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

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

### Equilibrium, Part 1

- Equilibrium is dynamic: the forward and reverse reactions are still proceeding, but since they happen at the same rate, there is no observable (net) change.
- There is a mathematical (numerical) expression for the Equilibrium state involving concentrations of the aqueous and/or gaseous species in a reaction. Solids and Liquids do not show up in the equilibrium expression.
- Equilibrium positions – thus Constants – change with temperature.

### Equilibrium, Part 1

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 1:32
- Introduction to Equilibrium
- Equilibrium Rules
- Example 1: Part A
- Example 1: Part B
- Example 1: Part C
- Example 1: Part D
- Example 2: Part A
- Example 2: Part B
- Example 2: Part C
- Reverse a Reaction
- Example 3

### AP Chemistry Online Prep Course

### Transcription: 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 H _{2} + O_{2} → 2 H_{2}O.*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 NH _{3} 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 H _{2} 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 H _{2}gas; this second one is the NH_{3} 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: N _{2} + 3 H_{2} goes to NH_{3}.*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 NH _{3} 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 NH_{3} (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, H _{2}, 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 NO_{3}^{-} (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 NO_{3}^{-}.*1188

*Well, K ^{+} cancels K^{+} (oh, again with the lines!); NO_{3}^{-} goes with NO_{3}^{-}; 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 CO _{2}, 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 NH _{3} gas + 7 O_{2} gas forms 4 moles of nitrogen dioxide gas and 6 moles of H_{2}O; 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 NO_{2} concentration, raised to the fourth power, over the NH_{3} concentration, raised to the fourth power, times the O_{2} 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, CaCO _{3}--this time, this is going to be a solid, and it is in equilibrium with calcium oxide, which is also solid, plus--interestingly enough--CO_{2} gas.*1497

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

*So, calcium carbonate decomposes into calcium oxide and CO _{2}.*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 CO_{2} concentration.*1523

*That is it--nice and simple.*1541

*This equilibrium expression only depends on the CO _{2} 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 CO _{2} 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 N _{2} + 3 H_{2} going to 2 NH_{3}.*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 NH _{3} concentration squared, over the N_{2} concentration times the H_{2} concentration cubed.*1811

*Well, that equals...the NH _{3} 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

*N _{2} 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 3.8x10 ^{4}; 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 NH_{3} going to 3 H_{2} +N_{2}.*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 H _{2} 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 3.8x10 ^{4}, 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 N_{2} (now we're going to change the coefficients) + 3/2 N_{2} goes to NH_{3}.*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 N_{2}...I'm sorry, it's going to be NH_{3}, divided by the concentration of N_{2} 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, N _{2} raised to the 1/2 power, times (oh, this is H_{2}) the concentration of H_{2} 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 3.8x10 ^{4} 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 N _{2} + 3 H_{2} going to 2 NH_{3} 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 N _{2} + 3 H_{2} going to 2 NH_{3} 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 N _{2} concentration was 1.000, and of course, concentration is in moles per liter.*2334

*The H _{2} concentration is at 1.000, and the NH_{3} 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) NH _{3} concentration squared, divided by the N_{2} concentration times the H_{2} 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 N _{2} concentration, an H_{2} concentration, and NH_{3} 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 N _{2}).*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

1 answer

Last reply by: Professor Hovasapian

Wed Jan 18, 2017 6:12 PM

Post by Magic Fu on January 16 at 05:42:30 PM

Hi, Professor H. Can you explain how kinetic and equilibrium are related?

1 answer

Last reply by: Professor Hovasapian

Fri Jun 3, 2016 6:49 PM

Post by Uriel Cordoba on May 21, 2016

in equilibrium, example 2, the answer is 0.03785, not 38000

1 answer

Last reply by: Professor Hovasapian

Fri Mar 25, 2016 10:23 PM

Post by Weipeng Sun on March 21, 2016

Hi Raffi. It seems to me that the third experiments in example 3 is a reverse reaction. Why its equilibrium constant isn't the reciprocal of the other two? Thank you!

1 answer

Last reply by: Professor Hovasapian

Thu Dec 3, 2015 12:39 AM

Post by Jeffrey Tao on November 27, 2015

How do chemists actually measure the concentration of the individual products? They are usually all mixed together, so it seems pretty difficult.

1 answer

Last reply by: Professor Hovasapian

Tue May 19, 2015 1:50 AM

Post by katie kim on May 18, 2015

Hello Professor Hovassapian, I have a question on the equilibrium concept. When a compound is enclosed in the equilibrium constant, like [NH3], what does the brackets [] represent? Does it just resemble the amount of ammonia, or the concentration, or something else?

1 answer

Last reply by: Professor Hovasapian

Thu Mar 13, 2014 11:40 PM

Post by Samiha Bushra on March 3, 2014

In Example three, in the second group of concentrations, isn't it a reverse reaction so shouldn't the Keq be 1/Keq of the first one?

2 answers

Last reply by: Professor Hovasapian

Wed Jan 22, 2014 4:11 PM

Post by Tim Zhang on January 16, 2014

Hello Pro H. Could you help me to solve this problem? A 25.0-g sample of garden compost was analyzed for chloride content. The sample was dissolved in water, and the chloride was precipitated as silver chloride. 1.58 g of dried precipitate was obtained. Calculate the mass percent of chloride in the sample. My answer for this 4.76% using (1.58*(AgmolMass/Ag+Cl))/25

1 answer

Last reply by: Professor Hovasapian

Fri Jan 3, 2014 4:06 PM

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)?

can show me the process to do that

1 answer

Last reply by: KyungYeop Kim

Mon Jun 3, 2013 8:03 PM

Post by Professor Hovasapian on June 3, 2013

Hi Kyung,

When a system is already at equilibrium, any disturbance causes the system to try to re-establish that equilibrium. So, when pressure is increased, the system will try to compensate by decreasing the pressure if it can, as much as it can...the only way to decrease pressure is by having less particles in the gaseous state hitting the walls of the container...so by shifting to the side with fewer mols, there will be fewer particles -- thus lower pressure.

If the pressure is decreased, the system wants to bring the pressure back up, so it moves toward the side with higher particles in the gaseous state -- more mols.

Hope that helps. Let me know...

Take care.

Raffi

1 answer

Last reply by: Professor Hovasapian

Mon Jun 3, 2013 6:22 PM

Post by KyungYeop Kim on June 3, 2013

I seem to never get how pressure affects equilibrium. I know their relationships, but I don't know why. Why is it that when pressure is increased, the equilibrium shifts to the side that produces the smallest number of molecules? Could you explain why number of moles matter at all, because I don't understand why more moles or less mores matter in the first place. Thank you, your lectures are very helpful.

1 answer

Last reply by: Professor Hovasapian

Wed May 15, 2013 1:31 AM

Post by Nawaphan Jedjomnongkit on May 13, 2013

Thank you for the great lecture but I still don't get the picture when it comes to example 1 part D where Keq depend on only [CO2], Let say if we have experiment like in example 3 which start from different concentration of reactant we will get different concentration of products, right? So how Keq still constant?

1 answer

Last reply by: Professor Hovasapian

Sun Mar 3, 2013 1:23 PM

Post by Abdihakim Mohamed on March 3, 2013

I love the way Professor teaches, straight to the point no time waist. Thank you very much.

0 answers

Post by Max Mayo on February 15, 2013

Picking out his trivial mistakes doesn't make you look like a genius... He's doing an effective job at teaching the material, and it's not like your input is doing anybody any good. The guy helped me make an A in gen chem last semester, so I would say his content grossly outweighs his mistakes.

1 answer

Last reply by: Professor Hovasapian

Mon Feb 11, 2013 3:46 PM

Post by oscar paniagua on February 11, 2013

Hello Professor,

There is a mistake in Example 2 Part A; you have H2=3.1x10^-1.

It should actually read as H2=3.1x10^-3. Otherwise while doing the calculations they do not add up to the answer of =3.8x10^4.

Cheers,