AP Chemistry Equilibrium: Examples

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

Today, we are going to continue our detailed discussion of equilibrium.0005

We are just going to be doing examples.0009

We have talked about what equilibrium is, the expression; we have actually done a fair number of problems so far, and problems that are reasonably complicated.0012

But, I wanted to do some more that go a little bit more in-depth, as far as--they have some slight variations; they are a little more detailed and require a little bit more care--for the purposes of showing you where things can possibly go wrong, but mostly just to get you really, really accustomed to these equilibrium problems.0020

It has been my experience, from all of my students who have had a solid grounding in equilibrium--the rest of chemistry is a complete breeze for them.0039

They know exactly what to do; they know where to go; I virtually didn't have to teach the class after this.0046

We hammered the equilibrium so much that they knew exactly what was going on; that is why this is important.0051

If you can get your head around this, and be able to handle at least a good 70 or 80 percent of these problems, the AP test should be absolutely a breeze for you, I promise.0057

OK, so let's jump right on in.0066

Our first example is going to be a synthesis of hydrogen fluoride from hydrogen and fluorine gas.0070

Let's go: Example 1: So, in a synthesis of hydrogen fluoride gas...0077

Now notice, I didn't say hydrofluoric acid, because this is hydrogen fluoride gas--it's in gaseous form; remember, an acid is something that is when you take the thing and you drop it in the water, and it dissociates into a hydrogen ion and fluoride ion, or hydrogen ion and any other anion.0093

That is when it is an acid--the acid is the H⁺; when HF is together, it is not an acid--it won't do any damage.0111

Well, I won't say it won't do any damage, but it won't do any damage as an acid--let's put it that way.0119

In the synthesis of HF from H₂ and F₂ at a given temperature, 3.000 moles of H₂ and 6.000 moles of F₂ are mixed (it might be nice if I actually knew how to spell--yes, that would be good) in a 3.0-liter flask.0124

So again, we have moles, and we have a 3-liter flask; this time it is not 1-liter, so we are going to have to actually calculate the concentration.0168

OK, the eq constant at this temperature is 1.15x10².0177

What are the equilibrium concentrations of all species?0196

Or I should say, "What is the equilibrium concentration of each species?"0202

Being able to handle these ICE charts is the real key to the majority of the second half of chemistry.0218

The ICE chart itself is always going to be a ubiquitous feature of the problems that we solve.0225

The difference is what the ICE chart looks like: acid-base problems, equilibrium problems, thermodynamic problems, whatever it is...the ICE chart itself is going to change, depending on what the problem is asking.0229

Being able to handle that--that is the real deal; that is when you know you actually know what is going on--when you know how to arrange the ICE chart as necessary; the rest is just math, basic algebra.0242

OK, so in the synthesis of HF from H₂ and F₂ at a given temperature, 3 moles of hydrogen and 6 moles of fluorine are mixed in a 3-liter flask.0255

The equilibrium constant is 1.15x10², or 115.0267

What is the equilibrium concentration of each species?0272

OK, so let's write our equation; like we said, that is our process: H₂ + F₂ goes to 2 HF.0275

We want to make sure that the equation is balanced; we want to write our equilibrium expression--it is going to be the concentration of HF squared (stoichiometric coefficient), over the concentration of H₂, times the concentration of F₂.0283

OK, now let's go ahead and calculate the initial concentration of H₂.0298

Well, it says we have 3.00 mol of H₂ in a 3.000-liter flask; so we have 1.000 Molar H₂.0304

That is what this little 0 here, down at the bottom is: H₀; it just means the initial concentration.0319

If you want to put an i, that is fine, too.0324

F₂, the initial concentration--it says we have 6 moles of fluorine gas, over (again, it's in the same flask, so it's) 3 liters, so our concentration is 2.00 Molar.0327

Well, Q--the next thing we want to do is, we want to calculate the Q in this case, the reaction quotient, to tell us what direction it is in.0343

There are a couple of ways to do this in this particular problem: we can just plug it in; the concentration of HF to begin with...well, there is no HF, so it's just 0...over 1, times 2, which is 0.0350

That is fine: you can do it that way, or you can just say, "Well, since there is no HF to begin with, I know the reaction is going to move in that direction; there is none of this yet--there is only this and this--so it has to move forward; there is nowhere else for it to go."0364

But, I think it is best to stick with the process, and just go ahead and put the numbers in, and do it that way.0382

OK, so this is 0, which implies that the reaction will move to the right.0388

Moving to the right means it will form product and it will deplete reactant.0397

H₂ and F₂ concentration will go down; HF concentration will rise.0402

Now, we can do our ICE chart: H₂ + F₂ (I'll give myself plenty of room here) goes to 2 HF; Initial, Change, Equilibrium.0407

We start off with 1 Molar of hydrogen; 2 Molar of fluorine; 0 Molar of HF.0421

This is going to deplete by x; this is going to deplete by a certain amount, x; this is going to increase by an amount, 2x, because of that 2.0429

That is the whole idea: 1, 1, 2.0440

We get equilibrium concentrations of 1.000-x, 2.000-x, and we get 2x.0443

Now, we plug in these equilibrium concentrations into this expression, and since we already know what that is, we solve the algebraic equation.0454

So, let's write: K is equal to...well, it's equal to the concentration of HF squared, so it's 2x squared, over 1.000-x, times 2.000-x.0472

That is our equation, and we know it is equal to 1.15x10².0494

They gave us the K; now, let's solve for x.0500

Well, this is just a quadratic equation; so you are just going to have to sort of do it.0503

You can...there are several ways you can do this: you can do it by hand, do the quadratic formula; you can go ahead and use your graphic utility, for those of you that have the TI-83s and 84s and 87 calculators.0515

You can handle these really, really easily; that is what I prefer; that is what I use; I have a TI-84--that is how I solve these.0527

Let's just go ahead and at least work out the algebra, and then we'll just write out the answer, presuming that we actually used a graphical utility or something like that to do it.0534

We end up with 4x²=1.15x10² (2.000-3.000x+x²), and we get 4x²=230-345x+115x².0546

And then, we end up with 111x² - 345 x + 230 = 0.0582

And again, when I put this into my graphical utility, and I solve, I end up with (again, this is a quadratic equation, so I have 2 values): the first x-value is 0.968 Molar, and the second root is going to be 2.14 Molar.0592

Now, we can't have both of these be true; there is only one equilibrium condition; one of these has to be true.0611

Here is how you decide: well, take a look at the original concentration--1 Molar of hydrogen gas, I think it was.0617

Well, 1 molarity - 2.14 molarity--that is going to give you a negative 1.14 molarity; you can't have a negative concentration, so that one drops out.0626

Let's do that in red; that one drops out--this is the x-value.0637

So, we have that .968 Molar is the x-value; we plug those back into the equilibrium concentrations in our ICE chart to get the following.0642

Our H₂ concentration is 1.000-0.968=3.2x10^-2 Molar at equilibrium.0653

Our F₂ concentration is 2.000-0.968, is equal to (let's get a better-looking equals sign than that) 1.032 molarity.0669

And our final HF concentration is 2x, so it's 2 times 0.968, is equal to 1.936 molarity (you know what, these numbers are getting strange again...1.936 molarity).0687

And my friends, we have our final solution.0709

Notice, we started off with 1 Molar of H; we end up with 3.2x10^-2; that means most of the H is used up.0714

That makes sense, because again, you are looking at a very, very large K_eq; the K_eq is 115--that means the reaction is very far to the right.0724

It favors the product formation, not reactants.0735

When it has come to equilibrium, virtually no reactants are left over.0740

The only reason that you have 1.32 moles per liter left over of the other reactant: because you started off with 2 moles per liter of that, and the stoichiometry of the H₂ to the F₂ is 1:1.0744

You have used up half of it; that is the only reason that it looks like this number is so big, compared to this number.0757

The K_eq tells you that it is mostly to the right; you started off with no hydrogen fluoride; you ended up virtually 2 moles per liter of hydrogen fluoride.0764

That is confirmed; so, the numbers match up; everything is good.0774

OK, let's see what is next.0780

Example 2: OK, this is going to be an example of an equilibrium problem where the K_eq is actually very, very small.0785

And again, depending on the equation, you might run into some rather complicated things, like cubic, quartic, quintal equations, which you can certainly handle with your graphical utility--it's not a problem--that is the great thing about having graphical utilities.0799

But, the method we are going to show is going to be a slightly simplified version, if you just want to do the math quickly.0810

And then, we will give you a way of checking to see whether any simplifications you made were actually viable--whether you could actually get away with it.0816

There might be situations where you might simplify something, and you can't get away with it.0824

We will give you a rule of thumb for doing that.0827

OK, so Example 2: Gaseous NOCl decomposes to form gaseous NO and Cl₂.0830

At 40 degrees Celsius, the eq constant is 1.4x10^-5; it is very small--that means there is virtually no product at equilibrium.0858

OK, initially, 1.0 mol of NOCl is placed in a 2.0-liter flask.0876

Here we go again: what is the equilibrium concentration of each species?0901

OK, let's write our equation: 2 NOCl decomposes into 2 NO + Cl₂.0923

Let's write our equilibrium expression--that is what we always do: it's going to be the concentration of NO squared, times the concentration of Cl₂, divided by the concentration of NOCl squared (is that correct?--yes, that is correct).0933

OK, so now we do that, and let's see what else we have.0955

Initial concentrations--we have to do initial concentrations.0962

The NOCl equals 1.0 mol over...it looks like a 2.0-liter flask: is that correct?--yes.0967

Let me circle my numbers: 2 (oops, let me use blue) liters, 1 mole; that is my K_eq; OK.0984

Let's go back to red; that is going to equal 0.50 moles per liter, and then I have my initial NO concentration, which is 0, and my initial Cl₂ concentration, which is also 0.0997

Therefore, Q is equal to 0 squared times 0, over 0.5, equals 0, which is definitely smaller--the K_eq is small, but it is still bigger than (the 1.4x10^-5 is still bigger than) 0, which implies that the reaction moves to the right.1015

Moving to the right means we are forming product.1042

We are depleting reactant.1048

OK, so let's go ahead and set up our ICE chart.1053

We write our equation: 2 NOCl goes to 2 NO + Cl₂; Initial, Change, Equilibrium.1056

We start off with 0.50 Molar and none of those; we are going to deplete this by 2x; we are going to form 2x, and we are going to form x.1069

Therefore, the equilibrium concentrations...just add them straight down: what you started with, what you lost, and what you end up with.1086

0.50 minus 2x; 2x; and x; now, let's go ahead and put it into our expression.1093

Our K is equal to 2x squared (that is the squared part--2x squared) times x to the 1 power, divided by that squared, (0.50-2x)².1105

That equals 1.4x10^-5; OK.1132

Well, all right, see now: it is getting a little complicated.1141

You are going to end up with 4x² times x; it is going to be 4x³; I don't know...you are certainly welcome to go ahead and solve this, just because you have a graphical utility; it's not a problem--you can do that.1144

But, let me give you an alternate procedure, which actually makes things at least a little bit more tractable mathematically.1156

Because we notice that the K_eq is very small, which means that the reaction is over here, mostly; that means not a lot of product; this is small, because that is small, meaning that there is not a lot of product--in fact, there is virtually no product at all...it is mostly NOCl still.1169

Well, because of that--because it is mostly NOCl still--that means that NOCl hasn't lost very much.1189

Very little of it has actually decomposed.1197

Well, very little of it has decomposed...well, we started with .5; that means x is probably really, really, really small compared to .5.1199

Because it is so small compared to .5, it is possible to take this term and just leave it out--ignore it.1212

Solve the problem, and then check to see whether it is actually valid or not that we did what we did.1224

So, this is how we do it: so again, to our first approximation, we are going to presume because this is small, it means that most of it is here; that means very little of this is going to decompose; in other words, very little of this is going to form.1233

Because this is small, we are saying it is so small compared to .5 that it is probable that the .5 minus the 2x is going to go unnoticed, so let me just ignore it and solve the easier problem.1246

OK, well, let's see what we can do.1256

We have: It's going to be 2x² times x, over 0.50 squared, equals 1.4x10^-5.1258

Well, what we end up with here is (what we are going to end up with, once we do all the multiplication): we are going to end up with: 4x³ is equal to 3.5x10^-6; x³ is equal to 8.75x10^-7; x is going to equal 9.6x10^-3.1273

We found a value of x, 9.6x10^-3; now, we want to check to see whether we were actually justified in ignoring it.1300

Well, if I take (yes, that is fine) 9.6x10^-3, and if I divide by the 0.50, and I multiply by 100, I am trying to see what percentage of the .5 this 9.6x10^-3 really is.1311

That is what I'm doing; I am trying to see how much of the .5 this is.1341

Well, as it turns out, it actually equals about 1.91%.1346

Now, we did, actually, 2x: so 2x would put us roughly at twice that; but, as you can see, the 1.91% is actually really, really small.1352

As it turns out, if you make an approximation of this nature, and the value that you get ends up being anywhere from about...anywhere less than 5 percent of your total value, it is actually (as a good rule of thumb) a valid approximation.1363

That means, virtually, the mathematics is not going to notice that you actually eliminated that.1379

So, because of the 1.91 percent, the x compared to the .5 (or, in this case, it's going to be 2x, but again, you are still going to be below 5%), you are actually pretty good.1384

And this allows us to sort of keep this value of x, as opposed to having to solve this entire equation, this cubic equation.1394

We just did it in a nice, simple way to get an answer, instead of having to use a graphical utility, and we got this value for x.1403

Given that value of x, we can go ahead and use it now to find our equilibrium concentrations.1411

Our NOCl concentration, final, is equal to 0.50, minus the 9.6...oops, this is going to be minus two times the 9.6x10^-3.1418

Our final NO concentration (I will go ahead and let you finish off the arithmetic here) is going to equal 2 times 9.6x10^-3.1442

Our final Cl₂ concentration is going to equal, well, just x: which is 9.6x10^-3.1461

The rest is just arithmetic, which I will leave to you--nice and straightforward.1473

Hopefully you are getting a sense of the general procedure and of the things that you have to sort of watch out for.1480

OK, so let's see: let's go ahead and close this off with an interesting type of problem; it should go pretty quickly, actually.1487

Let's do...yes, that is fine; we can start it on this page.1499

OK, Example 3: Calculate the value of the equilibrium constant for the reaction O₂ gas + oxygen gas goes to ozone gas, given reaction 1 (this is going to be sort of a Hess's Law kind of problem): NO₂ in equilibrium with NO + O, and K₁ is equal to 6.8x10^-49 (wow, that is really, really small) and that one, which is O₃ + NO goes to NO₂ + O₂; K₂ equals 5.8x10^-34.1508

Let's see what they are asking: they are asking you to calculate the value of the equilibrium constant for this reaction (let's do this in blue), given these two reactions and their corresponding equilibrium constants.1593

Well, we know from Hess's Law that if we want to find a final reaction that involves some of the reactants that we have, we have to rearrange them by either flipping them or multiplying them by constants.1607

In doing so, if we add all of the equations together, and then get our final equation--well, when we did ΔHs, when we did enthalpies, we just added the enthalpy; with Ks, with equilibrium constants, it is actually different.1621

When we add equations to get a final equation, what we do to equilibrium constants is: we actually multiply them.1635

So, let's go ahead and do this one.1643

In order to actually come up with this, I am going to flip Equation 1.1646

I am going to flip Equation 1, and that will give me: NO + O goes to NO₂.1654

Now, when I flip an equation, I take the reciprocal of the equilibrium constant, right?--because you are just flipping products and reactants.1664

That equilibrium constant, now, is 1.47x10⁴⁸--huge!1678

I am also going to flip the second reaction; I am going to make the reactants the products and the products the reactants; when I do that, I end up with NO₂ + O₂ in equilibrium with NO + O₃.1686

Well, this equilibrium constant--again, I flipped it, so I take the reciprocal of that, and I get 1.72x10³³.1702

Now I add these two equations; NO₂ cancels NO₂; NO cancels NO; I am left with O + O₂ goes to O₃; this was the equation that we wanted.1714

Now, in order to get the final equilibrium constant, I have to...I don't add these; I multiply them; it becomes K₁' times K₂'.1735

It equals 1.47x10⁴⁸, times 1.72x10³³, and I end up with getting some huge number, if I am not mistaken.1749

2.53x10⁸⁸: that is huge!--that means that any time oxygen gas and free oxygen atom come together, you will not find them separately!1766

This huge number tells me that the reaction is way to the right; there is no equilibrium here--not really.1782

Any time you have a bunch of equations, if you add them together to come up with a final equation, you multiply the equilibrium constants for all of the individual equations.1789

So, when we add equations to get a final net equation, our final K is equal to K₁ times K₂ times...all the way to K_n, if we had n equations; and that is it.1803

OK, so we have gone ahead and dealt with a fair number of problems in equilibrium.1837

Next time, we actually are going to continue our discussion of equilibrium; we are going to talk about Le Chatelier's Principle, and then we are going to do some more problems involving Le Chatelier's Principle.1843

And Le Chatelier's Principle, just to give you a little bit of a preamble, is basically just: If I have a system at equilibrium, if I stress that system out somehow--if I put pressure on it, meaning if I add this or heat it up or cool it down, what happens to that equilibrium.1854

So, I can shift the equilibrium; well, we know (well, I can tell you) that a system will always seek out equilibrium.1869

When I apply stress to a particular system, the system is going to respond by doing whatever is necessary to relieve that stress.1878

You know this from just being a human being: any time, any system that you apply stress to, the response of that system will be doing the things that relieve that stress.1885

Well, a chemical system behaves in exactly the same way, and it is a very deep, fundamental thing called Le Chatelier's Principle.1896

It is actually quite beautiful; so we look forward to seeing you next time.1903

Thank you for joining us at Educator.com and AP Chemistry.1907

We'll see you next time; goodbye.1910