For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

### Equilibrium, Part 2

- The equilibrium expression can be expressed in terms of pressure instead of concentration. A mathematical expression exists relating the two constants.
- A chart known as an ICE chart is used to solve most equilibrium problems. ICE stands for Initial, Change, Equilibrium.

### Equilibrium, Part 2

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
- Equilibrium 1:31
- Equilibriums Involving Gases
- General Equation
- Example 1: Question
- Example 1: Answer
- Example 2: Question
- Example 2: Answer
- Example 3: Question
- Example 3: Answer

### AP Chemistry Online Prep Course

### Transcription: 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 NH_{3} (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 NH_{3}, 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 H_{2} 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 NH_{3}, 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 N _{2}/RT to the first power, times the pressure of H_{2}/RT (that is concentration squared, concentration raised to the first, concentration cubed, right?)--just basic math.*0413

*And then, I have partial pressure of NH _{3}, squared, times 1/RT, squared, over the partial pressure of N_{2} times 1/RT, times the partial pressure of H_{2}, cubed, times 1/RT, cubed--so far, so good.*0443

*I have: partial pressure of NH _{3}, squared, over partial pressure of N_{2}, partial pressure of H_{2}, 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 NH _{3}, squared, over the partial pressure of N_{2} and the partial pressure of H_{2}, 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 H _{2}, 36.0 moles of CO_{2}, 24.0 mol of H_{2}O, 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: CO _{2} gas + H_{2} gas goes to carbon monoxide gas + H_{2}O 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 H _{2}O (everything is a gas), over the concentration of CO_{2}, times the concentration of H_{2} 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 H _{2}O: H_{2}O is 24 moles, so it's 24.0 divided by 2.*0950

*The concentration of CO _{2}: well, CO_{2} 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 H _{2} 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 PCl _{5} (phosphorus pentachloride) was placed in an evacuated flask so the pressure of pure PCl_{5} 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 PCl _{5} would be .5 atmospheres.*1221

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

*But, PCl _{5} decomposes according to: PCl_{5} goes to PCl_{3} + Cl_{2}.*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 PCl _{5} was placed in an evacuated flask so the pressure of pure PCl_{5} would be .5 atmospheres.*1310

*In other words, they filled it up with gas, and the pressure of the PCl _{5} gas was .5 atmospheres.*1316

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

*It decomposes into PCl _{3} and Cl_{2}.*1331

*Well, now there is an equilibrium that exists; now, you not only have PCl _{5} at .5 atmospheres; now, you have also produced some phosphorus trichloride gas and some chlorine gas, and there is also some PCl_{5} 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 (PCl _{5} gas, PCl_{3} gas, Cl_{2} 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

*PCl _{5} is in equilibrium with PCl_{3} + Cl_{2}.*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 PCl _{5} 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 PCl _{5} decomposes--there is no PCl_{3}, and there is no Cl_{2}; so this is our initial condition.*1544

*It's nice; well, a certain amount of PCl _{5} 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 Cl _{2}.*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--PCl _{5}--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 PCl _{3} and the Cl_{2}.*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 PCl _{3} is formed; 5.2 moles of Cl_{2} 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 PCl _{3} show up; .34 atmospheres of Cl_{2} show up; and .34 atmospheres of PCl_{5} 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 PCl _{3} gas, which, again, is x, which is 0.34 atm; and I have the partial pressure of the PCl_{5} 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 Cl_{2}, times the partial pressure of PCl_{3}, divided by the partial pressure of PCl_{5}, 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 PCl _{5} in equilibrium with PCl_{3} plus Cl_{2}; 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 NH _{4}Cl, was placed in an evacuated chamber, then heated; it decomposed according to: NH_{4}Cl, solid, decomposes into ammonia, NH_{3}, 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 NH_{3} 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

*NH _{4}Cl (write out the equation, and do it underneath; don't do it separately--you want to be able to keep everything straight) goes to NH_{3} + HCl.*2169

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

*OK, solid NH _{4}Cl--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 NH _{3} and HCl concentration--well, we started off with solid NH_{4}Cl; 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 NH _{4}Cl; 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 NH _{3} 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 NH _{3} 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 NH _{3}.*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 2.2^{2}.*2319

*They are different species; I promise you, if you write 2.2 ^{2}, 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 NH _{3} 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

1 answer

Last reply by: Professor Hovasapian

Wed Nov 26, 2014 12:32 AM

Post by Shih-Kuan Chen on November 26, 2014

Dear Professor,

Is it right to say that the ICE method discusses the partial pressures of each type of gas?

I, standing for "Initial Pressure," C, standing for "Change in Pressure," and E, standing for "Equilibrium Pressure?"

1 answer

Last reply by: Professor Hovasapian

Sun Jul 27, 2014 4:25 AM

Post by Jessica Lee on July 25, 2014

For 7:36 is it suppose to be a brackets or is it Parentheses ?

1 answer

Last reply by: Hong Zhao

Mon May 5, 2014 7:14 PM

Post by Hong Zhao on May 5, 2014

Why the initial concentration of PCL5 is 0.5? Thank you so much.

0 answers

Post by Winnie Hu on January 2, 2014

can somebody help my answers

0 answers

Post by Winnie Hu on January 2, 2014

the reaction 2NO2(g)=4NO(G)

has an equilipbrium constant of 4.5X10^3 at a certain temperature.

what is the equilibrium constant of 2N204(g)=4NO2(g)

0 answers

Post by Marian Iskandar on September 2, 2013

I love the ICE method...makes it so much easier to break down the problem, and work towards a solution. Great teaching, Professor!

1 answer

Last reply by: Professor Hovasapian

Wed Mar 20, 2013 2:46 AM

Post by Joseph Grosse on March 19, 2013

In example 3, we ignored the existence of the solid ammonium chloride in the flask when it came time to consider pressure.

I'm having trouble understanding that. According to the ideal gas law, pressure is dependent on the volume of the flask. So even if the solids are not directly contributing to the pressure of the gas wouldn't the space they are occupying restrict the volume available to the gases and so, affect the pressure?

Thank you.

Joseph

2 answers

Last reply by: okechukwu okigbo

Tue Jan 14, 2014 4:58 AM

Post by Gabriel Fuentes on March 11, 2013

For question 2 at 32:36 why is there a -1 delta n?

0 answers

Post by Abdihakim Mohamed on March 4, 2013

Thank u very much, you explain so great and to the point its not boring lessons, and I love it. The way we go straight to the point, and lot of samples, thank you again