, so you are decreasing the pressure.2317

But, you have 2 moles of particles on the right, and you have 2 moles of particles on the left; there is no place for the reaction to go, to offset the decrease in pressure.2330

It is just going to have a decrease in pressure; so there is no change.2341

OK, now let's do F: let's do a temperature increase.2347

The temperature is increased; OK...oh, I forgot to write...what is the ΔH here?--the ΔH of formation is 25.9 kilojoules.2354

So this is actually endothermic (because we need to know what the ΔH is, in order to decide what happens when we make a change to temperature).2366

So, if this is endothermic, that means that heat is one of the reactants--heat is one of the things required to actually move the reaction.2376

If temperature is increased (that means if I increase the concentration of heat), the system is going to want to offset by decreasing the concentration of heat that I put in; so, it wants to reduce this.2386

The only way to reduce this: the system must move in this direction; so it will shift to the right.2397

There you have it: Le Chatelier's Principle.2406

If you have a system at equilibrium, and you do something to mess with that equilibrium by either changing the concentration of a reactant or product, by changing the pressure of the system, or by changing the temperature of the system; the reaction is going to move in the direction that offsets that change.2410

It is that simple.2436

OK, thank you for joining us here at Educator.com, and thank you for joining us for equilibrium in chemistry.2438

This concludes our discussion of equilibrium.2445

Next time I see you, we will be talking about acids and bases.2448

Take good care; goodbye.2450