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

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

### Equilibrium: Reaction Quotient

- The Equilibrium Constant is a numerical measure of far a reaction has proceeded before it “stops”
- The Reaction Quotient, Q, is the same expression as the Equilibrium Expression, but contains concentrations at any given point.
- Comparing Q with K, we can decide in which direction the reaction has yet to proceed – forward or reverse.

### Equilibrium: Reaction Quotient

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 0:57
- Reaction Quotient
- If Q > K
- If Q < K
- If Q = K
- Example 1: Part A
- Example 1: Part B
- Example 2: Question
- Example 2: Answer
- Example 3: Question
- Example 3: Answer
- Steps in Solving Equilibrium Problems

### AP Chemistry Online Prep Course

### Transcription: Equilibrium: Reaction Quotient

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

*We are going to continue our discussion of equilibrium, and today we are going to introduce something called the reaction quotient.*0005

*As you will see in a minute, the reaction quotient is just like the equilibrium expression, except it's used at any given time during the reaction.*0011

*It actually will tell us in which direction the reaction must go in order to reach equilibrium.*0021

*Sometimes it's too far to the left; it wants to go to the right; sometimes the reaction is too far to the right; it wants to go to the left.*0027

*Sometimes, it is exactly at equilibrium.*0032

*So, this reaction quotient is going to be one of the fundamental things that we use, and we will actually see it over and over again as we continue to decide where our reaction is and in what direction it needs to proceed.*0035

*And, as you will see when we start doing the problems, it is going to describe the mathematics--what are we adding? What are we subtracting?--things like that.*0045

*So, anyway, let's go ahead and get started.*0055

*OK, so let's do a quick review; so let's do this in blue.*0059

*We'll take our equation that we have been dealing with, which is nitrogen gas, plus 3 moles of hydrogen gas, in equilibrium to form 2 moles of NH _{3} gas--ammonia.*0065

*OK, now we said that the equilibrium expression, K...we are no longer writing K _{eq}, remember--if we just do K, that will mean that we are talking about moles per liter, as opposed to K_{P}, which means we are definitely dealing in partial pressures, either in torr or atmosphere, as long as the units are consistent.*0079

*OK, so K is products over reactants, so it is going to be the concentration of NH _{3} squared, over the concentration of N_{2} (and again, I am using parentheses for concentration instead of the standard brackets, simply to make this a little cleaner, because my brackets tend to be a bit messy) and H_{2}^{3}.*0098

*So, this is our equilibrium expression; and now, just as a review for what this equilibrium expression tells us: depending on what this number is--if it's a really, really big number, well, notice: if it's a big number, that means the numerator is a lot bigger than the denominator.*0121

*That means that there is more product than there is reactant.*0137

*That means that the equation favors the products; that means that, at equilibrium, most of what is in that flask is going to be products.*0142

*We say it is really far to the right.*0151

*If this is a really small number, this ratio, that means that the numerator is small and the denominator is big.*0153

*That means that the reactants actually are favored in the equilibrium; so at equilibrium, you are going to find most of the stuff on the left-hand side; it's going to be mostly nitrogen and hydrogen, instead of the ammonia.*0162

*That is what this equilibrium constant is a measure of; it is a measure of the extent to which a reaction moves forward or doesn't move forward.*0173

*That is all it is; it is just a numerical value expressing that.*0181

*So, we can speak about it qualitatively--"It's far to the right; it's far to the left"--or we can be precise and quantitative--"It is far to the right, because the K _{eq} is 500," or "The K_{eq} is .001."*0185

*This tells us specifically how far to the left or right; so that is all that that is.*0197

*OK, now, as we said, there is also something called the reaction quotient, which we can use to tell us in which direction a reaction must go in order to reach equilibrium--in other words, where it is at that given moment.*0202

*OK, so let's go ahead and define our Q; this is called the reaction quotient.*0216

*It is the same expression as the equilibrium constant; so, in other words, it's the same thing as this (products over reactants, raised to their stoichiometric coefficients).*0229

*It's the same expression as K _{eq}, but concentrations/pressures are taken at any given moment.*0241

*So you know that this constant--equilibrium constant--it is the ratio of these concentrations, once the system has come to equilibrium.*0275

*Well, we can measure the concentration any time we want (of the NH _{3}, the N_{2}, the H_{2}, or the products, or the reactants), and we can put those into this expression, and we can see how far away from the equilibrium constant it is, and that will tell us whether it is too far to the left or too far to the right.*0283

*That is all we are doing here; so, for a general reaction, aA + bB going to cC + dD, we have that the reaction quotient, Q, is equal to the concentration of C raised to the c power, the concentration of D raised to the d power, over the concentration of A raised to the a power, the concentration of B raised to the b power.*0304

*It is exactly the same as the equilibrium expression, except these concentrations are at any given moment.*0330

*That is all; now, here are the criteria by which we decide where a reaction is--how far from equilibrium.*0337

*If Q is bigger than K, then the reaction will proceed (let's say...yes, you know, I think "proceed" is a good word; I wanted to use "shift," but I think "proceed" is better, or "move") to the left to reach equilibrium.*0346

*In other words, if Q is bigger than K, that means this is bigger than that, that means it has gone too far to the right; or it has not gone too far to the right--it is too far to the right.*0381

*In order for it to reach its equilibrium point (which, as we said, is a fingerprint for that particular reaction at a given temperature), it needs to move to the left; that means it needs to decompose product to form more reactant.*0396

*It needs to decompose this to form more this; it needs to move to the left--that is what that means.*0408

*So now, if Q is less than K, well, it's just the opposite: then, the reaction will proceed to the right to reach equilibrium.*0413

*At any given moment, if we take a bunch of concentrations of products and reactants and we stick it into this expression, we solve it, and it ends up being less than the K, that means there is too much of this and it needs to move in this direction to reach equilibrium.*0438

*That means this denominator is too big; Q is too small.*0452

*It needs to go this way, so it's going to proceed to the right to reach equilibrium.*0456

*So, in this case, reactants are depleting; products are forming.*0461

*And of course, last but not least, if Q equals K, well, you know the answer to this one: then the reaction is at equilibrium (here we go again with the stray lines; OK, that is nice).*0467

*That is it--nice and simple; a mathematical way to see where a reaction is and to see where a reaction is going.*0487

*OK, let's go ahead and do an example.*0496

*Example 1 (let's see): For the equation H _{2}O gas (you know, that's OK...well, I don't know; I want to sort of skip writing the gas, but I guess it's pretty important) + Cl_{2}O gas (oops, can't have a double arrow going--I need it to go that way and that way) forms 2 HOCl gas at 25 degrees Celsius, the equilibrium constant equals 0.0900.*0510

*OK, so: for the reaction H _{2}O gas + dichlorine monoxide, that goes to 2 HOCl gas (hydrogen hypochloride gas, OK?--this is not hypochlorous acid; it's in the gaseous state, so it's not aqueous, so it's not the acid--this is the hydrogen hypochloride), the equilibrium constant for this reaction at 25 degrees Celsius is .0900.*0561

*Notice, there is no unit here; we will get to that in just a minute.*0586

*Here is what we want to do: For the following concentrations, determine the direction (I don't want to run out of room over there) the reaction must go in, in order to reach equilibrium.*0590

*OK, so: For the following concentrations, determine the direction the reaction must go in order to reach equilibrium: a standard reaction quotient problem.*0637

*It is going to be basic; we are going to do it for most equilibrium problems, because we want to know where equilibrium is.*0645

*OK, so the first one: we have: A partial pressure of H _{2} O is equal to 200 torr, and the partial pressure of Cl_{2}O is going to equal 49.8 torr, and the P of HOCl is equal to 21.0 torricelli.*0651

*Now, our Q is equal to the partial pressure of HOCl squared, over the partial pressure of H _{2}O times the partial pressure of Cl_{2}O.*0680

*Well, that equals 21.0 squared (that is the HOCl), divided by the partial pressure of H _{2}O, which is 200, and it is 49.8; there we go.*0694

*Now, in this particular case, this is torr and this is torr squared; so the unit on top is going to be torr squared; this is torr; this is torr; it's going to be torr squared, so it's going to end up without a unit.*0715

*That is why this equilibrium constant doesn't have a unit.*0725

*It is because the torricelli, torricelli cancels with torricelli, torricelli, down at the bottom.*0729

*So, now, we get that Q is equal to 0.0443; so Q is equal to .0443; K is equal to .0900; clearly, Q is less than K, which implies that the reaction will move to the right to reach equilibrium.*0733

*In other words, reactant will deplete; product will form; it will move to the right to reach equilibrium.*0766

*It hasn't reached equilibrium yet; it's still moving to the right to reach eq.*0774

*That is it; that is all you are using the reaction quotient for--to tell you which direction it is going in.*0780

*OK, all right, let's see: so, let's do another one--another set of conditions.*0786

*This time, we will do: A 3.0-liter flask contains 0.25 moles of HOCl, 0.0100 mol of Cl _{2}O, and 0.56 mol of H_{2}O.*0796

*OK, notice: they give us moles, and they give us the actual volume of the flask.*0828

*Well, again, when we deal with these reaction quotients and equilibrium expressions, they have to be in moles per liter--in concentrations.*0832

*Let's just take each one and find the concentrations before we put them into our reaction quotient.*0841

*So, our concentration of HOCl is equal to 0.25 mol, divided by 3.0 liters; that is going to equal 0.0833 Molar (that m with a line over it means molarity; it's an older expression; you are accustomed to seeing it: capital M).*0846

*The Cl _{2}O: that is equal to 0.0100 moles, again divided by 3 liters, because it is in the flask; so, its concentration is 0.00333 Molar.*0875

*And finally, our H _{2}O concentration is equal to 0.56 mol, divided by 3.0 liter; and this concentration is 0.1867 Molar.*0895

*Now that we have the molarities, we can put them into our reaction quotient; the reaction quotient is going to be exactly the same thing.*0913

*Now, we have: Q is equal to the concentration, as we said, of HOCl squared, over the concentration of H _{2}O, times the concentration of Cl_{2}O, which is equal to (drop down a little bit here) 0.0833 squared, times 0.1867, times 0.0033; and we get 11.16.*0922

*Now, Q is 11.16; we said that K was equal to 0.0900; so, clearly, Q is much larger than K, which implies that the reaction will move to the right to reach equilibrium.*0968

*In other words, it is still moving--I'm sorry; not to the right--to the left!*0992

*The Q is bigger; yes, Q is bigger--it is going to move to the left.*0998

*Sorry about that; that means that there is too much product at this temperature, given the equilibrium, which is a fingerprint for that reaction; so, there is too much product; the product needs to decompose to form reactant.*1004

*It is moving to the left to reach equilibrium.*1019

*That is it; OK, now, again, notice that there are no units for that; there are no units for this particular equilibrium constant.*1028

*In fact, notice in the question: the question actually said, "Equilibrium constant equals"--it didn't say "K equals" or "K _{P} equals."*1040

*If you are given K _{eq}, K, or K_{P}, it will specifically mean that we are talking about pressures, or we are talking about moles per liter; but in this particular case, because there is no unit, that actually means that the K_{P} and the K are the same.*1050

*You remember the definition of the relationship between K _{P} and K; well, the fact that there is no unit means that there is no...if you look at the equation, Δn=0, so that RT that we had is 1.*1069

*So, when it says the "equilibrium constant," but it doesn't specifically specify whether it is a K or a K _{P}, well, it's the same thing--they are actually equal to each other, which is why we use the same number, .0900, with molarity and with the one that we just did, which was done in terms of pressure.*1085

*In both cases, we use the .0900; that comes from the fact that there is no unit that tells us that the K and the K _{P} are the same.*1103

*These are the little things that you have to watch out for; in other circumstances, when you do have a unit, you have to watch out; you have to actually (if you are dealing with molarity and you are given the K _{P}, you have to) either convert the K_{P} to a K, or you have to convert the molarities to pressures, if you can, depending on what the problem is asking.*1112

*Again, these are the sort of things; there is a lot that is going to be going on--there is a lot that you have to watch out for.*1132

*It isn't just "plug and play"--you don't just put numbers into an equation and hope things will fall out.*1139

*You have to understand what is happening; this is real science, and real science means conversions, units, and strange things.*1145

*OK, so let's do another example.*1155

*Here is where we begin to actually explore some of the diversity of these equilibrium problems.*1160

*What we are going to do is: most of our learning is actually going to come through the problems themselves.*1165

*That is why we are going to do a fair number of these equilibrium problems; it's very, very important that you have a reasonably solid understanding of how to handle these things, because it's going to be the bread and butter of what you do for the rest of chemistry--certainly for the rest of the AP and the free response questions; the electrochemistry; acid-base; the thermodynamics.*1170

*It's precisely this kind of reasoning, and equilibrium is fundamental to it all.*1191

*That is why it comes before everything else does.*1196

*Equilibrium is chemistry; it is that simple.*1198

*OK, so let's do Example #2: let's do this one in red--how is that?*1203

*OK, Example 2: The question is going to be a bit long, but...let's see.*1211

*At a certain temperature, a 1.0-liter flask contains 0.298 mol of PCl _{3} and 8.70x10^{-3} mol of PCl_{5}.*1221

*OK, now after the system comes to equilibrium (comes to eq), 2.00x10 ^{-3} mol of Cl_{2} gas was formed in the flask.*1257

*Now, PCl _{5} decomposes according to the following: PCl_{5} decomposes into PCl_{3} + Cl_{2} gas.*1289

*What we would like you to do is calculate the equilibrium concentrations of all species and the K _{eq}.*1310

*OK, so we would like you to calculate the equilibrium concentrations of all the species (in other words, the PCl _{5}, the PCl_{3}, and the Cl_{2}) and we would like you to tell us what the K_{eq} is--what the K is.*1337

*OK, so let's read this again: At a certain temperature, a 1-liter flask contains .298 moles of PCl _{3}, and 8.70x10^{-3} moles of PCl_{5}.*1349

*After the system comes to equilibrium, 2.0x10 ^{-3} moles of Cl_{2} is formed in the flask.*1362

*Great! So, let's go ahead and start this off; so I'm going to go ahead and move to a new page so I can rewrite the equation.*1368

*We are going to do our little ICE chart here: Initial, Change, Equilibrium.*1378

*PCl _{5} decomposes into PCl_{3} plus Cl_{2}; our initial concentration, our change, and our equilibrium concentration--which is what we actually want here.*1383

*It's telling me that PCl _{5} was 8.7x10^{-3} initially, right?*1396

*8.70x10 to the negative...oh, let's do...yes, that's fine; OK; I can just do this down below.*1406

*Notice: they give us (well, here; let me do this over to the side)--with the PCl _{5}, they gave us 8.70x10^{-3} moles.*1420

*Now, that is not a concentration--that is moles, not moles per liter, but they said we have a 1.0-liter flask.*1436

*Again, this is one of the other things: we have to make sure to actually calculate the concentrations; so, in this case, it's going to be 8.70x10 ^{-3} moles, over 1.0 liters.*1443

*Well, because it's 1 liter--it's a 1-liter flask--the number of moles is equal to the molarity.*1457

*So, I can just go ahead and put these numbers here.*1462

*If this were not a 1-liter flask--if it were anything other than a 1-liter flask--I would have to actually calculate the initial concentration, and those are the values that I use in my ICE chart.*1465

*OK, so in my ICE chart, I'm working with concentrations, not moles.*1476

*OK, so we have: 8.70x10 ^{-3} molarity, and we said the PCl_{3} was 0.298, and there is no chlorine gas.*1482

*Well, they said that, at equilibrium, there is 2.0x10 ^{-3} moles per liter of chlorine gas.*1496

*So, chlorine gas showed up; so the change was: well, for every mole of chlorine gas that shows up, a mole of PCl _{3} shows up, and a mole of PCl_{5} decomposes, because the ratio is 1:1, 1:1, 1:1.*1506

*If 2.00x10 ^{-3} moles shows up, that means here, also, 2.00x10^{-3} moles shows up; here, it is -2.00x10^{-3} moles.*1526

*Again, we are using just basic intuition and what we know about the physical system to decide how the math works.*1545

*This is what chemistry is all about--this is the single biggest problem with chemistry.*1553

*There is nothing intuitively strange about any concept in chemistry--it's all very, very clear--it's all very, very basic as far as what is happening; there is nothing esoteric; there is nothing metaphysical going on.*1557

*It is just that...how does one change the physical situation into the math?*1571

*Well, this is how you do it; you have to know what is going on physically, and then the math should fall out; just trust your instincts.*1576

*One mole of this shows up; well, the equation says one mole of this shows up.*1583

*That means, if one mole of this shows up--that means one mole of this was used up; that is it.*1587

*So now, we do 8.7x10 ^{-3}, minus 2.0x10^{-3}, and we end up with 6.7x10^{-3}, which I am going to express as a decimal; so, 0.067.*1592

*And then, we have this one; when we add it together, we end up with 0.300; and here, we have 2.00x10 ^{-3}.*1606

*It is probably not a good idea to mix the scientific notation and decimals, but you know what--actually, it is not that big of a deal--it is what science is all about.*1619

*OK, so now, because we have the equilibrium concentrations, we have solved the first part of the problem.*1627

*At equilibrium, we are going to have .067 Molar of the PCl _{5}, .300 Molar of the PCl_{3}, and 2.0x10^{-3} Molar of the Cl_{2}.*1635

*Well now, let's just go ahead and put it into our equilibrium expression, and find our K _{eq}.*1648

*That is the easy part.*1654

*OK, so K _{eq} is equal to...it is going to be...the Cl_{2} concentration, times the PCl_{3} concentration, divided by the PCl_{5} concentration.*1656

*That equals 2.00x10 ^{-3}, times 0.300, divided by 0.067.*1671

*Yes...no; 6.7x10 ^{-3}...oops, I think I have my numbers wrong here; this is supposed to be...yes, I should have just left it as scientific notation.*1688

*You know what, I'm just going to go ahead and leave it as scientific notation.*1698

*I shouldn't mess with things; 6.7x10 ^{-3}, and we will write this as 6.7x10^{-3}, also.*1702

*6.7x10 ^{-3}: I ended up forgetting a 0; it was .0067; OK.*1713

*And then, when we solve this, we end up with...(let's see, what number did we get?) 8.96x10 ^{-2} Molar.*1720

*So, in this case, it does have a unit; this is a K _{eq}--this is not a concentration.*1732

*That is why I am not a big fan of units when it comes to equilibrium constants; as far as equilibrium constants are concerned, I think that units should only be used to decide about conversions.*1738

*Other than that, I think they should be avoided; but you know what, we will just go ahead and leave it there.*1753

*This is the K _{eq}; K_{eq} equals 8.96x10^{-2}.*1756

*That is actually a pretty small number; what does a small K _{eq} mean?*1761

*That means most of this reaction is over here, on the left; there are not a lot of products.*1765

*Most of it is PCl _{5}; that is what that means.*1769

*Don't let these numbers say that it is mostly this, because, remember: we started off with a certain amount of the PCl _{5}; we started off with a whole bunch of the PCl_{3}.*1773

*So, .298--it only went up to .3; that means it only went up .002; that is not very far.*1786

*So, don't let these equilibrium amounts fool you into thinking that the reaction went forward; it is this equilibrium constant which tells you the relationship between these three numbers under these conditions.*1795

*But, this is a fingerprint; this reaction at this temperature will not go very far forward.*1811

*Most of it is still PCl _{5}; that is what is going on--don't let this .3 fool you; it doesn't mean that it has formed that much.*1816

*You already started at .298; you only formed .002 moles per liter of the PCl _{3}--not very much at all.*1824

*It is confirmed by the K _{eq}.*1833

*OK, let's do another example; that is what we are here to do.*1835

*Let's see what we have.*1842

*Yes, OK; let me write it out, and then...well, you know what, I am going to actually start a new page for this one, because I would like to see part of the problem while we are reading; OK.*1845

*This Example 3: Now, carbon monoxide reacts with steam (this is H _{2}O gas) to produce carbon dioxide and hydrogen at 700 Kelvin (we don't need a comma there).*1856

*At 700 Kelvin, the eq constant (equilibrium constant) is 5.10.*1901

*Calculate the eq concentrations of all species if 1.000 mol of each component (each component means each species) is mixed in a 1.0-liter flask.*1914

*OK, our reaction is: they said: Carbon monoxide gas plus steam, which is H _{2}O gas, forms carbon dioxide gas, plus hydrogen gas.*1953

*That is our reaction, and it is balanced; so it is 1:1:1:1--not a problem.*1966

*OK, so let's read this: Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen.*1973

*At 700 Kelvin, the equilibrium constant is 5.10.*1978

*So, we have our (let me do this in blue now) 5.10.*1983

*Calculate the equilibrium concentrations of all species if 1.0 mol of each of the components (that means 1, 1, 1, 1 mole of each) is mixed in a 1-liter flask.*1987

*OK, now the first thing we need to do is: we are going to (and again, this is sort of something we are always going to do--this is the procedure) write the equation, which we have.*1999

*We are going to write the equilibrium expression, and then we are going to check the reaction quotient to see where the reaction is at that moment, to see which direction it is actually going to be moving in.*2010

*Let's write the equilibrium expression for this, which is going to be the same as the reaction quotient.*2019

*It is the concentration of H _{2}, times the concentration of CO_{2}, divided by the concentration of CO, times the concentration of water.*2026

*OK, and that is also equal to the reaction quotient.*2037

*The first thing we want to do is calculate the reaction quotient to see which direction it is moving in.*2041

*Now, they say they put in 1.000 mol of each component.*2048

*Well, it is sitting in a 1.0-liter flask, so basically, I can work with moles: 1 mole, divided by 1 liter, is 1 mole per liter, so the concentration of each species is 1 mole per liter.*2054

*Well, now let's calculate the Q.*2068

*The Q equals...well, it is 1 Molar of the H _{2}, 1 Molar of the CO_{2}, divided by 1 Molar of the CO and 1 Molar of the H_{2}O; the Q equals 1.*2071

*Now, the reaction quotient Q, which is equal to 1, is less than 5.10 (remember, they gave us the 5.10, which is equal to K).*2088

*When Q is less than K, that means the reaction wants to move forward to produce more product, in order to reach equilibrium.*2096

*That means it hasn't reached equilibrium yet; it is still moving forward--it is producing more product to reach equilibrium.*2105

*That means carbon monoxide and H _{2}O gas are being depleted, and for each amount that these are depleted (because the ratio is 1:1), an equal amount of CO_{2} and H_{2} are being formed.*2113

*Now, we can do the actual equilibrium part of this problem.*2124

*So again, these "1 Molar"--that came from where we started at that moment.*2128

*At any given moment, I stick in 1 mole of each in a 1-liter flask, and let me find out what this value is.*2133

*It is 1; it is less than the equilibrium constant; that means it is going to move forward, to the right.*2140

*OK, so now let's do our equilibrium part, and we do that by doing our ICE chart, so let me rewrite CO + H _{2}O (I tend to rewrite things a lot--sorry about that; I hope it's not a problem--I'm sure you have different ways of doing it yourself--as long as each of these is here...) goes to CO_{2} + H_{2}.*2146

*OK, we have an initial concentration; we have the change; and we have our equilibrium concentrations, which is what we are looking for.*2173

*Our initial concentrations are 1.000, right?--that is how much we started with.*2181

*We stuck each of those in a flask.*2186

*Now, this reaction quotient tells us that the reaction is moving in that direction to reach equilibrium.*2189

*It is moving in that direction; that means CO is disappearing.*2196

*H _{2}O is also disappearing by an amount x--that means a certain amount is decomposing--a certain amount of CO is being lost, is being converted.*2202

*Well, since it is moving to the right, and this is 1:1:1:1, that means this is +x; that means CO _{2} is forming for every 1 mole of this that is disappearing.*2214

*This is also +x; I hope that makes sense.*2226

*Our equilibrium concentration (at equilibrium, once everything has stopped, a certain amount of CO has been used up, so our equilibrium concentration) is going to be 1.00-x.*2231

*A certain amount of H _{2}O has been used up; that is 1.00-x.*2242

*A certain amount of CO _{2} has formed, so it is going to be 1+x.*2248

*A certain amount of H _{2} has formed: 1+x.*2252

*These are our equilibrium (oh, wow, that is interesting; look at that--let's get these lines out of the way) expression, but notice: now, we have x, so we need to actually find x.*2258

*Fortunately, we can do that: we can plug it into the equilibrium expression, right?--because the equilibrium expression is a measure of these concentrations at equilibrium; that is what these values are.*2274

*We stick it in here; we know what the K _{eq} is--it's 5.10; and we solve for x.*2286

*That is it; it is just an algebra problem.*2291

*So, let's go ahead and do that: so K, which is equal to 1.000+x, times 1.000+x (that is the CO _{2} and H_{2} concentrations), divided by the CO and H_{2}O concentrations, which at equilibrium are 1.000-x, 1.000-x, and we know that that equals 5.10.*2294

*OK, so now let's just handle this algebraically.*2329

*This is (1.000+x) squared, over 1.000 (oops, too many 0's) minus x, squared, equals 5.10.*2333

*Now, I know that I said earlier (I think a lesson or two ago): when you are writing out the equilibrium expression, don't put the square--don't square it immediately--if two of the things are the same.*2353

*That is different than what I am doing now; I wrote the expression as each species separately, so that I can see that I actually have four species in my equilibrium expression.*2363

*Here, now, I am just dealing with the math.*2373

*Once you have actually written it out, then you can go ahead and write it like this to deal with the math.*2375

*Now, it's (1+x) ^{2} over (1-x)^{2}.*2380

*That is fine; but when you initially write the expression, don't cut corners; write down everything.*2384

*We want to know that the expression actually consists of four terms, not two terms, each squared.*2389

*OK, and now we just...well, we have a square here and a square here, so we'll just go ahead and take the square root of both sides.*2395

*We end up with 1+x over 1-x, equals 2.258, and then we multiply through to get 1.000+x=2.258-2.258 times x (I'm hoping that I am doing my math right here).*2406

*We end up with 3.258x equals 1.258, and x is equal to 0.386 Molar.*2440

*We found x; x is .386 Molar.*2458

*Now, if you go back to your ICE chart, it didn't ask for what x was; it asked for the final concentration.*2464

*Well, the final concentrations were the 1-x and the 1+x for those four species.*2471

*So, the CO concentration (carbon monoxide concentration), which also equals the H _{2}O concentration, is equal to 1.000-x, 1.000-.386, equals 0.613 Molar.*2476

*The carbon monoxide and the water are at .613 molarity.*2501

*Now, the CO _{2} concentration, which also happens to equal the H_{2} concentration, is equal to (OK, let's see if we can clean this up a little bit; I'm not going to have these stray lines driving us crazy all day)...CO_{2} (I just really need to learn to write slower; I know that that is what it is) equals H_{2} concentration, equals 1.000+x, equals 1.000+0.386, =1.386 molarity.*2505

*OK, and my friends, we have done it; we have taken a standard equilibrium problem, and here is how we have approached it.*2551

*1) Write the equation; this is chemistry--chemistry always begins with some equation; don't just go into the math.*2564

*Look at the equation--the equation gives you all of the information that you need--in fact, it will tell you everything you want.*2570

*2) Write the equilibrium expression--write the K expression.*2576

*Write it out explicitly--don't count on the fact that you will know how to do it later on when you are ready to plug things in.*2583

*Write it out; don't cut corners; doing things quickly is not impressive--doing things correctly is impressive.*2589

*3) Find Q, the reaction quotient; find Q to decide which direction the reaction is going in--to decide the reaction direction, if any.*2598

*That is how we knew...in this problem that we just did, the reaction quotient was less than the equilibrium constant--so that means the reaction was actually moving forward.*2614

*Well, that is how we knew that the reactants get the -x and the products get the +x.*2623

*If it were the other way around and the reaction were moving to the left, that means the products would get the -x and the reactants would get the +x.*2628

*You have to do this; you have to do the reaction quotient--very, very important.*2637

*After the reaction quotient, well, you set up your ICE chart: Initial, Change, Equilibrium concentration.*2642

*Once you get a value for the equilibrium concentration, you put the eq concentration expressions (in other words, the concentrations that you calculated there--the equations/expressions) into the K _{eq} expression.*2652

*Then, last but not least, you solve, depending on what they want.*2685

*If they want x, you stop there; if they want equilibrium concentrations, you take the x value; you plug it back into the equilibrium expressions; and you add and subtract until you get your equilibrium expressions.*2689

*If they want something else, they will tell you that they want something else.*2701

*Again, this process is what you are always going to be doing; this is the algorithm, the general, broad-strokes algorithm.*2705

*Write the equation; write the expression; find the Q; set up your ICE chart; put the equilibrium concentrations in; and then solve it.*2714

*Within this are the different variations that make up the different number of problems, which seem to be infinite (I understand completely).*2723

*That is all you are doing here.*2731

*OK, thank you for joining us here at Educator.com for our continuation of AP Chemistry in equilibrium.*2734

*In our next lesson, we are actually going to do more equilibrium problems, because again, this is a profoundly important concept; you have to be able to have a good, good, solid, intuitive understanding of what is going on.*2740

*So, we will do some more practice.*2751

1 answer

Last reply by: Jeffrey McNeary

Mon Jul 4, 2016 4:25 PM

Post by Jeffrey McNeary on July 4, 2016

for example 1 part A, you found Q using pressure. You then compared Q to K. My question is this: Why did you not convert Q into a equilibrium constant that used molarity by using the equation: K=Kp(RT)^-n? If you had done this, Q and K would both be in terms of molarity. However, in the problem, it seems like you are comparing pressure to molarity. To me this seems like you are comparing oranges to apples instead of apples to apples.

1 answer

Last reply by: Professor Hovasapian

Tue Dec 2, 2014 2:34 AM

Post by Shih-Kuan Chen on November 30, 2014

Dear Professor,

Just a curious question that has nothing to do with this lecture: Does the AP Chemistry test cover Organic Chemistry?

2 answers

Last reply by: Professor Hovasapian

Sat Aug 2, 2014 3:09 AM

Post by David Restrepo on July 30, 2014

In example 3, if the Q > K then would the +x be on the left side of the equation?

Thank You!

1 answer

Last reply by: Professor Hovasapian

Sun Jul 27, 2014 4:19 AM

Post by Jessica Lee on July 26, 2014

Why can't you just write 1 instead of 1.000, do you need that zeros? Example 3

2 answers

Last reply by: Tim Zhang

Mon Mar 17, 2014 8:41 PM

Post by Tim Zhang on March 16, 2014

I don't understant why a change of reactants are always negative and of products are positive, in such a reaction that can proceed reversibly.

1 answer

Last reply by: Professor Hovasapian

Sun Mar 16, 2014 11:20 PM

Post by Tim Zhang on March 16, 2014

In equailibrium, to calculate the concentation of a reaction the inital values start in Molarity. Does this mean the changes(C) and finial value (F) should all be in the unit of Molarity?

When a question tells you the changes in moles, Do you have to convert to molarity ?

2 answers

Last reply by: joebert binalinbing

Wed Feb 5, 2014 1:15 AM

Post by joebert binalinbing on February 3, 2014

for example 2 what happen if the ratio is 2 to 2 to 1. please help thanks

3 answers

Last reply by: Professor Hovasapian

Tue Sep 3, 2013 2:29 AM

Post by Marian Iskandar on September 2, 2013

For example 2, 8.7x10^-3 - 2.00x10^3 = 0.0067 (6.7x1^-3), not 0.067. Slight error, but thought I'd point it out.