AP Chemistry Equilibrium, Part 2

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

We are going to continue our discussion of equilibrium--in my opinion, absolutely the single most important concept, not just in chemistry, but in all of science.0004

Equilibrium is the thing that all systems tend to--this notion of balance; they don't like being off balance (too much of this, too much of that).0013

All systems tend toward equilibrium, and chemical systems are no different.0023

Last lesson, we introduced the notion of an equilibrium expression, this K_eq where, beginning with any sort of concentrations of a particular reaction, once the system has come to equilibrium, we measure those concentrations at equilibrium.0027

We put it into the equilibrium expression, and it ends up being the constant.0043

Different equilibrium conditions=different values; but the relationship among those values is a constant.0047

Today, we are going to continue that discussion, and we are going to introduce a variation of the expression for reactions that involve a gas.0054

As it turns out, when you have a reaction that involves a gas, you can either work in moles per liter (like we did last time--you are certainly welcome to do so by taking the number of moles in a flask and the volume of the flask, and dividing to get your concentration in moles per liter) or, as it turns out, pressure--you can work with pressures, because pressure is actually a measure of concentration.0062

We will show you how, mathematically; it is actually very, very nice.0087

Let's get started.0090

OK, so now, let's just go ahead and write, "For equilibriums involving gases, the K can be expressed in terms of P, pressure."0112

What that means is that I can either express my equilibrium expression with moles per liter concentrations or with pressure.0135

I'm going to write the two expressions, and then we will see how they are actually related, because there is a relationship between the two, which allows us to go back and forth, depending on what the problem is asking.0142

Because sometimes, it might be easier to deal with pressure--sometimes, easier with concentration; it just depends.0153

So again, we are going to stick with our tried-and-true nitrogen, plus 3 moles of hydrogen gas, going to 2 moles of ammonia.0158

We know that the actual K_eq expression in concentrations is: the concentration of ammonia squared, over the concentration of nitrogen times the concentration of hydrogen cubed.0170

Well, as it turns out, I can also express it this way: and now, I put a P down as a subscript to the equilibrium expression--P for pressure.0189

K_P--it's the same thing, and it equals the partial pressure of NH₃ (partial pressure means the pressure just of that gas in the container, because you have three gases in a container: one gas has one pressure, on has another pressure--we call those the partial pressures), squared (so again, everything is the same; it's products, over reactants, raised to the stoichiometric coefficients; the pressure of NH₃, squared...), because this is a 2, divided by the pressure of nitrogen gas, raised to a 1 power, because this stoichiometric coefficient is 1, times the partial pressure of H₂ cubed, because its stoichiometric coefficient is 3.0198

So now, let's talk about how these two are related.0240

Well, whenever we talk about gases, the one equation that we talk about is the PV=nRT expression.0244

Let me just erase a little bit of this and give me a little more room.0251

Pressure times volume equals (let me make my V a little more clear) the number of moles, times the gas constant, times the temperature in Kelvin.0254

Let's rearrange this a little bit.0267

I'm going to write P=nRT/V.0269

Now, I'm going to take two of these and combine them: n/V times RT: just a little bit of mathematical manipulation--I can do this.0275

I can take the denominator and put it with one of the numbers in the numerator.0286

Now, notice what I have: what is n/V?--n is the number of moles, and V is volume in liters; so, as it turns out, P and n/V are related by a constant, RT.0289

As it turns out, pressure is an alternative form of representing concentration in moles per liter.0308

n/V is just moles per liter, so this says, if I have something which is a certain number of moles per a certain number of liters, if I multiply that by RT, I actually get the pressure.0314

Therefore, instead of concentration, I can just put the particular pressure.0326

That is all this is: in dealing with gases, it is often difficult to...not difficult to deal in concentrations; it tends to be easier, simply by nature of gases, by measuring pressure.0330

That is all this is; so, whenever we have an equilibrium condition that involves some gases, we can use its pressure, because pressure is just an alternative form of concentration; that is it.0342

Heads, tails--it's just another way of looking at concentration.0353

Now, let's see what the actual relationship is between these two values.0357

OK, so P is equal to n/V(RT); so P is equal to concentration times RT (we will just use C as concentration, instead of writing it as n/V) or, we can write C=P/RT: concentration is equal to pressure over RT.0361

This is just standard mathematical manipulation.0383

So now, this is concentration; we are going to rewrite the expression here, and we are going to put concentration back in to see how it is related here.0386

We will do: K_eq is equal to...it's pressure over RT, the concentration: so we get the partial pressure of NH₃, over RT, squared.0397

So all I have done is: I have just taken this expression, put it into here, and divided by pressure of N₂/RT to the first power, times the pressure of H₂/RT (that is concentration squared, concentration raised to the first, concentration cubed, right?)--just basic math.0413

And then, I have partial pressure of NH₃, squared, times 1/RT, squared, over the partial pressure of N₂ times 1/RT, times the partial pressure of H₂, cubed, times 1/RT, cubed--so far, so good.0443

I have: partial pressure of NH₃, squared, over partial pressure of N₂, partial pressure of H₂, cubed, times 1/RT, squared, over 1/RT to the fourth power.0474

All right, and now, I am going to have...this 2 is going to cancel that; I'm going to end up with (let me write it one more time--well, two more times, actually): P_NH3, squared, over P_O2, times the P_H2, cubed, times 1/(1/RT, squared) (because this cancels two of those, leaving that) and 1/(1/RT squared) is RT squared.0497

So, it equals the partial pressure of NH₃, squared, over the partial pressure of N₂ and the partial pressure of H₂, cubed, times RT, squared.0542

Well, this is the K_eq; that is this expression.0562

This expression right here is the K_P (RT)².0567

So as it turns out, K_eq equals K_P, in this particular case, times RT squared.0572

There is a relationship between expressing it with concentrations and expressing it with pressures.0584

The relationship is that the K_eq equals K_P (for this particular reaction--we'll do the general one in a minute), times RT squared.0590

I can go back and forth between the two; so, if I'm working with K_P and I want K_eq, in concentrations, I can do that.0599

If I have K_eq and I want K_P, I can do that by multiplying by the square of RT.0605

OK, so now, let's do the general version; I just wanted to show you where the math came from.0612

For the general expression, aA + bB in equilibrium with cC + dD, our relationship is the following.0618

From now on, I'm not going to put the K_eq; I'm just going to put K; whenever you see K with no subscript on it, it just means moles per liter, concentration; K equals K_eq.0630

So, K is equivalent to K_eq; when we speak of P, we will write K_P; that means we are dealing with pressures.0641

When we just see a K, without this eq, they are the same--simply to avoid writing the eq over and over again.0651

As it turns out, the K is equal to K_P, times RT to the negative Δn, where Δn equals the sum of the coefficients of the products, minus the sum of the stoichiometric coefficients of the reactants.0659

We can write it that way; or, another way that it is written is in terms of K_P.0681

K_P is equal to K times (RT)^Δn; either one of these is fine.0686

If you have K, you can find K_P; if you have K_P, you can find K by just using this expression; that is it--nothing more than that.0696

Δn is this coefficient plus that coefficient, minus that plus that; that is all.0705

Let's just do an example, and everything will make sense.0711

Example 1: OK, at 427 degrees Celsius, a 2.0-liter flask contains 40.0 moles of H₂, 36.0 moles of CO₂, 24.0 mol of H₂O, and 11.8 mol of CO, carbon monoxide, at equilibrium.0718

So, at equilibrium, we measure 427 degrees; the system has come to equilibrium; we measure in a 2-liter flask; we find that we have this many moles of each of these species.0766

Now, the reaction is as follows: CO₂ gas + H₂ gas goes to carbon monoxide gas + H₂O gas.0777

All of these are gaseous species: that means all of them are involved in the equilibrium expression.0797

Now, our task here is to find K and K_P; so, find the equilibrium constant expression, in terms of moles per liter, and find the K_P constant, in terms of partial pressure.0804

Well, we have the relationship, so we can find K, and then we can find K_P, based on this--no problem.0824

OK, so now, this is our first example of a real equilibrium problem, in the sense that we really have to watch everything that is going on.0833

No two problems are going to read the same way; we can't follow an algorithmic procedure for solving every problem.0842

Physical systems--now, it just depends; different things can happen--you can have different data.0850

It can be worded in a certain way; you have to be able to extract the information that is necessary.0858

Yes, there are certain things that are universal, that you can always count on, but you have to watch every single little thing.0865

Notice here: they are giving you the volume of a flask, and they are giving you the amounts in moles; so you actually have to calculate the moles per liter, before you put it into the equilibrium expression, because the equilibrium expression requires that it be in concentrations and not in moles.0872

You have to sort of watch for that.0888

Let's write the equilibrium expression, K; we will do K first.0892

It is going to equal the products over the reactants, raised to their stoichiometric coefficients.0895

This is a balanced reaction, so everything is 1:1:1:1.0900

We have the concentration of CO, times the concentration of H₂O (everything is a gas), over the concentration of CO₂, times the concentration of H₂ gas.0904

Well, the concentration is the number of moles per volume; our volume is 2 liters (I'm going to go ahead and do this in red), and our moles are 40, 36, 24, and 11.8.0915

So, CO--the concentration of CO, moles per liter of carbon monoxide--is 11.8; so this is going to be 11.8 moles per 2 liters.0932

I take the number of moles and divide by the liters; it has to be concentration.0945

Times the concentration of H₂O: H₂O is 24 moles, so it's 24.0 divided by 2.0950

The concentration of CO₂: well, CO₂ is 36 moles, 36.0 moles; it is sitting in a 2-liter flask, so its concentration is 36/2, or 18 moles per liter.0959

And now, the H₂ is 40 moles in a 2-liter flask.0973

Now, you might think to yourself, "Well, wait a minute; 2, 2, 2, 2: don't the 2's cancel?"0977

Yes, in this case they cancel; that is because all of these coefficients are 1, 1, 1, 1.0981

But, you can't guarantee that all of the coefficients are 1, 1, 1, 1, so you can't just use the mole values in here.0989

The equilibrium expression explicitly requires that you use concentrations, moles per liter, so let's just use moles per liter.0996

Yes, they will cancel, but at least we know what is going on; we won't lose our way.1003

That is what is important: write down everything--don't do anything in your head; don't cut corners; don't take shortcuts.1007

I promise, it will go badly for you.1013

OK, when we calculate this: 0.197; that is what we wanted--we wanted the K.1015

Now, we want the K_P, because that is the other thing that we have to find.1026

Well, we know that K_P is equal to K times RT^Δn.1029

Well, what is Δn?--Δn is adding the product coefficients and subtracting the other coefficients.1036

Well, Δn is (let me go ahead and write it here) equal to 1+1 (is 2), minus 1+1 (2), is equal to 0.1045

So, in this case, Δn is 0.1056

K_P is equal to K, times RT, to the Δn, equals K, times RT to the 0 (anything raised to the 0 power is 1); is equal to K.1058

So, in this case, K_P is equal to 0.197...in this case; and the only reason it is this way is because everything is in a 1:1:1:1.1079

2 reactants, 2 products: 1+1 is 2; minus 1+1 (is 2) is 0; that is it.1091

Watch what you are doing very, very carefully; do not cut corners on equilibrium--do not cut corners ever, in any problem that you do.1098

Write everything out; it's very, very important.1104

OK, let's see: now, let's do a slightly more complicated problem, and again, this is going to be an example of taking information that is written in a particular problem and reasoning it out.1107

It isn't just the math--the math is actually pretty simple once you know what is going on.1125

That is the biggest problem with chemistry, or physics, or anything else: it's not how to turn it into math; it is, "What is going on, so that I know which math to use?"1130

That is the real issue, and no two problems are the same; no two problems will be worded the same; you cannot count on that, especially at this level.1139

OK, so Example 2 (and again, we are going to do a lot of problems, like I said, from here on in--from equilibrium all the way through at least electrochemistry, because this is the heart and soul of chemistry, not to mention the free-response questions that you are going to face on the AP exam)...1148

OK, so the question is a bit long: A sample of gaseous PCl₅ (phosphorus pentachloride) was placed in an evacuated flask so the pressure of pure PCl₅ would be 0.5 atmospheres.1172

So, a sample of gaseous phosphorus pentachloride was placed in an evacuated flask so the pressure of the pure PCl₅ would be .5 atmospheres.1221

We stuck it in there so that it would be .5 atmospheres.1229

But, PCl₅ decomposes according to: PCl₅ goes to PCl₃ + Cl₂.1232

The final total pressure in the flask was 0.84 atmospheres at 250 degrees Celsius.1259

Calculate K at this temperature.1287

OK, so let's make sure we understand exactly what this problem is asking.1298

Problems are going to be very, very specific; do not read into it anything that is there--read exactly what is there--very, very important.1302

Don't cut corners.1308

A sample of gaseous PCl₅ was placed in an evacuated flask so the pressure of pure PCl₅ would be .5 atmospheres.1310

In other words, they filled it up with gas, and the pressure of the PCl₅ gas was .5 atmospheres.1316

The problem is: "But, PCl₅ decomposes"--in other words, the PCl₅ they put there at .5 atmospheres--all of a sudden, it starts to come apart.1322

It decomposes into PCl₃ and Cl₂.1331

Well, now there is an equilibrium that exists; now, you not only have PCl₅ at .5 atmospheres; now, you have also produced some phosphorus trichloride gas and some chlorine gas, and there is also some PCl₅ gas left over.1334

This is an equilibrium expression; so now, it is becoming a little bit more complicated.1347

Now, you have three things in the flask, when you only introduced one.1351

We measure the final pressure of the flask, and it comes out to .84 atmospheres at 250 degrees Celsius.1354

Calculate the K at this temperature (K, we said, is K equilibrium, moles per liter).1363

They have given this to us in pressures; so, the first thing we have to do is: we have to find K_P and then calculate K.1369

So, we want to find (well, if it's OK, I'm not going to write out everything; we know what we are doing)...so notice: they didn't write and say "Find K_P."1379

They said, "Find K"; but the problem as written, since we are dealing with atmospheres and gases (PCl₅ gas, PCl₃ gas, Cl₂ gas)...we are going to deal with pressures, and then from pressures, we are going to use the RT and Δn expression that we just worked with to find the K.1387

Make sure that you understand what it is that they are asking for.1404

Don't just stop by getting the K_P.1407

OK, let's see how we are going to do this.1411

We are going to introduce something called an ICE chart; and this ICE chart is something that we are going to use for absolutely the rest of the time that we discuss this.1414

All equilibrium problems--acid-base problems, further aspects of acid-base equilibria, solubility-product equilibria, electrochemistry--we are going to deal with these things called ICE charts.1422

This is sort of an introduction to them, and it is a way of dealing with what is going on in the problem.1433

If you understand these, everything should fall out naturally.1438

OK, so let's write the equation.1442

PCl₅ is in equilibrium with PCl₃ + Cl₂.1445

ICE stands for Initial concentration, before anything happens; C stands for Change--what changes take place; and E stands for Equilibrium concentrations.1452

Well, it is these equilibrium concentrations that go into the equilibrium constant expression, right?1464

That is what we said: the K_eq: those values that we put in there are concentrations at equilibrium.1469

I means Initial; C means change.1475

Even if you are not sure what to do in a problem, just start, and just write down what you know; write down what is happening; eventually, the solution will fall out.1482

The single biggest mistake that kids make is: they think that they are supposed to just look at a problem and automatically know what is going on.1491

Even I don't just look at a problem and know what is going on!1497

After all of these years of experience, I have to sit there and stare at it, and sometimes just see where I am going.1500

ICE chart is a great place to start with equilibrium problems.1506

It will give you a sense of what is happening; then, you can put the math together.1511

Don't ever feel that you have to just know what is happening; you are extracting information.1514

And...sorry about that--E stands for equilibrium.1520

OK, so our initial concentration of PCl₅ was .5 atmospheres.1528

Remember what we said: pressure and concentration--they are the same; they are just different sides of the same coin, so we can deal in pressures the same way we deal with concentrations.1534

We start with 0.5 atmospheres, where, before anything happens--before PCl₅ decomposes--there is no PCl₃, and there is no Cl₂; so this is our initial condition.1544

It's nice; well, a certain amount of PCl₅ decomposes.1555

Well, look at our equation; it's 1:1:1.1561

For every 1 mole that decomposes, 1 mole is produced, and 1 mole is produced of the Cl₂.1563

We can say that, if -x is the amount that disappears (or, again...concentration and pressure are the same thing)...so, if the pressure drops--PCl₅--by a certain amount, that means it has to increase here by that same amount for each.1571

For every 1 atmosphere that drops, that means a certain amount has been used; well, that means that a certain amount has been produced of the PCl₃ and the Cl₂.1591

That is what the stoichiometry tells me.1598

That is why I have -x, +x, +x; I hope that makes sense--because, again, when this decomposes, this is forming; that is what is going on.1600

So, if 1 mole of this decomposes, 1 mole of this is formed; 1 mole of this is formed.1610

If 5.2 moles of this decomposes, 5.2 moles of PCl₃ is formed; 5.2 moles of Cl₂ is formed.1614

We don't know how much is formed yet, so that is why we use x: -x here, +x, +x.1621

Now, we add: the initial plus the change gives us the equilibrium condition.1627

At equilibrium, I have 0.50-x.1633

Here, I have 0+x is x; 0+x is x; so at equilibrium, this is how much I have.1637

Now, they want us to find K; well, what other information do they give us?1645

They give us the fact that our total pressure is equal to .84 atmospheres.1651

Well, our total pressure is the pressure at equilibrium: this plus this plus this.1657

The total pressure is the sum of the individual pressures; so, we write: 0.50-x+x+x (write it out; don't do it in your head first) =0.84 atm.1665

Well, this -x cancels with that x, and I'm left with 0.50+x=0.84; this is simple arithmetic--there is nothing hard about this.1682

x=0.84-0.5; it equals 0.34 atmospheres; look at that.1693

I have just solved for x, x, x; I did it; that is nice.1701

I found the value of x: .34 atmospheres of PCl₃ show up; .34 atmospheres of Cl₂ show up; and .34 atmospheres of PCl₅ is lost.1710

So now, we can actually find our K.1724

That is the best part, which is ultimately what we want; so we are solving for K_P here.1727

So now, I have partial pressures: I have the partial pressure of chlorine gas, is equal to x, which is 0.34 atm; I have the partial pressure of the PCl₃ gas, which, again, is x, which is 0.34 atm; and I have the partial pressure of the PCl₅ gas, which is 0.50-x (that is the equilibrium) minus 0.34, which is 0.16 atm.1732

Now, I can put these values into my equilibrium expression.1767

I know what my equilibrium expression is; my equilibrium expression is K_P equals the partial pressure of Cl₂, times the partial pressure of PCl₃, divided by the partial pressure of PCl₅, each raised to the first power, because the stoichiometric coefficients are 1.1772

Well, that equals (let's go down here) 0.34, times 0.34, divided by 0.16; I do the multiplication, and I get 0.72; this is my K_P; K_P equals 0.72.1792

They didn't ask for K_P; they asked for K, which means they asked for K_eq.1815

Well, the relationship between K and K_P is the following.1822

K is equal to K_P times (RT)^Δn, if that is correct; let me double-check; K_P equals...-Δn.1826

OK, so we have...let's see...yes, OK; so now, what is Δn, first of all?1847

Δn is equal to...well, we take the...again, let's write out the equation, so we have it on this page.1862

We have PCl₅ in equilibrium with PCl₃ plus Cl₂; Δn is equal to 1+1-1; 2-1=1.1871

So, K is equal to K_P (which is 0.72), times RT...OK, so here is where we have to be careful; R is not...well, the R that we are going to be using here is 0.08206; that is liter-atmosphere/mole-Kelvin; times temperature in Kelvin; it has to be in Kelvin, because the unit of this is liter-atmosphere...1887

You know, let me write out the units here, so you can see it.1920

This is liter-atmosphere per mole-Kelvin; so the temperature has to be in Kelvin.1925

At this particular temperature, add 273; you get 523 Kelvin to the -1 power (negative Δn; that was the relationship).1935

You put in K_P; R is .08206; temperature in Kelvin is 523; negative Δn...Δn was 1; negative 1.1948

We do the math, and we end up with 0.017.1957

There you go; in this particular problem, there was a lot going on; we handled it by just sort of stopping, taking a look, and making sure we wrote everything out.1966

We introduced this thing called an ICE chart; we write out the equation on top; underneath, we write the initial concentrations, we discuss the changes that take place, and we add to get our equilibrium.1976

From there, we take a look at what the problem is asking.1988

Sometimes, they might give us a K, and they might ask for a particular concentration.1991

I take those equilibrium concentrations, and I put them into my expression, and I solve that way.1996

In this case, they gave me a total pressure so I could find x.2001

I used that to find the K; it just depends on what they are asking.2005

This is why we are going to do a lot of different types of problems for these equilibrium and so on...at least through electrochemistry.2009

OK, so let's do another example.2018

Example 3: We have: Solid ammonium chloride, solid NH₄Cl, was placed in an evacuated chamber, then heated; it decomposed according to: NH₄Cl, solid, decomposes into ammonia, NH₃, gas, plus hydrogen chloride gas.2027

It's not hydrochloric acid; it's hydrogen chloride gas.2086

Now, after heating, the total pressure was found to be 4.4 atmospheres.2090

Our task is to calculate the K_P, the equilibrium constant with respect to pressures.2119

OK, well, let's write out the equilibrium expression first; it's always a great thing to do--write everything out.2125

Write out the equilibrium expression; write out the ICE chart; and then, just see where you go from there.2132

So, the equilibrium expression here, based on this equation--well, we have a gas; we have a gas; and we have a solid.2137

Solids don't show up in the equilibrium expression, so in this case, it is just these two.2143

The coefficients are 1, so we have: the K_P is equal to the partial pressure of NH₃ gas, times the partial pressure of HCl gas.2148

That is it; if we find the partial pressures, we are done--we plug them in, we multiply them, and we are done.2157

OK, now let's do our ICE chart.2163

Do our ICE chart: always do it this way.2165

NH₄Cl (write out the equation, and do it underneath; don't do it separately--you want to be able to keep everything straight) goes to NH₃ + HCl.2169

Our initial concentration; our change; and our equilibrium concentration...2183

OK, solid NH₄Cl--we don't care; it doesn't even matter, so we just put lines there; it doesn't show up in the equilibrium expression--it doesn't matter.2188

Our initial NH₃ and HCl concentration--well, we started off with solid NH₄Cl; so, initially, there is none of these.2198

The change--well, a certain amount shows up; that certain amount is what we want.2208

So, +x+x, right?--so again, you have to be able to see what the question is saying.2214

It is telling you that you start off with NH₄Cl; it decomposes--when something decomposes, that means it is going away; the products are showing up.2220

That is why you have a +x and a +x here.2229

This + goes here; it's not that +.2235

0+x is +x; 0+x is +x; well, they are telling me that the total pressure in this flask is 4.4 atmospheres.2239

Well, which gases are in the flask?2251

Well, the solid is a solid; that doesn't matter--that doesn't do anything for the gas.2256

The gases in here are NH₃ gas and HCl gas.2261

So, I basically have: x+x=4.4; they are telling me that the total pressure in there is 4.4 atmospheres; that has to be made up of the amount of NH₃ gas and the amount of HCl gas.2265

Well, that is even, because they are forming a 1:1 ratio.2283

So, 2x is equal to 4.4; x is equal to 2.2; well, 2.2--there you go; that is the partial pressure of HCl and the partial pressure of NH₃.2287

I have 2.2 and 2.2 (2.2, not 2.4...oh, numbers and arithmetic!); these numbers are the one that I put back in here.2307

So, K_P is equal to 2.2, times...OK, now watch this: even though 2.2 and 2.2 is the same, please don't write 2.2².2319

They are different species; I promise you, if you write 2.2², and if somewhere along the way you get lost and you have to come back, you will spend five minutes trying to figure out what happened, because again, stoichiometric coefficients--they show up in the equation, so write them separately; write 2.2 times 2.2.2332

Even though they are the same, they represent different species.2352

Don't mix them up.2355

We multiply that; we get 4.84; 4.84--that is the K_P, and we double-check: "Calculate K_P"--that is what we wanted; we're done; that is it.2357

So, they gave us a certain amount of information; they gave us an equation to work with; we wrote down the K_P expression; we wrote down the equation.2370

We wrote an ICE chart (Initial, Change, Equilibrium--ultimately, it is the equilibrium that we are concerned with, because the system has come to equilibrium).2380

We followed it; we get x and x.2388

They tell us that the total pressure is 4.4; well, the total pressure is the sum of the individual pressures.2392

There are only 2 gases in here (the NH₃ and the HCl); each one is x.2397

2x=4.4; x=2.2; at equilibrium, 2.2 atmospheres is hydrogen chloride gas; 2.2 atmospheres is ammonia gas.2401

You plug that into the equilibrium expression; we multiply, and we get 4.84: a standard equilibrium problem.2413

OK, this sort of gets us going with the types of problems we are going to be dealing with with equilibrium.2419

We are going to systematize this, and this whole idea of the ICE chart--our problems are going to start to become a little bit more complex, but this whole idea of using an ICE chart, writing the equilibrium expression, and then seeing what they want, based on what they give us--the ICE chart itself is going to be different.2425

That is what is going to change.2443

The approach does not change.2445

OK, so thank you for joining us here at Educator.com for AP Chemistry and equilibrium.2448

We will see you next time; goodbye.2452