AP Chemistry Equilibrium, Part 1

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

Today, we are going to start on what I consider to be probably the most fundamental, the most important, topic of the entire chemistry curriculum.0004

It is the one thing that shows up absolutely everywhere, in all areas of science.0012

It is the concept of equilibrium.0016

Starting from now, we are going to spend...I am certainly going to talk about it in great detail, and in great depth, and we are going to do probably more than the usual number of problems.0018

The reason is because these equilibrium problems (and, as we move on, acid-base and further aqueous equilibria and thermodynamics and electrochemistry and things like that)--the nature of the problems is such that there is no real algorithmic procedure to attack the problems with.0031

I mean, there is--there is certainly a set of things that you can do--but each problem is different, and it is very, very important to actually understand the chemistry.0049

We want the chemistry to lead the mathematics, not the other way around.0058

We don't just want to jump into a problem and start throwing equations together; we want to know what is going on.0061

So, I just wanted to give you that heads-up before we begin; this particular set of topics that we are going to talk about today is going to be more of just introductory, getting you comfortable with the notion of equilibrium, trying to wrap your mind around the concept of equilibrium.0067

Then, with the next lesson and subsequent lessons, we will really dive in and really get into the problem-solving aspect of this.0081

With that, let's go ahead and get started.0088

OK, so let's take a look, initially, at a reaction like this: the reaction of hydrogen and oxygen to form water.0094

2 H₂ + O₂ → 2 H₂O.0102

Now, notice: we wrote this arrow as just one arrow to the right.0107

Well, when this reaction runs forward (so when we put hydrogen and oxygen in a container and then we ignite it), all of the hydrogen and all of the oxygen end up turning to water.0110

We say that the equilibrium of this thing is very, very far to the right.0122

It is true that, at the end, there is a little bit of hydrogen left, and a little bit of oxygen left, but for all practical purposes, there is none left.0125

All of it is completely reacted, which is why the arrow goes in one direction.0132

Most reactions, however--remember, when we talked about kinetics, we said that as a reaction starts to move forward, some of the products start to show up, and some of those products start to break down, and they go back the other way to form reactants.0137

Well, let's take a look at a reaction like this.0150

Nitrogen plus 3 moles of hydrogen gas goes to 2 moles of ammonia.0154

So, if we put nitrogen and hydrogen gas in a container, and we let it start to react under a given set of conditions, it is going to move forward, and it is going to form the ammonia.0161

But, what is going to happen is: after enough ammonia starts to form, there is a certain point at which the ammonia starts to break down, and it starts to decompose into hydrogen gas and nitrogen gas.0171

Well, there comes a point where the forward reaction and the back reaction happen at the same rate.0181

Because they are happening at the same rate, we can no longer measure which one is really moving forward or backward.0186

The reaction is still talking place, so it is still a "dynamic" equilibrium (and I will write that in a minute).0192

But, the idea is that now, the forward and the reverse reaction are taking place at the same rate; that is what we call the equilibrium condition.0198

The system has reached a point, a natural point, for that reaction.0207

So, an equilibrium condition is something that is sort of like a fingerprint for that particular reaction.0212

In a minute, we will give a mathematical expression for that fingerprint, which is specific to that reaction.0218

That is it; it is that simple; equilibrium is when the forward and the reverse reactions are happening at the same rate, and we represent it as a double arrow.0226

Most reactions fall under this category.0234

In fact, all reactions do, but some reactions are so heavily forward, or so heavily in the back, that we just go ahead and, for all practical purposes, we assume that everything has reacted or nothing has reacted.0236

We say the equilibrium is far to the right (meaning all products), or we say it is to the left (all reactants--nothing has happened).0249

So, let's go ahead and actually write this down.0257

Equilibrium is the state where the forward and reverse reactions (rxns) happen at the same rate.0265

They happen at the same rate, so that there is no net reaction movement.0294

It is as if the reaction has come to a stop; there is no more change.0310

In other words, no more NH₃ forms; no more nitrogen diminishes; no more hydrogen diminishes.0313

They reach values that are stable--constant.0319

Let's sort of see what this looks like graphically.0322

We'll make an extra-long graph here; and this axis--let's say we have concentrations; so we'll start with, let's say...no, let's go up here.0328

Let's start with the H₂ concentration; so, it's going to look something like this.0342

OK, and this is just the reaction coordinate--time, in other words--so we'll just go ahead and put "time" here.0350

As time proceeds, the hydrogen concentration drops, drops, drops, drops, and then it levels off; it stays some place.0356

Well, now the nitrogen concentration--it is going to drop, drop, drop, drop, and then it is going to level off.0362

Then, of course, there is the ammonia concentration; if, after a certain amount of time, we take measurements of the ammonia concentration, it is going to rise, rise, rise, rise, and then it is going to level off.0372

Well, this point--where it levels off--that is the equilibrium point; so this is a graphical representation of what equilibrium looks like.0384

Let's stay we start with a flask; hydrogen and nitrogen are in there; and notice, the hydrogen diminishes (right?--it's being used up); the nitrogen diminishes; it's...0394

Let me label these; I'm sorry--so this first one is the H₂gas; this second one is the NH₃ gas; and this third one is the nitrogen gas; OK.0405

The nitrogen diminishes, diminishes, diminishes; and notice that the hydrogen--the slope here is actually steeper than the slope here, which makes sense, because hydrogen is being used up three times as fast as the nitrogen.0424

At any given point, initially, the slope of the hydrogen is going to be three times the slope of the nitrogen.0439

Here, the slope is positive, because there was no ammonia to begin with; but ammonia is forming, so again, this is concentration (that is what this axis is).0446

We are taking measures of concentration; the concentration of ammonia is increasing, increasing; but there comes a point where all three of them reach a value that no longer changes.0455

They just sort of stop; everything is constant at this point; that is the equilibrium condition.0465

The reaction is still going, forward and backward, but it is happening at the same rate, so there is no net movement of the reaction.0469

So, let us write here: Equilibrium is a dynamic process; in other words, it means that it is a dynamic equilibrium.0476

"Dynamic equilibrium" means that what is happening is still happening; it isn't that the reaction has come to a stop, except that it is happening at the same rate, so we don't notice a net process taking place.0492

It is different than sort of a static equilibrium, where...for example, like a rock that is poised some place, let's say out in Utah or something; the rock is not moving, so all of the forces on that rock are balanced.0505

But that is it; there is no...everything is balanced, but nothing is actually happening.0520

In chemistry, in chemical equilibrium, the reaction is still moving forward and backward; it always has to, because molecules are slamming into each other all of the time; but there is no net notice of actually any change.0524

That is what this represents graphically.0538

OK, now, we come to the good part: we have a way of actually representing this equilibrium condition mathematically.0540

So, let's go ahead and describe, for a general reaction that looks like this: aA + bB (and again, from now on, most of the time, we're going to be using the double arrows, because we are going to be talking about equilibrium conditions) to cC + dD.0550

Well, as it turns out, most of science works like this; you run an experiment; you collect a bunch of data; and hopefully, the data--the relationship among the numbers--the data that you have collected--is sort of clear mathematically.0569

But, often that is not the case.0588

Often, what you have to do is use just trial and error--and this is exactly how they did it, back in the old days.0590

They collected a bunch of data; so, for example, they will put a certain concentration of nitrogen, a certain concentration of hydrogen, and they will come back to it after a certain amount of time (once it has reached equilibrium), and they will measure the concentrations, and they will just collect data for different concentrations.0594

Sometimes they will start with all nitrogen and hydrogen--no ammonia; sometimes they will start with only ammonia--no nitrogen and hydrogen; sometimes they will start with a whole bunch of one and a little bit of the other two--all kinds of different things.0613

And then, they measure, at equilibrium, what the concentrations are.0625

Well, once you have these numbers, you have to try to find a way to see if there is some mathematical formula, some equation, that actually fits those numbers.0628

When it comes to data, often we try to find a constant.0638

In other words, is there a relationship among the different values at equilibrium that stays the same?0641

And, as it turns out, there is, and it is called the equilibrium constant--the equilibrium constant expression.0649

So, a reaction of this type, where you have these reactants--A, B, C, and D capitalized are the species; a, b, c, and d small are the coefficients--the equilibrium expression looks like this.0654

We write K_eq; we will also write K; we might not always write "eq," but any time you see a K something, with some subscript, it is an equilibrium kind of constant.0667

It is equal to the concentration of C, raised to the c, the concentration of D, raised to the d; over the concentration of A, raised to the a, and the concentration of B, raised to the b.0678

What that means is that, at equilibrium, if I measure the concentrations of A, B, C, and D, and if I take the concentration of C, raise it to its stoichiometric coefficient, multiply it by the concentration of D, raised to its stoichiometric coefficient (and remember, concentrations--these brackets always mean moles per liter--always), over the concentration of the reactants (A raised to its stoichiometric coefficient and B raised to its stoichiometric coefficient)--as it turns out, this number stays constant.0696

Later, towards the end of this lesson, actually, we will see a wonderful example of that, where we start with a bunch of different concentrations, and the concentrations are different at equilibrium.0727

In other words, the equilibrium position is different, but the constant is the same for every single particular experimental set of conditions.0738

The equilibrium constant is a mathematical representation of the equilibrium; it is a fingerprint for that reaction, at a given temperature.0746

The K is temperature-dependent; so, when you see these problems, it will often give you a specific temperature.0756

It changes with temperature, which makes sense--because, if you remember the Arrhenius equation, the Arrhenius equation says the rate of a reaction is dependent on temperature.0760

Oftentimes, the equilibrium will change as rates change.0769

So, that is it; this probably the single most important thing; if you don't take anything away from chemistry, take this away from chemistry.0774

We wait for a system to come to equilibrium; we measure the equilibrium concentrations and the relationship of the concentrations of the individual species at equilibrium.0782

That is what is important: this is at equilibrium; we cannot emphasize that enough.0791

We will talk about another quotient, which has the same form, but it's not at equilibrium; it is called the reaction quotient; but we use that to decide which way the reaction is moving at a given moment.0795

But, this is at equilibrium; so when you see a K_eq or a K (oops, we don't want these random lines)...you know what, let me just leave it as K_eq for the time being (and let me correct these lines; I don't want you to think that I am cancelling anything out here; this is D...), this is at equilibrium; very, very important.0807

OK, so let's just do a couple of examples.0838

Well, actually, before we do that, let's do some rules regarding the equilibrium constant expression, the K_eq.0841

So, some rules--very, very simple; #1: Pure liquids and solids: if you have a pure liquid or a solid as part of the equation, they don't show up in the equilibrium expression: "as reactants or products, are not included in the K_eq expression."0849

OK, so if you have a solid or a liquid in reactants or products, it is not included in the equilibrium expression.0898

2: Reactants or products which are aqueous (in other words, basically, ionic compounds)...I'll put aqueous ionic, because I think it's important, because we can certainly have covalent compounds that are aqueous, like sugar, which actually do show up in equilibrium expressions, because they are not ionic; so, aqueous ionic...must be expressed in free ionic form.0906

"Must be expressed"--and this is often where kids get into the most trouble, when they are putting together equilibrium expressions.0943

They don't really stop and realize that, when they are looking at a molecular formula, the molecular formula gives the combined ion, like silver nitrate or sodium sulfate.0950

Well, silver nitrate and sodium sulfate, in solution, are free ions, silver ion and nitrate ion.0961

It isn't silver nitrate that shows up in the expression; it's actually the silver ion, the nitrate ion, the sodium ion, the sulfate ion; so it's very, very important.0967

We'll do an example in just a minute to make this perfectly clear.0975

So, reactants or products which are aqueous ionic must be expressed in free ionic form.0978

But you know this already, because you know that, when you are dealing with ionic compound mixtures, you want to write the net ionic reaction, not just the molecular equation, because you have some spectator ions.0984

Spectator ions are ones that don't participate in the chemistry; we want only the things that participate in the chemistry.0993

In other words, we want the net ionic reaction...free ionic form.1000

OK, so let's do an example.1007

Example 1: Write the equilibrium expression (K_eq) for the following.1012

OK, we have A: so, we have: N₂ + 3 H₂ goes to NH₃.1034

And we say "goes to" simply out of habit; again, we are talking about equilibrium now--from now on, it is always going to be some kind of an equilibrium that we are talking about, unless otherwise stated.1046

So, this is gas, and this is hydrogen gas, and the formation is NH₃ gas.1056

Well, the K_eq expression for this is going to be (this is a gas--it's not a liquid or solid--so it does show up in the equilibrium expression)...we have: the concentration of NH₃ (oops, this is a 2 here) raised to the 2 power, because that is the stoichiometric coefficient.1063

It is products over reactants; again, the K_eq is always products over reactants.1082

Then, we have (in the reactant side) the nitrogen gas concentration, raised to its stoichiometric coefficient (which is 1); and then we have the hydrogen gas concentration, H₂, raised to its stoichiometric coefficient.1088

This is the equilibrium expression.1104

At equilibrium, if I measure the ammonia concentration, the nitrogen concentration, and the hydrogen concentration; and if I plug them in to this number; no matter what set of conditions I start off with, the ratio (this expression) will be constant.1107

It is a fingerprint for that reaction at a given temperature; it will not change.1123

OK, B: Here we have an example of an ionic situation: If I take a solution of potassium iodide, and if I mix it with a solution of silver nitrate, the question is: what is going to be the equilibrium expression when I have silver iodide (solid, because that is going to be the precipitate) and potassium nitrate?1129

So again, we are looking at an ionic situation; we are looking at this double-displacement reaction, where the potassium goes with the nitrate and the silver goes with the iodide.1153

The silver iodide drops out and falls to the bottom of the flask as a solid--that is the precipitate--and these stay aqueous.1163

In other words, they are dissolved as free ions.1170

Let me go ahead and write the total ionic equation for this; it's a nice review.1174

Potassium iodide is soluble (and this is balanced, by the way--we have to balance it before we do anything else--standard procedure in chemistry: always balance)1179

K⁺ + I^- (free ions), plus silver⁺, plus NO₃^- (free ions) goes to silver iodide (that is a solid; so a solid--the whole idea behind a precipitate is that it sticks together--water doesn't dissolve it; the bonds between the silver and the iodide are stronger than the bonds that the water might have...the solvation), plus K⁺, plus NO₃^-.1188

Well, K⁺ cancels K⁺ (oh, again with the lines!); NO₃^- goes with NO₃^-; and we are left with a net reaction of the following.1224

We are left with: silver ion, plus iodide ion, going to silver iodide, solid.1234

This is aqueous; this is aqueous.1244

Now, here is the interesting thing about it: we said that solids don't show up in the equilibrium expression; therefore, in the numerator, we just put a 1; that is it.1247

This iodide is in solution; it is aqueous; this silver ion is aqueous; so the equilibrium expression for this (let me actually rewrite the equation, since we are on a new page here: so...): I have: Ag⁺ + I^- goes to AgI, solid.1259

The K_eq expression, equilibrium constant expression--there is nothing over on the product side; so again, it's products over reactants; this is solid--it doesn't show up.1279

It is the only product; so nothing shows up at all in the numerator.1289

On the bottom, we have Ag⁺ to the first power, and we have I^- to the first power.1293

So, it is the net ionic reaction that you use for your equilibrium expression.1300

It's very, very important to do this in any kind of situation that involves some sort of soluble ionic compound.1305

Now, over on the product side, you might have gas formation; you might form water, and you might form CO₂, for example, if you end up mixing a carbonate with an acid.1311

You have to account for all of this; it is the net ionic reaction that plays a part, that gives you the particular equilibrium expression.1324

OK, let's do: 4 NH₃ gas + 7 O₂ gas forms 4 moles of nitrogen dioxide gas and 6 moles of H₂O; and let's just say that this is liquid.1333

So again, we have gas, gas, gas, liquid; liquids don't show up in the equilibrium expression.1363

Therefore, the K_eq for this is going to be...1368

Oh, here is another thing: now that we are talking about equilibrium, I personally get a little...there is a lot of notation involved in this, as you can tell.1373

You have brackets; you have symbols; you have charges; you have lines separating things; you have K_eq; there is a lot of writing.1382

I personally...concentrations are always written with brackets, but since we are talking about equilibrium, and often we will continue to talk about equilibrium, and we are always going to be writing some kind of K: K_eq, K, KP, whatever it is--we are always going to be talking about some equilibrium expression--instead of brackets, I prefer to use parentheses.1391

I hope that that doesn't confuse you.1414

Again, my parentheses here, in our discussion of equilibrium, always represents concentrations.1416

It is just so I can sort of make the writing a little bit faster, because my brackets tend to be a little sloppy.1423

So, K_eq is going to be the NO₂ concentration, raised to the fourth power, over the NH₃ concentration, raised to the fourth power, times the O₂ concentration, raised to the seventh power.1427

Notice, this did not show up in the numerator; it is a liquid; it does not show up.1446

Pure liquids, pure solids--pure liquids and solids--a pure liquid or solid is different that an aqueous solution.1452

You just show that in an aqueous solution, you have free ions floating around; that is not pure--that is just some ions floating around in solution; they are dissolved.1459

By "pure liquid," I mean like, for example, if the reaction said NaCl liquid; that doesn't show up...because NaCl--you can melt salt, and you can actually have a pure liquid salt, but that doesn't mean it actually shows up in the equilibrium expression.1467

So, differentiate pure liquid from aqueous; aqueous--yes; pure liquid--no.1485

OK, let's see what else we have here.1491

D: let's see another example--we have: Calcium carbonate, CaCO₃--this time, this is going to be a solid, and it is in equilibrium with calcium oxide, which is also solid, plus--interestingly enough--CO₂ gas.1497

You don't often see this--solids and gases in the same thing.1514

So, calcium carbonate decomposes into calcium oxide and CO₂.1517

It is a reversible reaction.1521

Well, the K_eq for this: well, this is a solid; we don't care about it; this is a solid--we don't care about it; we have just this, so as it turns out (and this is a balanced equation, so it's not a problem), it just equals the CO₂ concentration.1523

That is it--nice and simple.1541

This equilibrium expression only depends on the CO₂ concentration at a given temperature.1543

In other words, at a given temperature, no matter where I start, with this and this and this, at some point, the CO₂ concentration will always be the same; it will always end up at the same point.1550

It is a constant; it is a fingerprint for this reaction at a given temperature.1563

OK, now, in case you are wondering why it is that solids and liquids actually don't show up in equilibrium constant expressions, it is the following.1572

I can take a gas, and I can change the concentration of it by changing the pressure, volume, temperature...things like that.1580

I can condense it or expand it; the concentration changes; I can change the volume.1586

A solid, whether it is, let's say, 1 gram of sodium chloride or 150 grams of sodium chloride--well, concentration-wise, concentrations of liquids and solids don't change, because the concentration is a measure of the amount per volume.1592

Well, in a solid and liquid, the amount per volume is a constant; so, when something is a constant, it doesn't show up in an expression because it is constant.1609

It is part of the equilibrium constant, if you will; it is part of it...it doesn't show up.1619

That is the reason why: gases and aqueous ions--yes; their concentrations can change.1624

For example, I can take one liter of a liquid; I can drop in a certain amount of salt and dissolve it; now, I have a certain concentration of sodium ions--let's say it's .5 moles per liter.1630

Well, I can drop in more salt; I haven't changed the volume--I have changed the amount.1646

So, now, the concentration has changed; this is why free ions and gases show up in the equilibrium constant expression; their concentrations are not constant.1650

But, for solids and liquids, I can't actually change the concentration of them--they are constant.1661

OK, let's do another example here.1666

Let's continue with using our nitrogen and hydrogen to form ammonia.1669

We have Example 2: we have N₂ + 3 H₂ going to 2 NH₃.1677

OK, for this reaction, the following equilibrium concentrations were recorded at 128 degrees Celsius.1691

So again, most of these equilibrium things--they take place at a certain temperature, because K_eq is temperature-dependent.1718

All right, so for this reaction, the following equilibrium concentrations were recorded at 128 degrees Celsius.1723

What this means is that, at 128 degrees Celsius, we measured the amount of ammonia, hydrogen gas, and nitrogen gas that there was in a flask, and here is what we discovered.1729

We discovered that the ammonia concentration is 3.1x10^-2 Molar (moles per liter--actually, you know what, let me actually write out "moles per liter").1740

I have never really cared for that capital M sign; I actually like to have all of my units absolutely apparent.1755

OK, the nitrogen concentration was measured to be 8.5x10^-1 moles per liter.1761

The hydrogen ion concentration was 3.1x10^-1 moles per liter.1772

This is an equilibrium condition.1781

In a given flask, I measured the concentration of the three species involved in this reaction, and these are the concentrations that I measured.1784

This is at equilibrium--"the following equilibrium concentrations."1792

Now, we want to calculate the K_eq, OK?1796

Well, we know what the K_eq is; we know that it is the products over the reactants, raised to their respective stoichiometric coefficients.1802

So, we have the NH₃ concentration squared, over the N₂ concentration times the H₂ concentration cubed.1811

Well, that equals...the NH₃ concentration is 3.1x10^-2 moles per liter squared (I'm going to leave off the units; I hope you don't mind--units are not altogether that important for equilibrium concentration calculations, because they are going to change, depending on these exponents; so we don't really worry about them all that much).1821

N₂ concentration is 8.5x10^-1.1841

And then, 3.1x10^-1 cubed.1847

When we multiply all of this out and divide, we end up with 3.8x10⁴; that is the K_eq at that temperature for that particular reaction.1853

That is it--nice and simple: measure the concentrations and put it in there--simple, basic math--it's only arithmetic.1873

OK, now, let's calculate...this is...so now, we'll do Part B.1880

Now, let us calculate the following equilibrium: Calculate K_eq for 2 NH₃ going to 3 H₂ +N₂.1889

Notice, this is the same reaction, except it's reversed.1906

So now, I'm going to calculate the equilibrium based on ammonia being the reactant, and hydrogen and nitrogen being the products.1909

Well, again, it is just the K_eq--by definition, it is always the same; it is always going to be products over reactants.1917

So now, we have the H₂ concentration cubed, times the nitrogen concentration, divided by the ammonia concentration squared.1924

Well, you notice that this is just the reciprocal of your other K_eq that we found.1936

Let's call this K_eq...let's call it K prime...how about K_eq prime?1940

That is this one; well, this is equal to 1 over the K_eq; it's just the reciprocal, which makes sense--when we reverse a reaction, we just take the reciprocal of it to find...1949

OK, so it's going to be 1 over 3.8x10⁴, and this one ends up being...let's see what we get; we get 2.6x10^-5.1960

OK, and now we will do one more.1976

Part C is: Calculate the K_eq for this one: 1/2 N₂ (now we're going to change the coefficients) + 3/2 N₂ goes to NH₃.1980

Basically, we have taken the standard balanced equation, the one with the integral coefficients--the 1, the 3, the 2--and we have multiplied everything by 1/2.2002

We have divided everything; it is still balanced, but now, because the equilibrium expression is dependent on the stoichiometric coefficients, it is going to change a little bit.2014

So now, the K_eq expression is equal to N₂...I'm sorry, it's going to be NH₃, divided by the concentration of N₂ to the one-half power...2023

OK, this is not going to work; these lines all over are getting in the way.2043

All right; it always seems to happen at the bottom of the page--it's very interesting.2046

OK, N₂ raised to the 1/2 power, times (oh, this is H₂) the concentration of H₂ raised to the 3/2 power.2051

Well, if you notice what this is--essentially, what we have done...so let's call this K double prime...this is equal to our initial K_eq that we found up there for the standard, balanced reaction--this is equal to K raised to (oh, what is wrong with these lines!?) 1/2.2067

It is as if we have taken...not as if; we actually have; we have taken the equilibrium expression for the original balanced equation, the ones with the integral coefficients, and we have raised it to the power of the thing that we multiplied the equation by.2095

Going from the original equation to this equation, we have multiplied everything by 1/2.2112

Well, that means take the original K_eq and just raise it to a power of 1/2.2117

In other words, take the square root of it; that is what these fractional powers mean.2122

Something to the 1/2 power means to take the square root of it.2126

And, when we do that--when we take 3.8x10⁴ to the 1/2 power, we end up with...oh, you know what, I didn't even do the calculation; that's OK--don't worry about it; just put it in your calculator and plug it in.2130

So that is it; this is the original that we got; we multiplied the equation by 1/2 to get a new equation with new coefficients--still balanced; the K_eq changes accordingly.2150

The K_eq changes by simply taking the original and taking the square root of it.2160

So now, let's write down the take-home lessons of all of this.2165

If you reverse a reaction, that means the new K_eq is the reciprocal of the original K_eq, which makes sense.2167

If you reverse a reaction, you switch products and reactants; now what is on the bottom comes on top, and what is on the top goes on the bottom.2196

Now, if you multiply a balanced equation by a factor m, then your new K is equal to K raised to the power of m; that is it.2203

We are just saying that if you change the equation--do anything to it--the equilibrium expression also changes.2238

That is all that is going on here.2243

OK, let's do our final example, and we will sort of wrap up this basic introduction to equilibrium.2245

OK, Example #3: The following data were collected for N₂ + 3 H₂ going to 2 NH₃ at a particular temperature--the temperature is not specified.2254

OK, so here is what it looks like: let's go ahead and do (no...yes, that is fine...you know what, I am actually going to start the data on another page).2288

So, the following data were collected for N₂ + 3 H₂ going to 2 NH₃ at a particular temperature.2305

And now, I will go ahead and give you the table: this is the experiment; this is the initial concentration; the equilibrium concentration; and the actual K, which we know the expression for.2310

So, we have three experiments: we have one experiment: the N₂ concentration was 1.000, and of course, concentration is in moles per liter.2334

The H₂ concentration is at 1.000, and the NH₃ concentration was 0.2346

Our equilibrium concentration, when we measured it, is 0.921, 0.763, and 0.157.2353

And the K ends up (so this is...now the K is equal to) NH₃ concentration squared, divided by the N₂ concentration times the H₂ concentration cubed; that is our equilibrium expression.2366

The K ends up equaling 6.02x10^-2.2386

When we did Experiment 2, we found the following.2393

We started with an N₂ concentration, an H₂ concentration, and NH₃ concentration, and this time we did 0, 0, and 1.000.2397

I'll go over and discuss what all of this means in just a moment.2410

We ended up with 0.399, 1.197, 0.203.2413

And, when we calculated K, we get 6.02x10^-2.2424

Notice, they are the same, even though these are not.2431

And we did one more experiment, and we started off, this time: nitrogen, hydrogen, and ammonia.2434

We did 2.00; we did 1.00 and 3.00.2444

We ended up with (oops, no, we don't want these stray lines)...we did 2.59; we did 2.77; and 1.82.2451

And, when we...6.02x10^-2.2465

OK, so here is what is going on: we did three experiments.2470

The first experiment: what we did is, we measured initial concentrations; so we put in a flask 1 mole per liter of nitrogen gas, 1 mole per liter of hydrogen gas, and no ammonia.2474

We let the system come to equilibrium; we came back once it reached equilibrium--once we realized that these values are not changing; everything is constant--and we measured them.2486

It turns out that the nitrogen concentration was .921; hydrogen concentration was .763; and the ammonia concentration was 0.157.2497

We put these values, because they are equilibrium concentrations, into the equilibrium expression: ammonia squared over nitrogen times hydrogen cubed (let me rewrite the nitrogen to make it a little more clear here--we are dealing with N₂).2506

And what we did was got the number 6.02x10^-2.2520

That is fine; equilibrium concentrations--we stick it into the expression; we get a certain number.2525

Now, a whole different experiment: this time, the initial concentration of the nitrogen and hydrogen was 0.2529

We just put in ammonia in the flask, and we let the system come to equilibrium.2536

In other words, the ammonia started decomposing; now, nitrogen and hydrogen are forming.2542

At equilibrium, when the concentrations were constant and nothing was changing anymore, we measured the concentrations.2546

Well, now, the nitrogen is .399; hydrogen is 1.197; .203; completely different equilibrium conditions--completely different; we started differently, and we ended at different concentrations.2552

But notice, when we put these numbers into this expression, 6.02x10^-2: the same number shows up.2564

Hmm, interesting pattern.2572

Experiment #3: This time, we start with a little bit of everything.2574

We start with 2 moles per liter of nitrogen, 1 mole per liter of hydrogen, and 3 moles per liter of ammonia.2578

Well, we let the system come to equilibrium; all the concentrations are now constant--no longer changing.2585

We measure it; we get this, this, this; we put this into the equilibrium expression, and what do you know--6.02x10^-2.2591

This is proof that these things right here--these are equilibrium conditions; equilibrium conditions depend on the particular experiment that you are running at any given moment.2599

These individual numbers are different among the experiments, but the relationship (based on the equilibrium expression) of the products raised to their coefficients (stoichiometric coefficients), divided by the reactants raised to their stoichiometric coefficients--that number stays constant.2611

This is very profound: any time you have some data, and data which should not have anything to do with each other, when you mix and match and when you keep getting the same number over and over again, you have something very, very special there.2629

That is the whole idea; much of science is based on trying to elucidate some sort of a constant--what stays?--and this is what it is for the equilibrium condition.2642

At this temperature, this reaction--this is a fingerprint of that reaction.2652

Equilibrium does not change; equilibrium conditions change; the equilibrium constant doesn't change.2661

That is why we call it a constant; it is a fingerprint for that reaction.2667

OK, so last but not least, let's distinguish between the equilibrium constant and the equilibrium position.2672

These are equilibrium positions that vary, depending on the experiment.2681

This is the equilibrium constant, based on this expression; it does not change--it is not variable.2685

So, we speak about constants, and we speak about conditions--very, very simple; very important.2692

OK, so hopefully this has given you a sense of what equilibrium is.2698

Again, a profoundly important topic: those of you that go on into biology, biochemistry, biophysics...things like that--the equilibrium condition (or physics--any kind of science)...all systems tend toward some kind of an equilibrium.2704

It is the reason we are alive, because our bodies are maintaining--constantly are trying to maintain--a certain equilibrium.2722

In fact, if you want to give sort of a broad definition of disease, a broad definition of disease is a deviation from equilibrium; that means something is going wrong.2729

The body is trying to bring things back to equilibrium; equilibrium is where everything wants to be--the point of lowest energy.2738

We can express that mathematically, with chemical reactions, with these expressions.2746

It is actually quite amazing that it is this simple.2750

Thank you for joining us here at Educator.com and for AP Chemistry.2753

We will see you next time for a continuation of equilibrium; goodbye.2756