*Hello, and welcome to Educator.com; and welcome to the first lesson of AP Chemistry.*0000

*Today, I'm going to start by discussing the naming of compounds, because we have to know what it is that we are dealing with in order for us to jump into the chemistry.*0005

*So, let's get started.*0013

*OK, so this is, of course, the periodic table, and I'm just going to do a quick overview of the periodic table, just to get a sense of where we need to be, because this periodic table is what you are going to be using all the time--it's going to be your primary reference in chemistry--as you already know.*0017

*Over here, on the left, we have this first column; these are called the alkali metals.*0033

*Actually, you know what?--let's sort of break it up broadly.*0040

*Generally, everything from about right here onward--everything that is blue, and this orange right here--so the whole left side--these are all metals.*0044

*Up here, starting with boron, silicon, arsenic, tellurium--everything here and up to the right--those are going to be nonmetals.*0053

*So, as you can see, most of the periodic table--most elements (naturally occurring elements) consist of metals.*0060

*These over here are called the alkali metals; the second column is called alkaline earth metals; these here, in the middle--these are called the transition metals.*0066

*These over here are called the noble gases, basically because they are unreactive.*0078

*These are called the halogens.*0084

*These are the standard nonmetals.*0086

*The only thing on the left that is a nonmetal is hydrogen.*0088

*Now, let's talk about some of the diatomic gases.*0091

*There are certain gases (well, certain compounds; not just gases, because bromine is actually a liquid)--certain compounds that occur...*0096

*The periodic table lists actual atoms; astatine, iodine, bromine, chlorine, fluorine, oxygen, nitrogen, and hydrogen--they occur as diatomic gases.*0105

*In other words, there are two atoms that make a molecule of that in the natural, normal state.*0116

*What you are breathing right now is nitrogen and oxygen gas; not nitrogen and oxygen atoms, but there are two atoms stuck together to form a molecule.*0121

*That is really all that you have to know.*0132

*These numbers up on top (the ones right below the symbols): those are the atomic numbers--that is the number of protons and electrons in a neutral atom.*0135

*This number, right down below it (which, for many things, is almost double the atomic number)--that is the atomic mass; that is an average of all of the masses of the isotopes, depending on the frequency of occurrence of the particular isotope.*0144

*Again, if any of this is strange--things or words like isotope and stuff--I would definitely urge you to look through it in your books, but for all practical purposes, it won't come up all too often.*0162

*So, I just wanted you to get a quick overview; and now, we can jump into actually naming the compounds.*0174

*Because, before we do anything else--many of the problems are not going to actually have equations for us, or symbols for us; they're just going to give us names of compounds.*0179

*We have to know what it is that we are writing, so we can write an equation--all chemistry begins with a chemical equation.*0187

*OK; let's start off, quickly, by talking about what ions are.*0193

*Ions are atoms that have lost or gained an electron--that is it; lost or gained one or more electrons.*0201

*Your atoms (your naturally-occurring atoms) are neutral; in other words, they have the same number of protons and electrons, which makes them electrically neutral.*0223

*If I add to those, or if I take away from those, all of a sudden they become charged particles.*0231

*Positive ions are called cations, and negative ions are called anions.*0237

*So, an example of that would be something like this: Mg*^{2+} is a magnesium atom that has lost two electrons.0252

*Because an electron is the thing that is being added or subtracted, when we add (electrons are negatively charged), it carries a negative charge.*0259

*So, for example, O*^{2-}; that is an atom that has gained two electrons, so now it's carrying two extra negative charges.0268

*Magnesium lost two electrons, so now it has two more positive charges than it does negative charges; that is why we put the *^{2+}.0277

*Fluorine is *^{-}; potassium is ^{+}; aluminum is ^{+3}; iron(II) is that; there is also an iron(III)--and we'll talk about that in a minute--why it is that certain transition metals actually can lose more than one electron, and they have different oxidation states.0285

*That is it--ions, and the first thing that we're going to discuss, of course, is ionic compounds: simple binary ionic compounds, where the cation is made of one element and the anion is made of the other element.*0306

*Let's go ahead and get started.*0319

*OK, so now let's go to ionic, and we're going to go from the symbol to the name.*0323

*We need to be able to go from the symbol and be able to name the compound, and we need to be able to go backwards (from the name to the symbol).*0331

*So, from symbol to name--we'll do that first.*0338

*NaCl...now, if you look at your periodic table, you're going to notice that the alkali metals on the left, the alkaline earth metals (which is the second), and then on the right (if you ignore the noble gases), you're going to have the halogens, and then you're going to have the next group and the next group, working to the right.*0344

*As it turns out, the metals always lose electrons; nonmetals always gain electrons.*0365

*So, when we're talking about ionic compounds, it's always going to be the metal that is going to be positive, and it's going to be the nonmetal that is going to be negative.*0376

*Now, what we do when we put together ionic compounds is: we are just putting them together in such a way that we actually cancel the charge.*0385

*But again, we will get to that in just a minute, when we're dealing with name-to-symbol.*0394

*So, sodium...this NaCl, the name for this is sodium chloride--that's it.*0399

*Anytime you see something like this, basically, what you do is you take the name of the metal (sodium); you take the base of the nonmetal, and you add -ide to it; that's it.*0404

*How about MgBr*_{2}; OK, this is just magnesium, and this is bromine, so it becomes "bromide."0422

*That's it; it is always consistent--it's the name of the cation first (the positive, or the name of the metal, first), and then the name--the root of the nonmetal, plus -ide.*0436

*All binary ionic compounds are named like this.*0446

*How about something like this: Al*_{2}O_{3}; don't worry about what the numbers mean yet--we'll get to that in just a minute.0451

*This is aluminum oxide.*0460

*And then one more for good measure: K*_{2}S; this is potassium sulfide.0465

*That's it; very, very nice.*0477

*OK, so now we're going to do to ionic, and we're going to go from the name to the symbol.*0479

*OK, so if I see something like calcium oxide, how do I symbolize that?*0489

*Well, here is where we have to actually look at the charges; and the charges come from their arrangement on the periodic table.*0499

*The alkali metals are in the first group on the left; when they react with nonmetals, they actually lose one electron; they're in the first group.*0506

*The things in the second column (like magnesium and calcium)--they lose two electrons when they react.*0517

*Well, if we go to the right-hand side of the periodic table (ignore the noble gases), the halogens--they actually gain one electron when they react.*0525

*The next column over (next to the halogens), the one where oxygen and sulfur are--they gain two electrons.*0538

*Then, if you go one more, you're going to end up with the nitrogen and phosphorus column; those gain three electrons.*0544

*So, when you're putting ionic compounds together that consist of a metal from the left and a nonmetal on the right, you look at their charges.*0553

*In the case of calcium, calcium is a 2+ charge, because it's in the second column.*0560

*Oxygen has a 2- charge, because it is the second column over, ignoring the noble gases; it has a *^{2-} charge.0566

*What we're trying to do here is: we need to combine these in such a way that the charges actually cancel.*0574

*Here, the positive 2 and the minus 2 cancel, so the symbol is just CaO.*0580

*In other words, I just need one calcium atom and one oxygen atom to make calcium oxide; the charges balance.*0585

*OK, now let's do something like sodium sulfide.*0594

*Sodium sulfide: well, sodium is in the first column, so it has a +1 charge; sulfur has a -2 charge; it's in the second column over from the halogens (again, ignoring the noble gases).*0603

*Well, this is a +1 and this is a -2; in order to balance this charge, this +1 needs to be +2 to balance this -2.*0615

*Therefore, we need 2 sodiums; so, the symbol becomes Na*_{2}S.0623

*I need two sodiums for a total 2+ charge to balance the -2 charge.*0628

*And again, all ionic compounds have to be neutrally charged--their charges have to balance.*0632

*Let's do something like aluminum iodide.*0641

*Aluminum is in the 1, 2, third column over on the periodic table, when you actually skip the transition metals (we're going to get to the transition metals next--they are handled a little differently).*0652

*So, from your perspective, the first column, the second column...you skip the transition metals, and you go to the other column, where aluminum is.*0661

*That is a 3+ charge, so everything in that column carries a *^{3+} charge.0670

*Al*^{3+}; and iodine is a halogen, so it has a -1; well, how many iodines do I need to balance the 3+ charge? I need three of them.0675

*So, this becomes AlI*_{3}, aluminum iodide.0684

*Let's try calcium phosphide.*0690

*Calcium phosphide is Ca*^{2+}, because calcium is in the second column; phosphorus is in the third column over, so it carries a negative 3 charge when it's reacting with a metal.0701

*Well, how do I balance 2+ and 3-? I can't do it directly, so I need the least common multiple of these.*0711

*The least common multiple is 6.*0717

*In order to make 6 positive charges, I need 3 calciums; and 6 negative charges--I need 2 phosphoruses.*0720

*So, this becomes calcium phosphide; that is it.*0729

*You're just taking a look at the charge on the anion and the cation, and you're arranging them--taking specific numbers of them and putting them together so that the charge is 0--that's all you're doing.*0732

*We'll finish it off with aluminum sulfide.*0743

*Aluminum sulfide: OK, we said that aluminum was 3+, and sulfur is in the column with oxygen, so it is a 2-; so again, we have to make 6, so we end up with Al*_{2}S_{3}, aluminum sulfide; that is it.0750

*OK, now we'll do polyatomic anions.*0765

*Polyatomic anions: again, you have seen them before; polyatomic anions are anions that are made of more than one atom.*0776

*Some examples would be something like ClO*_{3}^{-}; that is chlorate; SO_{4}^{2-}; that is sulfate; maybe PO_{4}^{3-}; that is phosphate.0788

*Let's see...ClO*_{2}^{-}; that is chlorite; and one of the things you are going to notice with the polyatomics--they generally tend to end in -ate or -ite--most of the time.0808

*There are some exceptions, like, for example, the NH*_{4}^{+}; that is called ammonium.0820

*Now, of course, you have seen lists of polyatomic ions; and there is a page, a list, in your book, of polyatomic ions; or, you can go on the Internet, and you can find a list of polyatomic ions--some short, some long--and it lists all of the polyatomic ions that are available.*0828

*These are treated exactly the same way as the regular atomic anions, in the sense that you are treating them as a whole.*0846

*When you name them, you just basically use the entire name of the anion; so these actually work out really, really easily.*0855

*So, let's go from symbol to name first.*0863

*Symbol to name: for example, if I saw something like NaClO*_{3}; well, Na is sodium, and ClO_{3}...when I look it up on the polyatomic ion chart, it is chlorate; so the name is sodium chlorate--very, very simple.0870

*If I have something like KClO, well, K is potassium, and if I look up ClO, it is going to be hypochlorite.*0891

*So, it is potassium hypochlorite.*0902

*If I had Ca(CN)*_{2}, well, Ca is calcium, and CN--when I look it up, it is cyanide; so it is calcium cyanide.0908

*So it's very, very simple--you are handling it the exact same way: the name of the metal first, and then the name of the polyatomic anion.*0923

*Again, these are treated as a whole; it's like--just think of it as one unit that happens to have a specific charge on it: it could be -1, -2, -3...that's it.*0929

*And again, there is a whole list of these atomic...you will use them enough times so that you'll eventually memorize them, or perhaps you already have them memorized.*0940

*OK, so now let's go the other way; let's go from the name to the symbol.*0949

*Now, let's go from name to symbol.*0955

*Let's do potassium phosphate.*0960

*If you see "potassium phosphate," well, potassium is in the first column; so, when it reacts as an ion, it is K*^{+}.0963

*When you look up "phosphate," it is PO*_{4}^{3-}; we are doing the same thing.0977

*We just need to put them together in such a way that the charges balance.*0982

*In order for the charges to balance, we need three potassiums in order to balance the three negative charges; so we write it as K*_{3}PO_{4}: potassium phosphate.0986

*How about something like aluminum nitrate?*0999

*Well, aluminum has a 3+ charge; nitrate, when you look it up--it is NO*_{3}, and it carries a -1 charge.1008

*I need three nitrates to balance the 3+ on the aluminum, so it becomes Al, and whenever you need more than one polyatomic, you of course put it in parentheses, so it is written as (NO*_{3})_{3}.1016

*Now, this three--that means you have three NO*_{3}s; so it's three nitrogens, nine oxygens.1030

*Now, let's do one more: magnesium hydroxide.*1037

*OK, magnesium is a 2+; hydroxide, when you look it up--it is a -1; you need two of these to balance the 2+ charge.*1047

*So, it is Mg(OH)*_{2}; good--nice and easy.1054

*OK, so now we're going to move on to ionic compounds involving transition metals.*1060

*Ionic compounds involving transition metals: OK, now let's take a look at two compounds, just to start with.*1069

*Let's take a look at Fe*_{2}O_{3}, and let's take a look at FeO.1084

*Well, if I ask you to just name these the way we have been doing, you take the name of the metal; you take the name of the anion, add -ide to it, and you will get iron oxide.*1091

*Well, it's true--this is iron oxide; however, there is this other compound, that is FeO; this is also iron oxide.*1101

*These are two entirely different compounds with completely different chemistry; how do we differentiate between the two?*1109

*As it turns out, transition metals (all of those things in the middle of the periodic table)--they can actually lose different numbers of electrons--the same atom.*1114

*For example, you can have iron 2+, iron 3+; you can have manganese 2+, manganese 4+, manganese 6+...so these things--we actually have to specify, in parentheses, in the name, how many electrons it has lost--in other words, its charge.*1124

*So, in this case (and here is how you actually end up doing it), let's take Fe*_{2}O_{3}.1140

*Fe*_{2}O_{3}; well, what do we know about oxygen? Oxygen, when it reacts, always carries a -2 charge.1149

*There are three oxygens, so the total charge is -6.*1156

*Well, this -6 has to be balanced by a +6, because this is a neutral compound, right?--it's neutral.*1160

*This +6 charge is divided among two irons; that means that each iron is carrying a +3 charge; so what we do is: we write III in Roman numerals, in parentheses, right after the iron.*1167

*So, it's not just iron oxide; it's iron (III) oxide.*1183

*This thing in parentheses tells me the charge on an individual iron atom.*1186

*Now, if you want, you can write it as...it depends on your teacher; I actually use Arabic numerals instead of Roman numerals, so I just write iron (3), like this: iron (3) oxide; it really doesn't matter.*1191

*Some teachers like Roman numerals--it's more traditional; some teachers actually don't really care; but obviously, depending on your teacher...if they want Roman numerals, give them Roman numerals.*1204

*You don't want to be losing a point here and a point there for silly reasons like that.*1215

*Now, this one--how about iron oxide: which iron is this one?*1220

*Well, oxygen is -2; there is only one of them, so it's -2; that means it has to be balanced by a +2.*1224

*This +2...there is only one iron here, so this is actually iron (2) oxide.*1231

*So, when we are naming transition metal compounds, we have to specify the charge on the transition metal.*1236

*The charge is different, depending on how it has reacted: we express that with these parentheses.*1243

*Let's do a few of them.*1249

*Let's do Co*_{2}(CO_{3})_{3}.1252

*Now, again, these transition metals--they can react with individual atoms, nonmetals, or they can react with polyatomic ions.*1267

*Here, we have a cobalt that has reacted with a carbonate.*1275

*Carbonate--when you look it up, you know the name--it's carbonate; the charge on carbonate is 2-.*1279

*The charge on carbonate is (you know what, I'm going to do this a little lower, so I can...let me put it over here...Co*_{2}(CO_{3})_{3}) -2.1287

*There are three of them, so it gives us a total of -6.*1300

*This -6 has to be balanced by a +6, because cobalt carbonate is neutral; and this +6 is actually divided by...there are two cobalts that are carrying that total of 6 charge, so each cobalt has a +3.*1303

*So, we name this cobalt (3) carbonate.*1321

*That is it--very, very straightforward.*1327

*Let's do one more: let's do Mn(SO*_{4})_{2}.1330

*When I look up SO*_{4})_{2}, or if I know it, I know that SO_{4} is sulfate; it has a -2 charge; there are two sulfates, therefore the total charge is -4.1336

*That is going to be balanced by a +4, because this is neutral.*1345

*There is only one manganese, so that one manganese is carrying the entire 4 charge.*1348

*So, this becomes manganese (4) sulfate.*1352

*That's it--very, very straightforward.*1361

*OK, so now, let's go from name to symbol.*1364

*Name to symbol: let's say we have something like palladium (2) acetate.*1371

*OK, these are really, really simple; they are handled exactly the same way: you are given the charge on palladium--it's right there in parentheses; it tells you that it's a positive 2.*1382

*Acetate, when you look it up--it is going to be (I'm going to symbolize it as AcO) a negative 1 charge, so palladium acetate...well, this is -1; this is +2; which means you need 2 of these to balance that +2 charge.*1390

*So, you end up with Pd(AcO)*_{2}; that's it.1403

*Let's try chromium chloride.*1415

*Let's give ourselves a little bit more room here; I don't want to use up too much here.*1419

*Chromium (6) chloride: OK, chromium (6)--that is Cr*^{+6} (or ^{6+}; it doesn't matter--I actually generally tend to write it with the number first).1425

*And then, chloride is a -1 charge; well, we need 6 of these to balance that.*1440

*So, what we get is CrCl*_{6}.1445

*That's it; that is chromium (6) chloride.*1449

*If it were chromium (4) chloride, it would be CrCl*_{4}; chromium (2) chloride--CrCl_{2}; nice and easy.1451

*So now, we have done ionic compounds; now we're going to name covalent compounds.*1466

*Naming covalent compounds: a covalent compound is a compound that consists of a nonmetal-nonmetal bond.*1473

*Ionic compound--it always involves a metal with a nonmetal; covalent is nonmetal-nonmetal.*1488

*Nonmetal-nonmetal bond: OK, so now let's go from symbol to name.*1499

*If we have something like PCl*_{5}; all right, so phosphorus and chlorine--they are both nonmetals.1510

*When we name these, the -ide ending stays the same, but what we use is: we need to specify the actual numbers here now--how many (oops, let me get rid of these lines--so PCl*_{5}) of each atom there is; and we use the prefixes mono-, di-, tri-, tetra-, penta-, hexa-, hepta-, octo- or octa-, nona-, and deca-; so, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.1518

*This would actually be called phosphorus (actually, I think that is spelled with a **u*; you know what, I can't even remember now) pentachloride.1563

*Now, notice: when this first atom only has one atom in it, we don't use the mono- prefix; we just say phosphorus pentachloride, carbon dioxide.*1582

*However, if we had something like this: CO; if the subsequent atoms after the first have just one atom--this is carbon monoxide.*1595

*We use the mono- for subsequent atoms; we don't use it for the first atom.*1608

*That is carbon monoxide; and let's see--let's try N*_{2}O; N_{2}O is dinitrogen monoxide.1614

*You might have something like N*_{2}O_{5}, which is dinitrogen pentoxide (you could say pentaoxide, but...it's up to you; I say pentaoxide; you could just drop the *a*, and do pentoxide).1629

*Again, very, very simple--just specify the number of atoms that you have; that is all you are doing.*1649

*OK, so now, let's go from name to symbol.*1654

*Name to symbol: let's do nitrogen trifluoride--very, very simple.*1662

*Nitrogen trifluoride--just put them in: NF*_{3}; probably the easiest thing in the world.1671

*Let's see; let's try dichlorine monoxide; that is going to be dichlorine, Cl*_{2}, monoxide, O: Cl_{2}O; that's it.1677

*OK, now let's go ahead and finish up with a discussion of how to name acids.*1693

*Naming acids: Acid...basically, you are going to have some anion, any anion, balanced by an appropriate number of hydrogen ions.*1702

*Now, acids are interesting; hydrogen is in the first column; it's right above the alkali metals.*1734

*When it loses an electron, it loses one electron completely, so it becomes H*^{+}; well, H^{+} reacting with some anion (which is maybe a 1-, 2-, 3-)--they're going to combine.1739

*Acids are a little different, because...some people consider them ionic compounds; some people consider them covalent compounds; acidic behavior is more about behavior.*1758

*We will, of course, talk about acids in detail a little bit later on (in fact, in a couple of lessons), and then again in the middle of the course.*1768

*But right now, we just want to worry about naming them.*1775

*An acid is always going to have the H first; so if you see something like this: CH*_{4}; that is not an acid.1778

*But, if you see something like HCl, that is an acid.*1785

*Just by convention, again, the cation (which, in this case, is the hydrogen) comes first.*1789

*Let's go from symbol to name.*1794

*Let's say that we have something like HBr.*1798

*All right, now this is called hydrobromic acid.*1802

*OK, here are the rules.*1812

*When an acid does not have...well, actually, I'm going to do this a little bit differently.*1817

*Let me write one more, H*_{2}S, so that you can see the pattern: hydrosulfuric acid.1836

*When we are dealing with a binary acid...binary means that it's a certain number of H's, and the anion is actually just a single atom--bromide, fluoride, sulfide, iodide, things like that); when it's just those two, it's a binary acid.*1849

*Now, if it doesn't have oxygen in it--there is no oxygen--then the prefix is hydro-, and then it ends with an -ic; so, hydrobromic, hydrosulfuric...if you saw HCl, this is going to be hydrochloric.*1869

*Again, notice, we take the root...the root...the **root*, and we add -ic to it, and that is always consistent.1905

*If you have a binary acid that doesn't have any oxygen in it, that doesn't have any polyatomic ions, it is always named the same way.*1913

*So, let's say HI; this is going to be hydroiodic.*1923

*It's always a hydro- prefix and an -ic ending, always; there is no exception.*1932

*Now, let's do something else; let's say if you see an acid like this: H*_{2}SO_{4}, or if you see HClO_{3}, or if you see H_{2}SO_{3}, or let's say HCm...no, that's going to be a couple of different things...let's say HNO_{3}.1942

*OK, these acids actually have oxygen in them; these are named in a different way.*1969

*The ones that have oxygen in them are generally going to be associated with a polyatomic ion: notice, this is sulfate; this is chlorate; this SO*_{3} is sulfite; this NO_{3} is nitrate.1975

*You might have HNO*_{2}; this is called nitrite.1989

*OK, here is how these are named.*1995

*You don't use the hydro- any time you have an acid that has oxygen in it.*2000

*This is just the root of the element that makes up the polyatomic ion, and if the name ends in -ate--if a polyatomic ion name ends in -ate--you use the -ic ending; if the polyatomic ion name ends in -ite, you use the -ous ending.*2004

*This one would actually be called sulfuric acid.*2030

*Again, it's just a balance of charge: SO*_{4} is just 2-; H is +1; that is why you have 2 hydrogens.2039

*That is all you are doing: you need to still balance the charge--this is still treated like an ionic compound.*2048

*This is sulfuric acid; well, if I look up ClO*_{3}, that is chlorate; it ends in -ate, so the -ate becomes -ic; so it becomes chloric acid.2053

*This one, SO*_{3}, is sulfite; therefore, this becomes sulfurous acid (remember, if it ends in -ite, the -ite changes to -ous; if it ends in -ate, it ends in -ic).2067

*Anything with an oxygen in it does not have the hydro- prefix.*2082

*This right here is nitric acid, because it's from nitrate.*2087

*This HNO*_{2}...NO_{2} is nitrite; therefore, this is nitrous acid.2095

*That's it; that is how acids are named.*2106

*This is going from symbol to name; if we go the other way around...let's say, for example, you had something like chlorous acid.*2108

*Chlorous acid...well, this -ous ending is telling me that I am dealing with a chlorite anion; therefore--I know chlorite is ClO*_{2}^{-}, so--I know that H is +1; they balance; so the symbol for this is HClO_{2}; that's it.2122

*Let's do a couple more: let's say if we said...how about chromic acid?*2146

*Chromic acid: well, it ends in -ic, so it comes from a polyatomic ion that ends in -ate; as it turns out, it is chromate, CrO*_{4}.2155

*It is 2-; well, H is +1; how many H's do we need to balance the 2- charge? We need 2 of them.*2166

*So, the symbol is H*_{2}CrO_{4}; that's it.2173

*Let's say you had something like...how about phosphoric acid?*2182

*Phosphoric acid: well, phosphoric acid ends in -ic, so we know that we're dealing with the phosphate anion.*2191

*The phosphate anion is a 3- charge; H is a +1 charge; so I need 3 of these: H*_{3}PO_{4}.2200

*That is the symbol for phosphoric acid.*2209

*OK, so now we have talked about how to name compounds, how to go from the symbol to the name and from the name to the symbol.*2211

*This is going to be, of course, ubiquitous throughout the course.*2217

*You absolutely have to be able to understand how to name the compounds.*2221

*On the AP exam, sometimes the symbol will be given; often, in the free response section, the symbol will not be given.*2225

*You have to come up with the symbol--not only the symbol, but you have to come up with the equation; in fact, one of the last sections of the free response is a series of reactions, and they're just going to give you the names.*2232

*They're going to say, "This reacts with this; what happens?"; you have to know the symbol and you have to know the reaction; so naming--if you can't do naming, unfortunately, nothing else will work...so there is that.*2246

*OK, so I'm going to close off this discussion with a problem in percent composition.*2258

*Again, this is sort of just a general review of the basic techniques of chemistry, that I'm just going to go over quickly.*2264

*One sort of standard problem is percent composition; in other words, if I have some compound, like C*_{6}H_{12}O_{6}, which is glucose, what percentage of it, by mass, is carbon? What percentage of it is hydrogen? What percentage of it is oxygen?2270

*I'm just going to do one of those problems, just so you see the general process--as a review.*2289

*Percent composition by mass (and generally, they'll be talking about mass; if they talk about anything else, they'll specify what they mean): OK, so calculate the percent composition of each element in Mg(NO*_{3})_{2}.2297

*This is magnesium nitrate.*2335

*Well, let's see what we have.*2339

*We need...basically, when we are calculating percent composition--as you know, a percent is a part over the whole, so we're going to take the mass of magnesium over the whole, the mass of nitrogen over the whole, and the mass of oxygen over the whole.*2344

*Let's see what we have: magnesium is 24.31 grams per mole, and then we have 2 nitrogens--that comes to 28.02 grams per mole (and a gram per mole is the unit of molar mass, which I'm hoping you're familiar with), and 6 oxygens gives us 96 grams per mole.*2358

*OK, so the total mass is going to be 148.33 grams; therefore, the percent magnesium is equal to 2431 divided by 148.33, times 100, and you end up with 16.39%.*2388

*That means, of this 148.33 grams, about 16 or 16 and a half percent of it is magnesium; that is all that means.*2420

*Percent nitrogen: it equals 28.02 divided by 148.33, times 100, and you end up with 18.89%.*2428

*That means, of that mass, about 19% of it is nitrogen.*2445

*Then, the percent oxygen; you can either calculate it, or you can just add these two and subtract from 100; and you're going to get the same answer.*2449

*96 divided by 148.33, times 100...and you end up with 64.72%.*2459

*So, in magnesium nitrate, the majority of the mass is actually occupied by oxygen.*2470

*OK, thank you for joining us here at Educator.com for our first lesson of AP Chemistry.*2478

*We'll see you next time; goodbye.*2482

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

*Today, we're going to be discussing kinetic-molecular theory and properties of real gases.*0003

*Up until now, we've been discussing the ideal gas law, PV=nRT; well, the real gases actually behave well at low pressures and high temperatures.*0008

*When the pressures start to get high (let's say above 3, 4, 5, 6 atmospheres), and the pressures start to drop, volume starts to drop--at that point, the gas behavior starts to deviate from the ideal.*0023

*Towards the end of this particular lesson, we will discuss about how we adjust for that--how van der Waals adjusted for that.*0037

*Let's go ahead and get started.*0043

*We're going to run through the kinetic-molecular theory first.*0047

*So, again, what we are about to list right here--the four or five axioms of the kinetic-molecular theory--are precisely that: they are axioms; they are observations that we have made, and they are assumptions that we are making.*0050

*Based on those assumptions, we can start to build a theory, and hopefully have it correspond with what we observe empirically (in other words, what we get from experimentation).*0062

*Let's just list them out.*0073

*Let me see: our first axiom: The kinetic energy of an ideal gas (and again, we are talking about ideal gases' behavior--toward the end, we will discuss the deviation from ideal behavior, but the kinetic-molecular theory applies to ideal gases) is directly proportional to the temperature.*0076

*What that means, in terms of an equation, is: KE (the kinetic energy--the energy of motion of the molecules of the gas) is equal to 3/2 RT.*0118

*Now, R is the gas constant; however, here it is not equal to .08206 liter-atmosphere per mole-Kelvin.*0131

*In this particular case, because energy is expressed in Joules, and T here is in Kelvin, this R has to be Joules per Kelvin per mole; so, as it turns out...let's see: 8.31...it's going to be 8.31 Joules per mole-Kelvin.*0139

*And again, it's just a way of making the units work out; that is all it is; it's the same constant, R, the Rydberg constant; it's just so that it actually works out in terms of the appropriate units.*0163

*T is the temperature in Kelvin.*0172

*OK, now, also, what I'll put down here (this is a very, very important relationship--notice, it establishes the relationship between the energy, or the temperature; that is really what is going on here--so temperature is really just a measure of the average kinetic energy of the molecules)...*0182

*Most people confuse temperature and heat; temperature and heat are not the same thing.*0200

*Temperature is a measure of the motion, random motion, of the molecules; heat is that energy that actually flows from something hot in the direction of something cold.*0204

*They are two different things; temperature does not measure heat; it measures kinetic energy--it measures the random motion of the molecules, the vibrations, the bouncing around of each other.*0214

*OK, let's say there is one other little mathematical thing I want to put down here: the average kinetic energy of a gas particle (this one is not altogether that important, but I might as well just put it down here) equals 1/2 mass, times the average velocity squared.*0223

*Again, in this particular case, mass has to be in kilograms.*0249

*When velocity is expressed in meters per second, we get our kinetic energy in Joules.*0255

*That is why we need mass in kilograms.*0261

*OK, the second assumption of the kinetic-molecular theory is that the particles are so small (the particles of gas that are bouncing around) compared to the distance between the particles that the volume of the molecules is negligible.*0264

*In other words, molecules are very, very tiny; and when they're bouncing around in a gas, they are really very far apart from each other; so essentially, you can think of them as volume-less point particles that are just bouncing around.*0308

*Now, obviously, when we squeeze, when we drop the volume, increase the pressure, lower the temperature--when the volume gets really, really tiny--now the volume of the molecules actually plays an important role.*0318

*We can't really make this assumption anymore; we have to adjust this; but, for kinetic theory--ideal behavior--we just ignore the volume of the actual particles themselves.*0329

*They do have volume; they're just not important.*0339

*Three: the third axiom is: Particles exert no forces on each other.*0345

*In other words, they are just bouncing around randomly; one has nothing to do with the other; they don't stick together; they don't repel; they just bounce off of each other in perfectly-elastic collisions.*0364

*Now, that is not the case--we know for a fact that it doesn't happen that way--but for ideal behavior, we can make this assumption.*0377

*For all practical purposes, it behaves this way at low pressure.*0383

*OK, now the fourth one: The particles are in constant motion (this makes sense), and the collisions (oops, this random stray line all the way across the page--let's get rid of it here) with the walls of the container are the cause of exerted pressure.*0388

*So, if I have some gas in a closed vessel, and I measure the pressure, the pressure of that gas is coming from the particles--the gas molecules or atoms--bouncing and hitting the walls of the container.*0444

*That is where pressure comes from; it's just like a ball that hits a wall--just imagine billions and billions and billions of balls hitting a wall--well, that wall is going to feel it!*0457

*So, these are the assumptions of the kinetic-molecular theory of an ideal gas.*0466

*Now, from this, we can go ahead and deduce some things.*0473

*I'm not going to go ahead and derive any of these equations; I'm just going to throw them out, because, again, we're concerned with using the equations--not necessarily where they come from.*0477

*You can go ahead and follow the derivation in any one of your textbooks--they are either in the appendix or in the actual text itself--so I'll let you look at them if you want to; it certainly helps if you do.*0484

*If not, no harm--no foul--we're just going to be able to use them; that is what we're going to do.*0496

*The root is equal to (let's see) velocity and 3R...actually, that's going to be for the whole thing...3KT over m: so, the root mean square speed--just think of it as the average speed at which a particle is moving--is equal to 3 times K times T times m.*0502

*Now here, K is something called Boltzmann's constant, a very, very, very important constant--probably the single most important constant in physics (at least in my opinion--other people would say that the Planck constant is--they are closely related, in fact).*0540

*Boltzmann constant: that is 1.38x10*^{-23} Joules per Kelvin.0560

*T is temperature in Kelvin, and m is mass in kilograms.*0573

*It is the mass of an individual particle in kilograms; that is why we use the small m instead of the capital m.*0582

*Now, let's go ahead and write this same thing as 3RT/M, where R is, as we said before, the Rydberg constant: 8.31 Joules per mole-Kelvin--and that actually happens to equal Boltzmann's constant times Avogadro's number, 6.02x10*^{23}.0587

*So, this is also a really interesting relationship to keep in mind--that R, the Rydberg constant, equals the Boltzmann constant times Avogadro's number.*0623

*8.31 Joules per mole-Kelvin equals 1.38x10*^{-23}, times 6.02x10^{23}.0632

*This is where you end up getting R from.*0640

*M, the capital M, is the molar mass.*0643

*This just gives me average--the root mean square speed of gas particles.*0649

*OK, so let's throw a couple of other definitions out there.*0657

*Essentially, what I'm going to be doing is just sort of laying out these things--what they are--defining them, and then, once I have them on the page, I'm going to go ahead and use them to solve some problems.*0660

*But, I just wanted to lay them out as they are, as opposed to laying one out, doing a problem...I'm going to save the problems until the end.*0669

*There is something that we call effusion, and effusion is nothing more than the passage of a gas through a tiny opening--that is it.*0677

*So, if I have a balloon and I poke a little needle hole in it, the gas from the inside of the balloon effuses out.*0696

*The rate of effusion is just the rate at which it actually comes out.*0702

*So, it's going to be a certain volume per unit time, like, let's say, 10 milliliters per second.*0708

*That means 10 milliliters of air are escaping for every second; that is all it is--a rate is just an amount over a unit time.*0713

*But, you know this already.*0721

*Well, somebody by the name of Graham discovered that the rate of effusion is inversely proportional to the molar mass of the particular gas.*0723

*Or, we can write it as: The rate times the molar mass is equal to a constant.*0736

*Well, for two gases under similar circumstances--for the same temperature and pressure--they are going to equal the same constant.*0750

*Therefore, what you have is something like this: you have: The rate of the first gas, times its molar mass, equals the rate at which the second gas effuses, times its molar mass.*0760

*This is how I like to use it; however, it is totally equivalent to (and more often than not, you will see it written like this): The ratio of the rates--rate 1 over rate 2--is equal to the molar mass of 2 over the molar mass of 1.*0774

*It really doesn't matter: you can write it this way; you can write it this way--it's a totally personal choice; most books will write it this way, because they like to have ratios of the same thing.*0794

*The rate of one over the rate of the other equals the square of the molar mass of one over the square of the molar mass of the other...absolutely the same thing.*0801

*OK, now let's define something called diffusion.*0812

*Diffusion just means one gas mixing with another.*0816

*If I open up a vial of pure ammonia, and if I put it on the table, let's say 5 feet away from me, it's going to take a little while, but eventually, I'm going to actually smell the ammonia.*0826

*Well, the ammonia is mixing with the air--the oxygen and nitrogen--the 70%...air is nitrogen and oxygen; it's mixing with the air, and diffusion is just the extent to which it mixes.*0837

*The rate of diffusion is how quickly it actually mixes.*0851

*Effusion and diffusion are actually closely related; you can actually use the same equation for both.*0855

*Again, we will talk more about this when we actually do a problem; it will make more sense; but I just want to throw out the meanings.*0861

*So, effusion: how quickly a gas escapes from a tiny opening; diffusion is how a gas actually mixes with another gas--how quickly it penetrates that other gas.*0866

*OK, now let's go ahead and talk about real gases, as opposed to ideal gases.*0879

*Well, we know that the ideal gas law is PV=nRT; I am going to rearrange this, and I am going to write it as nRT/V, and here is the reason that I am going to do this.*0888

*We said that the ideal gas law works for low pressures--low pressures where the gas molecules are flying around; they're really far apart from each other; the individual volumes don't matter; but, if I were to all of a sudden take a volume, where the individual gas particles really fall apart, and they are floating--they don't matter--and I increase the pressure, by increasing the pressure, I reduce the volume.*0902

*So now, I have dropped it down to something like this, where the same number of particles...now the particles don't have as much space as they did before.*0930

*So now, the volumes of the individual atoms and molecules start to matter.*0944

*What ends up happening is that this initial volume...when we use this particular volume, now, because there are so many particles, and the volume of the particles is actually a fair percentage of the total volume--now, the volume available for the particles to move around in is not the same as--is actually lower than--the ideal volume, by assuming that volumes don't matter.*0952

*So, if I assume I have a point--if I have a particle and another point--there is a certain volume that is available to it; but now, if this point is an actual volume--occupies volume--it's taking away volume, so there is less room for this other particle to move around.*0979

*We have to make an adjustment to this side by reducing the volume.*0993

*We write this as P=nRT/V-nb; and I'll talk about what b is in just a second (n is just the number of moles).*0997

*This is a volume adjustment; it is saying that, as we increase the pressure or decrease the volume or lower the temperature (which also decreases the volume), now the volume of the individual particles matter, and I have to make an adjustment for the volume.*1011

*There is less volume available for the particles of gas that are initially there to move around in.*1024

*It is not V; it's less than V--that is why it is V-nb.*1030

*OK, now, the next adjustment is this: we assumed, in the kinetic-molecular theory for ideal gases, that the particles exert no forces on each other.*1034

*They have no attractive force to each other.*1042

*Well, as it turns out, particles do have an attractive force to each other, and in fact, the more polar the particular molecule (like, for example, water molecule)--they're going to stick together.*1044

*Because they stick together, the collisions that they experience are not elastic, so the pressure that we actually measure is going to be less than the pressure that it would be under ideal conditions, because now you have fewer particles actually bouncing around and hitting the containers, because more of these particles are actually sticking together.*1055

*There is loss of energy, if you will, in some sense; the total energy of the system is conserved, but individually, there is sort of a loss; so the pressure that we measure is actually going to be less than the pressure...I'll say that the pressure observed is going to be less than ideal pressure.*1078

*That is why I put the "obs" here.*1095

*So, this factor is going to be a, times n over V squared; this is the pressure adjustment.*1097

*This is the volume adjustment; this is the pressure adjustment.*1107

*Now, I'm going to rearrange this again, and I'm going to write it as P+a times (n over V)*^{2}, times V-nb, equals nRT.1110

*This is PV=nRT, but because of real gas behavior, I have made adjustments to the pressure and the volume for real gas behavior.*1126

*Now, this a and b are called van der Waals constants, and we have calculated different constants for different gases.*1135

*You can see them in any chapter on gases in a chemistry book.*1154

*They have them for (most of them list maybe 10 or 12) the most common gases--methane, oxygen, nitrogen, hydrogen...things like that.*1157

*That is all it is; these are just numbers that you put in there; n is just the number of moles of particles that you are dealing with, and this is a better representation of how gases behave at high pressures and low temperatures--in other words, small volumes, because volume matters now.*1165

*That is all this is; this is just PV=nRT, adjusted for real gas behavior.*1183

*OK, let's go ahead and jump into our examples, and I think a lot of this will start to make sense.*1189

*So, our first example is going to be: Calculate the root mean square velocity of atoms in a sample of methane gas (which is CH*_{4}) at 40 degrees Celsius.1194

*So, calculate the root mean square velocity of atoms in a sample of CH*_{4} at 40 degrees Celsius.1227

*Basically, how far is your average atom flying around at?*1231

*OK, well, let's (let me see--we know what we're going to deal with...so...) just use our equation: root mean square speed is equal to 3 times R times T, over the molar mass; square root.*1236

*Well, T (temperature) is in Kelvin; 40 degrees Celsius becomes 313 Kelvin; we have to make sure to work in Kelvin.*1253

*Also, remember molar mass: molar mass has to be in kilograms per mole--not grams per mole, so we know that oxygen is 16 grams per mole, but oxygen is going to be .016 kilograms per mole; that is what is really important here.*1264

*These problems are not difficult; the difficulty is going to be remembering to work in the appropriate unit, so that we actually get our answer in meters per second.*1280

*OK, so methane is CH*_{4}; C is 12; there are 4 H's--that is 16; so, 16 grams per mole becomes 0.016 kilograms per mole.1287

*And now, when we put these numbers in, we get 3, and we said R is 8.31, not .08206, so 8.31 Joules per mole-Kelvin, and the temperature is going to be 313 Kelvin, and we have 0.016 kilograms per mole; all of this under the square root sign.*1304

*Now, when we do the mathematics, as far as the numbers, we're going to get 699 meters per second.*1335

*Now, I want to show you where the meters per second comes from.*1341

*Kelvin cancels with Kelvin; mole cancels with mole; now, what we end up with, as far as units, is Joules over kilograms.*1344

*All right, here we go: J over kg; well, the Joule is kilogram-meter*^{2} per second^{2}; that is the unit--force times distance, a newton times a meter, gives me a Joule.1355

*So, the unit of a Joule is a derived unit; it's kilograms-meters*^{2} per seconds^{2}, over kilograms.1374

*Well, kilograms cancels with kilogram, leaving us meters*^{2} per seconds^{2}; then, when I take the square root of that, I get meters per second.1382

*So, the units work out--very, very important.*1391

*R has to be 8.31 Joules per mole-Kelvin; temperature has to be Kelvin; molar mass has to be in kilograms per mole--very, very important.*1394

*OK, let's move on to a second example here.*1403

*We have...let's see...the problem says: The effusion rate of an unknown gas is found to be (we can measure this) 32.50 milliliters per minute.*1407

*In other words, 32.50 milliliters are leaking out of a hole every minute.*1438

*That is all that is; it's a rate--an amount per time.*1444

*Now, under identical conditions, the effusion rate of oxygen gas (O*_{2}) is found to be 31.50 milliliters per minute.1448

*Is the gas (the unknown gas) methane, carbon monoxide, nitrogen monoxide, carbon dioxide, or nitrogen dioxide?*1482

*Well, let's use what we know.*1500

*We know that Graham's law says that the rate*_{1}, times the molar mass of 1, equals rate_{2}, times the molar mass of 2.1502

*Well, we know the rate of the unknown gas--we measured it--that is 32.50 milliliters per minute (I'll go ahead and leave the unit off--it's not that important).*1514

*I don't know what its molar mass is; I would like to know that, because, when I know that, I can just compare it to the molar masses of my choices and pick the right one.*1526

*Well, I know the rate for the second one is 31.50 milliliters per minute--that is the oxygen--and its molar mass is going to be...it's O*_{2}, so it's not 16; it's 32--32 grams per mole.1534

*In this case, we can use the 32 grams per mole; it doesn't have to be in kilograms per mole, because, again, the ratios cancel out.*1550

*It is OK, as long as the units are the same.*1557

*That is it; so we have that the square root of the molar mass is going to end up being 5.48, and then, when we square both sides, we get a molar mass of about 30 grams per mole, and when we compare it, it looks like nitrogen and oxygen--nitrogen is 14; oxygen is 16; that is 30.*1561

*So, our gas is nitrogen monoxide.*1584

*That is it: Graham's law; rate is related this way; you could put it in that other form, where you have rate*_{1} over rate_{2} = molar mass of 2 over molar mass of 1.1588

*I like it this way, because everything is consistent--1, 1, 2, 2--but it is your choice.*1599

*Let's do another problem here.*1607

*This one says: Calculate the pressure exerted by 0.600 moles of nitrogen gas in a 2.0 liter vessel at 35 degrees Celsius, using a) the ideal gas law, PV=nRT, and b) van der Waals equation (the van der Waals equation was that adjusted one, the P-a over n/V squared--that one).*1614

*We want to compare them to see what we are looking at--to see how closely, actually, ideal behavior and non-ideal behavior is for this particular situation.*1664

*Well, 35 degrees Celsius--that is a pretty high temperature, in terms of Kelvin.*1673

*2 liters volume is a reasonable volume; and it is .60 moles.*1678

*Well, let's just sort of see what happens.*1683

*OK, so a) PV=nRT: we rearrange; we get nRT/V, and we just plug the values in.*1686

*We get 0.600 (is our number of moles); R is .08206.*1695

*And remember, when you are dealing with the ideal gas law, you have to use the .08206 for R; when you are dealing with issues of Joules and things like that, root mean square speed...that is when you have to use the 8.31 Joules per mole-Kelvin.*1701

*The problem itself--if you just stop and take a look at what units you want--they will tell you which R you are actually going to use.*1719

*Then, 308 Kelvin should be for 35 degrees Celsius, and our volume is 2.0 liters.*1726

*We end up with 7.58 atmospheres; that is pretty high pressure.*1735

*OK, now let's use pressure + a, times (let me write it as n*^{2} over V^{2}...no, you know what...I'm just going to leave it as (n/V)^{2}).1743

*Notice this n/V, by the way: it's the number of moles over a volume--it's a concentration.*1759

*The pressure is actually contingent on the concentration, which makes sense; it has to do with: the more molecules you have, the more heavily concentrated, in a given volume--the pressure is going to increase.*1764

*So, just recognize that; that is why I didn't do n*^{2} over V^{2}.1774

*I left it that way so that you could see that it is a concentration term.*1779

*...V-nb=nRT.*1782

*OK, and then, when we rearrange, we get P=(nRT/(V-nb))-(a(n/V)*^{2}).1788

*Now, the pressure adjustment constant (van der Waals constant) for nitrogen gas is 1.39, and (the units are irrelevant; you can certainly look up the units if you want--they are not altogether important) B*_{N2} is equal to...the b, the volume adjustment parameter, is 0.0391.1807

*When we put all of these values in here, we end up with 7.67 for this first term, and we end up with 0.1251 for this second term, which gives us a total pressure of 7.56 atmospheres.*1833

*So notice, under the ideal gas law, we have 7.58 atmospheres; using the real gas behavior, we have 7.56 atmospheres.*1853

*7.56 and 7.58 are virtually the same, so under these conditions, we are welcome to go ahead and use the ideal gas law.*1862

*Which--just by looking at this--35 degrees Celsius is a pretty high temperature; 2 liters--it's a pretty big volume, actually, for .6 moles of gas, so you can pretty much say to yourself, "You know what? I'm just going to go ahead and use the ideal gas law; I don't want to use the van der Waals equation--it's not important."*1870

*This confirms that that is the case.*1889

*If the conditions were different, you might get a lot of deviation.*1892

*OK, so we talked about the kinetic-molecular theory; we have talked a little bit about root mean square speed, average kinetic energy of a gas sample...we have talked about effusion and diffusion and done some problems.*1896

*In the next lesson, we're going to actually sort of tie it all together and do some regular gas problems.*1908

*Thank you for joining us at Educator.com, and we'll see you next time--goodbye!*1914

*Hello, and welcome back to Educator.com.*0000

*Last lesson, we finished off the discussion of gases with real gases and the kinetic-molecular theory.*0002

*This lesson, we're just going to sort of tie it all together; we're just going to do some practice problems.*0008

*We're just going to work on the entire range of gas problems.*0013

*We're not going to do 1,000 problems, but we're going to do a fair sampling, and it will be the type of problems that you will see on the AP exam, and the type of thinking that you're going to have to do, and a little bit of the manipulations.*0017

*So, let's just go ahead and get started--that is the best way to make sense of anything.*0029

*The first example that we're going to do is the following (let's see...go ahead and write 1 here)--the problem is as follows.*0036

*The density of a gas was measured to be 1.25 grams per liter (now, notice: we have grams per liter--it's still a mass per volume; the actual, individual units don't really matter; for gases, grams per liter is pretty standard, simply because gases don't weigh very much).*0046

*So, we measured this density at 29 degrees Celsius and 1.75 atmospheres.*0073

*OK, we want to know: What is the molar mass of this gas?*0082

*A pretty standard problem--it's often how we find the molar mass of a gas.*0089

*What is the molar mass of the gas?*0093

*Now, I can guarantee you that this problem, in one form or another, will show up on the AP exam--perhaps even twice; I guarantee it; it always does.*0098

*So, let's see; let's talk about how we're going to approach this.*0106

*Well, let's just look at the definition of molar mass: they are asking for a molar mass, so what is molar mass?*0110

*Let me use red ink here.*0115

*Molar mass is equal to the mass in grams, over the number of moles; that is it--that is all molar mass is.*0118

*Let me rearrange this equality; let me write it as mole equals mass over molar mass.*0129

*Now, let me use a capital M for molar mass, but I'll go ahead and leave mass itself as...*0139

*And you know this already--you know, when you're finding the number of moles, you take the mass of something and you divide it by the molar mass; you get the number of moles.*0145

*Well, mol is n in the ideal gas law, so let's go ahead and put--wherever we see n, let's put mass over molar mass; how is that?*0152

*We write PV=nRT; that means PV=mass, over molar mass, times RT; it equals--and then we drop the V down; we solve for P; we get P=mass, times R, times T, over molar mass, times volume.*0163

*Now, let me switch some things around here; I'm going to bring the V and put it under the m, under the mass, and I'm going to write mass over V, volume.*0190

*I'm going to separate it out from everything else, and I'm going to write RT over (no, wait...PV over mass...oh, I'm sorry; I made a little mistake here); I drop down the m; I drop down the V; I actually want to drop...I'm looking for molar mass, so that is what I want to solve.*0201

*I want to bring the M up here, drop the V and the P down here...there we go: it looked like there was something wrong.*0229

*And then, I'm going to combine the mass and the mass over V; I'm going to write that as a unit and write RT/P (so this is the same thing--we just played with some variables).*0237

*Well, what is mass over volume?*0247

*You know that it is density; so, there you go--the molar mass is equal to the density (which--I used a Greek small letter delta), RT, over P.*0250

*That is how you find the molar mass.*0266

*If you had it the other way--if you had the molar mass and you wanted the density--you just rearrange it; that is what is nice about this--you just rearrange your variables, based on the units that they are made of, and you can come up with all kinds of different variations of the particular equation at hand.*0268

*That is often how you do it in laboratories.*0282

*So, let's go ahead and put our values in.*0285

*We have the density, which is 1.25 grams per liter; we have R, which is .08206 liter-atmosphere per mole-Kelvin, and then we have the temperature, which is 29 degrees Celsius, which is 302 Kelvin (again, we always work in Kelvin).*0287

*Then, our pressure, it says, is 1.75 atmospheres.*0312

*OK, that is it: so we get: atmosphere cancels atmosphere; Kelvin cancels Kelvin; liter cancels liter; what we'll end up with is g on top, mole on the bottom--grams per mole--molar mass; our units work out.*0317

*Our final answer is 17.7 grams per mole; that is it.*0333

*We took the PV=nRT; we solved the molar mass thing for moles; we put it in; we rearranged, and we get the molar mass equals the density, times R, times T, divided by P.*0339

*This equation right here will show up on the exam, in various forms; you will see it in the multiple choice, and more often than not, chances are pretty good (I would say 80 to 85%) that you will see it in the free response section or the essay section (the part that asks you for some descriptive--where you are not really doing anything mathematical, but you're describing what is going on).*0350

*So, it will more than likely show up.*0370

*OK, so let's go to another example; that is what we are here to do.*0375

*2: we have: Equimolar quantities of nitrogen gas and nitrogen dioxide gas are put into a closed vessel at constant temperature (so, closed vessel--fixed volume; constant temperature; we're good).*0382

*This is going to be an example of a qualitative problem that you see in the free response section, but they are not going to ask you to do any calculations, but they are going to ask you to reason things out.*0419

*So, this is a typical, typical example of what you will see on there.*0428

*Part A (and it comes in multiple parts): A: Which gas has the larger partial pressure?*0432

*These kind of questions want to test your qualitative understanding of what is going on.*0445

*In other words, you might be able to solve a problem mathematically, because you have sort of seen it a thousand times, and you know how to fiddle with it; but if you don't actually know what is going on, you won't be able to answer these questions.*0449

*So, these questions are probably the hardest, in the sense that they are testing whether you can reason things out, based on what you know chemically.*0459

*So, you have to know the chemistry; it is not just the math.*0467

*The larger partial pressure: OK, well, all right; we know that the partial pressure is equal to the number of moles, times R, times T, over V; it's just the ideal gas law for a mixture of gases.*0471

*We know that the total pressure is equal to the sum of the partial pressures; each partial pressure is equal to the number of moles of that particular gas, times RT, over V.*0487

*Well, b (a and b are just the nitrogen and O*_{2})--that equals the number of moles of b, times RT, over V.0496

*It is equimolar--that means the same number of moles.*0504

*That means the number of moles of nitrogen gas equals the number of moles of nitrogen dioxide gas.*0508

*Therefore, the partial pressures are equal.*0513

*The partial pressure of nitrogen gas is equal to the partial pressure of nitrogen dioxide gas.*0516

*You have reasoned it out; there it is.*0523

*Part B: here it is interesting: Which has the greater density?*0526

*Which has the greater density: well, in the problem that we just did, we said that the molar mass is equal to the density, times RT, all over P; let's just solve for the density now, instead of the molar mass.*0538

*So, let's go ahead and convert this to: Density equals P times M, over RT; in other words, P, R, T...those are fixed.*0553

*Molar mass; if the molar mass is higher, the density is higher--that is what this equation tells me; I just rearrange an equation.*0565

*Well, of these two--the nitrogen gas and the nitrogen dioxide gas--which has a higher molar mass? Nitrogen dioxide.*0572

*Therefore, that implies that NO*_{2} has the higher density.0579

*This equation tells me so.*0585

*This is the kind of stuff that you have to do.*0589

*Much of the multiple choice section is also going to be stuff just like this: reasonably simple--you can reason it out in a number of steps--no real math involved (or if there is math, it's simple numbers--2, 4, 6, 8); but there it is.*0593

*It's qualitative understanding; that is what is important.*0605

*Any of your professors will tell you that, if they had a choice between you understanding things qualitatively or quantitatively, your qualitative understanding is actually a little bit more important.*0611

*Math will always come; qualitative understanding doesn't always come; or, if it does, it comes a lot slower.*0620

*OK, C: let's see: Which has a greater concentration?--interesting.*0628

*Which has a greater concentration: Well, what is concentration?*0638

*Concentration--the unit of concentration is defined as (when you see :=, it means "is defined as") number of moles per liter.*0649

*Well, they have the same number of moles--their equimolar amounts were put in--and they are in the same volume flask, so concentration of 1, of nitrogen gas, equals the concentration of NO*_{2} gas--nice!0661

*OK, let's see what is next.*0678

*Which has the greater average kinetic energy?*0682

*OK, which has greater average kinetic energy: in other words, on average, which collection of atoms or molecules are actually moving faster?*0687

*Well, kinetic energy--remember what we said?--3/2 RT is average kinetic energy.*0702

*It is directly proportional to temperature.*0716

*Temperature is constant; therefore, the average kinetic energy of the two gas samples is the same.*0721

*That is kind of extraordinary: different gases, different molar masses--their average kinetic energy is actually the same.*0727

*It is a function of temperature; that is it; it's not a function of anything else.*0734

*OK, E: Which has the greater average molecular speed?*0740

*Energy and speed: not necessarily the same here.*0745

*Which has the greatest average molecular speed: well, U*_{rms} is equal to 3RT over M; so now is where that molar mass thing comes in.0748

*So, nitrogen--the higher the molar mass on average, the slower the molecule.*0779

*Nitrogen dioxide is heavier than nitrogen gas; therefore, nitrogen gas is lighter; therefore, nitrogen gas is faster.*0786

*So, that implies that nitrogen gas is faster--moving around faster.*0792

*OK, now, F: Which gas will show greater deviation from ideal behavior?*0803

*This one is a bit challenging.*0825

*Think about these two molecules: nitrogen gas, N*_{2}, is nonpolar; NO_{2} is a bent molecule--it is polar.0830

*Because it is polar, there is greater charge distribution, which means that the individual molecules of NO*_{2} are going to attract each other more.0847

*Because they attract each other more, the effect that they have on the pressure, the volume, the relationship that we had to adjust in the van der Waals equation, is greater.*0856

*So, NO*_{2} will demonstrate greater deviation from ideal behavior, precisely because it is more likely to experience that deviation because of the pressure adjustment--the volume adjustment.0864

*That's a typical problem that you will see.*0881

*OK, let's see: number 3: A mixture of neon and argon gas at a total pressure (P*_{T}) of 1.4 atmospheres--if there are twice as many moles of neon as argon (so we have twice as much neon as we have argon gas), what is the partial pressure of argon gas?0884

*So, we have a mixture of neon and argon gases; the total pressure in there is 1.4 atmospheres; if there are twice as many moles of neon as argon, what is the partial pressure of the argon?*0940

*OK, this is a mole fraction/partial pressure kind of thing.*0951

*We said that the mole fraction...remember, we said it is the number of moles of the part, over the total number of moles of the whole, which is also equal to, as far as pressures are concerned, the partial pressure of argon, over the total pressure.*0956

*Well, let me rearrange this.*0973

*The partial pressure of argon is equal to the mole fraction of argon, times the total pressure; I know the total pressure--it's 1.4--I should be able to figure out the mole fraction here, and that should give me my answer.*0976

*OK, well, let's see if we can't figure this out.*0988

*Let me use this version of it here; yes, I think I will use this version of it here, instead.*0996

*The mole fraction is also equal to the moles of argon, over the total number of moles, and it says that the moles of argon, let's say, is x.*1004

*OK, so the moles of argon is x; well, the moles of neon--it says that I have twice as many, so it's 2x.*1022

*Therefore, the mole fraction of argon is going to equal x over the total (which is x + 2x); or, if I wanted to do just 1 over 1+2, that is fine.*1029

*That is equal to the partial pressure of argon, divided by 1.4.*1041

*OK, well, x over (x + 2x) is equal to x over 3x; the three x's cancel; that leaves me with one-third, equals the partial pressure of argon over 1.4; I multiply 1.4 times one-third, and I get that the partial pressure of argon is equal to 0.47 atmospheres--which makes sense.*1049

*It's twice as much; there are 3 total; so therefore, it's going to be one-third of that 1.4.*1077

*Again, the mathematics sort of bears it out, based on reasoning it out.*1085

*Mole fraction--just plug it in; everything will fall out.*1089

*Let's see what we have: let's see, #4: we have: 2 moles of hydrogen gas, plus 1 mole of oxygen gas, gives 2 moles of water gas at a constant pressure and volume.*1097

*If T*_{i} is the initial temperature before the reaction takes place (meaning here on the left), what is the final temperature after the reaction takes place (in other words, here on the right)?1125

*Before and after: those are the important words here.*1156

*Be very careful; read the question very carefully.*1159

*Well, there is a change in the system here, right?--so we're going to use the P*_{1}V_{1}/T_{1}n_{1}= P_{2}V_{2}/T_{2}n_{2}.1162

*But, they say that pressure and volume are constant, so they basically drop off.*1175

*That leaves me with 1/T*_{1}n_{1}= 1/T_{2}n_{2}.1180

*Another way of rewriting this is just T*_{1}n_{1}=T_{2}n_{2}.1188

*Again, it is important to be able to see where all of this comes from.*1196

*The equation doesn't just drop out of the sky; you can rearrange it; you can fiddle with it to get what it is that you need.*1199

*Now, you just sort of put them all in.*1204

*We solve this for T*_{2}; we want our final temperature--that is T_{2}.1207

*That is equal to the initial temperature, T*_{i}, times n_{1} over n_{2}--the total number of moles to begin with, the total number of moles that you end with.1211

*Well, T*_{i}...the total number of moles that you begin with is 2+1: 3 moles.1224

*The total number of moles that you end with is 2; so our answer is...that is it; that is the expression.*1233

*The initial temperature, times 3, divided by 2, will give me the final temperature--and it's nothing more than an application of the ideal gas law, with pressure and volume completely ignored.*1240

*All it is is the temperature and the number of moles.*1250

*Before the reaction, the number of moles is 3 moles of gas particles; after the reaction, the number of particles is 2 moles; that is what we have to watch out for.*1253

*Let's see: we'll do a quick multiple choice one here.*1267

*Which of the following conditions would be most likely to cause deviation from ideal behavior in a gas?*1272

*We have 1) Low pressure, 2) Low volume, third choice is Low temperature, fourth choice is High temperature, and our (oops, can't have these random lines here; let me put temp there) 5)--we have High pressure.*1306

*OK, so our choices are: 1 only; 2 and 3 only (let's see); 2, 3, 5; 1 and 4; and the final choice is 5 only.*1347

*Well, which of the following conditions would be most likely to cause deviation from ideal behavior?*1380

*So, deviation from ideal behavior has to do with high pressure; high pressure or low temperature causes the volume to be small.*1386

*A small volume, a high pressure, low temperature: these things tend to induce deviant behavior.*1396

*1 only (low pressure): no, that is not it.*1406

*B: 2 and 3 only--well, low volume: yes, that will do it; low temperature: yes, low temperature will induce a low volume; that will do it.*1414

*High temperature: no; but there is also high pressure--high pressure will also cause a low volume, so this says "2 and 3 only," but 2 and 3 and 5 also work, so it isn't B.*1425

*We look at C (2, 3, and 5): yes, that works.*1437

*Let's check our others, just to be on the safe side.*1441

*1 and 4: 1--no (low pressure); 4 (high temperature)--no, that is definitely not it.*1443

*5 only: well, 5 works (high pressure)--yes, that is certainly viable--but it's not that one only; it's also 2 and 3 (low volume, low temperature, high pressure).*1450

*The idea is the "low volume" part.*1460

*Low volume comes from high pressure; low volume comes from low temperature.*1463

*So, that is what you have to keep in mind; so deviant behavior comes when you have low volumes.*1468

*OK, so that gives you a sampling of the type of problems that you are going to see on the AP exam.*1475

*Some of them are from a multiple choice section; some of them are from what they call the essay section; some from the free response section.*1479

*This is pretty typical of the type of thing that is going on.*1487

*They are not altogether difficult mathematically; it's just that they require a qualitative understanding.*1489

*It is very, very important that you know the chemistry.*1494

*Being able to do the math is nice, but, as you will discover as we go on in the course, a lot of the mathematics tends to be very, very long; it's not a one- or two-step thing.*1498

*If you understand the chemistry, the math will follow; but just being able to do the math doesn't mean that you will be able to reason out the chemistry, because the math can lead you astray.*1506

*As you see, if you miss some of the units, you are going to be all over the place; numbers are not going to make sense.*1515

*Qualitative understanding is what we seek in science.*1520

*Math will always be there; qualitative understanding will not always be there.*1523

*OK, thank you for joining us here at Educator.com; we will look forward to seeing you next time--goodbye.*1528

*Hello, and welcome back to Educator.com.*0000

*Welcome back to AP Chemistry.*0002

*We're going to start, today, on our unit discussing thermochemistry, and there are going to be about three or four lessons discussing thermochemistry--chemistry related to heat.*0004

*Today, we're going to talk about energy, heat, and work--some of the basic, fundamental concepts--and eventually, we will get more into the chemistry of it.*0015

*Let's just jump in and get started.*0023

*The first thing that we want to definitely make sure you know is that temperature and heat are not the same thing.*0026

*So, temperature and heat are not the same thing; this is probably the single biggest misconception.*0032

*OK, so temperature is a measure of the average kinetic energy of the particles in the sample that you are taking the temperature of.*0048

*Heat is the thing that actually flows when there is a temperature differential.*0078

*In other words, heat is the thing that flows from a hot object to something that is colder.*0085

*Or, it doesn't necessarily need to be an object; any time there is...let's say you have some hot thing, or something that is at a higher temperature; it doesn't have to be hot; there just has to be a difference in temperature.*0089

*So, if something is 30 degrees Celsius, and let's say the air surrounding it is 10 degrees Celsius, heat is that energy, if you will, that flows from the higher-temperature object into the lower-temperature object.*0099

*Heat is the energy that actually flows; what temperature does is actually measure the random motions of the molecules.*0113

*There is a difference between the two; they are related, but they are not the same thing.*0121

*Heat is the thing that flows from hotter to colder--we'll say objects (objects could also just be the air)--due to the temperature differential.*0126

*Differential, as you know, just means the difference.*0162

*Heat is a form of energy; it is a very disordered form of energy.*0166

*OK, so heat--profoundly important; thermochemistry is going to be all about the heat.*0179

*Work is another form of energy.*0187

*Clearly, energy can come in different ways; you can have light energy; it could be heat energy; work is another form of energy; all of these things are energy.*0199

*They are represented by the same unit, and the idea, essentially, of major areas of science, is: How do we convert one form of energy to another form of energy?*0208

*That is really what is going on.*0217

*Energy...work is another form of energy, and we define it as Work=Fd, which is force times distance.*0221

*So, the best way to think about that is: if I apply a certain force to an object, and if I move it a distance d, well, the work is going to be the force that I apply, times the distance that I actually move it.*0232

*Now, notice: work, the way we discuss it in science, is not the same sort of way of everyday usage of the word "work."*0243

*I can stand and I can push against a wall, and I'll definitely start sweating, and I'll definitely start getting tired, eventually.*0252

*But, the wall isn't moving; I'm applying a force, but the wall isn't moving.*0258

*Force is a positive value, but distance is 0; the wall isn't moving.*0261

*Scientifically, there is no work actually being done; yes, I am doing work, in the sense that I am actually expending energy, but no actual work is being done by the force.*0267

*So, there is a difference: when we say "work," in chemistry, in physics, in engineering, we're talking about a force that causes something to move.*0277

*OK, so let's look at some units.*0287

*Force equals mass times acceleration; that is Newton's second law.*0290

*So, if you want to give a certain body of a certain mass a given acceleration, you have to apply a certain amount of force to it, and the relation between these three variables is F=ma.*0297

*Well, mass is in kilograms; that is the standard unit of mass.*0306

*Acceleration is in meters per square second.*0313

*Well, this unit right here--the kilogram-meter per square second--this is called a newton.*0317

*A newton is a unit of force, symbolized with an N.*0323

*Now, we just said that work equals a force times a distance; well, force, we just said, is a newton; and distance is in meters; so now, we have this unit of a newton-meter.*0327

*This unit, which is a (well, let me go ahead and actually write out everything so you see it)...*0346

*We said that a newton is a kilogram-meter per square second, and then we multiply it by a meter, which is there; so this is the force; this is the distance; and we end up with kilogram-meter squared per second squared, and this is called a Joule.*0355

*This is our standard unit of energy, the Joule.*0377

*If you want, you can think of it as kilogram-meter squared per second squared, or you can think of it as kilogram-meter per square second, times meter--a newton-meter.*0381

*So, a newton-meter is a Joule, or if you want to break down the newton and sort of put some things together, you can get kilogram-meter squared per second squared.*0390

*It depends on the problem you are working on; sometimes you will just deal with newtons; sometimes, you will need to break it up.*0399

*As you go on in science, you will realize that you will need to sort of break up more of these things, in terms of units, to deal with certain constants.*0404

*Constants--depending on the problem, you might have to use different constants, which are the same constant, but in different units, they have different values.*0412

*OK, so the biggest piece of advice I can give is: watch your units.*0421

*If we're talking about a mass, make sure that kilogram matches with gram; or, if something is in two different units, make sure that kilograms, meters, seconds, Joules, newtons...that everything matches; otherwise, your values will be wrong.*0426

*OK, so let's talk about the two ways to actually transfer energy.*0442

*The two ways to transfer energy--to transfer energy from one object to another, from one thing to another, from one environment to another, from the outside to an inside, to transfer energy: we have two ways of doing it.*0448

*We can transfer it as heat or work, or both.*0465

*That is it; so, in some sense, this is really simple.*0473

*We're going to have some sort of a system, and we want to either give energy to it or take energy away from it; well, there are only a couple of ways we can do that.*0477

*We can either put heat into it or take heat out of it, or we can do work on it, or it can do work on us.*0485

*So, let's be a little more specific here; we'll draw a picture, and we're going to start talking about systems and surroundings.*0491

*OK, so I have a system; a system is defined as the thing that we are interested in.*0500

*I know that that is kind of a vague definition--"thing that we are interested in"--it will make more sense as we do some problems and as we talk about it more.*0513

*Well, for chemistry, most systems are going to be the reaction: the reaction, and maybe the vessel that is containing it...a solution of hydrochloric acid plus sodium hydroxide--that is your system...the beaker plus the water; that is the system.*0525

*The surroundings--that is everything else.*0540

*That is everything else; so, very, very intuitive--nothing strange is going on here.*0545

*We said that we can transfer energy in two ways: we can do it via heat, or we can do it via work.*0550

*This is our system; this is (I'm sorry, I should put the system in here)...I'll call this sys, and out here is the surroundings.*0557

*So now, this is the boundary, and it's always going to be something; there is going to be some sort of a boundary that separates your system from your surroundings.*0575

*Now, we're going to transfer heat in and out; we're going to transfer work in and out.*0583

*We have to pick a point of view.*0588

*In chemistry, we always look at the system's point of view; it's not always the case; there might be other circumstances, maybe in engineering, where you need to look at the surroundings' point of view; but in chemistry, we always look at it from the system's point of view.*0591

*From the system's point of view, when heat is added to a system (OK, so we say heat in), that heat is positive.*0602

*When heat is given off of a system, that heat is negative; in other words, the system is losing heat--it's going from a certain value of heat, it's losing heat, and its final heat is actually less than that.*0617

*So, energy and heat is flowing out; so heat is negative.*0631

*That is the sign convention; if heat is put into the system, that means more energy; so it has a certain amount of energy; we put more heat in; it goes up; that is positive.*0635

*We are looking at it from the system's point of view; notice, it would be the reverse if it were the surroundings' point of view.*0643

*The other way to transfer energy is work.*0648

*Now, the system could do work on the surroundings; the surroundings could do work on the system.*0651

*We will be a little more specific about what that means, but the same sign convention applies.*0657

*If I do work on the system--work **on* the system--that is positive work.0662

*That means I have put energy into the system via the work that I have done to it.*0676

*If the system does work on the surroundings, that is going to be negative work.*0680

*This is called work done by the system.*0687

*So, in this case, these prepositions are very, very important; so, when you read your questions, makes sure you read them carefully, because it will say "work done on the system" or "work done by the system."*0693

*It won't tell you whether it is positive or negative; you have to decide.*0701

*So, work done by the system--that means the system used its energy to do something to the surroundings, and in the process of doing it, it transferred its energy to the surroundings.*0706

*Now, energy is depleted.*0715

*Work is a little more abstract than heat; with heat, you have a pretty good intuitive sense--you know when something gets colder or gets hotter.*0718

*Work is a little different, but again, we will do some problems, and we'll get used to it.*0725

*So, two ways to transfer energy: heat and work; work into the system is positive; heat into the system is positive; heat out of the system is negative; work out of the system is negative.*0730

*Again, from the system's point of view--we are always looking at it from the system's point of view.*0744

*Whenever a process actually gives off heat (the system gives off heat), we call it exothermic.*0749

*It makes sense, right?--exo, outside.*0755

*And then, heat in is called endothermic--an endothermic process; that means the system is taking in heat.*0758

*These are very, very important words, and we will use them in the context of our problems.*0767

*OK, so let's look at just a couple of quick examples.*0773

*If I take methane gas, and if I combust it (burn it in oxygen), I produce two moles of CO*_{2}, 1 mole of CH_{4}; I need 2 moles of oxygen; I produce 2 moles of CO_{2} and I produce 2 moles of water, and I produce heat.0776

*As you know, any time you burn something, it gives off heat; you feel the heat radiation; that is giving off energy as heat.*0797

*This is an exothermic process.*0804

*Burning, combustion, is an exothermic process.*0810

*If I take nitrogen gas, and if I react it with oxygen gas, in order to produce a couple of moles of nitrogen monoxide gas, as it turns out, I have to add heat to the system.*0813

*So, what happens: if I have a vessel, and if I actually ran this reaction, what would happen is: the vessel itself would get cold--I would feel the vessel get cold--and the reason is, now the reaction--in order for it to work--needs energy put into it.*0826

*So, it takes heat away from the glass (let's say it's in a glass container); it sucks heat out of the glass container, and because the heat from the surroundings is not actually going into the glass fast enough, when I touch the glass, I feel cold.*0843

*So, it isn't that the glass got cold; it is that the heat was pulled out of it, bringing it to a lower temperature.*0856

*That is what is going on.*0863

*This is an endothermic process.*0864

*The system evolves heat and gives it up.*0870

*This system pulls in heat--needs the energy in order to perform its functions (in this case, the systems are the reactions).*0873

*OK, now let's go ahead and put some mathematics to this.*0883

*We said that there are two ways to transfer energy: heat and work; well, the change in energy of a system is going to be the heat that goes in or out of the system, plus the work that goes in or out of the system.*0890

*Again, heat and work are just two different types of energy.*0909

*So, if a system starts in one state of energy, and it goes to another state of energy, a couple of things have happened.*0912

*It has either lost or gained heat, lost or gained work, or some combination of both.*0920

*This is our fundamental equation; this is actually a statement of the first law of thermodynamics--that energy can neither be created nor destroyed; that any energy change is going to take one form or the other: heat, work, or both.*0926

*But, energy is conserved; that is what this equation is saying.*0942

*OK, so these heat and work thermodynamic quantities--they have two parts.*0947

*They have a magnitude, and they have a sign.*0957

*Well, again, we dealt with the sign issue here; with thermodynamics, oftentimes we have to stop and think about the physics of what is going on, in order to decide what the sign is going to be.*0963

*Often, we know what the magnitude is, but we have to make sure to keep the sign straight; otherwise, the mathematics will not work out.*0974

*All right, now let's fiddle with this equation; let's talk a little bit more specifically about this thing called work.*0982

*OK, so a common type of work in chemistry (in fact, the most common in general) is work done by a gas.*0991

*So, it's either going to be...well, work done by a gas: notice, here is that "by" again; that is expansion; and when I draw the picture in a minute, it will show you what I mean.*1015

*Or, work done to a gas, which we call compression.*1030

*Let's say I have this container, and it has a moveable piston that can move up and down; this is a chamber that has the volume.*1044

*So, I can push this piston down and squeeze this volume--make it smaller--or I can pull the piston up and make the volume bigger.*1056

*In this particular case, if there is a gas in here, and it expands, that means the gas is actually pushing against the external pressure of the atmosphere.*1064

*It is doing a certain amount of work; if I actually push the piston down, and the volume gets smaller, this pressure is higher than this pressure; I am actually pushing this down--I am doing work on the system here.*1073

*Let's take a look at what this looks like mathematically.*1088

*We just said that the change in energy is equal to q + w, the heat that goes in and out of a system plus the work that is done on or by the system.*1091

*OK, q: well, we said that work is equal to force times distance; great--simple substitution.*1101

*We'll leave q alone, and now, here is what I'm going to do: I know that there is a relationship between force and pressure and area; as it turns out, if I take the force of something, and if I divide it by the area (remember, back when we talked about gases, we said that pressure is equal to the force per unit area, so...) I have this F here; if I divide by A, in order to retain this, I have to multiply by A, so I basically have multiplied this F/d by A/A, right?*1111

*I did this in order to manipulate this F a little bit.*1144

*Now, I have F/A, and I have distance times area.*1147

*Well, this thing right here--I see it in the profile, but really, it is just a cylinder.*1153

*Let's see what we get here: we have q; well, force over area is pressure, and distance times area is...well, distance is...let's say that this force here is a pressure that is being applied to this piston.*1160

*The area is, of course, the area of this circular part; so there is a certain pressure being applied; well, the area times the distance...the distance is the distance that this piston actually goes down or up, right?*1186

*So, this is the distance; well, area times the distance, as far as the circular cylinder is concerned, is just volume.*1204

*You end up with: Change in energy equals q; P is constant, so the only thing that is changing here is, from this to this, whether we either press down or pull up.*1215

*Pressure...PΔV; this is pressure times the change in volume.*1233

*This is what we mean by pressure, volume, and work.*1238

*So, when we push down on a gas, we are actually doing work on the gas; we are doing work to the system.*1242

*When the gas expands and pushes the piston up, then the gas is actually pushing against an external pressure, and it is actually doing work on the surroundings.*1251

*This final equation...this is often how we will do it: we're going to be dealing with some gas in a container, and this is going to be the expression that we use: The change in energy of the system is going to be the heat that flows in or out of the system, plus the change in volume, based on a certain amount of pressure.*1266

*This pressure is always going to be the external pressure.*1291

*It is an external pressure that pushes down, and squeezes it, and makes the volume smaller; or, it's going to be the pressure against which the gas pushes in order to make it bigger.*1294

*In either case, it is a pressure that is pushing against the boundary here.*1306

*So, it is always an external pressure--that is what this P is, not the pressure of the gas.*1311

*OK, so now let's talk about some signs.*1318

*All right, this is going to be important; so now, we're going to actually talk about this term right here, and we're going to relate...so we know that work and pressure times volume are equivalent in terms of magnitude whenever we are doing something like this.*1322

*So now, we want to talk about what sign convention we are going to use.*1339

*All right, so let's just write "Work = PΔV"; so, work equals pressure times the change in volume.*1345

*If I start at a certain volume, and if the gas expands, that means the gas inside is pushing against an external pressure; well, the final volume is bigger than the initial volume.*1351

*Final volume, minus initial volume, is going to be bigger than 0.*1369

*Well, in this particular case, it is bigger than 0; the pressure is bigger than 0; here, this work is going to be positive; this number is going to be positive.*1375

*Pressure times the change in volume, for an expansion, is positive; but, because it is expanding, it is the system that is doing work on the surroundings; therefore, the work is negative, because again, we said we are looking from the system's point of view.*1383

*If the system is pushing against this piston and expanding, it is doing work on the surroundings.*1398

*In other words, work is leaving the system.*1404

*Therefore, work is negative; so, by putting this negative sign, we also account for the other direction.*1406

*If the outside pressure is pushing down on the piston, and making the volume smaller, that means the final volume is smaller than the initial volume.*1413

*Therefore, ΔV, which by definition is final minus initial, is less than 0.*1425

*Well, if ΔV is negative, and P is positive; this negative sign--negative times negative--makes the work positive, which works with the fact that now the surroundings are doing work on the system.*1430

*The work is positive, because the system is now taking in energy in the form of work.*1444

*It is positive from the system's point of view.*1451

*The energy of the system is increasing.*1453

*So, this is the actual relationship: work is negative pressure times the change in volume (change in volume always being final volume minus initial volume--which is the definition of Δ--final minus initial).*1456

*OK, so with this little bit of a background, let's just jump into the problems, and I think we can start to make sense of some of this.*1473

*Let's do Example 1: So, we want to calculate the change in energy for an endothermic process in which 29.6 kilojoules of heat flows (and notice: we said endothermic), and where 13.7 kilojoules of energy is done on the system.*1482

*OK, so we want to calculate: we say that, for an endothermic process, 20.6 kilojoules of heat is flowing.*1540

*Endothermic means heat is coming into the system; that means heat is positive.*1545

*"Where 13.7 kilojoules of energy is done on the system": on the system means work is flowing into the system--that means work is also positive.*1551

*So, we write our equation: change in energy equals q + w; q is heat; w is work.*1559

*We said an endothermic process, so it's positive, so it's going to be positive 20.6 kilojoules; so notice, the problem will give you a magnitude; it won't necessarily tell you what direction it is going.*1569

*If it said exothermic, then I would have to use -20.6.*1582

*If this said "by the system," I would have to use -13.7.*1585

*But here, it's endothermic, so heat is positive, and then the work is also positive 13.7, because it is done on the system.*1589

*Kilojoules...this is a simple arithmetic problem.*1597

*So, we end up with 34.3 kilojoules of energy.*1600

*I put heat into it in the amount of 20.6 kilojoules; I did work on the system by 13.7 kilojoules; the total amount of energy that I imparted to the system is a positive 34.3 kilojoules of energy.*1608

*Oh, by the way, I should let you know: oftentimes, another unit of energy is something called the calorie; it's an older unit of energy, and it is defined as the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius.*1626

*The calorie is a unit of heat; it's a unit of energy, because that is what heat is--it's just energy.*1640

*So, as a conversion factor, one calorie equals 4.184 Joules.*1646

*We are probably not going to run into calorie too much; we may or may not; we'll find out; but just so you know...*1653

*You have heard of the word "calorie"; you hear about it all the time; that is what it is: 1 calorie is that many Joules--it's a unit of heat; it's a unit of energy.*1658

*OK, so Example 2: What is the work associated with the expansion of a gas from 0.750 liters to 1.760 liters at constant external pressure of 13 atmospheres?*1667

*The gas is in a container; there is a pressure of 13 atmospheres on the outside; and again, pressure is always going to be external.*1724

*On the outside, 13 atmospheres is pressing on it.*1732

*But, the gas is actually expanding from the inside, and it goes from .750 liters to 1.760 liters.*1736

*In other words, it is pushing out against the surroundings.*1746

*It is doing work on the surroundings, which means that energy is flowing out of the system and into the surroundings, which means that it is actually going to be negative.*1750

*Well, work equals negative pressure, times change in volume, equals negative 13 atmospheres; change in volume--we have 1.760 liters, minus 0.750 liters, and when I do this, I end up with -13.13 (I don't know if you can read that) liter-atmosphere.*1761

*OK, so notice: atmosphere, liter; this is not exactly a unit of energy that you have seen, although it is a unit of energy; it just happens to be different units, and I will show you what this is.*1809

*The conversion factor here is this: 1 liter-atmosphere is equal to 101.3 Joules of energy.*1822

*I'll show you in a minute where this actually comes from.*1833

*But again, it just goes back to the point: units are very, very important; just because you end up with some unit that seems strange doesn't necessarily mean that it is strange.*1835

*So, all of the math that we have done is correct; work is equal to pressure times change in volume; pressure is in atmospheres; volume is in liters; this is actually a unit of energy--it's a unit of work.*1844

*It is actually equivalent (by a conversion factor) to a Joule.*1858

*So, don't let this throw you off.*1863

*Now, let's go ahead and do the conversion.*1865

*When we multiply this out, we end up with 1330.4 Joules, or (can't have these stray lines floating around--let's erase this) 1.33 kilojoules.*1867

*There we go.*1887

*OK, so let me see what I have here.*1890

*OK, so really quickly...you know what, let's skip the unit conversion part; if you can, you are welcome to look it up, but the liter-atmosphere--one liter-atmosphere is equal to 101.3 Joules, so that conversion will always be there for you.*1898

*Let's just keep going on with these examples.*1918

*All right, another problem here: Now, a balloon is inflated from 4.00x10*^{6} liters to 4.50x10^{6} liters by the addition of 1.4x10^{8} Joules of heat energy.1922

*So, I add some heat to the system, and I add 1.4x10*^{8} Joules of heat.1972

*Now, if the balloon expands against a constant pressure of 0.95 atmospheres, what is the ΔE for the process?*1981

*OK, so let's just draw a quick picture here: we have this balloon--obviously, we are talking about a really, really large balloon; we pump in heat--pumping in heat energy--so heat is going to be positive.*2019

*This balloon is going to expand, which means that it is going to push out against the .95 atmospheres; so now, the balloon is actually doing work on the surroundings.*2034

*Work is going to be negative.*2043

*Now, let's work it out.*2049

*ΔE is equal to q plus w; we have worked out our sign convention--heat is going in; work is going out, because it is expansion.*2052

*We also know that work is equal to -PΔV, so let's go ahead and calculate our work.*2062

*The work is equal to -0.95 atmospheres (that is the external pressure; and again, pressure, volume, work; PΔV; that P is the external pressure against which you are pushing, or that is pushing on you, or pushing on the system), times 4.50x10*^{6}, minus 4.00x10^{6}.2076

*We end up with (when we do this mathematics) -475,000 liter-atmospheres; and then, when we multiply by 101.3 (oops, that is not kilojoules--the conversion factor is 101.3 Joules per liter-atmosphere), we end up with a total of -4.8x10*^{7} Joules of work.2103

*Well, we just (oh, no, we definitely don't want that...OK) said that ΔE equals q + w; we calculated w--that is -4.8x10*^{7} Joules--so now ΔE equals q plus that.2141

*Well, q, positive--it is 1.4x10*^{8} Joules (you know what, I am always confused about whether I should write it out or do the letters; for years it has been like this--OK, I'll just do the letter), minus 4.8x10^{7} J, so our change in energy is 9.2x10^{7} J.2163

*Notice what has happened here: energy flowed in as heat.*2196

*Work ended up expanding the gas; the gas ended up pushing against the external atmosphere; so energy came in as heat; energy left as work.*2202

*The total energy change for this process was 9.2x10*^{7} Joules; this is a positive value, so there is now more energy in the system that stayed as heat than went out as work.2214

*That is all this says--the fundamental equation; this is the first law of thermodynamics.*2231

*The change in energy is equal to the heat that goes in and out of a system, plus the work that goes in and out of a system.*2235

*Energy itself can be transformed, but it cannot be created or destroyed; it always has to be accounted for.*2241

*OK, thank you for joining us here at Educator.com.*2247

*We'll see you next time for some more thermochemistry; goodbye.*2250

*Welcome back to Educator.com; welcome back to AP Chemistry.*0000

*Today, we're going to continue our discussion of thermochemistry.*0003

*We're going to talk about enthalpy and Hess's Law.*0007

*I have to, before I begin...I wanted to discuss--just take a couple of minutes to talk about--thermodynamics and thermochemistry, and a lot of the terms that are sort of being thrown around, and some of the equations.*0010

*Back in the early...well, not early days; in the turn of the century, there is a saying about thermodynamics; a very famous thermodynamicist says this; he said, "None of us really understands thermodynamics; we just get used to it."*0025

*Now, of course, that is not completely true; we understand it; but to a large extent, a lot of it really is true.*0037

*Thermodynamics is a very, very strange thing; heat behaves in very difficult ways--very unusual ways.*0043

*These terms, like enthalpy and heat and energy and work, and pressure-volume...I know that a lot of that is very, very odd; it's difficult to wrap your mind around; it's difficult to sort of get a good, intuitive feeling for what is going on.*0049

*My recommendation for dealing with that is: don't really worry about it too much.*0065

*A lot of the comfort that comes from dealing with thermochemistry and thermodynamics, aside from some of the stuff like today's (it's actually not that bad)...a lot of it just comes from familiarity.*0069

*You will be seeing it over and over and over again, and you will be doing the problems over and over again, so that you will get more of a sense.*0079

*So, if you don't understand it the way you do other things, I wouldn't worry about it too much; that is just the nature of thermodynamics.*0085

*Again, just to throw something out there to make you put your mind at ease: it's like that for everybody.*0092

*With that, let's go ahead and get started.*0099

*OK, so let's start with a definition.*0103

*We're going to define enthalpy: enthalpy (which we use the symbol H for, and you will understand in a minute, when we actually equate it to heat) is equal to the energy of a system, plus the pressure of the system, times the volume of the system.*0108

*So, let's say I had some container of gas; it's going to have a certain energy that is associated with it; it's going to have a certain pressure that is associated with it and a certain volume that is associated with it.*0132

*By definition, that is the enthalpy; if I take the energy, plus the pressure, times the volume, I get the enthalpy.*0142

*Now, like most thermodynamic properties, we don't really know what absolute enthalpies are; the only thing we can actually measure (which is science: in science, we measure things) are changes in enthalpy.*0149

*So, ultimately, we are going to be concerned with ΔH; and, in fact, all of thermodynamics is going to be concerned with the Δ of some properties.*0162

*Later on in the course, we will talk about ΔS; we will talk about ΔG, which is free energy entropy; and H is enthalpy.*0169

*We are concerned more with Δ, and, a little bit later in this lesson, we will show why the Δ is important...in any case, just so you know...*0177

*The definition of enthalpy is the energy plus the pressure times the volume of the system.*0189

*Now, let's recall what we did last time: we said that the change in energy of a system is equal to q, plus the work, which was -PΔV.*0194

*Now, let me rearrange this; let me bring this -ΔV over to this side; q equals the change in energy, plus PΔV--now, let me just set that aside for a minute.*0209

*Now, let me come over here, and let me take my definition of H as equal to E, plus PV, and now I'll do ΔH.*0222

*Well, ΔH, which is final minus initial, equals Δ of this, which is the ΔE plus Δ of PV.*0236

*Well, in this particular case, if we keep the pressure constant, that means we can pull this out of the Δ; we end up having ΔH=ΔE+PΔV.*0247

*Notice what we have: we have that q, which is the heat that is transferred, is equal to the change in energy plus the pressure times the change in volume (constant pressure).*0266

*And we have that ΔH, just based on the definition, equals ΔE, plus pressure, times Δvolume.*0276

*These are the same; q equals ΔH at constant pressure.*0282

*So, this is a constant pressure situation here, OK?--at constant pressure (which is pretty much what we are doing--what we do most of our chemistry in, at a constant atmospheric pressure), q equals ΔH.*0293

*In other words, the enthalpy is nothing more than the heat that is transferred--the energy that flows as heat.*0307

*In other words, if a certain reaction takes place, and it gives off a certain amount of heat or it takes in a certain amount of heat, that heat is equal to the enthalpy.*0314

*So, I can talk about the enthalpy by just referring to the heat; all I have to do is deal with the heat--it happens to be the same as the enthalpy, under constant pressure conditions.*0325

*That is why constant pressure is really, really nice, because under those conditions, I don't have to deal with enthalpy; I just deal with the heat; they are equivalent.*0335

*Heat is a very easy thing to measure; you just run a reaction, and you measure how hot something gets or how cold something gets; that is your enthalpy.*0343

*It is a thermodynamic quantity that is related to heat; under constant pressure conditions, it is the heat--it is equivalent to the heat; so that is really nice.*0351

*This is the important thing to know.*0362

*Heat and enthalpy are equivalent in magnitude under constant pressure conditions.*0366

*OK, so when we talk about the enthalpy of a reaction, ΔH*_{rxn}, well, that is equivalent to the heat of the reaction.0370

*We will often talk about the heat of the reaction.*0380

*OK, now, ΔH of the reaction is equal to...actually, you know what, no; I'm not going to...I'll save this for the next time; there is no need to confuse us with any information that we don't need right away.*0392

*OK, so let's just do a quick example to get a feel for what is going on--a little bit of stoichiometry.*0408

*Example 1: When 1 mole of C*_{6}H_{12}O_{6}, which is glucose, is fermented to ethanol at a constant pressure, 67 kilojoules of heat is released.0415

*The system releases heat; heat is flowing out of the system; it is negative.*0458

*How much heat is released when 7.6 grams of glucose is fermented?*0467

*1 mole of glucose is fermented to ethanol at constant pressure; 67 kilojoules of heat is released--exothermic.*0490

*The enthalpy is negative; now heat and enthalpy is the same--constant pressure.*0499

*How much heat is released when 7.6 grams of glucose is fermented?*0505

*OK, so let's write out what this looks like to get a sense of how we sort of start these problems, and what it looks like, notationally, for a chemist.*0509

*So, C*_{6}H_{12}O_{6} is fermented to 2 C_{2}H_{5}OH (ethanol--this is regular drinking alcohol), plus carbon dioxide gas.0518

*And we often write the ΔH over here as -67 kilojoules; so we see that heat is released; ΔH is negative; it is an exothermic process.*0534

*That is what it means: exothermic--ΔH is negative; ΔH is negative--it is exothermic; it is giving off heat, which means, in addition to this product and this product, one of the other products is that much heat.*0543

*That is why we write it; imagine heat as that third product that also comes out of the reaction.*0556

*This heat comes from the bonds in the carbon, hydrogen, and oxygen; that is where it is coming from.*0563

*OK, let's draw a little energy diagram, so you see what is going on here.*0569

*C*_{6}H_{12}O_{6}: this is H, enthalpy (heat at constant pressure), and this is just the reaction coordinate.0576

*The reaction coordinate just means that the reaction is proceeding in that direction.*0588

*Well, there are some...here we have the C*_{2}H_{5}OH, ethanol, plus our CO_{2}.0594

*Now, what this means--this -67 kilojoules--as it turns out, thermodynamically, the reactants--there is more heat in these bonds; when going from glucose to ethanol as CO*_{2}, the amount of energy in these bonds is actually 67 kilojoules less.0605

*So, this difference right here, from this point to this point, is the 67 kilojoules.*0627

*Because, again, energy cannot be created or destroyed, the energy in these bonds is returned into the energy of these bonds.*0634

*But now, I have an excess amount of energy; what am I going to do with it?--well, the reaction just releases it as heat.*0642

*That is what this says; it is going to a lower heat--this excess heat is what is given off.*0647

*That is why it is -67; this is what it actually looks like.*0655

*The products are thermodynamically at a lower energy.*0658

*OK, so let's go ahead and calculate: well, C*_{6}H_{12}O_{6} is 180 grams per mole; we have 7.6 grams of it, times 1 mole--180 grams; that gives us 0.0422 mol, and I hope that I did my arithmetic correctly.0663

*Well, I have 0.0422 mol, and it is telling me that it releases 67 kilojoules per mole (that is what the problem says).*0694

*It is a simple arithmetic problem.*0705

*-2.8 kilojoules of heat is released.*0708

*2,800 Joules of heat is released with 7.6 grams of glucose.*0723

*7.6 grams of glucose is not very much; it's a handful--not even a handful; it is 2,800 Joules; that is a lot of heat.*0729

*There is a lot of heat in those bonds; that is why the human body metabolizes glucose--it breaks it down, not into alcohol and CO*_{2}--it breaks it down completely into carbon dioxide and water, and all of the energy that is released--the body uses that energy to produce a molecule called adenosine triphosphate, and it is the adenosine triphosphate that runs the body.0743

*That is our energy currency, and it all comes from the energy that is stored in the bonds of C*_{6}H_{12}O_{6}.0767

*Well, just 7.6 grams produces 2,800 kilojoules of energy!*0773

*You can imagine the amount of glucose we actually take in, in the form of carbohydrates and other things; the body requires a lot of energy to run.*0778

*All right, now let's talk about something called a state function.*0790

*Let me...all right, let's define a state function.*0798

*A state function is a property (you could call it a state property; I don't know why they call it a state function, but it is a property) that does not depend on the path taken to achieve that state.*0803

*OK, so let's say I have something here and something there; these are two states; I can get from this state to this state--I can either go this way, directly, or I can go this way and come back; I can go this way, this way, this way, this way, this way, this way.*0846

*Now, as it turns out, there are certain properties that are not state functions, like heat and work.*0862

*So, for example, if I went from here to here, I would have to do a certain amount of work.*0872

*Clearly, if I went from here to here to here to here to here to here, I am doing more work.*0876

*But, as it turns out, energy (which, as we said, is equal to heat plus work, neither of which is a state function)--as it turns out, energy is a state function.*0880

*As it turns out, it doesn't matter how I get to the final state--all that matters is where I started and where I ended up.*0894

*That is why we are concerned with ΔS: ΔH, ΔG--those are state functions; enthalpy is another state function.*0900

*So, we said that enthalpy is equal to the change in energy, plus the pressure, times the change in volume, right?*0907

*Well, this is a state function; pressure is a state function--it doesn't matter how I get there; at a certain point, and at the pressure that I start and end up with, it is just a certain pressure; it doesn't matter how I get there.*0916

*Volume: it doesn't matter how I go from 3 liters to 5 liters; I can go up to 18 liters, drop down to .1 liter, and then go up to 5 liters; I have still just gone from 3 to 5--the net effect is the same.*0929

*So, ΔH is also a state function.*0942

*It is a state function because it is the sum of two state functions.*0944

*Energy is a state function despite the fact that neither of these is a state function.*0948

*This is also quite extraordinary, that that is the case.*0953

*OK, so as far as chemistry is concerned, now: chemistry--if we start with certain reactants, and we want to end up with certain products, well, as far as the enthalpy is concerned, it doesn't matter how I get there.*0957

*I can get there in 2 steps; 15 steps; 147 steps.*0974

*Now, yes, there are areas of chemistry where we are concerned about the steps, but as far as a thermodynamicist is concerned, all he cares about is the enthalpy at the beginning and the enthalpy at the end.*0978

*It is a state function; it doesn't matter how you get there; all that matters is that you get there.*0988

*It is the two states that matter; that is all that matters.*0993

*Because of that, we can actually take a reaction that we are interested in, and perform it, and if we want to find the enthalpy of that reaction, well, if we have enthalpies of other reactions that we can use to get to our final reaction, we get our final enthalpy.*0996

*OK, so now, this is the idea of Hess's Law.*1014

*And, rather than talking about it or writing about it, the best thing to do is just do a problem, and of course, it will make sense.*1021

*So, let me write out Hess's Law here.*1026

*Well, I won't write it out; we'll just do an example, and it will make sense.*1032

*OK, we want to find the ΔH for the reaction of sulfur, plus oxygen gas, going to sulfur dioxide gas.*1036

*In other words, there is a certain heat of reaction associated with this; either it absorbs heat to create SO*_{2}, or in the process of creating SO_{2}, it releases heat.1059

*I want to find the enthalpy--the heat of the reaction.*1069

*Well, how do I do that?*1072

*OK, so we want to find ΔH for the reaction; well, as it turns out, we just happen to know that we have a couple of reactions at our disposal.*1074

*We know that, if I take sulfur plus three-halves oxygen gas, in the process of creating sulfur trioxide gas, we happen to know that the ΔH of that is -395.2 kilojoules.*1088

*We also happen to know that sulfur dioxide gas, 2 moles of that, plus oxygen gas, goes to 2 sulfur trioxide gas, and we happen to know that the ΔH of that equals -198.2 kilojoules.*1110

*So, we have this reaction that we know the ΔH for; we have this reaction that we know the ΔH for.*1134

*Hess's Law says it doesn't matter how we get to our final reaction, if we can come up (oops, no, we don't want these stray lines here) with a way of manipulating these equations (switching them, multiplying them by coefficients, a lot like you do for linear equations in linear algebra or algebra courses that you have taken).*1140

*If we can fiddle around with them and add the equations to come up with a final equation--this one that we want--well, we will just add the ΔHs, and we will get the final enthalpy of the reaction, because again, ΔH is a state function; it doesn't matter how you get there, as long as you get there.*1166

*So, let's see what we are going to do.*1184

*So, how can I fiddle with these equations in order that, when I add them vertically, I end up with this equation?*1186

*OK, well, let's see; let's reverse...let's see, what can I do?*1194

*I am trying to create SO*_{2} gas; so notice that SO_{2} here is on the right-hand side, but in these equations, SO_{3} is on the right-hand side.1201

*The only equation that has SO*_{2} in it is over here; I want to reverse it, and there is one SO_{2} here, but there are 2 SO_{2}, so I am going to flip equation 2, and I'm going to divide it by 2.1212

*OK, so let me...actually, let me rewrite everything again, so that we have it on one page.*1228

*We have S + O*_{2} going to SO_{2}; that is the reaction that we want; and we are given S + 3/2 O_{2} goes to SO_{3}; ΔH equals -395.2 kilojoules.1241

*Our second equation (we'll call that #1) is 2 sulfur dioxide gas, plus O*_{2}, goes to 2 SO_{3}, and the ΔH for that is -198.2 kilojoules.1262

*OK, so we said we want this equation from these equations.*1279

*We are going to flip this equation, #2; so, we are going to reverse #2 and divide it by 2.*1284

*When we reverse it and divide it by 2, this 2 SO*_{3} comes to the left, and it ends up becoming just SO_{3}; this 2 SO_{2} ends up on the right, but becomes SO_{2}; and this O_{2} also ends up on the right, but it becomes 1/2 O_{2}.1299

*Now, what happens to the ΔH?--well, exactly what you think.*1315

*If you flip a reaction--if you flip an equation--you change the sign of ΔH; if you divide an equation by 2, you divide the ΔH by 2.*1318

*So now, the ΔH is no longer -198; it is going to equal +99.1 kilojoules.*1327

*And now, we leave the...here, our other equation is S + 3/2 O*_{2} goes to SO_{3}, so S is on the left--there is one S on the left--so let's leave that one alone.1337

*We have S + 3/2 O*_{2} going to SO_{3}; that ΔH stays the same: -395.2 kilojoules.1351

*Now, we just add straight down.*1367

*Everything that is on both sides--if there is something on the left and something on the right, they cancel.*1369

*SO*_{3} on the left; SO_{3} on the right; it cancels.1375

*S comes down; that is taken care of.*1379

*I have 3/2 O*_{2} on the left; I have 1/2 O_{2} on the right; 3/2 minus 1/2 is equal to...well, there you go: O_{2} (two halves)...+ O_{2}.1383

*O*_{2} is taken care of; now, the only thing left is the SO_{2}.1400

*There you go; I have the final equation that I wanted by messing around with the equations for which I did have information.*1405

*Now, all I do is: I just add the ΔHs.*1413

*When I add the ΔHs: +99.1, minus 395.2; we get -296.1 kilojoules.*1416

*I used reactions that I knew, manipulated the equations, made the appropriate changes to the enthalpy, and I added straight down, added straight down; now I have the enthalpy for this reaction.*1426

*This is Hess's Law; I can use reactions that I do know to find a reaction that I want.*1439

*OK, let's do another example here.*1447

*Let's do this on a new page.*1454

*Calculate the ΔH for the synthesis (which means formation) of di-nitrogen pentoxide gas, N*_{2}O_{5}, from its elements.1461

*OK, elements--so, the reaction that we want is nitrogen gas, plus oxygen gas, goes to N*_{2}O_{5} gas.1489

*Now, we have to balance this: so, we have 5 and 2, so we'll put a 5 here; we'll put a 2 here.*1505

*We'll put a 2 here; now it's balanced.*1512

*This is the equation that we want; OK, now here are the equations that we have at our disposal.*1515

*Equation #1: we have H*_{2} + 1/2 O_{2} goes to H_{2}O; the ΔH of that is equal to -285.8 kilojoules.1521

*Our second equation is: N*_{2}O_{5} + H_{2}O goes to 2 HNO_{3}, which is nitric acid; the ΔH of that is -76.6; exothermic, exothermic.1540

*3: we have 1/2 N*_{2} + 3/2 O_{2} + 1/2 H_{2} goes to HNO_{3}; the ΔH of this is equal to -174.1 kilojoules.1565

*So, our task is to take these three equations, manipulate them by multiplying by coefficients, reversing them, and then adding them straight down and arranging them in such a way so that, when they add, they add to the final equation.*1585

*OK, so let's see what we have; what can we do?*1604

*I notice I have N*_{2}O_{5} on the right, and here I have N_{2}O_{5} on the left, so let me just leave that one alone for now.1608

*Let me see: I have 2 N*_{2} on the left, 5 O_{2}; I have 2 N_{2}, and the only equation here that has N_{2} in it is this one, so I'm going to multiply equation #3 by 4.1623

*So, multiply #3 by 4, and what I end up with, when I multiply the third equation by 4--I get 2 N*_{2} + 6 O_{2} + 2 H_{2} goes to 4 HNO_{3}, and the ΔH also gets multiplied by 4.1638

*Negative...so this is multiplied by 4 minus 696.4 kilojoules; so whatever I do to the equation, I do the same thing to the enthalpy.*1673

*OK, now let's see what I have; HNO*_{3}...I have 4 HNO_{3}, but I don't have any HNO_{3} here, so I need 4 HNO_{3}s on the left.1683

*I have 2 HNO*_{3}s here, so I'm going to flip this equation and multiply it by w.1695

*So, we'll flip 2 and multiply by 2.*1699

*When I flip this and multiply by 2, I get 4 HNO*_{3} goes to 2 N_{2}O_{5} + 2 H_{2}O, and now the ΔH for this...I have flipped it, and I have multiplied it by 2, so now this is a positive 153.2 kilojoules.1711

*Now, #1--let me see: #1 equation: H*_{2}; I have...what do I have?...I have 2 H_{2}s on the left; I have an H_{2} here; I need to cancel the H_{2}s.1741

*I have 2 H*_{2}Os on the right; I have an H_{2}O on the right here; so I have used equation 3 and used equation 2; I need to get this on the left, and I need to multiply it by 2, so I'll do the same thing.1755

*I'll flip #1 and multiply by 2, and I end up with the following equation.*1765

*I get 2 H*_{2}O goes to 2 H_{2} + O_{2}; and again, the ΔH is going to be +571.6 kilojoules.1776

*And now, I should end up with what I have; so let me see here; let me...we are concerned with this equation, this equation, and this equation; so let's see what cancels.*1797

*H*_{2}O; 2 H_{2}O, 2 H_{2}O; 2 H_{2}, 2 H_{2}, right?--this is on the left of the arrow; this is on the right of the arrow.1815

*They are on top of each other, but it is where they are on as far as the arrows are concerned.*1824

*I have 4 HNO*_{3}, 4 HNO_{3}; 2 N_{2}, so I'll bring that down; that takes care of the 2 N_{2}.1830

*I have 6 oxygens on the left; I have 1 oxygen on the right; so, 6 minus the 1 leaves 5 oxygens on the left.*1841

*So, that is taken care of and that is taken care of; now, the only thing left is the 2 N*_{2}O_{5}, which is on the right, so I am not adding it.1853

*That is on the right-hand side of the arrow; so I get 2 N*_{2}O_{5}, which is exactly what we want: 2 N_{2} + 5 O_{2} goes to 2 N_{2}O_{5}, exactly what we found.1866

*Now, let's just add the ΔHs straight down, and when we add them, we end up with 28.4; again, you might want to check my arithmetic; I'm notorious for bad arithmetic.*1880

*So, this reaction: nitrogen gas plus oxygen gas to form di-nitrogen pentoxide: it is positive 28.4 kilojoules, so this is endothermic.*1893

*That means, in order for this reaction to go forward, I actually have to add heat to it.*1906

*Or, if I leave it alone, and there are other circumstances where there is something happening--there is enough heat that it can pull from the surroundings--it will pull that heat from the surroundings in order to make this go.*1911

*So, there you have it--Hess's Law.*1924

*We want to find the enthalpy, the heat of a given reaction; well, if we have other reactions at our disposal that we know the heat for, and we can rearrange those equations in a certain way--we can fiddle with the enthalpies appropriately--add the equations; we will get the equation that we are looking for, and we will get the enthalpy of that equation.*1926

*OK, thank you for joining us today at Educator.com for the discussion of enthalpy.*1947

*We will see you next time; goodbye.*1953

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

*In the last couple of lessons, we have been talking about enthalpy and energy and heat--things like that.*0004

*Today, we're going to continue our discussion of enthalpy, and we are going to be talking about standard enthalpies of formation; and we are going to be able to use these standard enthalpies of formation to help us bypass Hess's Law.*0011

*When we are given a reaction that we need to find the enthalpy for, the heat of reaction for (whether it is given off or absorbed), we can just do it by reading off of a table of enthalpies that have been prepared for a whole number of compounds.*0024

*It's very, very convenient, because Hess's Law, although it is very convenient in the sense that we can manipulate equations and just add them--still, it requires a little bit of work.*0039

*With standard enthalpies of formation, we have a standard, and then we have some things tabulated, and we can just do some really simple arithmetic and get the answer that we want--very, very powerful.*0049

*So, with that, let's go ahead and get started.*0060

*As often with these, we are going to start off with a definition, because I think it always makes it a little bit more worthwhile.*0064

*So, definition: the standard enthalpy of formation (and, if you have not figured it out by now, when we say "enthalpy," we just mean heat--you can use heat of formation, enthalpy...I actually prefer to use the word "heat"; I don't really care for the word "enthalpy" myself; but there it is) of a compound is the change in enthalpy that accompanies (let me spell this a little bit more appropriately here) the formation of 1 mole of the compound from its elements, with all substances in their standard states.*0071

*OK, so the standard enthalpy of formation of a compound is the change in enthalpy that accompanies the formation of 1 mole of the compound from its elements, with all substances in their standard states.*0184

*That means, if I wanted to talk about the enthalpy of formation of liquid water, it would look like this.*0195

*H*_{2} gas (because hydrogen is a gas in its standard state); and standard state means 25 degrees Celsius, 1 atmosphere pressure--normal; the way you would find it on an average day.0205

*Plus O*_{2} gas (these are its constituent elements); they would form liquid, and the ΔH standard of formation--this is the symbol for it, ΔH_{f} for ΔH of formation--this little degree sign at the top means that it is standard conditions, 25 degrees Celsius, 1 atmosphere pressure; equals...some number.0218

*This is what it would look like: we calculate the ΔH for the formation of a compound, 1 mole of that compound.*0244

*So notice, this is not balanced; now it is balanced.*0251

*1 mole of the product...it is always like this; this is always going to be 1, so we don't really balance this equation in terms of whole-number coefficients.*0256

*Because this is 1, you will end up with fractions over here.*0265

*So again, it is a standard; we pick a standard, and in chemistry, the standard is the mole; so that is what it is.*0269

*Now, let's actually list the standard states.*0277

*We have listed a couple of them.*0279

*What we mean when we say "standard states": for a gas, it is 1 atmosphere; that is standard for a gas.*0281

*For a solution--if we are talking about a solution in standard state--we mean 1 Molar.*0291

*And again, this is an older notation, an m with a line over it; you are more accustomed to seeing the capital M; Molar, mole per liter.*0297

*OK, an element...the state of an element is the state the element takes at 1 atmosphere and 25 degrees Celsius, which is roughly room temperature.*0305

*An element...the standard state of aluminum is metal; the standard state of bromine is liquid; the standard state of mercury--liquid; the standard state of oxygen is a gas.*0328

*The standard state of sodium is the metal; that is it--an element.*0341

*OK, now you should notice: standard states--this 1 atmosphere, 25 degrees Celsius, 1 Molar solution--this is not the same as standard temperature and pressure that we talked about with gases.*0346

*That is 0 degrees Celsius and 1 atmosphere pressure; so don't mix the two--when we are talking about a standard state, we are talking about a particular state that that compound, that element, is taking.*0360

*We have listed that standard as: a gas under 1 atmosphere pressure; a solution, 1 Molar; an element, the state that it assumes under roughly room temperature (25 degrees Celsius) and 1 atmosphere pressure--in other words, your average day at sea level.*0373

*OK, now, once again (I can't reiterate this enough): enthalpies of formation are always given, **always* given per mole of product, because the product is the thing that we are finding the enthalpy of formation for.0389

*The enthalpy of formation is the amount of heat either generated (either given off) or absorbed when you form that product from its constituent elements.*0418

*We just did this one up here.*0431

*H*_{2} + 1/2 O_{2} goes to H_{2}O.0435

*Now, here is the nice part: the ΔH of the final reaction, the standard ΔH, the enthalpy of reaction (or the heat of the reaction--standard heat of reaction) is equal to the ΔH of formation of the products, minus the ΔH of formation of the reactants, when you add them together.*0440

*So, I didn't want to use a summation symbol; and this also includes...well, actually, you know what--let me go ahead and write out...instead of using the summation symbol, I'm going to write it out.*0481

*You are going to notice that I am not a really big fan of symbolism; I think it's nice, but I often think it gets a little bit in the way.*0497

*I will put "Sum of the ΔH formations of the products" on the right-hand side of the arrow, minus the sum of the ΔHs of formation for the reactants.*0504

*That is it; this is the standard definition of ΔH.*0529

*We use the tabulated...at the end of your book, in the appendix, you are going to see a list that is called "Standard Thermodynamic Data."*0532

*It is going to list a whole bunch of compounds, and you are going to have three columns in there.*0540

*The first or second--it is usually the first or second column--one of the columns, anyway--is usually the enthalpy, the ΔH.*0546

*It has a ΔH; it has a little degree--this zero degree--on it; and it has a little f.*0553

*It is given in kilojoules per mole.*0559

*Well, another column is going to be ΔG, which is free energy, which we will talk about later.*0562

*The other one is going to be S, entropy, which is in Joules per mole per Kelvin.*0567

*But again, those are other thermodynamic quantities that we will talk about a little bit later in the year, when we talk about equilibrium, spontaneity, and things like that.*0576

*Right now, we are just concerned with enthalpy.*0584

*That table in the back--that column that lists the ΔHs of formation--that is what you are going to use to tabulate this.*0587

*It will make more sense in a minute, when we do a problem.*0593

*OK, a couple of things to remember: let's see, the first thing to remember is that the ΔH of formation for elements is 0, because you are not forming the elements; the elements are already there.*0597

*It's very, very convenient that it is 0; it comes in very handy.*0615

*OK, #2: ΔHs of formation are per mole of product formed.*0619

*So, when you come across a reaction, make sure the reaction is balanced--balance the equation, so that you have the appropriate stoichiometric coefficients.*0643

*Balance the equation, because again, we are using the ΔH values of formation, but they are per mole of product; but when you are using it in an actual reaction, you may have 3 or 4 or 5 moles, which means you have to multiply the ΔH of formation by 3 or 4 or 5.*0659

*And again, it will make more sense in a minute, when we do an example, which we are actually going to do right now.*0678

*Example 1: OK, ammonia is burned in air to form nitrogen dioxide and water; what is the ΔH of this reaction?*0686

*OK, so let's see what this says; write out the equation.*0721

*Always start with an equation; this is chemistry; there is always an equation somewhere.*0726

*OK, so ammonia is NH*_{3}; it is burned in air, which means it is going to take oxygen; and it forms nitrogen dioxide, NO_{2}, plus H_{2}O.0732

*When we balance this, we end up with 4; with 7; with 4; and with 6.*0746

*Now, this is our reaction; we want to find the heat of this reaction; we want to find the enthalpy of this reaction--ΔH*_{rxn}, the standard ΔH.0754

*We are going to use the ΔHs of formation that are tabulated in the back for this, for this, for this, and for this.*0767

*Don't mistake the two: the final heat of reaction is for the entire reaction; what we are doing is we are using the heats of formation that are tabulated for the individual pieces of this reaction--products and reactants.*0774

*So, we said that the ΔH of the reaction is equal to the sum of the ΔHs of formation for the products, including coefficients, minus the sum of the ΔHs of formation for the reactants, including coefficients.*0787

*All right, that equals--well, let me put a circle around the (oops, I actually wanted to use red; there we go--so that we have it here) equation; ΔH.*0824

*We are going to take the ΔH of formation of H*_{2}O, multiplied by 6; add it to the ΔH of formation of NO_{2}, multiplied by 4.0843

*From that, we are going to subtract the ΔH of O*_{2}, multiplied by 7.0851

*And then, subtract the ΔH of formation of NH*_{3}, multiplied by 4.0856

*Products minus the reactants--the sum of the ΔHs of formation of the products, minus the sum of the ΔHs of the reactants.*0864

*All right, so now, when we look up 34, 4...so which one shall we do first?...let's do...it doesn't really matter, so let's do the NO*_{2} first.0875

*I always like to do it this way; so we have 4 moles (that is a 4) times the ΔH of formation of NO*_{2} from its constituent elements.0888

*We look in the back at our thermodynamic tables; it is 34 kilojoules per mole; again, it is per mole of compound.*0900

*That is what a ΔH of formation is; it is a standard.*0907

*Plus 6 moles--that is where this comes from--that is where the 6 comes from; this, times the ΔH of formation of water, which is -286 kilojoules per mole.*0910

*From that, we are going to subtract 4 moles, times the ΔH of formation of NH*_{3}, which is -46 kilojoules per mole; plus 7 moles--that is where that comes from--times 0 kilojoules per mole, because again, this is an element--oxygen gas.0926

*We said that oxygen gas's ΔH of formation (for elements) is 0.*0953

*So, it's nice; these are not elements--these are compounds--so they have actual, finite values.*0958

*Well, when we do this, we end up getting -1396 kilojoules; is this exothermic, or is it endothermic?*0965

*ΔH is negative; that means it is giving off heat; that means, when you mix NH*_{3} and O_{2} gas, you produce three things: you produce nitrogen dioxide; you produce H_{2}O, and you produce heat.0976

*That is what this negative means; this is exothermic--it gives off heat.*0988

*Now, notice: this is in kilojoules; moles here have canceled; ΔH of formation is given in kilojoules per mole--we are multiplying by the stoichiometric coefficient in the balanced equation to give us a total for the reaction, as written in the balanced equation.*0994

*Moles cancel; the ΔH of reaction--standard ΔH of reaction--is equal to -1396 kilojoules.*1012

*ΔH of reaction is in kilojoules.*1024

*The ΔH of formation for the individual compounds--that is in kilojoules per mole of compound formed; that is why it is standard--we have standardized it to one mole.*1027

*OK, now let's see: let me show you what happened in terms of Hess's Law.*1040

*1/2 of N*_{2}, plus 3/2 H_{2}, goes to NH_{3}.1054

*The ΔH of formation is equal to -46 kilojoules per mole.*1063

*H*_{2} + 1/2 O_{2} goes to H_{2}O; the ΔH of formation is -286 kilojoules per mole.1071

*1/2 N*_{2} + O_{2} goes to NO_{2}; the ΔH of formation is equal to 34 kilojoules per mole.1083

*These are just the numbers that I used, that I looked up; and O*_{2}--that is just 0; the ΔH of formation equals 0.1095

*If I used Hess's Law, I would take these equations; I would manipulate them, flip them, turn them, multiply by things, and add straight down to get the final equation that I wanted.*1104

*Then, I would add the appropriate changes in the ΔH.*1114

*Well, a lot of that has already been done for you; again, the process--the steps don't matter.*1117

*All that matters is where you start, where you begin; because these values have been tabulated, I can just take the products, minus the reactants; that gives me my ΔH of reaction from the ΔHs of formation that are tabulated in the thermodynamic data in the back.*1122

*That is it; OK, let's do another example, and I think this will suffice to explain the use of ΔH of formation.*1140

*Calculate the ΔH of reaction, standard, for the thermite reaction (thermite--for those of you who don't know about thermite, it is the very, very, very powerful, very...well, you'll find out in a minute, as far as the ΔH, whether it is positive or negative).*1151

*The reaction is: 2 moles of aluminum metal, plus 1 mole of iron (3) oxide, goes to aluminum oxide, plus 2 moles of iron.*1176

*So, when you mix aluminum metal with iron oxide, you end up producing aluminum oxide; you end up releasing iron metal.*1201

*It's kind of amazing, actually.*1212

*So now, let's see what we have.*1214

*Let's go ahead and list some ΔH of formation values: the ΔH of formation of aluminum equals 0 (aluminum is an element); the ΔH of formation of Fe*_{2}O_{3} is equal to -826 kilojoules per mole.1218

*The ΔH of formation of Al*_{2}O_{3} is equal to -1676 kilojoules per mole: wow, incredibly exothermic!1245

*The ΔH of formation for iron--well, iron is an element: 0.*1260

*Good--so now, let us take 1 mole times -1676 kilojoules per mole, plus 0.*1268

*That is the aluminum plus the (yes, that is right...the products) iron, minus the reactants, which is 1 mole times -826 kilojoules per mole, which is the Fe*_{2}O_{3}, plus 0.1286

*When we do this, we get -850 kilojoules; that is 850,000 Joules.*1311

*Exothermic is an understatement.*1324

*Exothermic releases a lot of heat; in fact, it releases so much heat that the iron that comes out is not solid iron; it is actually liquid iron; the iron is melted.*1327

*OK, so let's see what we have here.*1341

*Do we need to do anything more?--no, that is it.*1347

*So, we found our ΔH of the reaction; it is equal to -850 kilojoules; notice, ΔH of reaction is in kilojoules, not kilojoules per mole.*1349

*Kilojoules per mole is ΔH of formation; that is what we used to get the ΔH of the actual reaction.*1361

*That is it; that is standard enthalpies of formation--nothing too difficult; I hope that made sense.*1369

*With that, I will go ahead and stop there.*1374

*Next time, we are going to talk about calorimetry, the transfer of heat and transfer of energy directly from one object to another.*1376

*Until then, thank you for joining us here at Educator.com; we'll see you next time; take care.*1385

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

*Today, we're going to be talking about calorimetry; we are going to be finishing off our discussion of thermochemistry.*0004

*We have talked about enthalpy, and a little bit about Hess's law, and things like that; we are going to finish it off with a discussion of calorimetry today.*0011

*So, let's just jump on in and see what we can do.*0018

*OK, so let's recall that, at constant pressure, the ΔH, the enthalpy, is actually equal to the heat of the reaction--which is very, very convenient for chemistry.*0022

*We do things at constant pressure so, when we talk about the heat and the enthalpy, they are the same.*0032

*Let's just write that down to begin with: recall that, at constant pressure, ΔH is equal to the q.*0038

*Remember energy: you can transfer energy to and from a system...or the change in energy of a system--the internal energy of a system, ΔE...there are two ways that energy can be changed.*0053

*It can either transfer as heat in or out of the system, or as work; in other words, you can do work on the system, or work out of the system.*0066

*So, q was heat; w was work, and it was usually pressure-volume work; and at the constant pressure, as it turns out, the enthalpy is equal to the heat of reaction.*0074

*We speak about the heat of a particular reaction: 2 H*_{2} + O_{2} goes to 2 H_{2}O; there is a certain heat that is given off--it is the enthalpy; it's written as ΔH=...negative for exothermic, positive for endothermic, things like that.0085

*Now, let us define something called the heat capacity.*0100

*No, I'm not going to write "definition"; I just said that it is a definition.*0106

*Heat capacity: the heat capacity is basically the heat absorbed by a material, divided by the increase in temperature.*0110

*For example, let's say you had absorbed 20 Joules of heat--some material, we don't know what it is--and the rise in temperature was, let's say, 4 degrees Celsius.*0134

*Well, of course, you know, when you do division, basically what you are doing with all division is taking the denominator to 1; so, this is going to be Joules per 1 degree Celsius.*0147

*That is the whole idea behind units; so 20 over 4--you get 5 Joules per degree Celsius.*0156

*That means, for every 5 Joules of energy that you put into this thing, the temperature of that thing is going to rise 1 degree Celsius.*0163

*That is a general definition of heat capacity: heat absorbed, divided by the increase in temperature.*0170

*OK, now, temperature is going to be expressed sometimes in degrees Celsius, degree Kelvin--the scales for degree Celsius and degree Kelvin are the same, so the actual gradation is the same; they just differ by 273; so it's not a problem.*0176

*Sometimes, you will see J/K; sometimes you will see J/ΔC.*0193

*All right, now different substances have different heat capacities, and this is what is kind of interesting.*0198

*Now, as a standard, in order for us to have something to discuss...as a standard, we choose 1 gram of a substance.*0204

*When we choose 1 gram of a particular substance to apply a certain amount of heat to, to see how much the temperature rises, it is called the specific heat--so not just the heat capacity, but the specific heat capacity.*0217

*So, as a standard, we choose 1 gram of a substance, and call it the specific heat capacity.*0228

*Now, heat capacity is Joules per degree Celsius; well, when we choose one gram of that, it's going to be Joules per degree Celsius per gram.*0252

*The unit for specific heat capacity: the unit is Joules per degree Celsius per gram of substance, which is often written as Joules over gram-degree Celsius; that is the unit of heat capacity.*0263

*OK, there is also molar heat capacity, and it is exactly what you think it is.*0283

*It is going to be the amount of energy that I need to...the amount of energy that one mole of something absorbs, and rises by 1 degree Celsius.*0295

*Or, if I want another way of saying it, if I want to raise the temperature of 1 mole of something by 1 degree Celsius, or 1 degree Kelvin, how much heat do I have to put into it?*0305

*1 gram; 1 mole; we can work with both; so the unit of that is, of course, Joules per degree Celsius per mole, which is equivalent to (the way you will see it is) mole-degree Celsius.*0314

*And again, this can be a K instead of a C; it just depends on what the problem is asking for.*0333

*Specific heat capacity/molar heat capacity: the amount of heat required to raise a particular amount of something 1 degree Celsius.*0337

*OK, now let's talk about calorimetry.*0347

*There are two types of calorimetry: there is constant-pressure calorimetry, and there is constant-volume calorimetry.*0354

*We're going to start with constant-pressure calorimetry.*0359

*Constant-pressure calorimetry measures heats of reaction (you remember that little rxn; it's a shorthand for "reaction"), or the enthalpies of reaction, for reactions occurring in solution--so in aqueous solution or some kind of solution; it doesn't necessarily need to be water as the solvent--it could be anything else--but something that happens in solution.*0366

*If I have hydrochloric acid, and I mix it with sodium hydroxide, and I stir it all up to...you know that it's an acid-base neutralization reaction...it is going to give off some heat.*0409

*Well, in calorimetry, what you are doing is: that reaction is taking place in solution, so the heat that that reaction either gives off or absorbs is going to be reflected in the temperature of the solution.*0422

*We actually use the solution to measure decrease in temperature, and because we know the particular heat capacity of water, it is an indirect way of measuring the actual amount of heat that is being given off, or released, from the particular reaction.*0436

*It will make more sense when we do a little bit of a problem.*0452

*Now, recall that energy is conserved; so, if a reaction gives off heat (exothermic), the solution gets warmer.*0456

*You know that from experience; if a particular reaction is taking place in solution, the solvent that is containing it is like some sort of a container, if you will.*0489

*The reaction itself gives off heat; well, the water absorbs that heat, so the heat given off is equal to the heat absorbed.*0497

*That is sort of the fundamental idea behind calorimetry.*0504

*Calorimetry is a device that measures heats given off or absorbed, using a particular medium that we know the properties of; and that gives us the heat of a particular reaction.*0507

*OK, so we measure the rise in the temperature or the decline in the temperature, and then we calculate heat absorbed or released.*0521

*OK, now, energy released equals energy absorbed.*0529

*If a system is releasing heat, the surroundings are absorbing that heat.*0544

*If a system is absorbing heat, it is the surroundings that are giving off that heat; it's a trade-off, and it happens at the boundary.*0548

*That is why we talk about systems and surroundings and universes.*0556

*OK, now, this is the particular equation that we are interested in: mCΔt.*0561

*q is the energy or the heat; so energy or heat...heat and energy are interchangeable.*0571

*This right here--this is the mass of the solution, or the mass of the particular thing that you are discussing--it could be a solid; it could be a liquid--its mass, generally in grams.*0582

*This C right here is the symbol for heat capacity--specific heat capacity.*0594

*Heat capacity: I'll say it's heat capacity, and I'll put "specific," because more often than not, it is specific heat capacity.*0601

*Remember, we are talking about a gram of something; this is why this mass is in grams.*0610

*This is going to be specific heat capacity, and this ΔT (I'm sorry, this shouldn't be a small t) is a change in temperature.*0614

*So, that is the change in temperature; now, mind you, it is not the beginning or the ending temperature--it is the actual change in the temperature.*0625

*It is the final temperature, minus the initial temperature--that is the definition of Δ.*0633

*It is the change in temperature; so, if a particular problem says the temperature went from 25 to 30, our ΔT in this case is 5.*0639

*Or, if they say (a problem could say) there was a temperature increase of 4.7 degrees, our ΔT is 4.7; so this is the change.*0648

*OK, so once again: mass, times heat capacity, times the change in the temperature, is equal to the energy; let's make sure that the units match here.*0658

*Mass is going to be in grams; the heat capacity is in Joules per gram per degree Celsius; and the change in temperature is going to be in degrees Celsius.*0667

*Degrees Celsius cancels degrees Celsius; gram cancels gram, leaving you Joules--sure enough, Joules: heat or energy is in Joules.*0679

*This equation, right here, is our fundamental equation for calorimetry.*0688

*Anything having to do with heat capacity, work, the heat absorbed or released by a substance (liquid, solid, whatever) is equal to the mass of that substance, times its heat capacity, times the change in temperature that it experiences.*0692

*It's fantastic: 1, 2, 3, 4: you have 4 variables--given any three of them, you can find the fourth.*0711

*Let's say you have the heat, the mass, and the change in temperature; you can calculate the heat capacity.*0718

*Let's say you have the heat capacity, the temperature, and the q; you can calculate the mass.*0723

*Whatever it is--like any equation, you can rearrange it to get what it is that you need.*0728

*You will run into different problems where you have to solve for each one of those, so it isn't just solving for one or the other; this is just how it is arranged.*0733

*You will rearrange it accordingly.*0742

*OK, well, let's just go ahead and do a problem, and I think it will start to come together.*0745

*Let's do an example: we have: 50 milliliters of a 1 Molar hydrochloric acid solution (and this m with a line on it--that is just an older symbol for molarity; it is just something that I'm used to; you will see the capital M--just as a reminder) is mixed with 50 milliliters of a 1 Molar sodium hydroxide solution, NaOH.*0755

*OK, temperature rises from 25 degrees to 31.9 degrees (this is Celsius).*0794

*Calculate the heat released...let's see...and the molar heat released.*0811

*OK, so 50 milliliters of a 1 Molar hydrochloric acid solution is mixed with 50 milliliters of a 1 Molar sodium hydroxide solution; the temperature rises from 25 to 31.9 (let's make this 31; sorry, it looks like a 37 here); we want you to calculate the heat released in this reaction when I mix these, and the molar heat released (in other words, the heat released per formation of 1 mole of whatever it is that is actually reacting here).*0836

*Let's just write a reaction; this is chemistry--you always want to start with a reaction, if you can, which...more often than not, you are going to be starting with some reaction.*0867

*We are putting together hydrochloric acid, and we are putting together sodium hydroxide.*0875

*We have H and we have OH; this is a neutralization reaction.*0880

*Any time you have an H and an OH, they are going to seek each other out, and they are going to form liquid water.*0885

*Of course, what you have left over is the sodium chloride.*0892

*Once you actually write it out, then you balance it; in this particular case, everything is balanced, so our reaction is HCl + NaOH → H*_{2}O, which I am actually going to rewrite a little differently, simply because I like to.0895

*It's up to you; you can write H*_{2}O; I have always been a big fan of writing it HOH, because I know that this H (let me do this in red)--this H comes from here, and this OH comes from here.0908

*I like to keep them separate, especially when I am balancing things that aren't 1 to 1; it makes it a little bit easier to balance.*0921

*So, HOH is perfectly acceptable--unless your teacher says otherwise, but hopefully, they will be OK with it.*0928

*50 milliliters of a 1 Molar--all right, so let's see what we have here.*0935

*Well, they want the heat released, so q=mCΔT.*0942

*Well, we need to find the mass of the solution; so the heat that is going to be released in this reaction (let me draw a little picture here)--this reaction takes place throughout this solution.*0952

*The solution is 100 milliliters of solution; well, the heat released into it is going to go into the water.*0963

*The water is going to rise in temperature; that is what is happening here.*0970

*The mass of the water--well, 100 milliliters; if you have 100 milliliters, times...water is 1 gram per milliliter, so you have a mass of 100 grams.*0976

*This mass has to be expressed in grams.*0988

*We have 100 grams of water; and again, heat released is heat absorbed.*0991

*Heat released by the reaction is heat absorbed by the water, by the solution.*0995

*Well, we have 100 grams of water; now, the heat capacity of water happens to be 4.18 Joules per gram per degree Celsius.*1000

*What that means is that, if I have 1 gram of water, and if I want to raise it by 1 degree Celsius, I have to put in (I have to heat it by) 4.18 Joules.*1010

*It is a very high heat capacity; that is what makes water the extraordinary substance that it does--one of the things is the fact that it has such a high heat capacity.*1019

*The change in temperature, now: 25 to 31.9--what is that--6.9 degrees Celsius?*1027

*OK, so now, Celsius cancels Celsius; gram cancels gram; we do our multiplication, and, if I am not mistaken, we end up with (if I have done my arithmetic right) 2884.2 Joules.*1034

*Now, you can express this as kilojoules if you want; that is going to be 28.9 kilojoules; but I'm just going to go ahead and leave it as Joules here--it's not a problem.*1054

*That is how much heat is released, based on everything that they gave us.*1062

*Rise in temperature: the heat released by the reaction is absorbed by the water; this is what calorimetry is about; usually, water is the particular...because we know the heat capacity of water, we can measure it, so this reaction released this much heat.*1069

*Now, this was part A; now we want to find the molar heat; OK.*1086

*This is: now we want to find the amount of Joules per mole of whatever it is we are discussing.*1093

*In this particular case, this particular reaction is a neutralization reaction.*1100

*The reaction that is taking place is: H*^{+} is getting together with OH^{-} to form H_{2}O.1104

*What we want to find is the amount of heat that is going to be released per mole of water formed.*1112

*That is what this is--as is: with 50 milliliters of a 1 Molar HCl, 50 milliliters of a 1 Molar NaOH, what we have is 28.9 kilojoules.*1122

*Now, we want to know what that is per mole of water formed.*1131

*OK, here is where we have to calculate how much is actually formed.*1135

*Well, 50 milliliters is 0.050 liters; times 1 mole per liter--that gives me 0.050 moles of HCl, which gives me 0.050 moles of H*^{+}, right?1141

*One molecule of HCl releases 1 hydrogen ion.*1159

*Now, we do the same thing for the hydroxide.*1164

*We have 0.050 liters of the sodium hydroxide, and it is also 1 Molar; liters cancel; that gives me 0.050 mol of OH*^{-}, because 1 mole of sodium hydroxide releases 1 mole of hydroxide.1167

*Well, this is nice: the equation is 1 to 1, so there is nothing limiting here; so every H is going to join with an OH--nothing is going to be left over, and we have 1:1:1.*1184

*We are not going to have .1 here; we don't add these; it says 1 mole of H*^{+} and 1 mole of OH^{-} goes to 1 mole of water.1197

*We're going to form 0.050 moles of H*_{2}O.1204

*OK, so now, we can go ahead and finish the problem.*1213

*We said that we ended up releasing 2884.2 Joules in this reaction.*1219

*Well, the amount of water formed was 0.050 moles; so again, when you divide, you are taking the denominator, and you are turning it to 1.*1229

*So, this becomes 57684 Joules per mole, which is equivalent to 57.7 kilojoules per mole(when the numbers start to get this large, generally they will express it in kilojoules)--per mole of water formed.*1244

*So, a particular set of conditions gives you the amount of heat, just in Joules.*1270

*Then, if you adjust that by putting the number of moles of the thing that you actually created, or that you are talking about, in the denominator, that gives you the molar heat released.*1276

*There you have it; OK.*1289

*Now, that was constant-pressure calorimetry; and again, constant-pressure calorimetry takes place in solution, because you are not really changing the pressure of the environment when you are pouring--when you are working in a solution.*1295

*The pressure doesn't change; the pressure is the atmospheric pressure on that solution, so everything works out really nicely.*1307

*Well, there is also something called constant-volume calorimetry, and so let's talk about that a little bit.*1314

*Constant-volume calorimetry: this takes place in something called a bomb calorimeter.*1324

*So basically, what a bomb calorimeter is (I'll give you a quick...): it is this container, and it is full of water, and inside the container, there is another container.*1336

*That is connected to the outside; and what we do is: we put some compound in there; we ignite it; and we burn it--we combust it completely--we blow it up, basically.*1353

*Now, because it is constant volume--the volume doesn't change--we're keeping a constant volume, the pressure rises; the heat rises; the heat escapes into the surrounding medium.*1365

*Well, what we have done with a standard: we have actually already calculated the heat capacity for the entire calorimeter, which involves the metal, any parts, the solution, the water--everything in there.*1382

*So, for constant-volume calorimetry, it isn't q=mCΔT; it is actually C*_{m}ΔT; this C_{m} actually includes (so I'll put it over here--q=mCΔT)...this C_{m} is the heat capacity of the entire bomb--of the calorimeter itself.1396

*It includes--it has already taken into account--the heat capacity of the water, the mass of the water, the heat capacity of the metal, the mass of the metal...everything is taken into account; so this one is actually going to be Joules per degree Celsius, or Joules per degree Kelvin.*1422

*This is slightly different; so, for constant-volume calorimetry, this is our equation.*1441

*The mass is accounted for, which is why they ended up putting it as a subscript on the heat capacity; so now, it's just a symbol reminding you that we are talking about constant-volume calorimetry.*1448

*The principle is the same: so you are just taking something--in this case, you are blowing it up; you are completely combusting it, seeing how much heat it releases or...well, yes, it's going to release heat.*1460

*The heat passes through the bomb into the solution; the temperature of the solution rises; we measure that temperature change, and then, we just use this.*1470

*Same thing; so, let's do an example.*1480

*OK, let's do this one in red.*1484

*Example (let's see): 0.1964 grams of quinone (which is C*_{6}H_{4}O_{2}) is placed in a bomb calorimeter; the bomb calorimeter is known to have a heat capacity of 1.56 (let me actually write this over here, so I can include the units)...C_{m} = 1.56 (not 7) kilojoules per degree Celsius.1488

*So here, they are actually expressing the specific heat in kilojoules per degree Celsius.*1553

*Now, the rise in temperature is (I don't know where these lines are coming from; OK) 3.2 degrees Celsius.*1559

*Now, the rise in temperature--this is the ΔT, right here.*1575

*Calculate (whoa, that was interesting; we don't want that; OK) the heat of combustion per gram and per mole of quinone.*1584

*We want to calculate the heat of combustion; we want to calculate the enthalpy of this combustion reaction.*1619

*Quinone--so, just to let you know, we have this thing quinone--I'm just going to write out the reaction with names.*1625

*Combustion: Quinone + O*_{2} → CO_{2} + H_{2}O.1632

*All combustion reactions are the same; O*_{2} is the other reactant; they produce carbon dioxide, and they produce water; this is a standard combustion reaction.1638

*We want to find out what the heat is per gram of quinone, and per mole of quinone.*1647

*OK, so we use our constant-volume equation, which is C*_{m}ΔT.1652

*Well, we know what C*_{m} is; it is 1.56 kilojoules per degree Celsius.1665

*We want to multiply that by 3.2 degrees Celsius.*1673

*What you end up with, when you multiply this out: you get 4.99 kilojoules.*1678

*OK, 4.99 kilojoules is released; now, A: per gram--so 4.99 kilojoules was released, and we said that there were .1964 grams of this quinone, so we take the 4.99 kilojoules, and we divide it by the mass, 0.1964 grams.*1686

*We want Joules per gram; when you divide by something, you are making the denominator 1.*1718

*So, you end up with 25.4 kilojoules per gram; that is how much heat is released.*1725

*If you have a gram of quinone--not a lot--and you combust it, it releases--the bonds in quinone release--25.4 kilojoules of heat; that is a lot of heat.*1734

*Now, per mole: well, we just have to now change this to moles.*1748

*Well, we know how to do that: we use the molar mass, grams per mole.*1752

*We go to 25.4 kilojoules per gram, times grams per mole, the molar mass; and now, we look up quinone--the molecular formula.*1755

*It ends up being 108 grams per mole; we end up doing the multiplication; we get 2743 kilojoules per mole.*1771

*108 grams of quinone--that is 1 mole--it releases 2,743 kilojoules (in other words, 2 million, 743 thousand Joules) of heat.*1783

*That is a very, very, very powerful molecule; there is a lot of energy stored in those bonds.*1796

*So again, basic calorimetry: constant-pressure calorimetry--let's just see--constant P: you have that the heat comes from mass, times the heat capacity, times the change in temperature.*1802

*And you have constant volume--Constant V--which is q=C*_{m}ΔT.1819

*The mass has been accounted for; it is just one calorimeter; you have the bomb; everything is accounted for.*1828

*Now, this equation could be used for anything; as far as calorimetry is concerned, the heat capacity--notice, we have been using this for water, 4.18 Joules per gram-degree Celsius.*1834

*But, water doesn't have to be the thing that I am heating up; this equation is perfectly valid for anything that you heat up, whether it is iron, concrete...anything at all; water, mercury...*1845

*Each different substance has its own heat capacity, so there you have it.*1857

*OK, let's close off with one more problem, and it will sort of tie everything together.*1864

*OK, a 28.2-gram sample of nickel at 99.8 degrees Celsius is placed in--is dropped in, if you will--150.0 grams of water at 23.5 degrees Celsius.*1872

*So, the nickel is at 99.8 degrees Celsius; we drop it in water that is at 23.5 degrees Celsius.*1909

*That is what this says.*1915

*After thermal equilibrium is reached (in other words, after we allow everything to settle down, where now no more heat is being transferred between the metal and the water)--after thermal eq... (oh, let me write it out, actually--after thermal equilibrium is reached), the temperature of the solution is 25 degrees Celsius.*1919

*What is C--what is the heat capacity for nickel?*1952

*So here, we are taking hot nickel and dropping it in water; the heat is released into the water, so the water rises in temperature.*1959

*Now, from that, we want to be able to measure the heat capacity of the nickel.*1969

*OK, let's draw a little schematic of what is actually going on.*1974

*OK, I have the nickel here at a temperature of 99.8; this is actually a good way to do this, because you can actually see it pictorially--what is going on.*1978

*We have the water, which is at 23.5 degrees Celsius; this is degrees Celsius; now, this one drops in temperature; this one rises in temperature; and at some point, when thermal equilibrium is reached, they are both going to be at 25 degrees Celsius.*1988

*That is where we ended up; that is the thermal equilibrium.*2005

*OK, now, energy that is lost by the nickel is energy that is absorbed by the water.*2010

*So, in other words, the heat that the nickel gives off is equal to the heat that the water absorbs.*2023

*They are equal to each other; OK.*2037

*I'm going to run this calculation one way, and then I'm going to do something slightly different--at the end, I'm going to talk about this little negative sign that is going to show up out of nowhere.*2041

*So, q of nickel is mCΔT, and q of H*_{2}O is mCΔT; so now, let me see; let's go ahead and put our values in.2049

*We have 28.2 grams of nickel; we are looking for C--that is what we want--and ΔT is 25-99.8 (I don't know if I should do it this way or not; OK, that is fine); 25 minus 99.8--that is equal to the mass of the water, which was 150, and the specific heat of water is 4.18, and the change in temperature is 25 minus 23.5.*2077

*25 minus 23.5: that is the ΔT for the water--it goes from 23.5 to 25, and this one goes from (the final minus initial) 25 from the 99.8; that is why we have set it up this way.*2124

*OK, so let's do 28.2C, times (25 minus 99.8), equals 940.5 Joules; when we run through this math, we multiply this by this, divide through; we should end up with .446.*2141

*OK, this is the heat capacity for the nickel; this is in Joules per gram per degree Celsius.*2177

*OK, we said that the (so let's stop and take a look at this) heat capacity is not a negative number; this is just a mathematical artifact here, based on how we did the problem.*2187

*When you are doing this particular kind of problem, what you end up having to do is--because the heat absorbed is equal to the heat released, well, as it turns out, because one is releasing heat, it is losing, and the other one is gaining that heat, so it's positive.*2200

*So, as it turns out, the sum of those is actually equal to 0; that is the whole idea.*2227

*From one perspective, the loss is equal to the gain; therefore, to actually make them equal, you really have to stick a negative sign in front of it.*2234

*Or, you can just run this particular calculation as-is, just leaving everything: mCΔT, mCΔT.*2245

*Because one is dropping in temperature and one is rising in temperature, the Δ (because it's defined as final minus initial)--one of these is going to be a negative number.*2254

*That is where this little negative sign shows up; this is an artifact of the mathematics, based on how you did it.*2264

*If you want to run this same problem and not have to worry about this sign at the end, you can just say the q of one equals negative (the q of the other), because that is what it is.*2273

*One of them is a positive quantity--it is being absorbed; the other one is a negative quantity--it is being released.*2287

*It just depends on your perspective.*2294

*As far as the mathematics is concerned, one is the negative of the other.*2296

*So, you can go ahead and put the negative sign in here, and then run this calculation; or you can just run this calculation, doing final minus initial for nickel, final minus initial for water--leaving everything as it is, based on the definition of Δ.*2300

*And then, if you end up with a negative sign, just go ahead and drop that negative sign; the negative sign is a physical feature of the problem--it doesn't necessarily reflect the actual mathematics of the problem.*2317

*Heat capacity is a positive capacity, so we will just take the .446.*2329

*As long as you keep that in mind with problems of this nature, where something is dropping in temperature and something is rising in temperature, the final is going to be--one of them is going to be--negative (the ΔTs).*2334

*We have to account for that negative; you can account for it here, in the beginning, or you can account for it at the end.*2345

*I hope that that makes sense; we may actually be returning to a particular problem like this, maybe, to talk about it a little bit more, because sometimes there is a little bit of confusion regarding this.*2353

*OK, thank you for joining us for AP Chemistry, and thank you for coming to see us here at Educator.com.*2362

*Take good care; goodbye.*2367

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

*We just finished talking about thermochemistry, and now we are going to move on to discuss a series of topics that are going to form, really, the most important part of chemistry.*0003

*Not that what we talked about wasn't important--I shouldn't say important; I should say the real core of what it is that you are going to be using in your future work.*0015

*We are going to be discussing--the first thing we are going to be starting with is kinetics.*0025

*We are going to talk about kinetics; after that, we are going to talk about equilibrium; after that, we are going to talk about acid-bases and some more aspects of acid-base chemistry.*0030

*And then, we are going to talk about thermodynamics in greater detail--not just enthalpy, but we are going to talk about free energy and entropy, and then we are going to round it out with a discussion electrochemistry.*0040

*Afterward, we will actually come back and discuss some of the topics like bonding and solutions and things like that.*0052

*But, I wanted to get to the heart and soul of what chemistry is--the things that you are going to be using--because we want to spend a fair amount of time with it and get really, really comfortable with it, over and over and over again.*0060

*Our first jump into this is chemical kinetics.*0073

*Chemical kinetics is the study of how fast something happens.*0078

*As it turns out with chemistry, you have thermodynamics, which tells you whether something can happen; and then you have chemical kinetics, which is to tell you how fast something happens.*0083

*You might have this wonderful, wonderful reaction that works thermodynamically, and it is the greatest thing you have ever seen, but it is just so slow that maybe it comes to completion after four thousand years.*0096

*There are plenty of reactions like that...so both are important: you have to know that a reaction actually works--will go forward without too much effort on your part--and that it will actually go forward in a reasonable amount of time, because there is the practical issue.*0106

*So, let's just jump in, and today we are going to talk about reaction rates, and we're going to introduce this idea of rate laws.*0119

*Much of our discussion today is just going to be getting you familiar with the context--that is what it is going to be about today.*0126

*OK, let's get started.*0136

*Let us start with a definition, like we often do.*0139

*No, I am not going to write the word "definition"; I am just going to write the rate.*0147

*Well, you all know what a rate is; it is the change in something over the change in time.*0153

*In this particular case, let's write ΔA over ΔT, where A, in the square brackets, is the concentration of a particular species (it could be any species--molecule, atom, whatever we happen to be on...ion...more often than not, it is going to be something in solution).*0160

*So, it's the change in concentration; so Δ is the concentration in moles per liter: molarity--pretty standard: M, and you have also seen it as m with a line on it; that is my particular symbol for it.*0183

*The rate is the final concentration of something, minus the initial concentration of something, divided by the time it took for it to go from the initial concentration to the final concentration.*0198

*Let's just take, as an example, a particular reaction: we'll write the balanced reaction, 2 moles of nitrogen dioxide decompose into 2 moles of nitrogen monoxide, plus a mole of oxygen gas.*0210

*Of course, all of these are gases, so let's go ahead and write that, just to be absolutely complete.*0227

*Now, I'm going to give a table: it's going to be a timetable, a table of time increments.*0233

*Then, I'm going to give you the concentrations of this, this, and this.*0239

*Then, we are going to see what is going on here.*0242

*Let's draw this out; so we have time t: 0, 50, 100, 150, 200, 250, 300, 350, and 400.*0245

*Now, we have NO*_{2}; and again, square brackets always means concentration in chemistry, in moles per liter.0265

*NO*_{2}, concentration of NO, concentration of O_{2}...so, 0.0100, 0.0079 (it's going to take me a couple of minutes to write this; I apologize), .0065 (I just want to give you as broad a picture as possible), 0.0055, .0048, and then these are 0 and 0...1, 2, 3, 4, 5; you know what, let's just stop with 5 of them.0274

*0.0021, .0011, 0.0035, .0018, 0.0045, 0.0023, .0052, 0.0026, and it goes on; but we have a pretty good idea here, so let's talk about what is happening here.*0312

*This is a particular reaction that was run, and it is a decomposition of nitrogen dioxide to nitrogen monoxide plus oxygen gas.*0347

*At different time intervals (at 50 seconds, then 100 seconds, then 150, and so on), I actually measure the concentration.*0354

*Well, at time 0, where we started, I started off with (let me use red) 0.0100 moles per liter of the NO*_{2}.0362

*Notice, this is 0 and this is 0; in other words, nothing has happened yet.*0373

*This is right at the beginning of the reaction.*0377

*So, 50 seconds later, I measure the concentration in the flask; I notice that I have 0.0079 moles per liter; there is 0.0021, 0.0011 of the oxygen.*0379

*That is all this is saying; I am just tabulating different concentrations of all three species at different times--50-second intervals.*0392

*100 seconds, 150...at 200, I have .0048--just about half; .0052, .0026.*0399

*Something you should notice here: notice that this column tends to be half of this column.*0410

*Well, it makes sense; if you go up here and you look at the coefficient, this coefficient is 1; this coefficient is 2.*0416

*It is telling me that, for every 1 mole of oxygen produced, 2 moles of nitrogen monoxide is produced.*0422

*These numbers confirm the stoichiometry.*0429

*This is a standard sort of kinetic data.*0432

*You are measuring concentrations, and you are taking time.*0438

*Well, let's go ahead and see what the graph of this looks like.*0441

*OK, so let me move over here; now, let's go ahead and draw this graph: it will be something like this.*0445

*This is the time axis, and this is the concentration axis; I'll just put brackets for concentration.*0453

*Let's see, let's do 1, 2, 3, 4...let's do 0.0100 here, and let's...so these are in increments of...0.0025, 0.0050, 0.0075; that is right--I know how to do math.*0464

*Then we have 1, 2, 3, 4, 5, 6, 7, and 8; OK, so now, we are going to take the data that we have, and we are going to graph it.*0491

*One of the graphs that we have looks like this: it goes down to about something like that; we have another one that goes up to about like that.*0507

*And then, we have another one which is about halfway--something like that.*0527

*OK, so this right here--this is the NO*_{2}.0533

*It starts off at 0.0100 concentration, and as time passes, it decreases; the concentration goes down, down, down, down, down.*0539

*Now, notice what is interesting to notice about this: this is not a straight line; this is a curve--it starts at a certain thing, but it actually starts to slow down--it starts to stabilize and become...OK, so that is interesting.*0547

*This up here--this graph--is for the NO.*0558

*As the reaction starts, there is none of it; the concentration increases as time goes by: boom, boom, boom, boom, boom, boom, boom.*0564

*Because the stoichiometric coefficient is 2 NO*_{2}, 2 NO, this line and this line are actually just copies of each other.0573

*As this one drops, that one goes up, and the crossover is going to be completely symmetric.*0581

*Now, this line over here--this graph--that is the O*_{2}.0587

*Notice, it is half of the NO, which works with the stoichiometry; for every 1 mole of O*_{2}, 2 moles of NO is actually produced.0591

*This gives a graphical representation of what we have.*0601

*Now, let's just find an average rate from, let's say, from t0 to t50; so, average rate that equals Δ...in this particular case, let's do ΔNO*_{2} over Δt.0604

*Let's take from 0 to 50--the rate from 0 to 50; it is going to be...Δ is final minus initial, so it is going to be 0.0079 (after 50 seconds, that is the concentration of the NO*_{2}--it has gone down), minus the initial, which is 0.0100.0630

*The time difference is 50 minus 0; what you end up with is -4.2x10*^{-5} moles per liter per second.0657

*So, remember these numbers right here: this is concentration--it's moles per liter.*0672

*That means that the nitrogen dioxide is actually decreasing, on average, between 0 and 50 seconds--the first 50 seconds of the reaction, it is decreasing at 4.2x10*^{-5} moles per liter per second.0676

*That is what this negative sign tells me--it tells me that it is decreasing; it is confirming this.*0689

*You also know this from the slope; the slope along that curve is negative.*0695

*OK, now, in kinetics, it is customary to express all rates as positive.*0700

*So, this negative 4.2x10*^{-5}--we will usually express it as a positive rate, so we will actually just negate it, so it will be something like this.0710

*Let me go back to blue here.*0721

*The rate is equal to -ΔNO*_{2}/Δt, equals -(-4.2x10^{-5}) moles per liter per second.0725

*Now, again, this is just a customary thing; when you get a negative rate, you know that something is decreasing; by now, chances are that you have had calculus or you are in calculus--you have a sense of what a rate is.*0744

*A negative rate means something is decreasing; a positive rate means that something is increasing.*0760

*Again, I myself don't particularly care for using positive rates--I like to keep things...I like to use the negative and positive signs, but it is customary, so I just thought that you should know that.*0764

*Symbolically, it looks like this.*0773

*This is the symbol that you will see: -ΔNO*_{2} when something is a reactant that is actually decreasing as you move forward in a reaction.0776

*OK, now let's see some average rates for different time intervals.*0786

*Let's do 0 to 50, 50 to 100, 100 to 150; so, let's put together another table here: the time interval and the average rate (which is -ΔNO*_{2}/Δt); we are just going to calculate rates for different time intervals.0791

*OK, and again, this is just a symbol; that is why this negative is here.*0812

*From 0 to 50, we already calculated as 4.2x10*^{-4}; again, we are keeping it positive.0815

*From 50 to 100, when we use the data, we end up with 2.8x10*^{-4}.0825

*So again, we take the concentration at 100 minus the concentration at 50, divided by 50, because 50 seconds passed.*0835

*That is what we are doing; and then, from 100 to 150, you have 2.0x10*^{-4}.0843

*This is going to be in (by the way, the units are) moles per liter per second.*0852

*If we go from 150 to 200, and then we will go from 200 to 250; we will get 1.4x10*^{-4}, and we will get 1.0x10^{-4}.0858

*So, at different times along the reaction (from 50 to the next 50, next 50, next 50), notice: the rate of the reactions, the rate of concentration decline--it is not constant.*0871

*That is confirmed from the graph, of course; if it were constant, you would have seen a straight line, but the graphs don't actually make a straight line--they actually slope down a little bit, so...different values.*0886

*OK, so not the same--the rate actually decreases.*0899

*Now, these are average rates: averages...sometimes we want the instantaneous rate at a given time.*0904

*Instantaneous rates: for an instantaneous rate, what you do is: you take the graph; you draw a tangent line at a particular point that you are interested in; and then you take two points on that line, and you calculate the slope.*0913

*The derivative is what you are calculating, but it is just the slope of the tangent line; so those of you who have had calculus--it is just--you know that that line...the slope represents the derivative of that graph, if you were to have an equation for it.*0936

*More often than not, you will not, because you are dealing with raw data.*0951

*You are going to take the data, make the graph, draw the tangent yourself, and then estimate.*0954

*So, for an instantaneous rate, we form the tangent line to the graph at (t, f(t)), or (t, concentration)--that is what f(t) is--it's the concentration--so maybe I should write that.*0961

*Time t, and the concentration of whatever species you have to be dealing with...then, calculate the slope.*0991

*That will give you an instantaneous rate: calculate the slope.*0999

*OK, so let's do something like that.*1007

*Let's take our graph, and we have a particular curve, and then we are going to draw a tangent line to that; let's say we choose 100--that is, we want to know what the instantaneous rate is; what is the rate of the reaction at t=100?*1011

*OK, well, you draw the tangent line; you pick a couple of points on that tangent line; you see where they hit the y-axis; and then, of course, you see where they hit the x-axis.*1034

*You take Δy/Δx (in other words, that is x*_{2}; that is x_{1}; y_{2}, y_{1}); you just take y_{2}-y_{1}/x_{2}-x_{1} from points that you pick from the line that you drew (not other random points).1050

*The point that it passes through--the tangent--is at the point that you are interested in; so, you will get (when we do this calculation based on the data that we have earlier--this is just a small portion of the other graph around 100)...the rate ends up being 2.4x10*^{-5}.1072

*And again, it is going to be moles per liter per second.*1094

*So, at 100 seconds, that is the rate of decrease of nitrogen dioxide, based on the particular data that we collected.*1097

*OK, well, notice: we did the reactants; we had NO*_{2} going to NO + O_{2}; balance it.1106

*Well, what if I wanted to express the rate with respect to the products, either nitrogen monoxide or oxygen gas?*1118

*I can do that, and it ends up looking something like this.*1126

*Rate of NO*_{2} consumption: well, I know that that equals the rate of NO production; it is also equal to twice the rate of O_{2} production.1130

*That is what the stoichiometry tells me--that the rate of this is twice as fast as this, because the ratio is 2:1.*1158

*Well, the symbols are -ΔNO*_{2}/Δt; this is a symbol for the rate.1165

*Again, we put a negative sign here so that everything is positive, because this Δ is going to be negative; so negative, negative makes this positive.*1173

*Well, that is equal to ΔNO/Δt; that is equal to twice the ΔO*_{2}/Δt.1182

*That is it; it doesn't really matter; more often than not, we tend to work with reactants--the depletion of the reactants.*1193

*If you wanted to, you can express it as the increase of the products, but again, most of the time, we'll be talking about (it's just sort of standard, at this point, to talk about) the depletion of reactants.*1202

*OK, now, let's talk about what it is that is actually going on here, and how it is that we can derive some sort of a rate law, some sort of a mathematical expression.*1217

*This is a symbol, and we have some data; how do we take some of this data and convert it into an actual, mathematical formula that expresses just how fast the rate is changing?*1228

*These right here give me rates of change of concentrations.*1241

*OK, if we take the reaction 2 NO*_{2} → NO + O_{2}, well, when you start the reaction, there is no nitrogen monoxide, and there is no oxygen gas.1248

*So, the reaction is just going to fall forward as fast as it can.*1265

*Well, in chemistry, many reactions...some go to completion all the way, but at some point, you actually do reach a point called equilibrium (which we will talk about later in more detail), but understand that this reaction is actually reversible.*1268

*At some point, enough nitrogen monoxide has formed, and enough oxygen has formed, so that now the reverse reaction starts to take place.*1282

*Well, so now you have the forward reaction of the decomposition of nitrogen dioxide, and you have the backward reaction, the reverse reaction, of the formation of nitrogen dioxide.*1290

*Things can get kind of complicated here.*1300

*So, what we do when we run these kinetic experiments is: we try to measure concentrations as close to t=0 as possible.*1303

*In other words, start the reaction, and immediately take all of the measurements that you need, in order to find the rate of the reaction, before this reaction has had a chance to get to a point where the reverse reaction becomes significant--where it starts to change the actual depletion of the nitrogen dioxide.*1314

*If I measure the rate of the depletion of nitrogen dioxide right at the beginning of that calculation, I don't account for how fast...I don't have to worry about when the reaction has run a little bit, and now the reverse reaction is going to affect the rate at which this depletes.*1333

*It changes things; it is very, very complex to deal with that mathematically and physically.*1353

*What we try to do is: we try to measure reaction rates right at the beginning of reactions.*1358

*When we do that, as it turns out, under these conditions, the rate is actually proportional to the concentration of the reactants, raised to certain powers.*1366

*So, in the case of NO*_{2}, as it turns out, the rate of the reaction--the actual rate of the reaction (this is just a rate of consumption and depletion; that is not the rate of the reaction) actually ends up being proportional to...raised to some power.1384

*The rate is equal to some constant, K, times the concentration of nitrogen dioxide, raised to some power.*1404

*Now, this K and this n have to be determined experimentally.*1416

*I can't just read it off from the equation.*1422

*Let me say this again; let me move to another page and write this again.*1425

*Now, as it turns out, the rate of a reaction is proportional to some power of the reactant concentrations.*1429

*If I have some species, A, going to B + C, the rate of the reaction, which is symbolized as -ΔA/Δt, is equal to some constant, K, times the concentration of A raised to some power.*1465

*This right here is what we call a differential rate law.*1486

*Now, what if you had something that had 2 reactants, A + B, going to...I don't know, C + D.*1491

*Well, now the rate law (and you can choose any one of these to actually symbolize it; let's just go with A again): -ΔA/Δt (this is just the symbol for the rate) is equal to some constant, K, times A to some power, times B to some power--because now, both reactants are involved.*1498

*Again, if you have 3, you will have 3 terms; it is just like that.*1528

*So again, the rate of a reaction is proportional to some power of the reactants' concentration.*1532

*Depending on whether you have one reactant or two reactants...if you have 1 reactant, you have 1 term like this; if you have 2 reactants, you have 2 terms like this...*1537

*This is called a rate law; specifically, it is called a differential rate law.*1549

*Now, again, this K, this n, and this m--they have to be determined experimentally; we have to run the reactions, and we have to work this out.*1554

*We will actually do this in the next lesson; we will actually talk about how to derive this rate law; it is actually quite nice--quite easy.*1562

*OK, now, let's write: these are differential rate laws.*1571

*We also have something called integrated rate laws, which we will also talk about in, not the next lesson, but the lesson afterward.*1591

*These are called differential rate laws.*1596

*It is when the rate of a reaction is expressed as some constant, times the concentration of the reactant, raised to some power.*1600

*If there is 1 reactant, that is 1 thing; if there are 2 reactants, you include those; if there are 3 reactants, you include all of those.*1609

*You would have A*^{n}, B^{m}, C^{s} power...something like that, times some constant.1615

*What is important is: the K, the n, and the m are determined experimentally.*1621

*You can't just read them off from the stoichiometry.*1625

*These coefficients in the actual equation have nothing to do with these numbers.*1628

*OK, now, let's see: let's actually go through a sort of basic example of determining a particular rate law for a particular reaction.*1634

*Let's start with a reaction: 2 N*_{2}O_{5} → 4 NO_{2} + O_{2}.1647

*OK, now let me just put in carbon tetrachloride; so this happens in carbon tetrachloride solution.*1661

*I'm actually going to...you know what, let me actually write the equation again up here.*1670

*2 N*_{2}O_{5} → 4 NO_{2} + O_{2}, in CCl_{4}.1679

*OK, so now we have some time measurements, and we have some concentration measurements: N*_{2}O_{5}...1692

*0, 200, 400, 600, 800, 1000, 1200, 1400; we have 1.00, 0.88, 0.78, 0.69, 0.61 (not 71), 0.54, 0.48, 0.43.*1702

*OK, so we have this set of data that we collected at different time increments; we measure the concentration of the N*_{2}O_{5}, and sure enough, it is decreasing.1750

*Now, when we graph this (I'll just do the graph right here), we are going to end up with something like...if this is 1, start off with 1; let's go ahead and put 0.2 down here, and let's put 400, 800, 1200, 1600...so, it's going to look something like this.*1763

*This data is just plotted on a graph: 400, 800, 1200, 1600--nice and simple; nothing particularly complicated.*1791

*This is time; this is concentration of the species; OK.*1805

*So now, what we want to do is: we end up actually...so now, I'm going to leave this here, and I'm going to move to the next page...when I take that graph, and I pick a couple of points, I can actually measure the concentration of N*_{2}O_{5} at particular points.1810

*When I do that, I have the following (I can actually measure the rate, the instantaneous rate; remember, you draw tangent lines at a given point; you measure the rate at that point; you measure the rate someplace else)...*1836

*So, at a concentration of 0.90 moles per liter, the rate ends up being 5.4x10*^{-4}.1857

*0.45: 2.7x10*^{-4}; now, notice what has happened here.1871

*At a certain concentration, I measured the rate of the reaction with a tangent line.*1884

*At another concentration, I measured the rate; but notice the relationship between these two--I did this specifically.*1891

*This is half of that; well, this is half of that; what happens here--the correlation--this also happens to be half of that.*1899

*So, when you have something like that, where when the concentration is halved, the rate is also halved (it won't always happen like this; this just happens to be this particular data that we did)--we arranged it so we can check the rate at concentration that is easily comparable.*1911

*.9; .45; this is twice that; this is half of that; as it turns out, when I halve the concentration, the reaction rate is also cut in half; that is a linear relationship--that is the whole idea of it.*1946

*If you double something, and the thing that you are measuring also doubles, that is a linear relationship.*1963

*If you double something, and the thing that you are measuring goes up by a factor of 4, that means it is a quadratic relationship.*1970

* 2, 4; 3, 9; here, you have halved it or doubled it, depending on which direction--how you want to look at it.*1977

*Here, you have doubled something; the difference has doubled; so it is a 1:1 relationship--that means it is a linear relationship: linear means an exponent of 1.*1987

*So, our rate law is: this is the symbol when we write the rate; in fact, oftentimes, you don't really need this symbol; I know that this symbol actually confuses a lot of students--it doesn't need to; I don't particularly care for it very much.*1998

*It is just the symbol--it means rate; for all practical purposes, you can actually leave this off and just write "rate =."*2018

*We said that the rate equals some constant (K), times the concentration of the reactant (N*_{2}O_{5}), raised to some power (n).2026

*Well, here, because the relationship is linear and I read it straight off, when a concentration is halved, the rate is halved; or when a concentration is doubled, the rate is doubled.*2037

*That is a linear relationship; that means n equals 1.*2049

*So, I don't need this; so this is the rate, equals some constant (K), times N*_{2}O_{5}.2055

*That means the rate of this reaction, based on the data, is actually some constant (K), raised to the concentration of the di-nitrogen pentoxide, raised to the first power.*2071

*This is the differential rate law for this particular reaction.*2083

*The most important thing to take away from this is that, under conditions where we actually measure rates, near points where the reaction is just starting, where we have a concentration and then another concentration: we have the rate at one concentration and the rate at the other concentration.*2101

*When we see how the concentration changes, and then we compare it to how the actual rate changes, that gives us the exponent of the particular concentration term of the reactant.*2128

*That n is called the order--the order of the reaction.*2140

*This is a first-order reaction; so the decomposition of di-nitrogen pentoxide is a first-order reaction.*2148

*Now, I'm going to go ahead and stop this lesson here; I just wanted to give you sort of a brief run-down about how this works.*2155

*Next lesson, we are actually going to get deeper into this, and we are going to calculate differential rate laws; we are going to calculate the exponents; and we are going to calculate the rate constants more systematically.*2161

*But, I just wanted you to see and get a sense of what this looks like and how things operate.*2173

*Thank you for joining us at Educator.com; we'll see you next time for a greater discussion of kinetics; goodbye.*2178

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

*Today, we are going to continue our discussion of reaction kinetics, chemical kinetics.*0004

*In the last lesson, we introduced rate laws--in particular, the differential rate law.*0010

*Let's just recap what that is, and we'll get into the actual method (or one of the methods) of determining that differential rate law for a particular reaction, given a particular set of data.*0015

*The method is called the method of initial rates; let's go ahead and get started.*0029

*We said that (you know, I think I'm going to use black ink now) the rate of a reaction, which is how fast it is going, symbolized by -Δ some species (reactant species), where this negative comes from...Δt is equal to some constant, times the product of the reactants (however many there are), raised to (A, B...let's say there are three reactants) certain powers.*0035

*These powers are called the order of that reactant.*0079

*n, m, s...the order of the reactant; this m, s, and n, and K, are experimentally determined.*0085

*The idea is this (we talked a little bit about it last time, but let's be more specific about it)--the idea is: we want to start a reaction, and we want to measure the rate of that reaction, the rate of depletion of this particular thing (the reactant) as quickly as possible, before any of the products have had a chance to build up and start working in reverse, because as product builds up, the reaction goes backwards.*0103

*That is the thing; chemistry, as it is going one way, is also going the other way; so, before it becomes too complicated, we want to see if we can measure the rate of the reaction--before anything else gets in the way.*0131

*Let's just write this down--so, the idea is this: we run several experiments with different initial rates...not initial rates; I apologize--initial concentrations.*0144

*We pick specific concentrations to start with--reactants and products--and we measure it immediately.*0171

*We start with different initial concentrations and measure the initial rates (that is what we are measuring--initial rates) before any products have an opportunity to build up and complicate matters.*0179

*OK, we then compare the rates of the different experiments...we then compare the rates among experiments...to see how the rate depends on the concentration (or specifically, the concentration change from one experiment to the other).*0228

*It depends on the concentration of a given reactant.*0272

*This will make a lot more sense when we actually do our first example, which we are going to do right now.*0277

*We are going to run a series of experiments--2, 3, 4, 5 of them--and we are going to modify concentrations (initial concentrations).*0283

*So, if I have an Experiment 1 and Experiment 2, and initial concentration 1 and initial concentration 2, I'm going to measure the rate; and depending on the concentration changes, I am going to compare the rates, and I'm going to see what the relationship is (if it doubles; does it go up by 4? does it go up by 6? does it go up by 19? 14.3?), and that is how I get those exponents.*0295

*That is how I get these numbers first.*0318

*Then, when I am done with that--once I get the orders of the reactions--then I can take any one of the data in the experiment and put it in here to find K.*0320

*Let's go ahead and do that; it will make a lot more sense.*0329

*OK, so let's take (you know what, I think I'm going to start with a fresh page here)...let's do the following reaction.*0332

*Let's take ammonium ion, plus the nitrite ion; when I mix those together, I am going to end up with nitrogen dioxide gas, plus 2 moles of liquid water.*0341

*Ammonium and nitrite ion forms nitrogen dioxide gas, which bubbles off; and I am left with liquid water.*0361

*Notice, I have two reactants--two reactants; I did three experiments on this; here is my data.*0372

*Experiment #1, Experiment #2, Experiment #3 (you know what, I probably don't need to make them this big--leave myself some room here--Experiment #2, Experiment #3)...*0390

*Now, I have NH*_{4}^{+} concentration; NO_{2}^{-} concentration; and I have initial rate, which is what I am measuring.0403

*Now, my first experiment: I started (remember, we are dealing with reactants here; we are measuring the rates before any products have a chance to build up)...0.100 moles per liter, and here I started with an NO*_{2}^{-} concentration of 0.0050 moles per liter.0418

*When I ran this experiment, right as it started, I measured the rate to be 1.35x10*^{-7}.0441

*Now, I ran a second experiment: 0.100; I left this concentration the same, but I doubled this one: 0.00...no, I doubled this concentration: 0.0100.*0450

*When I make a change from experiment to experiment, I change one species--I don't change everything.*0467

*The reason is because I need two experiments to compare; later, what I'm going to do is...notice, I have left this the same, and I have left this the same, but I have changed this; I'm going to use experiments 1 and 2 to compare, to find out how it's related to nitrite ion concentration--how the rate depends on this concentration.*0473

*I change one species at a time; I don't change all of them.*0493

*I left this the same and changed this one; when I do that, I ended up with a rate of 2.70x10*^{-7}.0496

*Now, Experiment 3: 0.200; now, I doubled the ammonium concentration, and I left this concentration the same.*0505

*When I do that, I ended up with a rate of 5.4x10*^{-7}.0517

*So, my first experiment; second experiment; third experiment: .1, .005, .1 (wait, I'm sorry, this is .1, not .01--I knew I had too many...I was going to say it looks a little odd); .100, and this is .100; that is right, .2.*0523

*OK, so .1, .1 initial; initial rate; and then, I do another experiment, and I start with .2 mol of this, .1 mol of that, and I measure an initial rate, 5.4.*0548

*Now, I can do my comparison.*0561

*OK, so here is where the good stuff starts.*0563

*Now, we said that the rate, which we will symbolize as -Δ (and I can pick any one--I am just going to go ahead and pick this one, so this is just a symbol); it's equal to some constant (K), times the NH*_{4}^{+} raised to some power (m--actually, let me use n, because I want to...); and the other is NO_{2}^{-}, raised to the power of m.0566

*So this is it; this was our assumption: under these conditions of initial rate, as it turns out, the rate is dependent on the product of the concentrations of the reactants, raised to different powers.*0601

*I write this, and now I am going to use this experiment to determine m, to determine n and determine K.*0615

*OK, now, Experiment 1 says that the rate is equal to 1.35x10*^{-7}.0622

*Well, the rate is equal to K, times the concentration of NH*_{4}^{+}, which in Experiment 1 is 0.100 to the m power, and the concentration of NO_{2}^{-}, which is 0.0050 to the n power.0641

*I just literally plug in these numbers into what I wrote down.*0662

*Experiment 2 says (I'll say this is rate 1, and this is rate 2, of course): rate 2 is 2.70x10*^{-7}, and that is equal to K times...now I use the values for Experiment 2: .100 to the n power, times 0.100 to the m power.0669

*OK, now I compare the two rates.*0698

*Rate 1, divided by rate 2, is equal to 1.35x10*^{-7}, divided by 2.70x10^{-7}, equals--this over that equals--K times 0.100 to the n, times 0.0050 to the m (there is a lot of writing here, but it is important to see everything), over K times 0.100 to the n, over 0.100 to the m.0704

*Now, look what happens: 1.35x10*^{-7}, over 2.7x10^{-7} (let me make this a little more clear here--this looks a little confusing...10^{-7})...this is equal to one-half.0742

*K cancels K; .100 to the n, .100 to the n--those cancel; what I end up with is .0050 to the m, over .100 to the m.*0758

*This is just .0050, divided by .100, to the m power.*0772

*Well (oh, I don't know where these lines are coming from!), it equals...this is just one-half to the m power.*0780

*Well, 1/2 equals 1/2 to the m power; that implies that m is equal to 1.*0796

*So, because m is equal to 1, that means the nitrite ion concentration, m, is 1; that means it is an order 1 in nitrite.*0807

*So now, I will do the same to find n; so I have already found m--now I am going to do the same thing for n.*0829

*This time, I am going to compare Experiment 2 and Experiment 3.*0836

*Let's go back to black.*0841

*OK, Experiment 2: the rate is equal to 2.7x10*^{-7}; it is equal to K times 0.100 to the n, times 0.100 to the m, over (oh, actually, no, not yet).0845

*OK, and then we will do Experiment 3: the rate is equal to 5.40x10*^{-7}, is equal to K times 0.200 to the n, times 0.100 to the m.0880

*Well, rate 2 divided by rate 3 (and you can do it in either order--you can do rate 3 over rate 2; it doesn't really matter) equals 2.70x10*^{-7}, over 5.40x10^{-7}, equals K times 0.100 to the n, 0.100 to the m, divided by K times 0.200 to n, 0.100 to the m.0900

*K cancels K; that cancels that; that is equal to 1/2; that is equal to 1/2, this time to the n power.*0939

*Well, 1/2 to what power is equal to 1/2?--that implies that n equals 1.*0949

*I found my differential rate law.*0956

*OK, now my differential rate law is this: my rate is equal to some constant (K), times the concentration of ammonium ion raised to the first power, times the concentration of nitrite raised to the first power.*0960

*The rate is first-order in each; the total order of the reaction is 2--you add up the orders.*0978

*So, m is equal to 1; n is equal to 1; their sum is equal to 2, so the overall order of this reaction is 2.*0986

*That is what I have done--I have compared rates to find this--but I am not done.*1001

*Now, I want to find K.*1004

*Now that we have our rate law, that the rate is proportional to the concentration of ammonium raised to the first power, times the concentration of nitrite raised to the first power, I can take any one of those experiments, 1, 2, or 3, plug in the values (I have a rate; I have this concentration; I have that concentration), and I can solve for K, and that is exactly what I am going to do.*1005

*I'm just going to go ahead and choose Experiment 2.*1030

*Experiment 2 says: the rate is 2.70x10*^{-7}, equals K times (oops, I don't have to actually do the brackets when I'm putting numbers) 0.100 raised to the first power, and the 0.100 raised to the first power.1034

*And now, I just go ahead and do my division.*1070

*OK, let me see here; .1; .01; minus and minus; and I end up with K is equal to 2.70 (please check my arithmetic; my arithmetic is notorious for being wrong) times 10 to the -5.*1079

*There we go; and the units, in this particular case, is liters per mole-second.*1098

*The reason is because we have a rate, which is moles per liter per second, and then you are dividing by moles per liter, dividing by moles per liter, so this K ends up with liters per mole per second.*1106

*K--this is not the unit for all K's; this is just the unit for K for this particular reaction, which is second-order--first-order in both of these.*1125

*OK, method of initial rates: we found the rate law, and then we used one of the experiments to find K, and then we plug it back in, so our final answer is: the rate equals 2.70x10*^{-5} times the concentration of NH_{4}^{+}, times the concentration of NO_{2}^{-}.1137

*That is our differential rate law for this reaction.*1164

*We are done.*1167

*OK, let's do another example.*1168

*This time, we are going to use three reactants.*1172

*This time, we will do bromate ion, plus 5 bromide ion, plus 6 hydrogen ion, forms 3 molecules of bromine plus 3 molecules of water.*1182

*3 reactants: our rate law is going to have 3 things in there.*1202

*OK, you want to determine the order of each reactant; you want to determine the overall order; and we want to determine the rate constant.*1207

*That is our task, OK?*1219

*Let's write it down: determine the order of each reactant, overall order, and rate constant.*1221

*OK, well, here is our experimental data...well, let's go ahead and write the rate law first--the general one.*1242

*The rate is equal to some constant (K); it is going to be that raised m power, Br*^{-} raised to n power, H^{+} raised to s power.1249

*This is the general form; we are going to find m, n, and s, and we are going to find K, based on the following data.*1268

*So, we have four experiments; let's see...*1274

*Experiment: BrO*_{3}^{-}, Br^{-}, H^{+}, and initial rate...1284

*All right, Experiment 1; 2; 3; 4; 0.10, 0.10, 0.10; initial rate is 8x10*^{-4}.1296

*0.20, 0.10, 0.10; notice, I have only changed one thing in this case--the concentration of the bromate: bromate, bromide, hydrogen ion.*1314

*We get 1.6x10*^{-3}--that is interesting; OK.1328

*0.20, 0.20, 0.10; this time, from here to here, we have changed the bromide.*1336

*We end up with 3.2x10*^{-3}.1344

*The fourth one: 0.10, 0.10, 0.20; we have gone back--that is the same; that is the same; that is what is different.*1349

*We end up with 3.2x10*^{-3}.1362

*OK, so in order to find m, which goes with the bromate ion, I'm going to use Experiments 1 and 2, because bromate changes from Experiment 1 to 2.*1369

*Let's go ahead and do that.*1384

*I'm just going to go ahead and write it out without writing absolutely everything.*1387

*So, rate 1, divided by rate 2, equals 8x10*^{-4}, divided by 1.6x10^{-3}, equals...well, the rate is equal to that whole expression--is equal to K times (and now, I'm going to stop using brackets; I'm just going to use parentheses; but I do mean concentrations) concentration 0.1 to the m; 0.1 to the n; 0.1 to the s, over K times 0.2 to the m, 0.1 to the n, 0.1 to the s.1390

*We have a whole bunch of cancellations; the only thing that doesn't cancel is this.*1444

*So, this number is equal to 1/2, equals 1/2 to the m power; that implies that m equals 1.*1450

*Therefore, the bromate power is 1; we are done with that.*1465

*Now, let's compare rate 2 and rate 3; in other words, Experiment 2 and Experiment 3.*1473

*Rate 2, divided by rate 3, equals 1.6x10*^{-3}, over 3.2x10^{-3}, equals K times .2 to the m, .1 to the n, .1 to the s, over K times .2 to the m, .2 to the n, .1 to the s.1480

*Cancel, cancel, cancel, cancel, cancel, cancel; we end up with 1/2 equals 1/2, this time, to the n power, which implies that n equals 1.*1517

*Well, n equals 1; that is the bromide ion concentration.*1530

*So, bromide is also order 1.*1533

*OK, now we will do rate 1 compared to rate 4.*1537

*Rate 1 is 8x10*^{-7} (is that right?); no, it's 8x10^{-4}, not -7.1545

*8x10*^{-4}; I was going to say: that is a little bit too much.1559

*8x10*^{-4}, divided by 3.2x10^{-3}, equals K times .10 to the m, .10 to the n, .10 to the s, over K times .10 to the m, .10 to the n (not s yet--.10 to the n--wow, symbols everywhere), .2 to the s.1565

*Cancel, cancel, cancel, cancel, cancel, cancel; we end up with 1/4 equals 1/2 to the s power.*1602

*Well, 1/2 raised to what power gives me 1/4?--that implies that s equals 2, so the hydrogen ion concentration has an order of 2.*1613

*There we have it: we have our rate, which is equal to (I'm going to go ahead and use one of them as a symbol) -ΔBrO*_{3}^{-}, over Δt, just a symbol, is equal to some constant, K.1625

*BrO*_{3}^{-} to the first power, Br^{-} to the first power, H^{+} to the second power: this is our differential rate law.1648

*You can just...that; this is just a symbol that says "the rate is."*1659

*The rate of depletion of bromate is equal to some constant, times this, this, this, squared.*1666

*That is what is going on.*1673

*Now, I take any one of the experiments; I put the values of the experiments in--that one, that one, that one--and I have the rate, because I calculated the rate--it's part of the data--and I solve for K.*1674

*So, I'm going to go ahead and take...oh, let's just take Experiment 1; it's not a problem.*1686

*OK, so I get 8.0x10*^{-4} equals K times 0.10, times (that is the bromate concentration) the bromide concentration--is 0.10, and this other one is 0.10 squared.1692

*When I solve for K, I get 8; in this particular case, it is liters cubed, over moles cubed, second.*1711

*And again, the unit doesn't really tell you all that much; it just tells you what is going on up here.*1722

*But, the real thing--this number is what is important.*1727

*The order of bromate is 1; the order of bromide is 1; the order of hydrogen ion is 2.*1731

*The overall order is 1 plus 1 plus 2; the overall order is 4.*1739

*The rate constant is 8.*1743

*OK, that is the method of initial rates: you write down "equals constant K, times the particular species, raised to..." and so on; it's usually not going to be more than 2 or 3; I think 3 is about the most that you are really going to get, as far as reactants are concerned.*1749

*Then, you take the data that has been collected: Experiment 1 has initial concentrations for each of the reactants, and a measured rate.*1768

*You compare the rate of one with the rate of the other by putting in values based on this equation, just like we did to derive those numbers.*1778

*Once you have the orders, you use them, plus one of the experimental values, concentration, concentration rate, to find K.*1787

*And now, you have your final differential rate law.*1798

*So, in this particular case, we get that the rate is equal to 8 times that, that, that: the rate of the reaction.*1802

*Now, I can plug in any concentration I want, randomly: OK, .26, 19, whatever; and based on this that I derived from experiment, I can tell you what the rate of the reaction is going to be--how fast it is going to be.*1818

*OK, so that was the method of initial rates; we did a couple of examples; next lesson, we are going to talk about the integrated rate law.*1836

*This was the differential rate law.*1843

*Thank you for joining us here at Educator.com; we will see you next time; goodbye.*1845

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

*In today's lesson, we are going to be talking about the integrated rate law and reaction half-life.*0004

*Last time, we discussed the differential rate law that says that the rate of a reaction is proportional to the concentration of reactant, to some power.*0009

*We used some concentration-time data to actually work out what that rate law was.*0021

*Now, what we are going to do is: we are going to use calculus; we are actually not going to do the derivations, but using the techniques of calculus, we can take that differential rate law; we can convert it to an integrated rate law; and then, instead of the differential rate law (which expresses rate as a function of concentration of reactant), we are going to express concentration as a function of time.*0027

*Let's jump in and see what we can do!*0049

*OK, so, as we said, the differential rate law expresses rate as a function of reactant concentration.*0053

*So, we had something like this: the rate equals (and there is this interesting symbol)--let's just use di-nitrogen pentoxide, N*_{2}O_{5}/Δt.0059

*And again, this is just a symbol expressing rate; and this negative sign is because, in chemistry, they prefer to have rates being positive.*0073

*Because a reactant is diminishing, the final minus the initial concentration is going to give you a negative value here, so a negative and a negative gives you a positive.*0080

*I wouldn't worry too much about this; what is important here is this thing--this K, times the N*_{2}O_{5}; and in this particular case, it would be some value of n, and we actually calculated it before in a previous lesson: n=1.0088

*So, this is a first-order reaction.*0105

*The decomposition of di-nitrogen pentoxide is a first-order reaction.*0107

*Well, the integrated rate law (this is the differential rate law) expresses concentration as a function of time, which is very, very convenient.*0112

*Given a certain amount of time--10 seconds, 50 seconds, 2 minutes--what is the concentration at that moment?--very, very convenient: as a function of time.*0139

*Now, the differential in differential rate law (or what we call just the rate law--when you hear "the rate law," they mean the differential rate law--they mean this thing)--the "differential" part comes from this symbol, this Δ; in calculus, they become differentials.*0149

*They become something that looks like this: d...let's say dA/dt.*0165

*When you fiddle around with this expression, and then you perform an operation called integration on it (the symbol for which is something like that), that is why it is called the integrated rate law; that is where these names come from.*0170

*If you go on in chemistry, you will understand--you will actually do the derivations; it is actually quite simple, but we will skip it for our purposes.*0181

*OK, so given a differential rate law, we can find an integrated rate law.*0187

*Or, if we have an integrated rate law, we can actually go backwards and find a differential rate law.*0193

*So, since we are going to deal with first-order reaction, let's write a reaction: aA decomposes into products; and the products themselves are not altogether important--what is important is that we have a single reactant on the left side of the arrow.*0198

*So, the rate is equal to (I'm going to skip the symbol; I'm just going to say) K, times the concentration of A, to the first power.*0216

*When we fiddle with this thing (actually, you know what, let me go ahead and write...it is actually pretty important: -ΔA/Δt; we might as well be consistent): K times the concentration of A to the first power; when we integrate this expression, we end up with the following.*0229

*We end up with: the logarithm of the concentration of A equals -K, times t, plus the logarithm of the initial concentration.*0257

*This (oops, try to get red ink here...red ink) is our integrated rate law.*0272

*For a first-order reaction, this is our differential rate law, and this is our integrated rate law, OK?*0281

*This says that the logarithm of the concentration at any given time, t, is equal to minus the rate constant, K (which is the same K here), times the time, plus the logarithm of the initial concentration.*0287

*Now, take a look at this equation here: this here (let's just call it y)...let's call this -K m; this t is your x, plus b.*0303

*So, this equation--the integrated rate law actually takes the form y=mx+b.*0315

*As it turns out, if you have some time and concentration data (which we will do in a minute), if you have a certain time and you are measuring concentrations at different times--if you actually plot, not the time and the concentration data, but if you plot the logarithm of the concentration, versus the time (the time on the x-axis, and the logarithm of the concentration); if you end up with a straight line, that actually tells you that this is a first-order reaction.*0322

*Just given some straight kinetic data (time, concentration), you plot the time on the x-axis and the logarithm of the concentration on the y-axis; if you get a straight line, that tells you that you have a first-order reaction.*0350

*That tells you you could write this and this; it automatically gives you the integrated and the differential rate laws.*0363

*That is what is so great about this particular one.*0370

*It is very practical, because again: at any given time, t, you can measure the concentration of your particular reactant.*0372

*OK, so let's go ahead and do an example, and I think it will make more sense, as always.*0381

*Let's go back to...no, let's keep it red; not a problem.*0388

*Our example will be: The decomposition of di-nitrogen monoxide...di-nitrogen pentoxide; I'm sorry...at a constant temperature (at a constant t) gave the following kinetic data.*0394

*OK, so we have--let's see: t is going to be in seconds, and then, of course, our concentration...*0426

*Now, if you don't mind, I actually tire of doing the whole brackets for concentration; I just like put parentheses for concentration; I'm going to sort of go back and forth between them--I hope you understand what it means.*0433

*When we are talking about kinetics, a parentheses around a species (like di-nitrogen pentoxide) is actually going to mean concentration.*0445

*Normally, it is true: a square bracket is concentration; I shouldn't be so reckless, but I am.*0453

*OK, so the concentration of N*_{2}O_{5} (and this is in moles per liter, as always--concentration is in moles per liter when you see it like that)...0460

*OK, so we are going to do 0 seconds, 50 seconds, 100 seconds, 200 seconds, 300 seconds, and 400.*0470

*The data we have is: we start with an initial concentration of 0.100--four decimal places; 0.0707; 0.0500; 0.0250; 0.0125; and 0.00625.*0480

*This is the kinetic data that we collected.*0505

*At different time intervals, we measured the concentration of reactant left in the flask, and these are the concentrations that we got.*0509

*Now, we want to know what the order of this reaction is, and we want to know what the rate constant is.*0515

*Let's go ahead and start.*0521

*Now, what we are going to do in order to find the order: we said that we are going to plot the logarithm of the concentration, versus the time (y versus x).*0524

*Whenever you see something versus something, it is always y versus x.*0534

*So, this is going to be the y-axis; this is going to be the x-axis...actually, the logarithm of this is going to be the y-axis.*0537

*Let's make a new table and draw a little line here; and this time, I'll do the table in blue.*0543

*It is going to be the same thing; the time is going to be the same thing--it's going to be 0, 50, 100, 200, 300, and 400.*0550

*Now, we are going to take logarithms of these numbers; so, when we take the logarithms of those numbers, we end up with the following data.*0560

*This is going to be ln of the N*_{2}O_{5} concentration in moles per liter.0569

*We end up with -2.3, -2.6, -3, -3.7, -4.4, and -5.*0575

*Now, we are going to plot this.*0589

*We have our y-axis and our x-axis: this is our time; this is going to be our logarithm of the N*_{2}O_{5} concentration.0592

*We will just put some numbers up here: so, -6, -5, -4, -3, -2; so, -6, -4, -2, and we will go with 100, 200, 300, 400; this is 100 seconds (oops, all of these lines are showing up again--this is always happening down at the bottom of the page--it's kind of interesting).*0604

*All right, so 100, 200, 300, and 400 seconds: when we plot this data, we actually end up getting something which is (let's go down to about right there; I guess that is pretty good), believe it or not--we actually do end up with a straight line, with all of these different points at various points.*0633

*Now, I should let you know: this is kinetic data; kinetic data is not always going to line up in an exactly straight line--it's not going to be that the line is going to go through every single point exactly.*0661

*But, it is going to be pretty good; you are going to get a linear correlation.*0672

*So, you should be able to draw a line; in other words, you are not going to get points all over the place.*0676

*You are going to be able to draw a line through many of the points.*0681

*Don't think that is has to be exact; real kinetic data, like real science, is not...doesn't fall into nice, perfect, clean square boxes.*0685

*There is a little bit of deviation; don't let that confuse you.*0697

*The idea is: when you plot this, you get something that is a straight line--pretty much a straight line.*0700

*OK, so because it is a straight line--because this kinetic data gives you a straight line--that implies that you have a first-order reaction.*0706

*That means that your rate law is K times the N*_{2}O_{5} to the first power, and that means your integrated rate law is (let's write it down here): the logarithm of the concentration of N_{2}O_{5} equals -K, times t, plus the logarithm of the initial concentration of N_{2}O_{5}.0722

*That is what is happening here: the initial concentration was 0.100; so we have identified that it is first order.*0757

*We can go ahead and write a differential rate law for it; we can write an integrated rate law for it; everything is good.*0769

*Now, the second part is: how do we find the rate constant?*0776

*OK, well, we have the straight line; the rate constant is (let me go ahead and move to the next page) the slope, the negative slope, of that line.*0779

*So, the slope equals -K.*0791

*So now, all we have to do: we take 2 points on that line--it doesn't matter which 2 points; now, when you are taking points on a line (on a straight line for a graph), you can only use points on that line.*0799

*We just said that real kinetic data doesn't always give you an exactly straight line; so you are going to have a best fit line, and you are going to do a least squares fit on that line.*0813

*Well, when you pick 2 points randomly on there, you don't pick 2 data points, unless those data points actually happen to be on the line.*0825

*If they are not, you pick points on that line.*0833

*There you go: you have your 2 points; you do Δy/Δx; set it equal to -K; and you solve for K.*0836

*When we take a couple of points on the graph that we just did in the previous slide, we end up with the following.*0847

*We end up with: -6.93x10*^{-3}, and the unit is per second, equals -K.0853

*Slope equals -K; negative equals negative; so K, the rate constant, equals 6.93x10*^{-3} per second.0867

*That is our rate constant.*0879

*What we did: given kinetic data (time and concentration), we plotted time on the x-axis, and the logarithm of the concentration on the y-axis.*0881

*If you get a straight line, it's a first-order reaction; if you don't get a straight line, it is not a first-order reaction; so it works both ways.*0892

*If you do get a straight line, the slope of that straight line is negative of the reaction constant, K.*0902

*OK, so now, let's move on to our second example, or actually a continuation of this first example.*0914

*Using the data and results from the previous example, find the concentration of N*_{2}O_{5} at t=150 seconds.0926

*OK, well, so now we have K; we have the integrated rate law, which is: the logarithm of the concentration is equal to -K, times t, plus the logarithm of the initial concentration.*0958

*What we are looking for is the concentration of N*_{2}O_{5} at 150 seconds.0979

*Well, if we look at our graph, or look at our data table, we have t at 100; we have t at 200.*0984

*We can't just split the difference to find the concentration in between there at 150.*0991

*You will see that in just a minute; but let's just solve our equation.*0996

*We end up with logarithm of the concentration of A (because we are looking for the concentration of A--this is the dependent variable here--this whole thing), equals -K (K was 6.93x10*^{-3}--let me make this a little more clear), times t (well, at 150 seconds), plus the natural logarithm of the initial concentration (the initial concentration was 0.1000).1000

*OK, so let's go ahead and do this: equals...well, when you do the math, you end up with: ln of the concentration of A equals -3.3425, and in a logarithmic equation, you solve by exponentiating both sides.*1036

*So, let's go ahead, and I'll just write "exponentiate," which means e to the ln of A equals e to the -3.3425, and you end up (well, the e and the ln go away) with a concentration of A=0.0353 moles per liter.*1059

*There you go: we used the kinetic data to find the integrated rate law, to find the reaction constant...the rate constant; and then we used that equation to find the value of a concentration at any given time (in this case, 150 seconds).*1093

*Now, let's take a look at some values.*1111

*The concentration of N*_{2}O_{5} at t=100 was 0.500; that is from the data table.1119

*The concentration of N*_{2}O_{5} at t=200, also from the data table, equals 0.250.1133

*OK, now, the concentration of N*_{2}O_{5} that we found, which was at t=150--it doesn't equal...it's between 100 and 200; it isn't this plus this, divided by 2; it isn't .375; it isn't right down the middle.1146

*That is the whole idea--that this relationship is logarithmic, so we found this to be equal to 0.0353, not 0.0375.*1165

*This is not right; hold on a second--I think we have some values wrong here; this is .1; this is .05; 0.0500; 0.0250; .0375.*1185

*It is not half of that; you actually have to use the integrated rate law to find the rate--it's this--the concentration--not that.*1204

*Be very, very careful; don't just think you can look at the data and say, "OK, well, half the time between 100 and 200, at 150, is half the concentration between .5 and .25, which is .0375"; that is not the case.*1214

*You have to use the integrated rate law, which gives you .0353; the concentration is actually less--more diminished than you would have expected.*1226

*OK, so let's go back here; now that we have taken care of the integrated rate law, we also want to talk about something called the half-life.*1236

*Let's give a definition.*1247

*The half-life of a reaction is the time required for the concentration of a reactant (in our case, the reactant) to reach 1/2 its original value.*1256

*In other words, when does A equal A*_{0}/2?1296

*When does the concentration of A equal A*_{0}/2?--when is half the reactant used up?1304

*That is sort of an important point; it is a line of demarcation; it gives us a standard--a reference, if you will--a reference point: half-life; it's pretty common.*1309

*We also speak about half-life in radioactive decay: how much time does it take for half of a particular nuclide, radioactive isotope, to completely disappear by half of its original amount?*1318

*So, in kinetics, we talk about the same thing: the half-life of a reaction is the time required for the concentration of a reactant to reach 1/2 of its original value.*1331

*In other words, concentration of A equals 1/2 of A*_{0}, the initial concentration.1339

*OK, well, the symbol is t*_{1/2}...half-life; the time.1345

*Now, let's go ahead and derive our half-life equation.*1359

*So, we have our equation, our integrated rate law: the logarithm of A equal -Kt, plus the logarithm of A*_{0}.1363

*Well, let's see: how shall we do this?*1378

*OK, let me go ahead and bring this over here; so we have ln of A, minus ln of A*_{0}, equals -Kt.1381

*Now, let me rewrite this logarithm thing as (yes, that is fine) logarithm of A, divided by A*_{0}, right?1397

*It is the property of a logarithm; the logarithm of something, minus the logarithm of something else, is the logarithm of that first something divided by the second something.*1408

*Equals -Kt; and now, we will use the fact that A=A*_{0}/2, right?1416

*We will go ahead and put this A*_{0}/2 in here.1424

*We get the logarithm of A*_{0}/2, over A_{0}, equals -Kt.1428

*The A*_{0}s cancel; we end up with ln of 1/2 = -Kt.1439

*We get that t is equal to ln of 1/2, over K.*1447

*So, the half-life--when half of the concentration has completely been depleted, the time that it takes for that to happen--is equal to the logarithm of 1/2, over the rate constant.*1454

*Take a really, really close look.*1469

*We can also write this another way: we can also write it as logarithm of 2 (I'm sorry, this is -K, right?--because the minus sign...) over K, and that just has to do with how we derive the equation.*1471

*Either one of these is fine.*1488

*This one or this one: you can use either one; it is just a question of: logarithm of 1/2 is a negative number, negative over negative is positive, logarithm of 2 is positive, so either one is absolutely fine; it's just that, the way I derived it, we came up with this one.*1490

*Now, notice what is important about this: the time it takes for a first-order reaction to be halfway complete, to be 50% depleted, does not depend on the initial concentration.*1507

*It doesn't depend on concentration at all; it just depends on a number, the logarithm of 1/2, and it depends on the rate constant.*1518

*So, it is a function of the rate constant; for a first-order reaction, the half-life does not depend on concentration; that is very, very important.*1525

*But, we will summarize this towards the end.*1534

*OK, so now that we have that, let's go ahead and do another example.*1537

*OK, a certain first-order reaction has a t*_{1/2} equal to 20 minutes--has a half-life of 20 minutes.1550

*That means it takes 20 minutes for a certain reaction to be halfway complete.*1568

*A: We would like you to calculate the rate constant.*1571

*B: We would like you to find out how much time is required for the reaction to be 75% complete.*1583

*OK, so we have a certain first-order reaction; it has a half-life of 20 minutes.*1610

*We want you to calculate the rate constant, and we want you to tell us how much time is required for the reaction to become 75% complete.*1614

*All right, well, part A: we have: t*_{1/2}=...let's just use the ln 2/K.1623

*We are solving for K here; we know the t*_{1/2}, right?1642

*So, K is equal to ln 2 over the t*_{1/2}.1645

*Well, ln 2 is a certain number, and t*_{1/2}; this is just ln 2 over 20 minutes; and what we end up with is 0.0346 per minute; that is the rate constant--nice and simple.1654

*So, you can find a rate constant from the half-life, or you can find a rate constant from the kinetic data--the slope.*1672

*Either one is fine; it just depends on what you have at your disposal.*1679

*Now, part B: well, 75% complete of the reaction--what does "75% complete" mean?*1684

*That means that your final concentration is only 25% of your original.*1693

*75% means that your final concentration of A is equal to 0.25 A*_{0} (that means you only have 25% of the initial concentration left; 75% of it is gone).1701

*We just plug that in to our integrated rate law.*1714

*So, let's write our integrated rate law again; we write it over and over again, until it becomes second nature.*1720

*-Kt + ln of A*_{0} (and you notice, I am getting, actually, very, very lazy; I am not even writing the parentheses anymore).1726

*We are talking about kinetics; we are talking about rate laws; we are talking about concentrations in moles per liter; I don't want to hammer certain things that are self-evident.*1735

*OK, so now, we said that A=.25 A*_{0}, so we write ln of 0.25 A_{0} = -0.0346t + ln, and our initial concentration was...actually, it doesn't even matter; it is just A_{0}.1746

*We know that it is 75% complete, so we don't actually have to have a number; we just know that this is .25% of that, because it is 75% done.*1776

*OK, so now, we move this over; we put them together; it looks like this.*1784

*The logarithm of 0.25 A*_{0}, over A_{0}, equals -0.0346t.1792

*These A*_{0}s cancel; we end up with ln of 0.25, divided by -0.0346, equals t.1803

*When we solve this on our calculator, we end up with 40 minutes; very nice.*1815

*Now, this another way to actually do this, based on the percentage that they asked.*1824

*They said 75% complete; 75% complete--that means 50% is gone (half-life), and then to 75% means that another half-life (half of the half) is gone.*1829

*So, you can think of it as the first 20 minutes, and then another 20 minutes, because again, the half-life of a first-order reaction is constant.*1842

*It doesn't depend on concentration; it only depends on the K.*1851

*In this case, they say it's 20 minutes; the half-life of any amount is always going to be 20 minutes for this reaction.*1856

*It means 20 minutes must elapse before half of what is left over goes away.*1864

*Then, half of that left over goes away--it takes another 20 minutes; half of that--another 20 minutes.*1869

*But, I think the best thing to do is just to use the equations, instead of sort of reasoning it out that way.*1874

*When you get a little bit more comfortable with it, that is fine; you can look at 75%, and you can say, "Oh, that is 2 half-lives; 2 times 20 is 40."*1879

*Just go ahead and plug it in, and everything will work out fine.*1887

*OK, so we have taken care of the differential rate law for a first-order.*1890

*We have talked about integrated rate law for a first-order.*1894

*We have talked about the half-life of a first-order reaction.*1898

*All of this, basically, comes from standard kinetic data, where you, at different points of time, measure the concentration of the reactant; you plot it, and then the plot will actually tell you whether you are looking at a first-order or a different kind of order reaction.*1902

*So, next lesson, what we are going to do is talk about second-order and zero-order reactions.*1919

*We will just continue with the same theme: new rate laws, new integrated rate laws, new formulas for half-life; and then, we will summarize everything.*1924

*Thank you for joining us here for AP Chemistry, and thank you for joining us at Educator.com.*1931

*We'll see you next time; goodbye.*1936

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

*Today, we are going to continue our discussion of integrated rate laws.*0004

*Last lesson, we introduced the integrated rate law for a first-order reaction, and also the half-life formula for a first-order reaction.*0007

*Today, we are going to talk about second-order reactions and zero-order rate laws.*0015

*So, let's just jump right on in.*0019

*OK, so again, we are talking about (in order to simplify matters for ourselves, just so we get a good sense of the kinetics) a single reactant that decomposes to products.*0024

*That is what we have been doing this entire time.*0035

*Our fundamental reaction that we have been working with is the following products.*0037

*OK, now, the rate law for a second-order is the following--this is the differential rate law; and again, when we say "rate law," they mean differential rate law.*0043

*When they mean integrated rate law, they will specifically say "integrated rate law."*0055

*So, the rate is equal to -Delta;A/Δt; this rate symbol, which is the differential part in differential rate law, is equal to some constant, K, times the concentration of A to the second power.*0060

*That is the differential rate law for a second-order reaction.*0079

*Now, when we integrate this particular function, we end up with the following.*0084

*Let me write; this is the differential rate law; now, the integrated, which relates concentration as a function of time.*0091

*It is going to be 1 over the concentration of A, is equal to Kt, plus 1 over the initial concentration.*0112

*For a second-order reaction, your integrated rate law says 1 over the concentration of A is equal to the rate constant, times the time, plus 1 over the initial concentration of A.*0124

*So again, in this particular case, we can set it up as y=mx+b, but this time, the y is 1 over the concentration--the data that we take.*0138

*That is what is important; and the slope of the line that we get is the rate constant.*0151

*Notice, this time it is a positive slope--positive K instead of negative K.*0158

*What we are going to do is: we are presented with some standard raw kinetic data; we are going to find...we have the time; we have concentration; we are going to also change those concentrations to 1/concentration, and we are also going to do logarithm of concentration, because now, we have to check; now, we have 2 different types of equation, 2 different types of reaction orders.*0161

*Now, we have to check to see whether it is first-order or second-order.*0185

*We have to have 2 columns: one with the logarithm of A, one with the reciprocal of the concentration of A.*0188

*And again, you will see that in just a minute, when we do the example.*0195

*So, now, this is the integrated rate law; let's go ahead and work out the half-life for a second-order reaction.*0199

*And again, remember: half-life means that the concentration of A is equal to the initial concentration, over 2; it is when half of it is gone.*0209

*We will go ahead and write; that implies using this equation with those values: 1 over A*_{0} over 2 equals Kt, plus 1 over A_{0}; this ends up being 2 over A_{0}, minus 1 over A_{0}.0219

*I flip this and move this over; it equals Kt; 2 over A*_{0} minus 1 over A_{0} is 1 over A_{0}, equals Kt.0247

*Now, I will divide through by K, and I get that t of 1/2 is equal to 1/K, times the concentration (the initial concentration).*0257

*This is the formula for the half-life of a second-order reaction.*0269

*Notice, in this particular (you know what, this is not too clear; let me write this out; I need to erase this, and let me write it bigger over to the side; I think it will be a little more clear; let me get rid of this thing; A*_{0}...so, we get t is equal to 1/K, times the initial concentration)...0274

*Now notice: in this case, the half-life of a second-order reaction not only depends on K, but it also depends on the initial concentration.*0302

*That was not the case with the first-order reaction; a first-order reaction does not depend on the concentration--it just depends on K; it's constant.*0310

*This is not constant; so the half-life--if you start with a certain amount, half of it is going to take a certain amount of time to get rid of.*0317

*Well, now that you have gotten through half of it, now that is a new initial amount.*0325

*In order to release half of that, that half-life is going to change; it is actually going to end up being longer.*0328

*So now, it depends on two things: the rate constant and the concentration.*0334

*There you go: we have that equation, which is the integrated rate law; we have this equation, which is the half-life; and, of course, we have this equation, which is the differential rate law.*0339

*These are the things that matter in kinetics: the differential rate law, the integrated rate law, and the half-life formula.*0354

*Pretty much everything else can be worked out from this information, and all of this, of course, comes from raw kinetic data: time, concentration.*0360

*OK, now, let's go ahead and do an example, because that is the best way that these things work.*0368

*Example: we have butadiene; butadiene dimerizes (and dimerization means that two molecules of something stick together); butadiene dimerization was studied, and the following kinetic data were obtained.*0377

*OK, so let's go ahead and write out the formula.*0417

*2 C*_{4}H_{6} turns into C_{8}H_{12}.0420

*OK, so now, we have our time value; I'm going to go ahead and write it as one big plot.*0429

*Our time value; we have our concentration of C*_{4}H_{6}, which is our reactant.0437

*Now, what we want to do here is: we want to find the rate constant; we want to find the order of the reaction; and we want to find the half-life (I'm sorry about that; I should have actually told you what it is that we are actually going to be doing here).*0446

*So, the first thing we want to do is: we want to find the order of this particular reaction, based on the data that we are given.*0459

*Then, we want to find out what the rate constant is; and then, we want to find out what the half-life of the reaction is, based on this data.*0467

*So, t (let me go ahead and write this down) is 0, 1000 seconds, 1800, 2800, 3600, 4400, 5200, and 6200.*0475

*And what we have is 0.0100, 0.0625, 0.0476...no, that is not right; that is 0.0100; so this is going to be 0.00625--the butadiene is diminishing--0.00476, 0.00370, 0.00313, 0.00270, and we are almost done--no worries--0.00241; and 0.00208.*0498

*OK, given this raw data, find the order of the reaction; find the rate constant; and find the half-life of this reaction, based on the fact that we started with .0100 moles per liter of this butadiene.*0552

*All right, so we need to check: is it first-order or second-order?*0569

*We need to plot both the logarithm of C*_{4}H_{6} versus time, and the reciprocal of C_{4}H_{6} versus time, to see which one of these gives us a straight line.0574

*If this gives us a straight line--the logarithm--it is first-order; if this one gives us a straight line, it's second-order.*0591

*That is how we do it.*0597

*OK, so here is what the data looks like.*0599

*Let's see; let's do logarithm first.*0602

*We get -4.605, -5.075, -5.348, -5.599, -5.767, -5.915, -6.028, and our last one is -6.175; reciprocals are a lot easier.*0604

*What we have is 100; 160; 210; 270; 320; 370; 415; and 481.*0639

*OK, so now we are going to plot (I'll do it in blue): we are going to plot this versus time, and we are going to plot this versus time, to see which one gives us a straight line.*0654

*Let's go ahead and take a look at what these plots look like.*0669

*I'm going to go ahead and put them next to each other; so, I'm not going to give you too much detail on these graphs--you can actually go ahead and plot them yourself on a piece of graph paper, or using a software like Excel or something like that.*0671

*What you are going to end up with, as it turns out--let's see here, this is going to be 200...yes; so let's do the logarithm of the butadiene versus time here; and let's do the 1 over the reciprocal of butadiene (which I'll just call A) over time here.*0687

*When I do this, the logarithm graph is something like this--definitely not a straight line.*0714

*When I do the reciprocal, I get a straight line; so I will have you confirm this, but this is exactly what happens.*0721

*So, because it is the reciprocal that gives us a straight line, it is a second-order reaction.*0729

*Second-order: so, we can go ahead and write the differential rate law: C*_{4}H_{6}/Δt equals the constant times C_{4}H_{6} to the second power.0737

*We know it now; it is a second power; we derived it from the graph now, so we have an order of 2.*0754

*Now, the next thing we want to know is: what is the rate constant?*0763

*OK, well, the rate constant--the best thing to do is to go ahead and (let's see, how shall we do the rate constant)...let's just go ahead and use the integrated rate law.*0767

*We know that the second order is: 1, over the concentration of A, equals -K (actually, you know what?--what am I saying?--let me just...write it out, so we have it)...equals **positive* K times t, plus 1 over A_{0}.0780

*This is y=mx+b.*0808

*We just graphed the reciprocal over that; we got a straight line, so now, what we want to do is: we want to pick two points on that straight line, take the Δy over the Δx of those two points...*0811

*And again, the points need to be on the line; if the data points that we have in the original data, when we calculated it, are on the line, that is fine--you can use those data points.*0825

*But, if they are not, make sure that you are using points that are on the line.*0835

*When we calculate Δy/Δx from this graph, based on the kinetic data that we derived, we end up with the following.*0838

*Let's see: let's use a couple of points, actually; so let's use Δy over Δx; it equals...as it turns out, a couple of the data points do fall on the line.*0848

*I pick the first and the last; so it's going to be 481 minus 100, over 6200 minus 0, which gives me 0.06145, and the unit is going to be liters per mole-second.*0862

*I know the units for the rate constants don't really make sense; they tend to change, depending on what order it is; it is actually the number that matters.*0882

*This is our rate constant; and now, we want to find t*_{1/2}.0889

*OK, well, t*_{1/2} of a second-order reaction is 1/K, times the initial concentration.0897

*It is equal to 1 over 0.06145, and the initial concentration was 0.0100; so, when we do that, we get 1627 seconds.*0909

*There you go: raw data: time and concentration; we calculated the logarithm of concentration; we calculated the reciprocal of the concentration; we plotted ln of concentration versus time for one graph; we plotted reciprocal of concentration versus time for a second graph.*0926

*We saw which one of those graphs gave us a straight line; in this particular case, it was the reciprocal versus time that gave us a straight line: second-order.*0947

*Because it is second order, I go ahead, and I can write the differential rate law, if I need it.*0955

*We need the rate constant; so I just found a couple of points on that straight line; I took the slope, y*_{2}-y_{1} over x_{2}-x_{1}.0961

*I ended up with the rate constant; and then, because I have the half-life formula, and I have the rate constant which I found, and I have the initial concentration (which is the initial concentration from raw data), I was able to calculate the half-life of this reaction--very, very straightforward.*0969

*Nothing altogether too complicated: again, differential rate law, integrated rate law, half-life--those three things are what is important, as far as the kinetics of a reaction are concerned.*0985

*OK, that was second-order reaction.*0996

*Let's talk about a zero-order reaction.*0999

*All right, so a zero-order reaction is exactly what you think it is; it says that the rate (well, let me actually write it in the top here)...*1001

*Zero-order reaction: well, the rate (which is equal to -Δ of some reactant/Δt) equals KA to the 0 power, which is K times 1, which is K.*1011

*In other words, the rate itself is constant; the rate doesn't change.*1032

*The integrated rate law: when we integrate this particular function, you end up with the following.*1039

*You end up with the concentration of A (you know what, I am not even going to write these anymore; they are driving me crazy), equals -Kt, plus A*_{0}.1045

*The concentration at any given time is equal to minus the rate constant, times time, plus the initial concentration.*1060

*Or, it equals the initial concentration, minus rate constant, times time; it's diminishment.*1067

*Again, y=mx+b.*1072

*So now, if you had some raw kinetic data, and you just plotted the concentration versus the time, and you got a straight line, that is a zero-order reaction; there you go.*1077

*Now, half-life: A=A*_{0}/2; so let's put it in here.1088

*A*_{0}/2=-K times t, plus A_{0}.1097

*A*_{0}/2 minus A_{0}=-Kt.1111

*-A*_{0}/2=-Kt.1117

*Therefore, t*_{1/2} equals A_{0} (these lines--they always show up in the most inopportune moments), divided by 2K.1125

*Let's do it in red.*1143

*We have this; we have this; and we have this.*1145

*For a zero-order reaction, the differential rate law says that the rate is equal to K, the rate constant; it is a constant.*1155

*The integrated rate law says that the concentration at any time A is equal to -Kt, plus A*_{0}.1163

*The half-life formula says that the initial concentration, divided by twice the rate constant, will give you the amount of time it takes for half of whatever you started with to disappear.*1169

*Notice, in this case: again, it depends on the initial concentration, and it depends on K.*1181

*Now, you might be wondering what a zero-order reaction is; where would you run across a zero-order reaction?*1187

*When would you have a reaction that doesn't actually depend on the concentration?*1192

*Zero-order reactions show up in places where...usually in things that involve catalysis; and, extended to the biological realm, that means enzymes.*1197

*So, for example, if you have--let's say just 10 enzymes, catalyzing a reaction; well, if all of those 10 enzymes are busy (meaning if they are full, doing what they are supposed to do), it doesn't matter how much more reactant you actually put in there.*1210

*It is called a substrate--the thing that the enzyme grabs onto and fills with; so it doesn't matter--once those 10 enzymes are full, you can't make the reaction go any faster, because it doesn't matter how much concentration--how much more of the reactant--you actually put in; the reaction rate is controlled by how many enzymes are actually doing the work and how fast they are doing it.*1229

*So, under catalysis conditions, the reaction rate actually depends on the catalyst, not so much the concentration of the reactant.*1252

*Now, this isn't always the case; we're just saying that, when you run across a zero-order reaction, that more often than not, it is going to be in a catalysis situation.*1260

*That is just a qualitative bit of information that you should know, if it happens to come up.*1268

*OK, so now, let us take a global view of what it is that we have done with zero-, first-, and second-order reactions.*1273

*Let's summarize what we have, and it will give us a good, nice one-page summary of reaction kinetics and how to deal with them.*1280

*OK, so this is going to be a summary for the kinetics associated with the reaction aA going to products.*1288

*So again, this entire time we have been talking about a single reactant decomposing into products.*1314

*It diminishes, and it forms something.*1320

*How can we figure out the rate?*1322

*All right, so here, let's go ahead and make ourselves a little table; and we will go ahead and put a line here, and we will say first-order, second-order, zero-order reaction.*1324

*OK, the differential rate law for a first-order reaction is: the rate is equal to K, times the concentration of A to the first power.*1345

*The rate is equal to K, times the concentration of A to the second power.*1360

*And, the rate equals just plain old K.*1366

*That is the differential rate law.*1369

*OK, our integrated rate law: we have, for a first-order reaction--we have: ln of A is equal to -Kt, plus (oops, I always do an h for ln; I don't know why) ln of A*_{0}; again, concentrations, kinetics--we are talking about concentrations.1371

*Now, the rate law for a second-order is reciprocal: 1 over A is equal to Kt, plus 1 over A*_{0}.1395

*And, this one is: A is equal to -Kt, plus A*_{0}.1408

*These are the integrated rate laws.*1414

*OK, a plot giving a straight line is going to be ln of A versus t, y versus x.*1416

*So, the plot of ln(A) for a first-order reaction--the plot of ln(A) versus t gives a straight line.*1440

*Here, it is 1/A versus t that gives a straight line.*1445

*Here, it is A versus t that gives a straight line.*1450

*And again, it is just the left-hand side of the integrated rate law: ln(A), 1/A...you check the data.*1456

*Slope: well, the slope is -K, K, -K of the particular line, if you happen to get one.*1464

*In other words, the slope is that; so you can find K; that is how you find K.*1478

*And, half-life: the t*_{1/2} is equal to ln(2) over K; notice, for a first-order reaction, it is kind of interesting; it does not depend on concentration at all; it is constant.1483

*The half-life for a second-order reaction is equal to 1/K times the concentration of A*_{0}.1503

*It depends on the initial concentration.*1512

*And, t*_{1/2} of this one equals the initial concentration over 2K.1515

*So, this is everything that we have done: first-order, second-order, zero-order reaction rates.*1521

*When we are given kinetic data, which involves time and concentration, we have to plot the logarithm of the concentration versus time, the concentration versus time, and 1 over the concentration versus time.*1528

*Whichever one of these graphs gives us a straight line, that is the order of the reaction.*1542

*When we have the order of the reaction (let's say, for example--boom!--we come up with second-order), we know the differential rate law; we can write the integrated rate law; we can find K from the slope; and we have the half-life formula.*1547

*We can tell you what the concentration is at any time; we can tell you what the concentration is going to be at any percentage that we want along the way, instead of the time--all of this information right here.*1561

*This is a summary of standard kinetic operating procedure.*1573

*We'll go ahead and leave it at that; next time, we will talk about temperature dependence of reaction rates, and talk about something called the Arrhenius equation, and activation energy, and things like that.*1581

*But until then, thank you for joining us for AP Chemistry and kinetics, and thank you for joining us at Educator.com.*1592

*Take care; goodbye.*1599

*Hello, and welcome back to Educator.com*0000

*Today, we are going to continue our quick review of the basics of chemistry before we jump into some solid AP work.*0002

*We're going to go over stoichiometry today.*0010

*I'm going to talk about several varieties of stoichiometry problems.*0013

*Stoichiometry, essentially, is about amounts; that is what it is.*0018

*In chemistry, the basic unit of amount is the mole.*0026

*This is something that I know you all have heard before, so really quickly: a mole is 6.02x10*^{23} particles.0031

*"Particles," because in chemistry, we can talk about atoms; we can talk about molecules; we can talk about ions, electrons, protons...whatever.*0042

*I tend to use just the generic term "particles"; and the problem itself will specify what particle we're talking about.*0049

*In one mole (so this is a conversion factor) of something, we have 6.02x10*^{23} particles.0057

*Now, the periodic table is arranged based on the mole; a molar mass, that you see on the periodic table--that is one mole of oxygen atoms, one mole of magnesium atoms, one mole of iron atoms--those weigh those particular numbers.*0065

*So, the mole is the standard unit that we use in chemistry--again, because we can't count individual atoms (they're too tiny), so we have to choose a unit, kind of like a dozen or a century--things like that.*0079

*Stoichiometry basically concerns conversion factors: setting up a bunch of conversion factors based on what you know versus what you want.*0093

*There is a series of steps to that.*0103

*So, I'm going to introduce this notion of how to deal with stoichiometry problems.*0106

*Because the mole is always what we're talking about, the mole is the central hub; you start from the mole and you go to the mole, or you pass through the mole getting to where you want.*0111

*One way or the other, you are going to have to use a mole.*0120

*The reason is: the equations in chemistry (like, for example, 2H*_{2}+O_{2} → 2H_{2}O, the reaction of hydrogen and oxygen gas to produce water)--well, these numbers here--the coefficients--the 2, 1, and the 2--they represent moles.0122

*What this says is, "Two moles of hydrogen gas need to react with one mole of oxygen gas in order to produce two moles of H*_{2}O."0143

*So, all equations--balanced equations--represent the standard that we use to work from. *0150

*It's all in moles.*0158

*Now, in any stoichiometry problem, any species that you're dealing with always has (this is the best way to think about it) moles (you know what, I'm going to actually draw this up here) of X.*0160

*Every species that is mentioned in a stoichiometry problem--something like that; from the mole, we can go to grams of X, or we can go to the number of particles of X.*0179

*Again, X might be a molecule; it might be an atom; it might be an ion; it depends.*0198

*Later on, we might need to go to liters of X.*0202

*But, they all have to pass through the mole.*0209

*Again, every species under discussion has one of these little things associated with it.*0213

*If a stoichiometry problem has two species that are mentioned, and there is some relationship among the two species (which we will get to in a minute), that species is also going to have something like this.*0218

*Let's call it mol y, and again, you can go to grams of y; number of particles of y; or maybe liters of y.*0231

*Now, here is what is interesting: the reason that I say that each species--you should write something like this: when you do something like this, this will give you a pathway to actually find out how to write the problem, stoichiometrically; how to solve it, mathematically.*0250

*This gives you a solution path.*0264

*Any time you are moving from one species to another--like, for example, in a minute we're going to sort of choose a couple of reactants and a couple of products and stoichiometric problems--movement this way or that way is based on the mole ratio, and these mole ratios are precisely the coefficients in the balanced reaction.*0267

*Now, this will make more sense in a minute, when we actually do a problem, and I'll show you the way to actually do this.*0286

*Hopefully, you are already good at this; but if not, this will be a good review and a little bit of practice.*0293

*Having said that, let's just jump into a problem, and I think it will make sense.*0299

*OK, so example number 1: How many Cl atoms are there in 18.50 grams of phosphorus pentachloride?*0304

*OK, let's see: so now, let's identify the species: one of the species is chlorine; one of the species is PCl*_{5}; and so, I'm going to write that.0333

*I'm going to write mol PCl*_{5}, and I'm going to write mol Cl; those are the two species.0344

*Now, what is it that I'm interested in?--I'm actually interested in the number of atoms, so I need the number of Cl atoms.*0354

*OK, what am I given?--I'm given grams of PCl*_{5}.0365

*I'm given grams of PCl*_{5}; I'm supposed to find the number of chlorine atoms.0375

*Here is how you do it: you go...this is the path that you follow, and each arrow represents a conversion factor.*0380

*I need to go from grams of PCl*_{5} to moles of PCl_{5}.0390

*From moles of PCl*_{5}, I need to go to moles of Cl; this is going to be the mole ratio.0394

*From moles of Cl, I can go to the number of chlorine atoms.*0400

*Here is the setup (actually, let me draw this a little bit higher, so that I can make some room): grams PCl*_{5}...0405

*This is why we say each species has this little setup here: when you set it up like this, you can actually go ahead and draw the pathway; so here, I'm going from grams of y, to moles of y, to moles of x, to number of particles of x.*0418

*That's it; I'm following a pathway, and I can go anywhere I want, but I'm always passing through moles; there is a mole ratio.*0434

*OK, you always write what you are given: 18.50 grams of PCl*_{5}.0441

*This is one conversion factor, two conversion factors, three conversion factors: I know I'm going to have three conversion factors--I don't even need to fill in the numbers yet.*0451

*I can just automatically do this.*0461

*Now, the thing that you are coming from is going to be the unit on the bottom; the thing that you are going to is going to be the unit on the top.*0463

*So, I'm going to have grams of PCl*_{5}, moles of PCl_{5}; well, how many grams in a mole of PCl_{5}?0471

*Well, the molar mass is 208.22 in one mole of PCl*_{5}.0481

*Now, my second conversion: what is the relationship between moles of PCl*_{5} and moles of Cl?0487

*One mole of PCl*_{5} releases 5 moles of Cl; that is just based on the molecular formula--it is PCl_{5}.0493

*All right, PCl*_{5}: one of these produces five of these.0501

*So, again, the thing that you are going to is on top--mole of Cl; the thing you are coming from is down at the bottom--mole of PCl*_{5}.0507

*Well, the relationship is: one mole of this produces five moles of that; that is the mole ratio; that is this step.*0524

*Now, I'm looking for number of Cl atoms in one mole of Cl.*0532

*Well, one mole of Cl--one mole of anything contains 6.02x10*^{23} (let me actually write "atoms").0544

*So again, here, the particle we're talking about is atoms.*0558

*When I do this multiplication, I end up with (OK, I can write it down here) 2.67x10*^{23} atoms of Cl; that is my final answer.0561

*Again, it's just based on this relationship; every species has a relationship; from mole, you can go to grams, to particles, to liters.*0578

*Then, among the species, there is a mole ratio--a relationship.*0588

*In this one, it was 1 mole of this produces 5 moles of that.*0592

*That is how you proceed through this network.*0595

*You have something that you are given; you have another thing that you want; you find a path through there.*0598

*If you're given this and you want that, you find a path through there, and it always passes through moles, because the mole is the basic unit of chemistry.*0603

*OK, we'll get to some more stoichiometry in just a minute.*0614

*The next part of the review: I'm going to cover an empirical formula problem.*0618

*Oftentimes, you are given a certain amount of data, and you need to find the empirical formula of a compound.*0623

*The empirical formula is the formula that gives the lowest number ratio of those things.*0630

*So, for example, if I had something like C*_{6}H_{12}O_{6}, which is glucose; and let's say I had another sugar, C_{12}H_{24}O_{12}; well, these are two different sugars.0636

*But notice, the base formula--I can actually divide it by 6 here; I end up with CH*_{2}O, and I end up with CH_{2}O, right?0656

*That is the lowest number ratio of atoms to each other.*0666

*In other words, all sugars have a ratio of one carbon to two hydrogens to one oxygen.*0670

*That is the empirical formula; it's a general umbrella formula that covers all of the molecules of a given compound.*0676

*Each individual compound has its own molecular formula.*0684

*These are molecular formulas; they are specific to the compound.*0687

*But, an empirical formula covers both; that's it.*0690

*So, let's go ahead and do an empirical formula problem.*0695

*Let's see...let's write this as example 2: all right, I have a compound that is 24.5% sodium, and 14.9% silicon, and 60.6% fluorine.*0698

*I want to know what the empirical formula is.*0722

*OK, here is how you handle it: because you are given percentages...percentage is based on 100; so, since they give us the percentages, let's just presume that we have 100 grams of something.*0724

*So, we can convert these percentages: 24.5% of 100 grams is 24.5 grams.*0735

*What I'm going to do is: I'm going to convert these to moles.*0741

*So, sodium: I take 24.5 grams, and I convert to moles: well, one mole of that is about 23 grams, so I get 1.07 moles of sodium.*0744

*Now, I do the same for silicon: silicon--I have 14.9%, which is 14.9 grams if we take a 100-gram sample; we're just doing this so we can make our math easy.*0760

*Well, one mole of silicon is 32.07 grams, so I get 0.46 moles of silicon.*0771

*Then, when we do fluorine, we have 60.6 grams of fluorine; one mole of fluorine is 19 grams, and I end up with 3.2 moles of that.*0782

*Well, now that I have it in moles, I divide by the lowest number of moles; so I divide all of them by 0.46, 0.46, 0.46, and when I do that, I end up with 2.3 here; I end up with 1 here, of course, and here I end up with 7.*0796

*I need whole-number ratios; so 2.3--in order to make this a whole number, I have to multiply by 3, which means I have to multiply everything else by 3.*0822

*This becomes 21; this becomes 3; and this becomes 7.*0832

*So, this is my empirical formula: I have: Na*_{7}Si_{3}F_{21}.0837

*That is my empirical formula.*0851

*It could be Na*_{14}Si_{6}F_{42}, or any other multiple, but the basic relationship, the ratio, of sodium to silicon to fluorine is 7:3:21.0853

*This procedure always allows you to find the empirical formula.*0866

*You change the percentage to gram, gram to mole for all of them; you divide by the lowest number of moles to standardize, so that you have at least one of them with a 1; and then you multiply by an appropriate constant to make sure that all of them give you whole numbers, because we can't leave it as 2.3; you can't have a fraction of an atom.*0872

*That is a standard empirical formula problem.*0891

*OK, so now we're going to take the next step, and I'm going to show you a molecular formula problem.*0894

*Once we find the empirical formula, if we happen to have the molar mass of a particular compound, we can find the actual molecular formula.*0901

*Let's do that one.*0908

*Let's go ahead and make this example 3: Menthol is composed of carbon, hydrogen, and oxygen.*0913

*1.005 grams is combusted to produce 2.829 grams of CO*_{2} and 1.159 grams of H_{2}O.0932

*If menthol is 156 grams per mole, what is its molecular formula?*0969

*Here, we're going to end up having to find the mass of the individual components--the carbon, the hydrogen, the oxygen.*0994

*We're going to use the empirical formula, and then, from there, we're going to use the fact that they give us the molar mass.*1003

*We're going to divide that molar mass, the 156, by the mass of the empirical formula to see how many empirical units actually go into the whole formula.*1010

*That is how we get the molecular formula.*1018

*Let's just jump right on in.*1021

*Notice, this didn't give you the percentages straight on; this is a slightly different problem--slightly more complicated.*1023

*Let's see: 2.89 grams of CO*_{2}; let's go ahead and...let me write this equation down: menthol, when you combust it, it's with O_{2}, and it's going to produce CO_{2} + H_{2}O.1029

*All right, now: 2 CO*_{2}...how many grams of carbon...OK, so now we have 2.829 grams of CO_{2}.1054

*One mole of CO*_{2} weighs 44 grams; now we're going to do...so we know that all of the carbon ends up in CO_{2}; all of the hydrogen in menthol ends up in H_{2}O; it's the oxygen that is split between the two.1077

*We need to find the number of grams of carbon, the number of grams of hydrogen; they give us the total number of grams of menthol that were combusted, so the balance of that is going to be the oxygen.*1092

*Then I can start my problem.*1104

*2.829 grams of CO*_{2}) times...again, this is a mole ratio: 1 mole of CO_{2} produces one mole of carbon, right?1107

*One mole of CO*_{2} produces one mole of carbon, times 12 grams of carbon per mole, and I end up with 0.7715 grams of carbon.1119

*Now, I have 1.159 grams of water times...1 mole of water is 18 grams: 1 mole of H*_{2}O produces 2 moles of hydrogen, right?1135

*Erase that: 2 moles of hydrogen--1 mole of this produces 2 moles of hydrogen.*1159

*Hydrogen is 1 gram per mole, so I get 0.1288 grams of (oops, this isn't carbon; this is) hydrogen.*1165

*Now, if I want to find the mass of oxygen, I take the 1.005 grams of menthol, and I subtract the amount of carbon, 0.7715, plus 0.1288, and I end up with a mass of 0.1047 grams of oxygen.*1182

*Now I have my masses of each, based on a combustion analysis.*1211

*I hope you understand how it is that I came up with that.*1216

*Now, I can start my empirical formula problem.*1218

*I have 0.7715 grams of carbon, times 1 mole over 12 grams; that gives me 0.06429--often, with these empirical formula problems, you want to carry out the decimals as far as you can, to get a good number.*1222

*Then, if I do hydrogen, I have 0.1288 grams, times 1 mole, which is 1 gram, and I get 0.1288--this is in moles.*1242

*Oxygen: I have 0.1047 grams; 1 mole of that is 16 grams of oxygen (again, we'll use the oxygen atom), and this is going to equal 0.0065438 moles.*1258

*Now that we have moles, we need to divide by the smallest, and it looks like the smallest here is this one.*1284

*When we divide by that, when we divide by 0.0065438, and we divide by 0.0065438, we end up with 10 here; we end up with 20 here; and we end up with 1 here.*1289

*So, the empirical formula for menthol is C*_{10}H_{20}O_{1}.1312

*Now, (OK, so here is where we have to...) what we do is: we take the molar mass of the compound, the 156 (let me actually write this out), and we divide it by the molar mass of the empirical formula.*1324

*When we do that, again, the empirical formula is the lowest number ratio, so in this particular case, we had C*_{10}H_{20}O_{1}; that means the carbon to hydrogen to oxygen ration is 10:20:1.1363

*Well, it might be C*_{20}H_{40}O_{2}; we don't know.1380

*So, we divide the mass of the empirical formula into the molar mass of the compound, which we do know.*1385

*This ends up being...they said it was 156 grams per mole, and the molar mass of the empirical happens to also be 156 grams per mole; therefore, the ratio is 1.*1391

*Therefore, menthol is C*_{10}H_{20}O_{1}.1405

*If this number were 3, it would be C*_{30}H_{60}O_{3}; that's it.1411

*This is the actual...not only empirical formula, but it happens to be the molecular formula for menthol.*1416

*This process will always work for molecular formula.*1422

*You run through it; you find the empirical formula; and then you divide the molar mass of the compound by the molar mass of the empirical formula, and that will give you the ratio--the number by which to multiply all of these little subscripts in order to get your final molecular formula.*1425

*Let's do a stoichiometric calculation based on an equation.*1443

*The fermentation of glucose produces ethyl alcohol and carbon dioxide according to the following equation: C*_{6}H_{12}O_{6}, and fermentation (so no oxygen involved), goes to 2C_{2}H_{5}OH + 2 CO_{2}.1450

*The question is, "How many grams of glucose are required to produce 5.2 grams of ethanol, C*_{2}H_{5}OH?1474

*So, how many grams of glucose do I need in order to produce 5.2 grams of C*_{2}H_{5}OH?1502

*OK, so let's take a look: the two species that we are talking about are glucose and C*_{2}H_{5}OH; we need to pass through moles.1508

*They want...we are given grams of C*_{2}H_{5}OH that we want; we want to find the number of grams of glucose.1519

*Here is the process: based on this equation (I'll write it like this), I need grams of glucose.*1526

*Grams of glucose comes from moles of glucose.*1538

*Moles of glucose I can get from moles of ethanol.*1543

*Moles of ethanol comes from grams of ethanol.*1551

*That is the process: I am given grams of this--I am given 5.2 grams of this; I have to go to the moles--from grams to moles, mole to mole (I use mole ratio--that one), and then from mole back to gram.*1559

*This is my path: I go here to here to here to here; that's it--I have 1, 2, 3 arrows; that is three conversion factors.*1574

*Let's go ahead and write this down; we always start with what we're given.*1586

*5.2 grams of EtOH (EtOH is just a shorthand for ethanol); that is 1, that is 2, that is 3; grams of EtOH here; I'm going to moles.*1590

*Moles of EtOH--notice, I'm not going to do the numbers until afterward--I have moles of EtOH here, because, again, they have to cancel, and it's also the thing that I'm coming from.*1605

*I'm going to moles of glucose, and this is moles of glucose; grams of glucose...*1617

*5.2 grams of ethanol; ethanol happens to be 46 grams per mole; the mole ratio--how many moles of ethanol to moles of glucose?*1629

*Moles of ethanol to moles of glucose: 1; that is why you need a balanced equation.*1638

*Glucose is 180 grams per mole; therefore, my final answer is 10.17 grams of glucose.*1645

*10.17 grams of glucose will produce 5.2 grams of ethanol--standard stoichiometry.*1655

*The two species were glucose and CH*_{3}OH; there is my 1 mole of glucose; mole of ethanol; these are the relationships: I need grams of glucose coming from grams of ethanol; this is my path.1664

*If you can't just do it automatically, draw it out; see the solution path, and the path will tell you what the mathematics looks like.*1681

*Let's finish off this lesson with a limiting reactant problem.*1692

*Limiting reactant problems are going to come up very, very often, so basically, it's just...a balanced equation is such that it says, "All of this will react with all of that; nothing is left over."*1697

*But in real life, it doesn't work like that; often, when we run a reaction, one of the reactants is going to be left over; one of them is going to run out.*1709

*The thing that runs out--when it runs out, the reaction stops.*1717

*It's a limiting reactant--it limits the outcome of the reaction.*1720

*It controls how much product you actually get, because when it runs out, again, the reaction stops.*1725

*Let's take a look at this equation here: 2 aluminum hydroxide, plus 3 moles of sulfuric acid, produce 1 mole of aluminum sulfate, plus 6 waters.*1732

*I'm given 0.350 moles of aluminum hydroxide, and I'm given 0.450 moles of sulfuric acid.*1756

*My questions are: a) which is the limiting reactant? part b) how many grams of aluminum sulfate (Al*_{2}(SO_{4})_{3}) can be recovered, theoretically?1769

*That is what we are calculating; we are calculating a theoretical yield, as if everything went right, theoretically.*1805

*And part c) is: How many moles of excess reactant are left over?*1811

*Let's do part a: what is the limiting reactant?*1831

*We need to find out which one of these runs out first.*1834

*You can choose either one to start with: basically, what you're going to do is...here is what it looks like: I'm going to go ahead and choose the aluminum hydroxide.*1837

*0.350 moles of Al(OH)*_{3}; the mole ratio between this and this is 2:3.1847

*Moles of Al(OH)*_{3} (this is the problem with chemistry--there is just so much writing as far as these symbols are concerned--the names!) to moles of H_{2}SO_{4}; I need to do a mole ratio calculation.1863

*It is 2 moles of this to 3 moles of that; this actually tells me that I require 0.525 moles of H*_{2}SO_{4}.1885

*In other words, if I have .350 moles of aluminum hydroxide, this calculation tells me that I actually need .525 moles of H*_{2}SO_{4} to completely react with it.1897

*Well, this is required; the question is: I require .525--do I have .525? I do not (let me do this in red).*1908

*I don't have .525--I only have .50; therefore, H*_{2}SO_{4} is limiting.1921

*Because it's limiting, it controls the rest of the reaction; this is the number that I use to find out how much of this I'm going to get; not this.*1932

*When this runs out, the reaction stops; there is going to be excess of this.*1940

*Therefore, this is what controls how much I have.*1944

*Let's go ahead and do part b, which is, "How many grams of aluminum sulfate can be recovered?"*1947

*All right, well, since this is my limiting reactant, that is the one that I take: 0.450 moles of H*_{2}SO_{4} times...well, what is the mole ratio here?--it's 3:1; 3 moles of H_{2}SO_{4}...this is not going to work...1954

*3 moles of H*_{2}SO_{4}...oh, these crazy lines; OK, let me try this one more time.1987

*Let's start from the top: 1 mole of aluminum sulfate comes from 3 moles of H*_{2}SO_{4}, and then we have 342.7 grams of aluminum sulfate per mole.2002

*There we go; this cancels that; that cancels this; my final answer is going to be in grams, and it's going to be =51.32 grams of Al*_{2}(SO_{4})_{3} (it's tedious writing all of those symbols, isn't it?).2035

*There you go; no, I'm not going to have...I do not want those lines to show up; therefore, I'm going to write this one more time.*2061

*=51.32 grams of Al*_{2}(SO_{4})_{3}; there we go!2073

*Based on .450 moles of H*_{2}SO_{4}, the most aluminum sulfate I can recover is 51.32 grams.2088

*The reason is: because that is what controls the reaction.*2096

*OK, now: this final problems says, "How many moles of excess reactant will be left over?"*2101

*This is our limiting, so this is our excess: we need to find out how much is used, and then subtract it from the amount that we started with, and that will give us the amount left over.*2107

*The amount used is, again, going to be based on the limiting reactant; so, part C: 0.450 moles of H*_{2}SO_{4}...2117

*Well, I have that 3 moles of H*_{2}SO_{4} requires 2 moles of aluminum hydroxide.2132

*I end up with 0.30 moles of aluminum hydroxide used.*2145

*0.350 moles-0.30 moles used (let me actually write these out: moles of Al(OH)*_{3} minus 0.03 mol used up) leaves me with 0.05 moles of aluminum hydroxide to end up with.2156

*You see what we did: once you know the limiting reactant, that controls the rest of the reaction.*2191

*.450 moles of H*_{2}SO_{4} uses up .3 moles of aluminum hydroxide.2196

*I started with .350; I subtract, and I'm left with .05 moles of aluminum hydroxide; that is my excess reactant.*2201

*OK, so in this lesson, we have done some basic stoichiometry coverage.*2212

*We have talked about empirical formula; we have talked about molecular formula; and we finished off with a limiting reactant problem.*2215

*That, and a previous lesson with naming, should provide sort of a good, general foundation for the rest.*2222

*Starting with the next lesson, we're going to jump straight into some good, solid AP chemistry.*2230

*Until then, thank you for joining us here at Educator.com; we'll see you next time; goodbye.*2234

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

*The last couple of lessons, we have been talking about reaction kinetics.*0004

*We have talked about the differential rate law; we have talked about the integrated rate law; and today, we are going to close off the discussion of kinetics with a discussion of activation energy and something called the Arrhenius equation.*0010

*You know from your experience that, when you raise the temperature on a given reaction, that the reaction tends to proceed faster; or, if you drop the temperature, that somehow things seem to go slower.*0022

*Like, for example, the whole idea of refrigeration is based on that fact: you drop the temperature, and the reactions that cause spoilage actually slow down.*0034

*So, not only does the rate depend on concentration, but the rate also depends on temperature.*0043

*The Arrhenius equation actually accounts for that dependence on temperature.*0048

*Let's just jump right on in.*0053

*OK, now again, we have seen that the rate depends on concentration; so, if we have some general equation like aA + bB → products, well, we know that the rate is going to be equal to some rate constant, times the product of A raised to some power, times the product of B raised to some power.*0056

*And, these n and m and n could be integers; they could be numbers; they could be anything.*0083

*Then again, let's reiterate that the m, the n, and the K are determined experimentally; these are not things that we can read off from the equation, the way we will do later on, when we discuss equilibrium.*0090

*Well, again, as it turns out, the rate is not only dependent on concentrations, but it is also dependent on the temperature.*0101

*So, as it turns out, the rate constant itself shows an exponential increase with temperature.*0108

*This constant of proportionality--yes, it is true that the rate is contingent, is dependent on A and B raised to some power; but in some sense, this K is a measure of that dependence.*0132

*As it turns out, for the temperature itself, the rate constant shows an exponential increase with temperature.*0143

*The rate constant depends on the temperature.*0149

*So now, we are going to introduce something called the collision model, and it is exactly what you think it is.*0153

*It accounts for this temperature dependence the same way that it accounts for the concentration dependence.*0160

*Let me just write this collision model; and the collision model basically says (I don't really need to write it down) that, in order for things to react, they basically need to come into close contact with each other.*0166

*And, since you know that molecules and atoms are sort of flying around, especially in the gas phase, at very, very high speeds--they don't just come close to each other; they collide.*0181

*And that is what it is; in order for a reaction to take place, they have to collide.*0191

*Well, so we know that, of course, the kinetic-molecular theory says that, as you raise the temperature, the average velocity of the molecules--of the atoms--of the particles increases.*0196

*Therefore, you have a higher frequency of collisions.*0210

*A higher collisions--more things collide; therefore, there is a greater chance of the reaction actually taking place.*0213

*And, that is all that is really going on.*0221

*Let's take a look at this.*0224

*Well, let's see: shall we write...let's write something for the collision model.*0231

*So, in order for a reaction to proceed, particles of reactants must come into close contact by colliding.*0236

*Now, I know this is not the most precise, rigorous definition; but it gives us a sense of what is going on, and that is really what is important; we want to understand the chemistry--we want to understand where these equations come from, if it actually makes sense--and this is perfectly fine.*0267

*OK, now, what is interesting is that, even though the temperature rises, and more particles are colliding--as it turns out, the frequency of collision goes up; but as it turns out, the rate itself seems to be not as high as it would be otherwise.*0286

*There are some other things going on; so, before we discuss those other things, let's just talk about what this man, Arrhenius, actually proposed.*0302

*Arrhenius proposed the idea of an activation energy.*0312

*The activation energy, if we want to define it: it is the threshold energy that the reactants must overcome in order to go to products--or in order for the reaction to actually move forward.*0318

*And now, we're going to draw a picture, and I think the picture is going to make it a lot more clear.*0353

*So, this is a standard energy diagram, and on the y-axis we have energy; and this thing is called the reaction coordinate.*0358

*The reaction coordinate is just a measure of...sort of time moving forward; it isn't time itself--it is saying how far forward the reaction is proceeding.*0368

*Let's go ahead and put the reactants over here; we'll put the products over here; as it turns out...*0378

*Now, thermodynamically, notice that the products over here are actually lower than the reactants.*0386

*Thermodynamically, this reaction is spontaneous; it will actually move forward.*0392

*But, that doesn't mean that it is going to move forward; there is this energy hill that has to be overcome--it has to get over this hump--in order for it to actually move forward.*0397

*This is what the activation energy is.*0408

*The reactants, the molecules...whatever it is--they have to slam into each other; they have to collide with enough energy to actually overcome this barrier.*0412

*Now, at a given temperature, not all of the velocities of the particles are uniform; it is not like all of the particles have one velocity.*0422

*There is a distribution of velocities, and only those velocities of the particles that are involved, that have enough kinetic energy for that kinetic energy to be transformed and equal to this activation energy (it's E with an a) will actually move forward.*0430

*That is why, when you raise the temperature--yes, it is true, you are actually causing more things to move faster and more things to collide; but not everything is going to move at enough speed to actually overcome this threshold barrier.*0450

*So, let's go ahead and draw a little...so this thing up here--we call it the transition state; we also call it an activated complex--the transition state.*0466

*So if we were to take something like...let's take a particular reaction; let's take 2 BrNO → Br*_{2} + 2 NO; so here, basically, two molecules of BrNO need to collide, and when they do collide, they end up releasing a bromine molecule and 2 nitrogen monoxide molecules.0478

*Well, in order for that to happen--it's true that they need to slam into each other with enough energy; if they actually overcome it--if they actually have enough energy to get over that activation energy and to form a transition state, well, you might think that it actually looks a little bit...something like this.*0514

*In order for this to happen, a bromine and a nitrogen bond has to break, and a bromine-bromine bond has to form; that is what is happening here.*0532

*So, we could think of it as BrNO and then another BrNO, and these dots are used for bonds that are breaking and bonds that are forming.*0544

*This thing would be the transition state; they would slam together, and as this bond is breaking (the BrNO bond), the Br-Br bond is forming.*0556

*This was sort of what it might look like; now, we don't know exactly what it would look like, but that is what we postulate.*0567

*We call this thing the transition state; there is a transition from reactants to products.*0574

*That is all that is going on here.*0579

*And, we will often see these energy diagrams in any number of contexts.*0581

*OK, now, let us talk a little bit about this distribution of velocities, so you see what is going on, exactly.*0589

*Now, we said that, at a given temperature, there is a distribution of velocity; some particles are moving very slowly; some particles are moving very, very fast; in general, they sort of average out, and the distribution looks something like this--like any other distribution.*0601

*It will be...OK, so this is the energy axis; and now, let's see...this distribution--let's call this at temperature 1; so let's say, at a certain temperature, there is a distribution of speeds; certain things are slow; certain things are fast.*0617

*Well, as it turns out, if the activation energy is here, that means only those particles have enough energy, have enough speed, to surmount that barrier.*0638

*All of the others do not: which is why, despite the fact that we raised the temperature, and we have a whole bunch of collisions--as it turns out, the sheer number of collisions that actually lead to a reaction is actually very small.*0652

*This is why it is very small: because, among the distribution of speeds (very slow, moderately slow...on average, where most molecules fall, and the very fast), it is only the very fast that have enough energy to overcome that activation barrier--the activation energy.*0665

*That is what this is.*0682

*Well, if we raise the temperature, we raise, on average, the velocities, and the distribution changes; the distribution looks like this now.*0683

*So now, notice that this is temperature 2; and the temperature 2 is higher than temperature 1.*0694

*Well, temperature 2 is higher than temperature 1, which means, on average, you have more things that are fast.*0701

*Well now, we have basically pushed, on average, all of the particles to a faster speed, but they still occupy a distribution: some are still slower than others; most are in the middle; some are fast*0708

*But notice, now you have a lot more (I'll do this in red) particles that have enough energy above the activation energy to actually move forward with the reaction, as opposed to (let's do this other one in black) this right here, the original temperature.*0721

*So, as you raise the temperature, you provide more particles with enough energy to overcome that activation energy barrier.*0750

*This is a qualitative description of why raising the temperature increases the reaction rate.*0762

*OK, now, Arrhenius proposed this: he proposed that the number of collisions with enough energy equals some percentage of the total collisions--right?*0770

*If you have 1,000 collisions, and let's say, of those 1,000 collisions, only 100 of them have enough energy--well, it's a certain percentage.*0798

*He quantified this; he actually came up with a mathematical formula: so, the total collisions (we won't worry about where these came from or how he derived this) e to the -E*_{a}/RT.0805

*So, this e*_{a}/RT is the fraction of the total collisions that have enough energy to move forward.0823

*And notice: this thing right here, the activation energy, shows up in this exponential.*0837

*R is the (let me rewrite this, actually: e to the -E*_{a}/RT) gas constant, which is not .08206--we are dealing with energy in Joules, so it is actually 8.31 Joules per mole-Kelvin.0843

*T is the temperature in Kelvin; so it's very, very important--you might be given degrees Celsius, but when you use this equation (and we'll sort of make it a little bit better in a minute), the temperature has to be in Kelvin, and R has to be 8.31, not .08206.*0868

*This E*_{a}--that is the activation energy, the energy that has to be overcome in order for the reaction to move forward--the energy that takes you up to the transition state--the energy that takes you up to the activated complex for it to go over the hump; it is the top of the hump.0888

*OK, now, as we said, even those particles with enough energy to overcome the barrier--still, we don't see the kind of rate change that we would expect.*0905

*Well, that is because there is something else going on.*0917

*Not only do these particles have to have enough energy in order to overcome the barrier--they actually have to be oriented properly.*0920

*In other words, they can't just slam into each other this way or that way or that way, and a reaction will take place; there is a specific way that they have to slam into each other in order for a reaction to take place.*0928

*So, there are two requirements (I'm going to put this down at the bottom here, and I'm going to do this in blue) for a successful reaction.*0939

*Two requirements for a successful reaction: the first is, of course, that the collision energy has to be greater than or equal to the activation energy, right?--we need to get over the hump.*0955

*Well, not only do we need to get over the hump, but the molecular orientation (well, I don't want to say "molecular"--well, yes, that is fine) must be such that it allows the reaction to take place.*0977

*We need an activation energy--the collision energy has to be greater than the E*_{a}--and the orientation of the molecules has to be such that it actually allows for the reaction to take place.1015

*That is why the number that we actually see, even though we have a bunch of particles that certainly have enough energy to overcome the barrier--still, only a fraction of those actually makes it past the reaction point.*1026

*So, we have a bunch of particles; only a fraction of them have a certain velocity, in order to activate, in order to get over the hump.*1039

*Of the amount that actually have enough to get over the hump, only a fraction of those have the right orientation.*1049

*So as it turns out, really, it is kind of surprising that reactions proceed the way they do, because really, ultimately, very, very few molecules actually satisfy these things.*1057

*But again, since we are talking about so many, we actually do see reactions take place.*1066

*OK, now, I am going to write a mathematical expression that quantifies everything that we have talked about.*1072

*Now, the rate constant is going to equal z, times p, times e to the (-E*_{a}, divided by RT).1084

*Now, let's talk about what these mean.*1096

*z is a factor that accounts for collision frequency--in other words, how often things collide.*1099

*p accounts for orientation, and it is always going to be less than 1; but that is fine--we are not really worried about that.*1114

*And of course, e to the (-E*_{a}/RT)--that accounts for the fraction of collisions having enough energy--in other words, greater than or equal to E_{a}.1127

*These three things: the frequency of collisions, the orientation, and those that actually have enough energy when they collide: all of these things contribute to the rate constant.*1161

*And, as we said, the rate constant increases exponentially with temperature.*1177

*All of these things are a function of temperature.*1182

*This is a constant; this is a constant; all of these things are essentially constants and factors.*1184

*It is this temperature that changes.*1189

*Now, we are going to write this as: K (which is the rate constant) equals Ae to the (-E*_{a}/RT).1192

*This A is called the frequency factor, and it accounts for these things--it accounts for the frequency of the collisions and the frequency of collisions that actually have the right orientation.*1203

*OK, this is called...this is the Arrhenius equation, right here.*1219

*This is the Arrhenius equation.*1231

*Now, like all things that tend to involve exponentials, let's go ahead and take the logarithm, and see if we can't come up with some linear relationship.*1233

*When I take the logarithm of both sides, I get the logarithm of K is equal to the logarithm of this whole thing; and the logarithm of a multiplication is the logarithm of one plus the logarithm of the other, so it is going to equal the logarithm of A, plus ln of e to the (-E*_{a}/RT).1240

*Well, ln(K)= ln(A), plus--the ln and the e go away, right, so you end up with -E*_{a}/RT.1264

*Now, let me rearrange this: I get ln(K)=...I'm going to pull out the -E*_{a}/R, because these are constants, times (oops, no, we want this to be very, very clear) 1/T, plus the logarithm of A.1283

*Now, notice what we have: we have (oh, these lines are driving me crazy) RT ln(A); let me actually do it over here.*1314

*We have y=mx+b.*1331

*The logarithm of the rate constant--that is the x; let me draw it down here: y=mx+b.*1336

*When we do y versus x, when we plot the logarithm of the rate constant against 1 over the absolute temperature in Kelvin, what we end up with is a straight line.*1350

*The slope of that straight line is equal to the negative of the activation energy, over R.*1362

*So, this gives us a way of finding the activation energy for a given reaction, when we measure the temperature and the rate constant.*1368

*The y-intercept actually gives us ln(A); well, when we exponentiate that, it gives us a way to find the frequency factor, A.*1375

*There we go; this is our basic equation that we are going to use when we are dealing with temperature dependence of the rate constant.*1386

*OK, now, let's see: let's go forward, and let's rewrite the equation again.*1397

*We have: logarithm of the rate constant equals minus the activation energy, over R, times 1 over the temperature, plus the logarithm of A (which is the frequency factor).*1405

*Thus, for a reaction which obeys the Arrhenius equation, the logarithm of the rate constant, K, versus 1 over the temperature, gives a straight line with slope equal to -A/R.*1422

*And again, R equals 8.31 Joules per mole-Kelvin, not the .08206.*1464

*Now, what is really, really nice is that most rate constants actually do obey the Arrhenius equation.*1472

*So, because that is the case, it actually ends up lending support to the collision model.*1477

*We derived this from the collision model; the fact that most things obey this is supporting the collision model--that is supporting evidence; so, our model is actually very, very good.*1484

*Now, let's go ahead and see if we can do an example; I think it's the best way to proceed.*1499

*Our example: the following data were obtained for the reaction 2 N*_{2}O_{5} decomposes to 4 NO_{2}, plus O_{2}.1507

*The following data was obtained: the temperature, which was in degrees Celsius, and the rate constant, which is in per second.*1537

*We have 20, 30, 40, 50, and 60; and these are rate constants, OK?--so this is per second, so we are looking at a first-order reaction.*1549

*2.0x10*^{-5}; we have 7.3x10^{-5}; 2.7x10^{-4}; 9.1x10^{-4}; and 2.9x10^{-3}.1562

*Notice, as the temperature increases, the rate constant increases; the rate is getting faster.*1582

*A higher rate constant means a faster rate.*1586

*OK, now, let's see: what is it that we want to do here?*1591

*We want to find the activation energy for this reaction.*1597

*In other words, how much energy do the N*_{2}O_{5} molecules have to have in order for this reaction to proceed properly?1602

*OK, well, let's go ahead and graph this data.*1610

*Now, I'm going to give a rough graph; it's going to be reasonably accurate, but the idea is to see what is going on here.*1614

*This is going to be 3; this is going to be 3.25; this is going to be 3.50; and then we have 6, 7, 8, 9, 10, 11, 10, 9, 8, 7, 6, so -6, -7, -8, -9, -10, -11.*1623

*OK, so these are...this axis, the vertical axis, we said, is the logarithm of the rate constant; and this is going to be 1/temperature.*1649

*So, let's go ahead and actually...so what we do is (now, notice: we were given K, and we were given T; what we have to do is) calculate 1/T and ln(K).*1667

*OK, so let me go ahead and just redraw everything here; let's do temperature in degrees Celsius; let's do temperature in Kelvin; let's do 1/T; and let's do ln(K)--so we can actually see the data, instead of just throwing it out there.*1681

*We have 20, 30, 40, 50, and 60; then, we have 293, 303, 313, 323, and 333.*1701

*Now, we have 3.41x10*^{-3}; 3.30x10^{-3}; 3.19x10^{-3}; 3.10x10^{-3}; and 3.00x10^{-3}.1718

*Then, we have -10.82; -9.53; -8.22; -7.00; -5.84.*1739

*So, again, our equation is 1/T and ln(K), which is why we took that data that we got, and we calculated the 1/T and the ln(K).*1753

*Now, this is what we plot: this on the x-axis, this on the y-axis.*1764

*When we do that, we end up with some line that looks like this.*1771

*Let's go over here; I'm just going to pick a couple of points: 1, 2, 3, 4, 5--we have 5 points, but I'm just going to pick a couple of them, because it is the point that I want to make that is ultimately important.*1776

*We go ahead and we draw a line through those; and again, this is kinetic data; everything is not going to lie on a straight line, but you are going to get a linear correlation.*1787

*You are doing the "best fit" line.*1798

*When you do that, you pick a couple of points on that line, and you calculate Δy/Δx.*1799

*Now, Δy/Δx, which, in this case, is equal to Δ of ln(K) (that is the y) over Δ of 1/T.*1808

*When you do this, you end up with -1.2x10*^{-4} Kelvin.1820

*Now, we said -1.2x10*^{-4}; that is the slope--well, the slope is equal to the activation energy over R, which is equal to the activation energy (which we are seeking) over 8.31 Joules per mole per Kelvin.1832

*When we multiply this through, we end up with an activation energy equal to 1.0x10*^{4} Joules per mole; there we go.1856

*We were given kinetic data, which consisted of temperature and rate constants; we used the logarithmic version of the Arrhenius equation; we found 1/T and ln(K); we wrote those values down.*1872

*We plotted that: ln(K) versus 1/T; we got ourselves a straight line; we picked a couple of points on that line; we found the slope; and we know that the slope is equal to negative of the activation energy, over the gas constant.*1885

*We solved basic algebra, and we ended up being able to find the activation energy of 1.0x10*^{4} Joules.1901

*That means that the particles--a mole of particles--need to have enough activation, and need to have 1.0x10*^{4} Joules (in order for this to actually proceed) per mole.1908

*That is all this is; that is all we did.*1921

*OK, let's see: what else can we do?*1926

*Let's do another example here.*1933

*This time, I will go ahead and...actually, before we do the example, let me give you a little preface to the example.*1935

*Now, instead of kinetic data, the activation energy can also be gotten from the values of the rate constant at only 2 temperatures.*1944

*In other words, we don't need a whole table of rate constants and their corresponding temperatures...or, temperatures and their corresponding rate constants.*1980

*As long as we have two temperatures and two rate constants, we can actually find the activation energy.*1989

*So, let's see how that is done.*1994

*Now, at temperature T*_{1}, we have rate constant K_{1}.1996

*The relationship is: the logarithm of K*_{1} is equal to minus the activation energy over R, times 1/T_{1}, plus the ln(A).2009

*Again, this is based on the equation: K is equal to A, times e to the (-E*_{a}/RT).2022

*This is the Arrhenius equation, and, when we take the logarithm of both sides, we get this version.*2032

*Well, for any K and T, that is the relationship.*2037

*And now, let's take another temperature; how about at T*_{2}?2042

*Well, at T*_{2}, we have a K_{2}; we have another rate constant, K_{2}.2047

*The relationship is: ln(K*_{2}) is equal to -E_{a}, over R (it doesn't change; it's the same reaction), this time times 1/T_{1}, plus ln(A).2053

*And again, A doesn't change; A is the frequency factor--it's a constant for that particular reaction.*2066

*Well, let me take this equation minus this equation.*2072

*So, I write: ln(K*_{2})-ln(K_{1}), which equals ln(K_{2}/K_{1}), is equal to...when I take this side and subtract this side, the ln(A)s cancel, and I end up with the following.2077

*-E*_{a}/R, times 1/T_{2}, minus 1/T_{1}.2097

*Now, in your chemistry books, you will often see this flipped, and you will see the sign change here.*2106

*All they have done is factor out a -1 from here, in order to get rid of this -1.*2112

*So, you will also see it as: E*_{a}/R, times 1/T_{1}, minus 1/T_{2}; it's up to you.2117

*I personally prefer this, because it doesn't change anything.*2128

*I think it is important that equations be written exactly as how they were derived, and that signs be left alone, simply so you can see everything, so everything is on the table.*2133

*When you start to simplify things, yes, you tend to make them look more elegant--and this is generally true of science; we tend to like our equations to look elegant.*2143

*But, understand something: that in science, and in mathematics, the more elegant something looks--that means the more that is hidden.*2153

*That is the whole idea: when something looks elegant--when something looks clean and sleek and simple--that means something is hidden.*2160

*In this particular case, what they have done is: they have hidden the negative sign, and they have flipped this around.*2165

*That is fine; this is reasonable straightforward--it is not going to confuse anybody too badly.*2170

*But, I think that if we are going to take something and subtract something else, we should leave things exactly as they are.*2174

*So, don't let this minus sign confuse you because it looks different than what you see in your books; it's the same equation; they just don't like minus signs--which is generally true of chemists; they tend not to like minus signs; I don't know why.*2180

*OK, so here we have this equation; if you are given two temperatures and two rate constants, you can calculate the activation energy.*2193

*Let's do an example.*2204

*We have the following reaction: we have: methane (CH*_{4}), plus 2 moles of diatomic sulfur, forms carbon disulfide, plus 2 H_{2}S gas (hydrogen sulfide gas).2208

*OK, now, we take a couple of measurements: temperature in degrees Celsius, and we calculated some rate constants for that.*2233

*That ended up being...let me see: we did it at 550 degrees Celsius, and we also did it at 625 degrees Celsius; we got 1.1, and we got 6.4; it makes sense--higher rate constant-faster rate; higher temperature-faster rate; so everything is good.*2248

*We want to find the activation energy.*2267

*Well, great; we have two constants, and we have two temperatures; let's use our equation that we just derived.*2271

*And again, you don't have to know that equation; as long as you know the Arrhenius equation, everything else you can derive from there, because you are just taking logarithms and fiddling with things.*2282

*That is why we are showing you the derivation--to show you that you don't have to memorize the equation; it is where the equation came from and what you can do with it.*2290

*OK, so let's take the logarithm of K*_{2}/K_{1} (well, you know what, let me write it again), is equal to (-E_{a}, over R)(1/T_{2} minus 1/T_{1}).2299

*And again, temperature is in Kelvin.*2317

*This is equal to the logarithm of 6.4 (see this number), over 1.1; is equal to -E*_{a}/8.31, times...now, 1/T_{2}; T_{2} is 625 degrees; that is 898 Kelvin...so it's 1/898, minus...T_{1} is 550; that is 823 Kelvin.2321

*When we solve, when we do this and this, we end up with the following: =1.7609, equals -E*_{a} × -1.22x10^{-5}.2354

*We end up with an activation energy of 144,195 Joules, or 144 kilojoules.*2372

*I am not a big fan of significant figures one way or the other, which is why it looks like this.*2386

*Numbers are probably going to be a little bit different, as far as you are used to in your book, but it is the process that is important, ultimately--not the significant figures.*2393

*Later, if you become an analytical chemist, then significant figures will be a bigger issue.*2400

*That is it; we have used the Arrhenius equation, which, again, says that the rate constant is equal to some constant (called the frequency factor), exponential, minus activation energy over RT.*2404

*From this one equation, we can do multiple things.*2422

*This expresses that the rate is dependent on temperature.*2427

*This is the independent variable; this is the dependent variable.*2431

*When we are taking kinetic data, we will often have temperature and a rate constant, and we can do things to that, based on how we fiddle with this equation.*2435

*OK, with the Arrhenius equation, this concludes our discussion of kinetics.*2447

*I want to thank you for joining us for this discussion, and thank you for joining us here at Educator.com.*2453

*We'll see you next time; goodbye.*2457

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

*In our last lesson, we concluded our discussion of kinetics, but I thought it would be a good idea to actually spend some time doing some practice problems, directly from the AP exam, and just to give you an idea of what it is that you are going to be doing on the exam, because ultimately, again, this is an AP chemistry course, and the final thing is the AP Chemistry test.*0005

*There is nothing altogether different as far as the example problems that we did, but I thought it would be nice to have some real, specific examples of what it is that you are going to see.*0025

*So, we're going to do a couple of multiple-choice and a couple of free-response questions, and things like that.*0035

*Let's just jump right in and see what we can do.*0040

*OK, so our first series of questions is going to be three multiple-choice questions, and it is going to be based on the following hypothetical reaction and the following choices.*0044

*OK, so let's see: we'll say: Consider the following hypothetical reaction.*0055

*Now, mind you, these multiple choice questions--you can't use your calculator in these, so the numbers are going to be very, very simple.*0072

*Often, you will be able to just look at something and see what it is that is going on, so if you find yourself having to do multiple strange computations, or if you think that you can't do it because of a computation, chances are that there is something wrong.*0078

*These multiple-choice questions--they test basic understanding: can you follow a logical train of thought, where the numbers are completely secondary?*0091

*It is the free-response questions where you can use your calculator to actually find specific answers.*0102

*Hypothetical reaction: we have: our reactant R, plus reactant S, goes to product T.*0109

*Now, the following are our choices.*0117

*A) Rate = K times concentration of R to the first power.*0124

*B) we have: The rate = K times the concentration of S to the second power.*0134

*Our C) choice is: The rate is equal to K, times the concentration of R to the first, the concentration of S to the first.*0143

*D) Our choice is: The rate is equal to K, times the concentration of R to the second power, the concentration of S to the first.*0152

*And E), our final choice, because you have ABCDE to bubble in: The rate is equal to K times R squared, S squared.*0163

*So, we have a hypothetical reaction, and we have these choices to choose from, based on the questions that we are going to ask.*0176

*The first question: #1: OK, it says: When R and S (in other words, when the concentrations of R and S) are doubled, the rate increases by a factor of 8 (so they are telling you this; they are telling you that they double the concentration of R; they double the concentration of S; the rate increases be a factor of 8).*0183

*The question is: What is the rate law?*0219

*A through E are our choices.*0228

*OK, so how do we deal with this?*0232

*Again, we should be able to do this reasonably quickly.*0235

*Well, they are telling us that they are doubling R and they are doubling S, and that the rate is increasing by a factor of 8.*0239

*Well, there are a couple of ways that we can do this.*0248

*I could say to myself, "Well, if I double the rate..." I can say by doubling something, I am just making it twice that; well, if the rate increases by a factor of 8, I can say, "Two to what power equals 8?"*0249

*But, I think that may be a little bit confusing, so let's go ahead and do this a more natural way.*0268

*Let's just go ahead and stick the doubled rate into these to see what happens.*0274

*Well, notice what happens: if I put in twice R into the first one, I get K2R; the 2 comes out--I get 2 KR.*0279

*Let me just do that; let's try choice A.*0293

*Try choice A: I'm a little bit longer on this first one, but it shouldn't take this long--I just want you to see the process.*0297

*Try choice A: Well, I get that the rate is equal to K, times twice the rate of R.*0304

*2 just comes out; so twice the KR.*0313

*Well, KR is the rate; so by doubling the rate, all I have done is multiply the rate by a factor of 2; that is not the one.*0319

*I do the same thing with S; so, if I try choice B, they are telling me that the rate is equal to K2S--just double the rate and put it in there; and this time, it's 2S*^{2}, right?--because our choice is S to the second power.0328

*Well, this is equal to 4 times KS*^{2}.0345

*So now, by doubling the concentration of S, I put it in the choice, and I run the mathematics, and I get that my final answer is 4 times K to the S squared.*0352

*Well, K to the S squared was the original rate; 4 times that--that is not it, either.*0363

*So, as it turns out, the only one of these, when I put it in, where I double R and I double S and I run the mathematics--the only thing that gives me an actual increase by a factor of 8--is choice D.*0369

*The answer is D, and the best way to see that--just in terms of by looking at it without having to run through this--is just: take the number 2--they double the rate--and put it in here.*0386

*2 squared is 4; times 2 is 8.*0398

*You get 8 on this side.*0402

*That is how you solved this problem: just put it in and see which one actually gives you what it is that they are asking for.*0404

*I hope that makes sense.*0411

*OK, under the same circumstances now, problem #2: When R and S are doubled (again, the concentrations are doubled), the rate increases by a factor of 2.*0414

*Well, they say that both R and S are doubled; so, if you go back to your choices, and if you look through your choices--if you put 2 in for R, 2 in for S, well, the only one that actually gives you double the rate is choice A.*0448

*So, A is the answer.*0469

*And again, if you don't see that: the rate they told me was K, times R to the first power.*0477

*Well, if I put in 2, they tell me R and S are doubled; in this case, this choice A is not dependent on S at all--it only depends on R; so when I put 2 in there, I get K times 2R equals 2 times K times R.*0492

*Well, 2 times R--that equals 2 times the rate; that means the rate is doubled; that is the only choice.*0513

*If I put it into the other choices, I get that the rate is increased by a factor of 4, by a factor of 8, by a factor of 16...so the only one that is doubled is choice A, so our answer is A.*0520

*That is how we do it; these are very, very quick--you just need to be able to know what to put where and what to multiply.*0534

*OK, well, let's take a look at #3.*0539

*This time, they say, R is doubled; so this time, they are only doubling R.*0543

*R is doubled; S is unchanged.*0552

*Well, let's write this a little differently.*0562

*This time, what we have is: When R is doubled (the concentration of R), and S is unchanged, the rate is unchanged.*0569

*OK, well, they tell me that R is doubled, and they tell me that S is unchanged, and they tell me that the rate is unchanged.*0608

*Well, if I go back to my choices and if I take a look--well, if R is doubled--notice that choice A, choice C, choice D, and choice E all have R in them.*0618

*So, if I stick a 2 in there, I am going to get an increase; there isn't going to be no change; but they tell me that S is unchanged.*0631

*Well, if S is unchanged, the only choice (which is choice B)--the rate on this one is equal to K, times the concentration of S squared.*0641

*Well, they are telling me that S is unchanged; well, if I don't change S, I don't change the rate.*0658

*S unchanged; rate unchanged; my only choice is B, because everything else has an R in it, and if the rate is doubled, that means there is going to be some sort of a change to the rate.*0663

*In this case, my only choice is choice B, under these circumstances.*0675

*I hope that makes sense.*0682

*OK, let's try a free-response question here.*0686

*No, actually, I'm sorry; let's try one more multiple-choice question.*0697

*OK, so now: For the hypothetical reaction A + B → C, based on the following data, what is the rate law?*0700

*OK, so we are going to give you some data here; and they give you some data: experiment, and then we have the concentration of A; we have the concentration of B; and then we have the rate.*0737

*Experiment 1, 2, 3; 0.20, 0.20, 0.40, 0.10, 0.20, 0.10; and we have 2.5x10*^{-2}; we have 5.0x10^{-2}; and we have 5.0x10^{-2}.0752

*Our choices are the following: K times the concentration of A; K times the concentration of A squared; C is K times the concentration of B; D is K times the concentration of B squared; and E: K times the concentration of A, times the concentration of B.*0778

*OK, so now, let's take a look at this data.*0804

*Whenever we compare two experiments, when we are given concentration data and rate information, whenever we compare two experiments, we need to find something where only one of the concentrations changes--because again, we are only dealing with one independent variable for anything that we compare.*0811

*So, in this case, when we change B from point 1 to point 2, when we double it, we ended up doubling the rate.*0828

*OK, so that means that it is going to be first-order in B.*0841

*The rate is K, so so far we know that it is first-order in B, because again, when you double something and you double the rate, that is first-order in that reactant.*0848

*Well, now let's take a look at Experiment 1 and Experiment 3.*0862

*Let me do this in red; now, we look at 1, and we look at 3.*0866

*Here, the B concentration stays the same, but we double up A and A.*0870

*Let me see here: .2, .4, .2, point...oh, wait; I'm sorry; our data is wrong.*0880

*It is not .1; this is .2; that is .20; OK, let's try this again, shall we?*0891

*Let me erase this, and let me put in blue again: this is .10, .20; this is .20; OK.*0900

*Yes, I think that is going to change some things.*0911

*OK, so now let's take a look at Experiment 1 and 2 again; that is fine; so .2, .2; that is the same; we double up the concentration of B, and then we end up doubling the rate; so yes, not a problem.*0913

*Our rate is...that part is the same; it is going to be first-order in B.*0927

*And now, let's see; now, let's look at Experiment 2 and Experiment 3.*0934

*.2, .2; B is the same; when I end up going from changing the concentration of A from .2 to .4, when I double that, the rate goes from 5.0x10*^{-2} to 5.0x10^{-2}; I double the concentration, but the rate didn't change.0939

*Therefore, the rate is not dependent on A at all; so we are done.*0956

*Rate is KB; our choice is C; that is it.*0961

*Any time you have data like this, you take a look at some doubling, some changing, concentration, and you see how the rate changes.*0969

*That will tell you the order of that particular concentration term in the differential rate law; and then, you just look through your choices.*0977

*Good; OK, let's see; let's try another one here.*0987

*#5: Reactant P underwent decomposition, and the concentration was measured at different times.*1001

*Now, based on the data, what is the rate law?*1038

*OK, so we have time data, and we have concentration of P data.*1055

*0 time (this is in hours, and the concentration is in moles per liter, as always): 0, 1, 2, 3; so we started off at 0.4, and an hour later, we measured the concentration; we have 0.2; an hour later, we have 0.1; an hour later, we have 0.05.*1064

*Now, our choices are the following.*1088

*First-order in P (I'm getting a little sloppy here); second-order in P; twice; D--we have half KP; and E--we have K.*1094

*OK, so they are saying that some reactant, P, underwent decomposition, and the concentration was measured at different times.*1132

*Based on the data, what is the rate law?*1140

*OK, take a look at the time: 1 hour, 1 hour, 1 hour.*1142

*The time increment is uniform; it's 1 hour.*1148

*Now, notice the concentration: after 1 hour, there is half of the initial concentration.*1151

*After another hour, there is half of this concentration.*1157

*After another hour, there is half of this concentration.*1160

*So, every hour, it is diminishing by half.*1163

*That should immediately tell you something; we are talking about half-life here; and what is interesting is that the time--the half-life for it to diminish by half, diminish by half, diminish by half--is uniform.*1168

*The t*_{1/2} is clearly 1 hour, and it is constant.1184

*Any time you have a constant half-life, you are talking about a first-order reaction.*1189

*Remember that the t*_{1/2} of a first-order reaction is equal to the logarithm of 2, over K, or the logarithm of 1/2, over -K.1199

*It is a constant; it doesn't depend on the concentration.*1217

*Because it doesn't depend on the concentration, it stays constant.*1220

*That is what this data is telling me.*1225

*One hour, one hour, one hour; half-life is constant in order for it to diminish by half, diminish by half, diminish by half.*1228

*So, that automatically tells me that I am talking about a first-order rate law; that is my answer.*1237

*My choice is A; I hope that makes sense.*1247

*OK, let's see what we can do here.*1254

*Let's see: OK, let's try a free response question.*1260

*This is #6: We have the following reaction: we have 2 nitrogen monoxide, plus chlorine gas, forming 2 moles of NOCl.*1264

*OK, we have some kinetic data; we run a series of experiments--we run 3 experiments--1, 2, 3.*1278

*We measure the concentration of nitrogen monoxide (initial concentration); we measure the initial concentration of Cl*_{2}; and we also measure...this time, we are measuring the rate of appearance of NOBr.1286

*That is fine; we can measure the rate of appearance--a rate is a rate; rate of appearance, rate of disappearance--it's the same; it's just that one is negative and one is positive.*1309

*We have: the following data were obtained: .02; 0.04; 0.02; 0.02; 0.02; 0.04.*1317

*We have 9.6x10*^{-2}; we have 3.8x10^{-1}; and we have 1.9x10^{-1}.1333

*Our first question to you is: what is the rate law?*1346

*What is the rate law?--OK, it should be easy enough.*1352

*We have some concentration data; we have rate data; we hold one of them fixed; we check the other one to see what is going on.*1357

*When we check Experiment 1 and 2, we notice that the chlorine concentration is the same.*1366

*We doubled the nitrogen monoxide concentration; when we doubled the concentration, we quadrupled the rate.*1372

*OK, I hope you see that: double, from .02 to .04, the concentration of nitrogen monoxide; this 3.8x10*^{-1} is four times 9.6x10^{-2}.1383

*OK, I hope that you see that; watch these exponents very, very carefully.*1395

*So now, double NO concentration means quadruple the rate.*1401

*This implies that it is second-order in NO.*1413

*2 squared is 4; that is where this 2 comes from here.*1423

*OK, now, let's take a look at Experiment 2 and 3.*1430

*When we look at 2 and 3, we notice that we have...actually, not 2 and 3; we will have to look at 1 and 3; I'm sorry.*1435

*1 and 3; and the reason is because .02...now, we're going to leave the NO concentration the same; we are going to double the Cl*_{2} concentration.1448

*So, in this particular case, when we double the chlorine concentration, we end up doubling the rate: 1.9x10*^{-1} is twice the 9.6x10^{-2}.1459

*We double the rate.*1476

*Well, that means--when you double something and the rate doubles, that means it is first-order.*1479

*Therefore, our rate law is equal to some K, times NO to the second power, times Cl*_{2} to the first power; that is our answer.1487

*That is it--nice and simple.*1504

*OK, let's do our next phase.*1507

*Part B: what is the value of the rate constant? Include units.*1513

*OK, what is the value of the rate constant? Include units: I do my units and my numbers separately; that is just something that I like to do.*1517

*You are welcome to do it any way that you like.*1545

*OK, so let's see: we have "What is the value of the rate constant? Include units."*1548

*OK, well, we just said that the rate is equal to K times NO squared, times the Cl*_{2} concentration.1553

*Well, just pick any one of the experiments; you have the rate; you have the NO concentration; you have the Cl*_{2} concentration; just solve for K.1565

*So, K is equal to the rate, divided by the NO concentration squared, times the Cl*_{2} concentration.1575

*Let's just take (I think I'm going to use Experiment 3) 1.9x10*^{-1}, divided by 0.02 squared, times 0.04; and when we do that, we end up with 1.2x10^{4}.1587

*That is the numerical value; now, let's do the units.*1615

*The rate: OK, the rate is in moles per liter per second, because that is how much was showing up, or that is how much was disappearing.*1619

*That is this up here; it is in moles per liter per second.*1635

*These are concentrations; you get moles per liter squared, times moles per liter; you get moles per liter cubed.*1638

*You get moles cubed over liters cubed.*1648

*Well, let's see what we have: this becomes 2; this becomes 2; that cancels that and cancels that; you flip that over, and you get liters squared over moles squared-seconds.*1652

*That is your unit; so your answer is 1.2x10*^{4} liters squared/moles squared-second.1672

*That is your answer.*1684

*So again, my recommendation, as far as kinetics is concerned, is: in one of the lessons (not the last lesson, but the one right before that), we did a summary of first-order reaction, second-order reaction, zero-order reaction.*1687

*On that summary, we discussed the differential rate law; we discussed the integrated rate law; we discussed the half-life; and we discovered the quality of the graph that gives us a straight line (whether it's the logarithm versus time; whether it's 1 over the concentration versus time; or whether it's just concentration versus time).*1704

*That is what you want to know, as far as kinetics is concerned.*1728

*If you understand that summary, if you understand that table and where each thing is coming from, all of these problems will fall out naturally.*1731

*OK, thank you for joining us here at Educator.com, and thank you for joining us at AP Chemistry.*1741

*See you next time; goodbye.*1747

*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

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

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

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

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

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

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

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

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

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

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

*Let's get started.*0090

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

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

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

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

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

*We know that the actual K*_{eq} expression in concentrations is: the concentration of ammonia squared, over the concentration of nitrogen times the concentration of hydrogen cubed.0170

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

*K*_{P}--it's the same thing, and it equals the partial pressure of NH_{3} (partial pressure means the pressure just of that gas in the container, because you have three gases in a container: one gas has one pressure, on has another pressure--we call those the partial pressures), squared (so again, everything is the same; it's products, over reactants, raised to the stoichiometric coefficients; the pressure of NH_{3}, squared...), because this is a 2, divided by the pressure of nitrogen gas, raised to a 1 power, because this stoichiometric coefficient is 1, times the partial pressure of H_{2} cubed, because its stoichiometric coefficient is 3.0198

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

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

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

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

*Let's rearrange this a little bit.*0267

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

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

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

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

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

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

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

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

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

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

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

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

*This is just standard mathematical manipulation.*0383

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

*We will do: K*_{eq} is equal to...it's pressure over RT, the concentration: so we get the partial pressure of NH_{3}, over RT, squared.0397

*So all I have done is: I have just taken this expression, put it into here, and divided by pressure of N*_{2}/RT to the first power, times the pressure of H_{2}/RT (that is concentration squared, concentration raised to the first, concentration cubed, right?)--just basic math.0413

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

*I have: partial pressure of NH*_{3}, squared, over partial pressure of N_{2}, partial pressure of H_{2}, cubed, times 1/RT, squared, over 1/RT to the fourth power.0474

*All right, and now, I am going to have...this 2 is going to cancel that; I'm going to end up with (let me write it one more time--well, two more times, actually): P*_{NH3}, squared, over P_{O2}, times the P_{H2}, cubed, times 1/(1/RT, squared) (because this cancels two of those, leaving that) and 1/(1/RT squared) is RT squared.0497

*So, it equals the partial pressure of NH*_{3}, squared, over the partial pressure of N_{2} and the partial pressure of H_{2}, cubed, times RT, squared.0542

*Well, this is the K*_{eq}; that is this expression.0562

*This expression right here is the K*_{P} (RT)^{2}.0567

*So as it turns out, K*_{eq} equals K_{P}, in this particular case, times RT squared.0572

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

*The relationship is that the K*_{eq} equals K_{P} (for this particular reaction--we'll do the general one in a minute), times RT squared.0590

*I can go back and forth between the two; so, if I'm working with K*_{P} and I want K_{eq}, in concentrations, I can do that.0599

*If I have K*_{eq} and I want K_{P}, I can do that by multiplying by the square of RT.0605

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

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

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

*So, K is equivalent to K*_{eq}; when we speak of P, we will write K_{P}; that means we are dealing with pressures.0641

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

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

*We can write it that way; or, another way that it is written is in terms of K*_{P}.0681

*K*_{P} is equal to K times (RT)^{Δn}; either one of these is fine.0686

*If you have K, you can find K*_{P}; if you have K_{P}, you can find K by just using this expression; that is it--nothing more than that.0696

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

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

*Example 1: OK, at 427 degrees Celsius, a 2.0-liter flask contains 40.0 moles of H*_{2}, 36.0 moles of CO_{2}, 24.0 mol of H_{2}O, and 11.8 mol of CO, carbon monoxide, at equilibrium.0718

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

*Now, the reaction is as follows: CO*_{2} gas + H_{2} gas goes to carbon monoxide gas + H_{2}O gas.0777

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

*Now, our task here is to find K and K*_{P}; so, find the equilibrium constant expression, in terms of moles per liter, and find the K_{P} constant, in terms of partial pressure.0804

*Well, we have the relationship, so we can find K, and then we can find K*_{P}, based on this--no problem.0824

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

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

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

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

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

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

*You have to sort of watch for that.*0888

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

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

*This is a balanced reaction, so everything is 1:1:1:1.*0900

*We have the concentration of CO, times the concentration of H*_{2}O (everything is a gas), over the concentration of CO_{2}, times the concentration of H_{2} gas.0904

*Well, the concentration is the number of moles per volume; our volume is 2 liters (I'm going to go ahead and do this in red), and our moles are 40, 36, 24, and 11.8.*0915

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

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

*Times the concentration of H*_{2}O: H_{2}O is 24 moles, so it's 24.0 divided by 2.0950

*The concentration of CO*_{2}: well, CO_{2} is 36 moles, 36.0 moles; it is sitting in a 2-liter flask, so its concentration is 36/2, or 18 moles per liter.0959

*And now, the H*_{2} is 40 moles in a 2-liter flask.0973

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

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

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

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

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

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

*I promise, it will go badly for you.*1013

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

*Now, we want the K*_{P}, because that is the other thing that we have to find.1026

*Well, we know that K*_{P} is equal to K times RT^{Δn}.1029

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

*Well, Δn is (let me go ahead and write it here) equal to 1+1 (is 2), minus 1+1 (2), is equal to 0.*1045

*So, in this case, Δn is 0.*1056

*K*_{P} is equal to K, times RT, to the Δn, equals K, times RT to the 0 (anything raised to the 0 power is 1); is equal to K.1058

*So, in this case, K*_{P} is equal to 0.197...in *this* case; and the only reason it is this way is because everything is in a 1:1:1:1.1079

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

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

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

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

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

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

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

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

*OK, so the question is a bit long: A sample of gaseous PCl*_{5} (phosphorus pentachloride) was placed in an evacuated flask so the pressure of pure PCl_{5} would be 0.5 atmospheres.1172

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

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

*But, PCl*_{5} decomposes according to: PCl_{5} goes to PCl_{3} + Cl_{2}.1232

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

*Calculate K at this temperature.*1287

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

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

*Don't cut corners.*1308

*A sample of gaseous PCl*_{5} was placed in an evacuated flask so the pressure of pure PCl_{5} would be .5 atmospheres.1310

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

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

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

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

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

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

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

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

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

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

*They said, "Find K"; but the problem as written, since we are dealing with atmospheres and gases (PCl*_{5} gas, PCl_{3} gas, Cl_{2} gas)...we are going to deal with pressures, and then from pressures, we are going to use the RT and Δn expression that we just worked with to find the K.1387

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

*Don't just stop by getting the K*_{P}.1407

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

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

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

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

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

*OK, so let's write the equation.*1442

*PCl*_{5} is in equilibrium with PCl_{3} + Cl_{2}.1445

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

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

*That is what we said: the K*_{eq}: those values that we put in there are concentrations at equilibrium.1469

*I means Initial; C means change.*1475

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

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

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

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

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

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

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

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

*OK, so our initial concentration of PCl*_{5} was .5 atmospheres.1528

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

*We start with 0.5 atmospheres, where, before anything happens--before PCl*_{5} decomposes--there is no PCl_{3}, and there is no Cl_{2}; so this is our initial condition.1544

*It's nice; well, a certain amount of PCl*_{5} decomposes.1555

*Well, look at our equation; it's 1:1:1.*1561

*For every 1 mole that decomposes, 1 mole is produced, and 1 mole is produced of the Cl*_{2}.1563

*We can say that, if -x is the amount that disappears (or, again...concentration and pressure are the same thing)...so, if the pressure drops--PCl*_{5}--by a certain amount, that means it has to increase here by that same amount for each.1571

*For every 1 atmosphere that drops, that means a certain amount has been used; well, that means that a certain amount has been produced of the PCl*_{3} and the Cl_{2}.1591

*That is what the stoichiometry tells me.*1598

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

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

*If 5.2 moles of this decomposes, 5.2 moles of PCl*_{3} is formed; 5.2 moles of Cl_{2} is formed.1614

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

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

*At equilibrium, I have 0.50-x.*1633

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

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

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

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

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

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

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

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

*I found the value of x: .34 atmospheres of PCl*_{3} show up; .34 atmospheres of Cl_{2} show up; and .34 atmospheres of PCl_{5} is lost.1710

*So now, we can actually find our K.*1724

*That is the best part, which is ultimately what we want; so we are solving for K*_{P} here.1727

*So now, I have partial pressures: I have the partial pressure of chlorine gas, is equal to x, which is 0.34 atm; I have the partial pressure of the PCl*_{3} gas, which, again, is x, which is 0.34 atm; and I have the partial pressure of the PCl_{5} gas, which is 0.50-x (that is the equilibrium) minus 0.34, which is 0.16 atm.1732

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

*I know what my equilibrium expression is; my equilibrium expression is K*_{P} equals the partial pressure of Cl_{2}, times the partial pressure of PCl_{3}, divided by the partial pressure of PCl_{5}, each raised to the first power, because the stoichiometric coefficients are 1.1772

*Well, that equals (let's go down here) 0.34, times 0.34, divided by 0.16; I do the multiplication, and I get 0.72; this is my K*_{P}; K_{P} equals 0.72.1792

*They didn't ask for K*_{P}; they asked for K, which means they asked for K_{eq}.1815

*Well, the relationship between K and K*_{P} is the following.1822

*K is equal to K*_{P} times (RT)^{Δn}, if that is correct; let me double-check; K_{P} equals...-Δn.1826

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

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

*We have PCl*_{5} in equilibrium with PCl_{3} plus Cl_{2}; Δn is equal to 1+1-1; 2-1=1.1871

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

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

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

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

*You put in K*_{P}; R is .08206; temperature in Kelvin is 523; negative Δn...Δn was 1; negative 1.1948

*We do the math, and we end up with 0.017.*1957

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

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

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

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

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

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

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

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

*OK, so let's do another example.*2018

*Example 3: We have: Solid ammonium chloride, solid NH*_{4}Cl, was placed in an evacuated chamber, then heated; it decomposed according to: NH_{4}Cl, solid, decomposes into ammonia, NH_{3}, gas, plus hydrogen chloride gas.2027

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

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

*Our task is to calculate the K*_{P}, the equilibrium constant with respect to pressures.2119

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

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

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

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

*The coefficients are 1, so we have: the K*_{P} is equal to the partial pressure of NH_{3} gas, times the partial pressure of HCl gas.2148

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

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

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

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

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

*OK, solid NH*_{4}Cl--we don't care; it doesn't even matter, so we just put lines there; it doesn't show up in the equilibrium expression--it doesn't matter.2188

*Our initial NH*_{3} and HCl concentration--well, we started off with solid NH_{4}Cl; so, initially, there is none of these.2198

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

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

*It is telling you that you start off with NH*_{4}Cl; it decomposes--when something decomposes, that means it is going away; the products are showing up.2220

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

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

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

*Well, which gases are in the flask?*2251

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

*The gases in here are NH*_{3} gas and HCl gas.2261

*So, I basically have: x+x=4.4; they are telling me that the total pressure in there is 4.4 atmospheres; that has to be made up of the amount of NH*_{3} gas and the amount of HCl gas.2265

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

*So, 2x is equal to 4.4; x is equal to 2.2; well, 2.2--there you go; that is the partial pressure of HCl and the partial pressure of NH*_{3}.2287

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

*So, K*_{P} is equal to 2.2, times...OK, now watch this: even though 2.2 and 2.2 is the same, please don't write 2.2^{2}.2319

*They are different species; I promise you, if you write 2.2*^{2}, and if somewhere along the way you get lost and you have to come back, you will spend five minutes trying to figure out what happened, because again, stoichiometric coefficients--they show up in the equation, so write them separately; write 2.2 times 2.2.2332

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

*Don't mix them up.*2355

*We multiply that; we get 4.84; 4.84--that is the K*_{P}, and we double-check: "Calculate K_{P}"--that is what we wanted; we're done; that is it.2357

*So, they gave us a certain amount of information; they gave us an equation to work with; we wrote down the K*_{P} expression; we wrote down the equation.2370

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

*We followed it; we get x and x.*2388

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

*There are only 2 gases in here (the NH*_{3} and the HCl); each one is x.2397

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

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

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

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

*That is what is going to change.*2443

*The approach does not change.*2445

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

*We will see you next time; goodbye.*2452

*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

*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 H*_{2} and F_{2} at a given temperature, 3.000 moles of H_{2} and 6.000 moles of F_{2} 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 1.15x10*^{2}.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 H*_{2} and F_{2} 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 1.15x10*^{2}, 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: H*_{2} + F_{2} 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 H*_{2}, times the concentration of F_{2}.0283

*OK, now let's go ahead and calculate the initial concentration of H*_{2}.0298

*Well, it says we have 3.00 mol of H*_{2} in a 3.000-liter flask; so we have 1.000 Molar H_{2}.0304

*That is what this little 0 here, down at the bottom is: H*_{0}; it just means the initial concentration.0319

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

*F*_{2}, 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

*H*_{2} and F_{2} concentration will go down; HF concentration will rise.0402

*Now, we can do our ICE chart: H*_{2} + F_{2} (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 1.15x10*^{2}.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 4x*^{2}=1.15x10^{2} (2.000-3.000x+x^{2}), and we get 4x^{2}=230-345x+115x^{2}.0546

*And then, we end up with 111x*^{2} - 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 H*_{2} concentration is 1.000-0.968=3.2x10^{-2} Molar at equilibrium.0653

*Our F*_{2} 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 H*_{2} to the F_{2} 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 Cl*_{2}.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 + Cl*_{2}.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 Cl*_{2}, 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 Cl*_{2} 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 + Cl*_{2}; 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 4x*^{2} times x; it is going to be 4x^{3}; 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 2x*^{2} 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: 4x*^{3} is equal to 3.5x10^{-6}; x^{3} 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 Cl*_{2} 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 O*_{2} gas + oxygen gas goes to ozone gas, given reaction 1 (this is going to be sort of a Hess's Law kind of problem): NO_{2} in equilibrium with NO + O, and K_{1} is equal to 6.8x10^{-49} (wow, that is really, really small) and that one, which is O_{3} + NO goes to NO_{2} + O_{2}; K_{2} 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 NO*_{2}.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 1.47x10*^{48}--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 NO*_{2} + O_{2} in equilibrium with NO + O_{3}.1686

*Well, this equilibrium constant--again, I flipped it, so I take the reciprocal of that, and I get 1.72x10*^{33}.1702

*Now I add these two equations; NO*_{2} cancels NO_{2}; NO cancels NO; I am left with O + O_{2} goes to O_{3}; 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 K*_{1}' times K_{2}'.1735

*It equals 1.47x10*^{48}, times 1.72x10^{33}, and I end up with getting some huge number, if I am not mistaken.1749

*2.53x10*^{88}: 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 K*_{1} times K_{2} 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

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

*Today, we are going to close out our discussion of equilibrium with a discussion of Le Chatelier's Principle in relation to equilibrium.*0004

*Now, that doesn't mean that we are just never going to talk about equilibrium again; we are--equilibrium is the central concept in chemistry and most sciences.*0011

*This was a general discussion about equilibrium--getting a little bit of practice in the techniques and how to subtract and add reactants to decide which way a reaction is going to go.*0019

*All of these things will show up when we discuss acid-base, solubility product, complex ion, equilibria...so, very, very important.*0031

*So now, we are going to discuss a qualitative method of deciding what happens when you actually (once a system is at equilibrium) place a stress on that system.*0040

*Your intuition is actually absolutely correct; when you place a stress on any system, the system is going to react in a way that it wants to offset that stress.*0050

*In chemistry, in an equilibrium situation, we call that Le Chatelier's Principle.*0059

*Let's jump in and get started.*0064

*Once again, it is a way of predicting the effect that a change in concentration or pressure or temperature has on the system that is already at equilibrium.*0067

*If we have a system at equilibrium, we can do a couple of things to it to stress it out: we can change the concentration of a particular reactant or product (in other words, add or subtract reactant or product); we can change the pressure of the system, either by directly changing the pressure or by changing the volume, or by adding an inert gas; or we can actually change the temperature (drop the temperature or increase the temperature).*0077

*When we do these things, the equilibrium position is going to change.*0102

*Now mind you, when we change concentration and when we change pressure, the equilibrium position might change, meaning there might be more or less of reactant or product; but the equilibrium constant does not change.*0106

*In other words, the equilibrium itself doesn't change; the position changes.*0120

*However, that is not the same with temperature, because, remember: the equilibrium constant is actually dependent on temperature.*0125

*So anyway, we will see more of that when we start doing some of the problems.*0131

*OK, so let's just go ahead and write out the principle as our fundamental thing to get started with.*0135

*The principle is: If a change is imposed on a system already at equilibrium (let me go ahead and not use that symbol; let me go ahead and just write out "at equilibrium"), the position of the equilibrium will shift in the direction that offsets the change.*0146

*Now, let me just describe what that means in terms of a chemical reaction.*0200

*So, let's just take aA + bB is in equilibrium with cC + dD.*0205

*Now, when we say it will shift in the direction that offsets the equilibrium, once the system actually reaches equilibrium, if I stress it out somehow, it will move in the direction to offset that change--the change that I have made.*0213

*What we mean is that it will either shift right or it will shift left; in other words, it will produce more product, or it will produce more reactant.*0224

*That is all that that means--shift equilibrium.*0232

*So remember, when we actually start a reaction, it moves in the forward direction; but eventually, when enough product develops, it starts to also move in the reverse direction.*0235

*At some point, the rate of the forward and reverse reactions is the same; that is what we call the equilibrium point--when the concentrations of this, this, this, and this are constant--when there is no more change.*0244

*That doesn't mean the reaction has stopped; it's a dynamic equilibrium--things are still going, except there is no net change.*0255

*So, a shift in equilibrium means will it move forward? Will it move back? And I can do things to push it forward and back.*0260

*That is essentially what chemical engineers do: they try to find ways to push reactions forward--the reactions that would not otherwise want to move forward.*0267

*How can we stress it out to make it go forward?*0277

*OK, the first...so we said that we can do a couple of things to it: we can affect, we can change (well, not change--we can--yes, let's go ahead and say change) the concentration by adding or subtracting; we can change the pressure of the system (that is another method by which we can affect the system), or the temperature.*0280

*Temperature, pressure, and concentration are the things that we can control, that we can actually do to a system at equilibrium.*0314

*When we effect these changes, what happens?*0322

*The first thing we are going to talk about is concentration.*0325

*Here we go: How does a system respond to a change in concentration?*0333

*OK, well, like this: If we add species X (and by species X it could mean any one of these--it could be A; it could be B; it could be C; it could be D)--I have this system here; it is in a reaction vessel; I can add or subtract any one of these species; here is how the system will react.*0338

*Well, if I add a species, X, well, the system is going to do what is necessary to offset the addition, which means (so, system offsets the addition) that it will consume X, which means it will deplete X, which means that the reaction will shift accordingly.*0361

*What "accordingly" means: it means away from X.*0411

*In other words (we will do an example, but I'll just do it here), if I have this system at equilibrium, and if I add B, that means I have pumped more B into here.*0419

*Well, the system wants to offset my addition of B, so it wants to reduce the concentration of B.*0429

*The only way to reduce the concentration of B is by actually using it up.*0435

*The only way it can use it up is by shifting the reaction forward to use up B.*0440

*Yes, it also ends up using up A, but the idea is that it is using up B.*0445

*It wants to offset what I have done to it: I have increased the concentration of B--it wants to push the concentration back down.*0449

*In order to push the concentration back down, it has to use it up.*0455

*In order to use it up, it moves the reaction forward; it produces more product, less reactant.*0459

*That is what it means; so in this case, we say that the equilibrium shifts to the right, towards the product.*0465

*Now, what happens when we subtract a species?*0471

*Let's say I have A, B, C, D in equilibrium, and let's say I siphon off...let's say O*_{2} gas is one of the products that is formed--what happens if I siphon off O_{2} as is produced?0475

*Well, the subtraction of X goes like this.*0487

*It is always the same: Subtract species X; well, the system will offset the subtraction...the system wants to offset the subtraction; that means it will...since I am subtracting X, it wants to do the opposite, which means it will produce X; that means that the reaction will shift in a direction which produces X.*0491

*It is that simple; you are just following the train of logic which produces X.*0538

*In the example of, let's say, aA + bB is in equilibrium with cC + dD, if I end up subtracting B (meaning if I pulled B away from a system already at equilibrium), well, the system is going to respond by wanting to increase the concentration of A.*0545

*In order to increase the concentration of A (in other words, produce A...I'm sorry, B), the reaction will shift in the direction that does that.*0564

*Well, the only way to produce B is to shift this way.*0572

*In other words, use up this and this so that more B is produced.*0575

*So, the reaction is going to shift to the left.*0579

*That is all you are doing here.*0583

*Let's do an example, a real-life example.*0585

*It's going to be Example 1: I have the equation 2 SO*_{2} + O_{2} goes to 2 SO_{3}.0592

*All of these are gaseous: sulfur dioxide, oxygen gas, sulfur trioxide.*0601

*Now, we want you to predict the direction the reaction will shift when its equilibrium is disturbed (so this is a system already at equilibrium).*0605

*One disturbance we are going to do is: we are going to add SO*_{2}.0633

*The second disturbance we are going to do: we are going to remove SO*_{2}.0640

*The third disturbance we are going to do is: we are going to remove O*_{2}.0646

*OK, well, so we have a reaction here; now, we want (the system is at equilibrium)--we add sulfur dioxide.*0652

*Well, if we add sulfur dioxide, the system wants to now reduce the sulfur dioxide--it wants to do the opposite of what we are doing.*0663

*We are adding; it wants to subtract; that means lower the sulfur dioxide concentration again.*0670

*In order to lower it, it has to use it up; in order to use it up, it is going to shift it to the right.*0675

*So, the equilibrium is going to shift to the right.*0680

*You can either write "shift to the right," or you can just do a right arrow; I tend to use arrows.*0682

*I like symbolics, when it works--when it is clear.*0686

*That is it; it is that simple.*0690

*What happens if I remove SO*_{2}?0691

*Well, if I remove SO*_{2} from an equilibrium mixture, the system is going to want to produce more SO_{2} to compensate for what I have taken away.0694

*In order to produce more SO*_{2}, it has to shift the equilibrium that way.0702

*That means product is going to decompose, O*_{2} and SO_{2} are going to form, until the SO_{2} level comes back to such that the equilibrium constant is brought back to where it is.0706

*So, in this particular case, it is going to shift to the left.*0720

*What happens if I actually remove...you know what, let's change this; I don't want to remove this; I want to remove SO*_{3}.0726

*Now, what happens if I remove SO*_{3}?0733

*Well, I go over here; SO*_{3} is on the product side; if I remove SO_{3}, if I pull it out of the flask, the system is going to want to compensate for that loss by producing SO_{3}.0735

*The only way to produce SO*_{3} is by pulling the reaction forward, depleting this, depleting that, producing SO_{3}.0748

*The reaction is going to shift to the right.*0757

*That is all that is going on here.*0760

*Find the stress that you are putting on the system, and find out what the system has to do in order to do the opposite of that stress--it's that simple.*0762

*OK, so now, let's go ahead and talk about stressing out the system via pressure.*0773

*Now, instead of changing the concentration of a given species, let's see what happens when we increase the pressure or decrease the pressure on the system.*0780

*So, this is going to be a change in pressure.*0790

*This is the second of the effects that we can run on a system at equilibrium.*0795

*Well, as it turns out, there are three ways to change the pressure.*0800

*OK, we can add or remove reactant or product--gaseous; OK, I'm going to put "gaseous" in; so again, when we talk about pressure, we are talking about gases.*0804

*So, if I want to change the pressure, one of the ways that I can do it is by adding or subtracting actual gaseous product--reactant or product, if one of those species happens to be a gas.*0823

*If it's not a gas, then I can't do anything about the pressure.*0835

*Add or remove--well, by adding or removing a reactant or product, I handled that the same way that I did just a moment ago; I'm just adding or removing a reactant or a product.*0837

*I handle it--it's the same as changing the concentration.*0849

*That is what I am doing; when I am adding or subtracting a gas, yes, I am changing the pressure, but what I am doing is actually changing the concentration, because remember: pressure and concentration are actually the same thing.*0852

*They are just variations of the same thing.*0862

*So that one--I handle it just by addition or subtraction of reactant, like we did a moment ago.*0866

*The other way to change it is by adding an inert gas.*0873

*If I pump in an inert gas (in other words, a gas that doesn't react with anything, that just takes up space), I have changed the pressure of the system; how does that system react when I have actually added an inert gas?*0877

*We will see the answer in just a minute.*0890

*3: I can change the volume.*0893

*Change the volume: so, you know PV=nRT; if the volume goes up, the pressure goes down; if the volume goes down, the pressure increases, all else being equal.*0896

*The relationship between volume and pressure is an inverse relationship.*0908

*So, these three ways are the ways that I can actually change the pressure of the system.*0911

*Now, let's see how each one reacts.*0916

*Well, 1: as we said, handle the same way as the addition or the subtraction, changing the concentration.*0918

*This is handled the same as a change in concentration, because that is it; that is all you are doing.*0929

*"As a change in concentration..." so we already know this one.*0938

*2: If we add an inert gas, as it turns out, adding an inert gas--it changes the total pressure, but it doesn't change the equilibrium, because it's not reacting with any of the species.*0943

*So, adding an inert gas changes nothing--no change.*0954

*Yes, total pressure will increase in the system, but because the gas is not really reacting with anything, it doesn't really shift the equilibrium one way or the other.*0964

*The same concentrations of reactant and product are in there, and the K*_{eq}, which is the product of the products, over the product of the reactants.0972

*Those concentrations don't change, so the K*_{eq} doesn't change.0988

*Well, nothing is changing; so there is no change.*0992

*3: OK, this one--there definitely is a change.*0996

*If I do a volume decrease, that is the same as a pressure increase; these double arrows--they just mean "implies"--it's following a train of logic.*1002

*Volume decrease is the same as a pressure increase; the system now wants to offset the pressure increase; "wants to offset" means it will seek out a way to decrease the pressure back.*1020

*That means it will shift in the direction of fewer gas particles.*1054

*I won't do an example just yet; I want to actually talk about a volume increase, and then I will do the example, and it will make sense.*1069

*So, let's follow the train of logic here: if I do a volume decrease for a system at equilibrium, well, if I decrease the volume, the pressure is going to increase.*1074

*If the pressure increases, we know that the system is going to offset that increase, so it is going to want to seek out a way to decrease the pressure back.*1084

*Therefore, it is going to shift in the direction that has fewer gas particles, because fewer gas particles means less pressure.*1091

*That is what the system is going to do.*1099

*Now, what happens if we have a volume increase?*1101

*OK, so a volume increase: well, it's going to be just the opposite.*1107

*A volume increase means a pressure decrease, so the system wants to offset by a pressure increase.*1110

*How does it do that?--well, it shifts in the direction of more gas particles.*1133

*More gas particles means higher pressure; so once again, if I increase the volume of the system, I decrease the pressure.*1149

*Well, by decreasing the pressure, the system needs to offset that decrease that I just made, so it is going to want to increase the pressure.*1158

*The only way that a system has to increase the pressure is by shifting the reaction in a direction that produces more gas particles.*1165

*So, if the products happen to have more gas particles, it will shift in the direction of products, which means it will shift to the right.*1173

*So now, let's do an example.*1180

*Oh, I should give one last note here: I'll put N.B. for nota bene; if reactants and products have equal numbers of gas particles, there is no change.*1184

*I can increase the pressure, decrease it...do whatever I need to; there is no change, because now, the system has nowhere to go.*1211

*There are equal numbers of particles on the reactant side and on the product side, so there is nowhere for the system to go.*1218

*The total pressure will increase or decrease accordingly; the system will stay exactly the same--there will be no shift.*1224

*In other words, it won't move to the right to create more product; it won't move to the left to create more reactant.*1230

*OK, so let's do Example 2; let's do this in red.*1236

*Example 2: we have: CH*_{4} gas + H_{2}O gas is in equilibrium with carbon monoxide gas, plus hydrogen gas.1247

*We have gaseous components everywhere--in reactants and products--we have 1, 2, 3, 4; OK.*1261

*Let's see: oh, I'm sorry, this is 3 H*_{2}; I was going to say, "There is something wrong here"; let's double-check: C, C, H_{2}O gas; we have H_{2}; that is 4; that is 6; that is 6; 1 oxygen; 1 oxygen; yes, OK.1270

*We have 1 mole of CH*_{4}, 1 mole of H_{2}O, 1 mole of carbon monoxide, and 3 moles of hydrogen gas.1288

*OK, the question is: What happens if we (hmm, temperature--interesting; OK) reduce the volume?*1295

*This system--let's say, all of a sudden, we pump all of this from a 1-liter flask into a 250-milliliter flask; we reduced the volume; what happens to the equilibrium position?*1319

*Again, the equilibrium constant doesn't change; that is the whole idea behind the shift; the constant wants to stay the same, so it has to shift in a direction that will keep it the same.*1333

*OK, what happens when we reduce the volume?*1344

*OK, so I'm going to do some symbolism here: a down arrow is decrease; and up arrow is increase.*1347

*A volume decrease implies a pressure increase; a pressure increase means that a system wants a pressure decrease.*1352

*In order to get a pressure decrease, I need fewer gas particles.*1368

*The only way to get fewer gas particles is to shift the reaction to the left.*1376

*The reason is: I have (let me do this in blue) 1 mole, 1 mole; I have 1 mole...I have 3 moles here.*1386

*The products have 4 moles of gas particles; here, I only have 2 moles of gas particles.*1396

*If I increase the pressure, the pressure will want to decrease to offset; the only way that the system has in order to decrease it is by shifting the reaction that way, by diminishing the product and producing more reactant.*1401

*4 moles, 2 moles; there are fewer moles of gas particles if the reaction goes this way.*1414

*In that case, it will actually reduce the pressure as far as it can to reach equilibrium.*1422

*That is all that is going on here.*1428

*OK, let's say the opposite: What happens if we just directly decrease the pressure?*1431

*Well, if we decrease the pressure, the system is going to want to offset it by increasing the pressure.*1445

*The system wants to increase pressure; the only way to increase pressure is to move the reaction in a direction that produces more gas particles.*1453

*So, it will shift to the right.*1465

*That is it; it's all you are doing--follow the train of logic; everything will fall out.*1468

*OK, now let's talk about the third change that we can make to a system at equilibrium, and that is going to be changing the temperature (heating it up, cooling it down); what happens to the equilibrium?*1474

*Now, remember: we said that the equilibrium position changes when we change the pressure or change the concentration.*1487

*In other words, the actual values of the concentrations change, but the equilibrium constant does not change.*1495

*That is the whole idea behind a constant.*1501

*With temperature, it is different, because the equilibrium constant is actually dependent on the temperature.*1507

*For most practical purposes, it is not really a problem, but you should know that the K*_{eq} is temperature-dependent.1514

*That is important; so, at a different temperature, you are going to have a different K*_{eq}.1525

*An equilibrium constant--yes, it is a fingerprint for that reaction, but it is a fingerprint for that reaction at a specific temperature, which is why the problems always give you what the temperature is.*1529

*It doesn't mean you have to use the temperature in a calculation; it just means "Know that it is at that temperature."*1540

*OK, so let's talk about what happens with temperature.*1547

*Temperature is actually really easy to deal with; let's do a quick review.*1555

*Exothermic means ΔH is (not equals) negative; ΔH is less than 0.*1560

*It also means that heat is released; heat is produced, in other words.*1577

*Endothermic is the opposite; that means the ΔH is positive (not equals, but is; the equals sign is used for numbers); in other words, heat is absorbed.*1591

*Or, a better way of saying it is "is required."*1610

*OK, so exothermic--now we will talk about how we deal with heat, as far as Le Chatelier's Principle is concerned.*1617

*Exothermic--heat is released; heat is produced; imagine that heat is a product--that is it.*1627

*Not only do you produce a chemical product, but you are also producing heat in the process.*1634

*Treat heat like a product; that is it.*1639

*Heat is a product, so you write it on the right side of the arrows.*1643

*Endothermic means heat is required; well, if heat is required, that means heat is a reactant--treat heat like a reactant--put it on the left side of the arrows.*1650

*Well, if I am writing heat as a product or a reactant, I can imagine it like a species.*1666

*If I increase the temperature of something, that means I am actually increasing the concentration of heat.*1679

*Now, I treat a change in temperature the same way that I treated an increase in concentration.*1686

*Well, let's write out what this looks like; let's do an example, actually.*1693

*Actually, you know what, let me just do a rough breakdown; so let me do aA + bB goes to cC + heat.*1696

*This is an exothermic reaction; I have written it; ΔH is negative.*1706

*I have written it as if heat is just another product; it is one of the things that is produced.*1711

*It produces a chemical, C, but it also produces heat.*1716

*Now, if I have this situation at equilibrium, what happens if I increase the temperature?*1720

*Well, if I increase the temperature, that means I am increasing the concentration of heat.*1726

*Well, if I increase the concentration of heat, the system wants to react by decreasing the concentration of heat; that means it wants to use up the heat--that means it is going to shift that way in order to use up the heat, to pull it away--just like the first part, when we did concentration.*1732

*If I do a temperature decrease on an exothermic reaction--well, with a temperature decrease, the temperature is going to want to offset by going up.*1749

*Well, the only way the temperature can go up is by producing more heat from the equation itself.*1759

*In order to produce more heat, the reaction has to shift to the right.*1764

*It is going to shift toward products, toward the production of heat--yes, and C, but of heat.*1770

*So again, when you are dealing with temperature, take a look to see which--whether your reaction is exothermic or endothermic.*1776

*If it's exothermic, write heat as one of the products; if it's endothermic, write it as one of the reactants.*1783

*And then, go back to treating it like it's just a reactant or a product that you are increasing or decreasing the concentration of.*1788

*So let's do an actual example.*1796

*We have: N*_{2} + O_{2} is in equilibrium with 2 NO; so nitrogen gas and oxygen gas in equilibrium with nitrogen monoxide.1799

*The ΔH for this process is 181 kilojoules.*1813

*I apologize; I tend to write my kilos with a capital K; I know it's probably bad practice, but...it's all right; it's not supposed to be, but...I'm going to be like that.*1818

*This is a positive 181 kilojoules; so this is an endothermic reaction.*1826

*OK, this is endothermic, so I'm going to write this reaction with heat as one of the reactants.*1831

*I am going to write heat, plus N*_{2}, plus O_{2}, is in equilibrium with 2 NO.1836

*So now, let's see what happens.*1845

*What can we do?...well, OK; if we increase (at part A) the temperature, what happens?*1852

*Well, if we increase the temperature, that means we are increasing the concentration of heat, because it is one of the reactants--it's an endothermic reaction.*1861

*It is going to want to decrease the concentration of heat to offset.*1869

*The only way to decrease it is to deplete it, which means it is going to push the reaction to the right.*1872

*So, the reaction will shift right.*1878

*In other words, shifting right means it will produce more nitrogen monoxide; in the process of producing more nitrogen monoxide, oxygen will deplete, N will deplete, and heat will deplete--it will use up the excess heat that I have put in there.*1881

*Well, if I do a temperature decrease--if I decrease the temperature--the system wants to increase the temperature back up.*1895

*The only way it has (the system has) to increase the temperature is by producing more heat.*1903

*The only way to produce more heat is by moving the reaction that way, to produce more heat, more N*_{2}, and more O_{2}.1908

*It is going to shift left.*1915

*That is all these shifts mean: is it going to move towards formation of more product or formation of more reactant?*1919

*Formation of more product means "shift right"; formation of more reactant means "shift left."*1926

*That is it; that is all that is going on here.*1933

*Let's do another example.*1936

*We have: 2 SO*_{2} + O_{2} going to SO_{3} (well, it's 2 SO_{3}; it's not balanced--and that is going to be 3 O_{2}; there, that is a little bit better).1937

*So, we have sulfur dioxide gas and oxygen gas, sulfur trioxide; I think I may have left the equation unbalanced earlier on; I apologize for that, if so.*1956

*So now, it says that the ΔH for this reaction is -198 kilojoules.*1966

*Well, a negative ΔH is exothermic; exothermic means it is releasing heat--it is producing heat--so I'm going to rewrite this as: 2 SO*_{2} + 3 O_{2} is in equilibrium with 2 SO_{3} + heat.1973

*Or, I can just write "energy"--and again, heat and energy are just the same thing.*1990

*So now, heat is a product; it is one of the things that is formed in this reaction.*1993

*Well, if I increase the temperature of this reaction at equilibrium--if I increase the temperature, the system is going to want to decrease that temperature back down.*1998

*The only way the system has to decrease the temperature is by shifting the reaction that way--using up heat and using up SO*_{3} to produce more reactant; so it is going to shift left.2008

*If I decrease the temperature, the system is going to want to offset by increasing the temperature.*2022

*The only way the system has to increase temperature is by producing more heat.*2029

*How does the system produce more heat?--it moves to the right to produce more sulfur trioxide and to produce more heat to compensate for the heat that I took away.*2032

*So, it ends up shifting right.*2043

*That is it.*2049

*So, we have a system at equilibrium, and there are a couple of things that we can do to it--well, three things that we can do to it.*2052

*We can add or subtract reactant or product; we can change the pressure of the system; or we can change the temperature of the system.*2059

*Each of those things causes the reaction to actually move in a direction to reestablish equilibrium.*2069

*That is the whole point: I have upset the equilibrium balance; the system needs to do what it can, if it can, to reestablish the balance.*2076

*That doesn't mean it can always do that; like we said, if we have 2 moles of gas on the reactant side and 2 moles of gas particles on the product side, and all of a sudden I either...let's say I change the pressure of the system by increasing it, well, there is no direction that the system can actually move in to reduce that pressure change, because both sides have 2 moles.*2085

*There isn't one side with a...so, basically, the system stays exactly where it is.*2108

*I have stressed out the system, but the system doesn't have a means of escaping, of alleviating that stress, so it stays there.*2112

*More often than not, it will--it will have an escape route; it will have an avenue by which to offset the stress.*2120

*But, it doesn't mean it will have; so you have to watch out for that.*2126

*OK, so let's do a final example here, and it will sort of be an all-encompassing example: What happens when we do all kinds of things to it?*2130

*So, let's have the equation example; let's do H*_{2} (I really, really need to write a little bit more clearly, or at least slow down; I know that I tend to write quickly--I apologize) gas, plus I_{2} gas, is in equilibrium with hydrogen iodide gas.2140

*Now notice: I didn't say hydroiodic acid; this is a gas--this is not aqueous.*2164

*When you take hydrogen iodide gas, and you bubble it in the water, that is when it dissociates into hydrogen ion and iodide ion.*2169

*That is when we call it hydroiodic acid.*2177

*This is just hydrogen iodide; it's not acidic--this won't hurt...well, I won't say it won't hurt you, but it won't hurt you as an acid.*2179

*OK, so now, what happens when H*_{2} gas is added?2187

*Well, if H*_{2} gas is added, the system is going to want to move in a direction that actually depletes what I added.2195

*It wants to offset it; in order to deplete H*_{2}, it has to move to the right, so it's going to shift right.2202

*I'm going to use symbols here.*2208

*Part B: What happens when I*_{2} is removed?2211

*Well, if I remove I*_{2}, the system is going to want to offset it by producing more I_{2} that was there.2217

*So, it is going to want to pull the reaction back this way, so it's going to shift to the left.*2223

*It is going to produce more reactant; yes, it will also produce more of this, but the idea is to produce more of this.*2227

*C: HI is removed.*2235

*Well, if I remove hydrogen iodide, the system is going to want to offset by creating more hydrogen iodide.*2241

*How do you create more iodide?--you shift the reaction to the right.*2247

*You don't do it; the system automatically shifts to the right (I'm sorry; I should be a little bit more precise).*2252

*You are putting a stress on the system; the system is reacting by moving to the right to offset the stress that you put on it.*2257

*Part D: Argon is added--argon gas.*2266

*Well, argon is a noble gas; it is inert; so we add an inert gas.*2271

*Well, if you are adding an inert gas, you are changing the total pressure of the system, but there is not really much that you can do; so, in this particular case, nothing is going to happen--no change.*2276

*OK, C, D, E: And...volume is doubled.*2296

*Well, if we double the volume, it is going to be sort of the same as the argon gas; even though it's inert...*2308

*Volume is doubled; that means the pressure is decreased, right?--volume is doubled; that means you are making the volume bigger, so you are decreasing the pressure.*2317

*But, you have 2 moles of particles on the right, and you have 2 moles of particles on the left; there is no place for the reaction to go, to offset the decrease in pressure.*2330

*It is just going to have a decrease in pressure; so there is no change.*2341

*OK, now let's do F: let's do a temperature increase.*2347

*The temperature is increased; OK...oh, I forgot to write...what is the ΔH here?--the ΔH of formation is 25.9 kilojoules.*2354

*So this is actually endothermic (because we need to know what the ΔH is, in order to decide what happens when we make a change to temperature).*2366

*So, if this is endothermic, that means that heat is one of the reactants--heat is one of the things required to actually move the reaction.*2376

*If temperature is increased (that means if I increase the concentration of heat), the system is going to want to offset by decreasing the concentration of heat that I put in; so, it wants to reduce this.*2386

*The only way to reduce this: the system must move in this direction; so it will shift to the right.*2397

*There you have it: Le Chatelier's Principle.*2406

*If you have a system at equilibrium, and you do something to mess with that equilibrium by either changing the concentration of a reactant or product, by changing the pressure of the system, or by changing the temperature of the system; the reaction is going to move in the direction that offsets that change.*2410

*It is that simple.*2436

*OK, thank you for joining us here at Educator.com, and thank you for joining us for equilibrium in chemistry.*2438

*This concludes our discussion of equilibrium.*2445

*Next time I see you, we will be talking about acids and bases.*2448

*Take good care; goodbye.*2450

*Welcome back to Educator.com; welcome back to AP Chemistry.*0000

*Last time we finished off our discussion of equilibrium, a very, very important topic, and now we are going to go on to probably the second central topic of chemistry, which is acids and bases.*0005

*They're very, very important in all aspects of chemistry, particularly in biochemistry, because much of the reaction that takes place in your body is, of course, acid-base reactions.*0016

*We are going to spend some time discussing acids and bases and writing reactions and talk about what is going on qualitatively, and then, we will also deal with it quantitatively, and we are going to introduce a new equilibrium constant, called the K*_{a} or the K_{b}.0026

*Now, fortunately, it isn't really anything new; it isn't new in the sense that it is a different way of writing the equilibrium constant.*0043

*It is just a different symbol for it; it is still exactly what you learned with equilibrium, and everything that we learned with equilibrium will again play the same part.*0049

*We will be doing ICE charts, so we will get plenty of practice in this.*0058

*If you feel like you don't have a solid grasp of equilibrium, acids and bases are a wonderful place to continue to practice that until you get a solid grasp of it.*0062

*Let's go ahead and get started with some definitions.*0071

*OK, what we are going to be using: there are a couple of models for acid-base behavior, and the primary model--at least, the one that is used most often--is something called the Bronsted-Lowry acid-base model.*0075

*Let's just talk about what that is; we'll just define it.*0087

*Bronsted-Lowry...the name itself doesn't really matter all that much--for historical reasons, I suppose it's nice to give credit to the people who deserve credit--but it is the model that is important: acid-base model.*0093

*Now, an acid is a hydrogen ion donor, and a base is a hydrogen ion acceptor--a very, very simple definition, and let's talk a little bit about what this means.*0109

*So, an acid is a compound that has an H*^{+} that can come off--that can potentially come off.0133

*I'll write "potentially come off"--in other words, it has a hydrogen ion to donate, if it needs to donate it.*0161

*More often than not, it will actually donate it--it will give up the hydrogen ion.*0170

*A base is something (I shouldn't say something--it's a compound--yeah, right, it's just "something")...is a compound that can potentially take that (not that; it will be that, but let's be as general as possible)...that can take an H*^{+} ion.0176

*That is it: so, a base is a compound that has a hydrogen attached to it, but the hydrogen can potentially separate from that compound, and just float around freely as hydrogen ion.*0221

*That is, in fact, the acid: when we speak about an acid, that is what an acid is: it is just free hydrogen ions floating around in solution--that is what does the damage.*0233

*A base is something that has the capacity to either take a hydrogen ion from something (an acid) that has it to give--in other words, rip it off--or, if there are free hydrogen ions floating around, a base can actually grab onto it.*0241

*That is it: so an acid is a hydrogen ion donor; a base is a hydrogen ion acceptor.*0260

*Acid-base chemistry--they come in pairs; when there is some acid, there is some base, usually; so we will see what that means in just a minute.*0266

*OK, there are two ways to write the reaction of an acid in water--or with water, I should say: to write the reaction of an acid with water.*0277

*And again, an acid is something that you are actually dropping in water, and once it is in water, it actually tends to come apart.*0304

*Here is what actually happens: we will just write H, and we will write A; so H is the H that actually comes off; this A can be anything--it's just a generic other part of the molecule.*0312

*...Plus H*_{2}O, goes to H_{3}O, plus A^{-}.0326

*What has happened is that the H has gone from this HA over, and attached itself to the H*_{2}O; it has actually attached itself to the oxygen, and it is H_{3}O^{+}, now.0335

*H*_{2}O is neutral; H^{+} is plus charge--that is why now, this has a plus charge; and because this was HA, it's a neutral ionic compound, but this H left its electron when it left (it left as H^{+}), it has an A^{-}.0346

*There is another way to write this reaction, which is the way that I am certainly accustomed to doing it, and in certain circumstances, it helps to write it this way as opposed to this way.*0365

*It is why we actually have two ways of doing it, because it depends on the problem and how convenient it is to write one way or the other.*0375

*HA...it's the actual dissociation reaction: when you put this in water, it becomes free ion, H*^{+} ion, and free A^{-} ion.0382

*This is a little bit more descriptive of what is really going on, in the sense that HA dissolves, like sodium chloride dissolves, and it breaks up into a hydrogen ion and the other ion (whatever the rest of the molecule is).*0393

*What is really happening is: there is an actual transfer.*0409

*This is the acid; it has the hydrogen to give up--it has the hydrogen to donate.*0413

*In this case, water is acting as the base; it has the capacity to take that hydrogen, or to accept that hydrogen.*0418

*So, when we write it this way, we actually talk about an acid conjugate base pair, which I will actually get to in just a minute; but I just want to sort of talk about these two equations, and how to represent them.*0426

*When we talk about an acid dissociating, or an acid reacting with water, we can either write it this way, or we can write it this way.*0443

*This H*_{3}O^{+} and this H^{+} are the same thing.0452

*This H*^{+} is just an H^{+}; when it is attached to an H_{2}O, we write it as H_{3}O^{+}, but the chemistry is the same.0456

*Now, let's go ahead and write the equilibrium constant for this.*0466

*The K*_{eq}--well, we said the K_{eq} is the concentration of product, concentration of product, divided by concentration of reactant, concentration of reactant.0470

*This is liquid water; it doesn't show up--remember, liquids and solids don't show up in the equilibrium constant.*0483

*This is going to be: the concentration of H*_{3}O^{+} in moles per liter, raised to its coefficient (which is 1), the concentration of A^{-}, raised to its coefficient (which is 1), divided by the concentration of HA; this is aqueous.0488

*OK, water doesn't show up; when we have an acid written like this, or (let me actually write this version of it: H*^{+}, A^{-}, over HA) when we are dealing with an acid, we don't call it K_{eq}; we call it K_{a}.0507

*This is the acid dissociation constant; it is the equilibrium constant for the reaction of an acid with water, or for the reaction of an acid dissolved in water.*0530

*When you dissolve it in water, we think about it as just a dissociation, like any ionic compound.*0543

*But what is really happening--it is actually reacting with water.*0547

*The acid is giving up the H; the water is acting as the base, accepting the H.*0550

*In some sense, when you look at this, this equation, this equilibrium--what it expresses, what it represents...it represents a competition between water and the conjugate base for this H.*0555

*In other words, does the H stay with the A, or does it go with the water?*0570

*If it goes with the water, that means the reaction is over on this side: H*_{3}O^{+} is formed.0574

*If the H doesn't want to leave the A--if it wants to stay with the A--that means most of the reaction is on this side.*0580

*That is what the K*_{a} measures.0586

*So, the equilibrium constant is called the acid dissociation constant.*0588

*It is the same as any other equilibrium, except it is representing what is happening when the acid reacts with water, when the acid is dropped into water.*0595

*That is it; so, from now on, when we are discussing acid-base chemistry, it is going to be K*_{a}.0605

*Later on, when we discuss bases, we are going to have an analogous equation, and it is going to be K*_{b}.0610

*But it is just an equilibrium constant.*0615

*And what does an equilibrium constant represent? It represents the extent to which a reaction has moved forward.*0619

*Once it reaches equilibrium, how much of this, how much of this, how much of this?--it's a measure of the extent of reaction.*0626

*How far forward has it gone? How far has it not gone?--that is all it is a measure of.*0634

*The higher the K*_{a}, that means the dissociation; that means it's to the right, because the numerator is a bigger number.0639

*OK, now, let's talk about polyprotic acids really quickly.*0648

*For acid dissociation, 1 H*^{+} leaves at a time.0656

*For example, if I have the acid H*_{3}PO_{4}, which we know of as phosphoric acid, it doesn't do this: H_{3}PO_{4}--it doesn't just dissociate into 3 H^{+}s plus PO_{4}^{3-}.0683

*It doesn't do that; that is not how acids behave.*0699

*This doesn't happen; what does happen is that 1 H*^{+} leaves at a time, so the dissociation of phosphoric acid is three steps, and it looks like this.0704

*H*_{3}PO_{4} + H_{2}O is going to be in equilibrium with H_{3}O^{+} + H_{2}PO_{4}^{-}.0722

*That is the first dissociation; there is some K*_{a} associated with that first equilibrium.0733

*And then, H*_{2}PO_{4} has two more Hs to give up; so, H_{2}PO_{4} (actually, you know what, I think I'm going to write it as just a regular dissociation--I'm sorry; I think it will be a little more clear) now gives up another one.0739

*It gives up 1 at a time: H*^{+}, and it becomes HPO_{4}^{2-}; there is an equilibrium associated with that.0762

*OK, and then we have HPO*_{4}; it has another H^{+} to give up, and now it's PO_{4}^{3-}; there is a third acid dissociation constant.0773

*And each one of these is actually different; these phosphoric acid dissociates to differing degrees--these are not the same number.*0787

*In other words, this equilibrium might be really far forward; this equilibrium may be really far this way; it just depends.*0796

*The key idea is: for a polyprotic acid (polyprotic just means having more than one hydrogen to give up), it gives them up one at a time.*0805

*Very, very important: for acids, it releases one at a time; it doesn't happen all at once.*0813

*So, H*_{2}PO_{4}: it doesn't just release both hydrogens--it releases one completely, and the second one comes off very little, in fact--so, just something that you should know.0819

*OK, so let's talk about acid strength.*0830

*Acid strength: Acid strength is defined by the position of the equilibrium constant--or by the position of equilibrium; in other words, which is measured by the equilibrium constant.*0837

*It's measured, it's defined, by the position of the equilibrium; and the equilibrium, we said, was some acid plus water, going to hydronium ion...or just H*_{3}O^{+}...0856

*We call it hydronium, by the way--let me just write that name down: H*_{3}O^{+} is called a hydronium.0874

*Anything with a positive charge tends to have an -onium ending; it's the same as H*^{+}--you can treat them the same.0882

*So, the acid strength is defined by the position of the equilibrium; if the equilibrium lies really far to the right, that means, if this acid has completely given up all of its Hs over to water, it is a strong acid.*0891

*So, the farther to the right the equilibrium is, that means it is a stronger acid.*0903

*We talk about a strong acid being something that actually does not want to keep its H's; it wants to give them up completely.*0914

*It wants to be A*^{-}, and it wants its H to be H^{+}.0922

*It does not want to be together.*0928

*OK, now, let's talk about...let's see; so, actually, let me write that again.*0930

*Let me say: farther to the right, stronger acid--so, when we talk about acid strength, we are talking about relative strength; it's always...we are comparing it to something.*0942

*Stronger acid--well, that means a higher K*_{a}, right?--because the K_{a} is products over reactants; well, farther to the right means there is more product and very little reactant--a really big numerator, a really small denominator, and a huge number--stronger acid.0958

*OK, now let's go back to blue here.*0976

*A strong acid implies a weak conjugate base.*0983

*Acids and bases come in conjugate pairs--"conjugate base"; we get the conjugate base by just pulling off the hydrogen from the acid.*0988

*So, let's write our equation again: HA + H*_{2}O is in equilibrium with H_{3}O^{+} + A^{-}.0999

*This is the acid; this is its conjugate base.*1010

*I pulled off the H; that leaves my conjugate base.*1021

*If I take a base (in this particular case, again, this is the acid--this has the H to give up; water is acting as the base--it is going to take that H, or accept it, if it wants to), when it accepts the H and it becomes H*_{3}O^{+}, this is called the conjugate acid.1026

*That is it; if you have an acid, you pull of the H; you are left with a conjugate base.*1048

*If you have a base, and if you add a hydrogen ion to it, you have its conjugate acid.*1053

*That is it--nothing more than that; no more, no less: don't read any more into that.*1060

*OK, so a strong acid means a weak conjugate base, which is exactly what you would expect.*1064

*A strong acid does not want to hold onto its H's; it will give them up completely; that means it has a weak conjugate base.*1070

*A weak conjugate base means that it doesn't want this H; that is the whole idea.*1077

*A strong base wants the H--will take the H; it's a competition.*1084

*A weak acid has a strong conjugate base.*1091

*A weak acid is one that does not want to give up its H's; and the reason it doesn't want to give up its H's is because its conjugate base is so strong that it literally holds onto its H's without letting go.*1104

*That is the idea; so, weak acid means conjugate base.*1116

*When you are thinking about this, you have to pick a perspective; you have to pick a point of reference; that is the whole idea.*1121

*We decide to take the reference point as the acid itself; we call it strong; we call its conjugate base weak.*1127

*That means the conjugate base doesn't want the H; it gives it up.*1134

*A weak acid is one that does not want to give up its H; that means that its conjugate base is very, very strong--it will not release its H.*1139

*So, it's really a question of perspective; you need a point of reference; and the point of reference that we have taken is the acid.*1149

*We talked about a strong acid and a weak acid: a strong acid has a weak conjugate base; a weak acid has a strong conjugate base.*1157

*OK, for strong acids, no K*_{a} exists...no K_{a} is listed, we should say.1172

*So, for example, hydrochloric acid is a very strong acid; its dissociation (it's not even in equilibrium)--it completely breaks up into free H*^{+} and free Cl^{-}.1184

*You won't find any of this at all.*1195

*Well, the K*_{a} for this is H^{+}(Cl^{-})/HCl.1198

*Well, since there is a whole bunch of this and a whole bunch of that, and virtually none of this, this denominator is really tiny; the numerator is really big.*1207

*So, you have a K*_{a} which is huge; in fact, the K_{a} is so huge that we can't even measure this.1216

*Because we can't measure the Cl concentration accurately, we don't even list a K*_{a} for it; it's just off the charts.1222

*Strong acids--full dissociation; that is the idea.*1228

*Once again, strong acid means full dissociation.*1233

*That means this H and Cl completely come apart; in solution, you will never find any Cl attached to an H.*1240

*It will free Cl*^{-} and free H^{+}.1249

*That is what makes it a strong acid; because there is so much H*^{+} floating around freely, that is what does the damage.1252

*So, when we say "strong acid," we are talking about a chemical property.*1259

*Strong acids do damage because there is so much H*^{+}; there is so much H^{+} because the H and Cl do not want to stay together--they want to be separate.1263

*Strong acid--full dissociation: that's very important.*1274

*OK, let's do an example to get a sense, or a feel for what is going on, for some numbers here.*1279

*So, Example 1: HF, which is hydrofluoric acid, has a K*_{a} of 7.2x10^{-4}.1290

*It's a small number; it's a weak acid--it's not very far forward.*1301

*Most of it is in this form; OK, so let me write the dissociation: HF goes to H*^{+} + F^{-}.1305

*When I measure this, this, and this, and I put these on top of that--the K*_{a}--I get this number.1314

*It's pretty tiny; a tiny K*_{a} means that most of this equilibrium will be found on this side, the reactant side.1320

*In other words, there is very little dissociation; most of it just stays HF.*1327

*Hydrocyanic acid, HCn--it has a K*_{a} of 6.2x10^{-10}--wow, really, really small.1333

*So, when I see HCn--when I drop some of that in water (that is Cn*^{-}), this would be the dissociation.1343

*This is so tiny that this is telling me that most of this reaction is over on this side.*1351

*I won't find a lot of free H*^{+} and Cn^{-}; I'll find some--I was able to measure it--but most of it is this.1356

*That is what the small K*_{a} means; a small K_{a} means it hasn't dissociated very much.1362

*A big K*_{a} means it has dissociated a lot more.1368

*Between these two, this is bigger; this is smaller; this is a stronger acid, because its K*_{a} is bigger--it has dissociated more.1371

*It is still a weak acid, but compared to hydrocyanic, it's stronger than hydrocyanic.*1383

*That is the difference; when we speak about the strength of acids, we are often going to be comparing; it's relative.*1390

*We are going to be talking about 2 or 3 or 4 species.*1396

*Now, the question is: which one has the stronger conjugate base?--that is the question.*1402

*Which acid has the stronger conjugate base?*1409

*Well, for HF, the conjugate base is--just take off the H; this is the conjugate base.*1421

*For HCn, it is the acid minus the H; that is the conjugate base.*1427

*Which has the stronger conjugate base? Well, the stronger conjugate base comes from the weaker acid, right?*1433

*Weak acid=strong conjugate base; OK, between these two, the weaker acid is the hydrocyanic acid.*1445

*In other words, it's a weak acid because it doesn't dissociate very much; that is confirmed with this really, really small K*_{a}.1456

*It doesn't dissociate very much; it's a weaker acid; it has a stronger conjugate base.*1464

*In other words, the base is so strong that it doesn't want to release its hydrogens; it wants to hold on to it.*1470

*That is what a strong base does: a strong base grabs onto its hydrogens and holds onto it.*1476

*A strong acid wants to give up its hydrogens; that is why we have strong acid, weak base--yes, stronger acid, weaker base; weaker acid, stronger base.*1482

*So, in this case, F*^{-} is a weaker conjugate base than Cn^{-}.1495

*That is it; we are using K*_{a} values to decide, depending on what species we are talking about.1509

*This K*_{a} means HF is a stronger acid than HCn.1515

*It means its conjugate base is weaker than Cn*^{-}; that is what is going on here.1520

*OK, so now, let's discuss water as an acid and a base.*1526

*Let's define something called amphoteric: an amphoteric substance is one that can behave as an acid or a base.*1538

*It should be "and a base," because it actually does both; "as an acid or a base"--in other words, something that can go both ways.*1561

*Well, let's look at the dissociation of water.*1571

*I'm going to write it slightly differently.*1575

*I'm going to write the first water: so now, we are talking about water as a species dissolved in water.*1577

*It's plain old water; I know it's a little weird, but think about it this way.*1583

*I'm going to write it as HOH + H*_{2}O.1587

*These are both water, but I wanted to emphasize what is happening.*1591

*This HOH, believe it or not, actually dissociates just a little bit; it breaks up into H*^{+} and OH^{-}.1596

*Well, this H goes over here to become H*_{3}O^{+} + OH^{-}.1602

*Another way of writing this is HOH, as free dissociation, without involving the water species, breaks up into H*^{+} + OH^{-}1611

*Well, let me rewrite it: HOH is in equilibrium with a little bit of H*^{+} and a little bit of OH^{-}.1622

*The K*_{eq} for this is equal to...well, this is liquid water; liquid doesn't show up in the equilibrium concentration; this is aqueous; this is aqueous; this is floating around, in other words, and this is floating around in the water, to a certain degree.1634

*It is equal to the H*^{+} concentration times the OH^{-} concentration--the products.1650

*Well, as it turns out, for water, we actually call it K*_{w}.1656

*We have done the experiment; and, as it turns out experimentally, we have actually measured, at 25 degrees Celsius...experimentally, at 25 degrees Celsius, the hydrogen ion concentration is equal to the hydroxide ion concentration; that is what this equation says.*1662

*For each H that is produced, one OH is produced.*1685

*They have the same concentration; it equals 1.0x10*^{-7} moles per liter.1688

*The concentration of H*^{+} and OH^{-} in standard, neutral, run-of-the-mill water, is 1.0x10^{-7}.1696

*Well, what does that mean?--K*_{w} is H^{+} times OH^{-}, so K_{w} equals 1.0x10^{-7} squared, equals 1.0x10^{-14}.1706

*Water has a K*_{a} of 1.0x10^{-14}.1723

*We give it a special name; we call it K*_{w} for water--it's very, very important that you know this.1728

*K*_{w} is 1.0x10^{-14} at 25 degrees Celsius--always.1734

*OK, what does this mean?--it does have a meaning.*1743

*This means (and here is where it gets really, really important): in **any*, in *any* aqueous solution, at 25 degrees Celsius, no matter what is in it--no matter *what*--no matter what is in it (in the solution), the product of H^{+} and OH^{-} must always equal 1.0x10^{-14}.1753

*OK, we need to stop and think about this, because this is really important.*1811

*Any aqueous solution, no matter what is in it--sodium chloride, potassium phosphate, the complex ion, magnesium hydroxide, phosphoric acid, permangenic acid--anything--if I were to measure the hydrogen ion concentration and the hydroxide ion concentration, they have to--the products have to equal--**have* to equal 1.0x10^{-14}.1816

*That is what this says; at any given moment, an aqueous solution (in other words, water--something that is in water)--the product of the hydrogen ion and the OH*^{-} ion concentration always 1x10^{-14}.1842

*That doesn't mean that each one is 1x10*^{-7}; that is just for neutral water.1858

*It just says that, if...so this is equal to the H*^{+} concentration times the OH^{-} concentration--that means, if the H^{+} concentration rises, the OH^{-} concentration has to drop in order to retain the constant value of 1.0x10^{-14}.1863

*So again, they don't have to equal each other; their product has to be a constant.*1883

*If one goes up, the other goes down; that is the relationship--that is the mathematical relationship.*1888

*OK, their product has to be 10*^{-14}; and it is going to be very, very important in just a minute.1894

*So, let's just...if the concentrations do equal each other, we call it neutral.*1899

*If the hydrogen ion concentration is bigger than the OH*^{-} concentration, we call it acidic.1910

*That is where we get the idea of "acidic"; so, when you say that orange juice is acidic, that means that, if you take some orange juice, and if you measure the concentration of H*^{+} versus the concentration of OH^{-}, the H^{+} concentration is going to be higher.1919

*But, the product of the two concentrations is still going to be 10*^{-14}.1934

*If the H*^{+} concentration is less than the OH^{-} concentration, we call that solution basic.1940

*That's it; OK, so let's do an example.*1948

*Nice, simple math: Calculate H*^{+} and OH^{-} (I'm really, really sorry about my handwriting; I know) concentration for the following solutions.1956

*OK, let's see: Oh, "at 25 degrees Celsius."*1982

*A: We have: 1.5x10*^{-5} molarity OH^{-}.1991

*Well, we know what the concentration of OH*^{-} is; it's right there--it's 1.5x10^{-5}.2000

*Well, we know that the H*^{+} concentration, times the OH^{-} concentration, at 25 degrees Celsius, equals 1.0x10^{-14}, so we just plug it in to find the H^{+} concentration.2010

*H*^{+} concentration equals 1.0x10^{-14}, divided by the OH^{-} concentration; I have just done a simple division.2026

*1.0x10*^{-14} (ten to the negative fourteen, not fifteen), divided by OH^{-}, which is 1.5x10^{-5}, and I get a concentration of 6.7x10^{-10}.2038

*Which one is a higher concentration? OH*^{-} is 10^{-5}; H^{+} is 10^{-10}; the OH^{-} is a bigger concentration--the solution is basic.2058

*It will still hurt you, so don't...it doesn't mean that a base won't hurt you; an acidic solution will hurt you--a basic solution will also hurt you.*2072

*OK, the idea is: we want a neutral solution.*2080

*By the way, blood pH is just about...we will get to that in a minute; sorry about that.*2083

*OK, so let's do another one; let's do a 2 Molar solution--2.0 Molar solution of H*^{+}.2091

*Well, again, we have the H*^{+} concentration: in this case, a 2.0 Molar solution, a solution that contains 2 molarity H^{+}; that is the hydrogen ion concentration.2101

*Well, we know that the H*^{+} concentration, times the OH^{-} concentration, equals 1.0x10^{-14}; therefore, the OH^{-} concentration equals 1.0x10^{-14}, divided by 2.0, which was that.2112

*And we end up with an OH*^{-} concentration of 5.0x10^{-15}--a very, very small number; in this case, the hydrogen ion concentration hugely, hugely out-values the OH^{-} concentration.2132

*This is acidic--I would say highly acidic; that is it; very, very dangerous.*2149

*OK, let's do another one here.*2158

*Example 3 (or, actually, I'm not sure if it's Example 3, but...I'll just do an example): At 100 degrees Celsius, the K*_{w} equals 5.13x10^{-13}.2163

*So remember: we said that, at different temperatures, the equilibrium constant is different.*2183

*At 25 degrees Celsius, the K*_{w} is 1.0x10^{-14}; but if I change the temperature to 100 degrees Celsius, now it has gone up to 5.13x10^{-13}.2187

*It has gone up.*2199

*The question to you is: Is this reaction, a dissociation of water into H*^{+} plus OH^{-}--is it endo- or is it exo-thermic?2202

*OK, well, let's see what happens here: at 100 degrees Celsius, they are telling me that the K*_{w} equals 5.13x10^{-13}.2227

*They want to know if this reaction is endo- or exothermic.*2240

*Well, let's see what we did: we increased the temperature, and by increasing the temperature, we actually pushed the reaction forward, because that is what this K*_{w} is telling us.2243

*This K*_{w} of 5.13x10^{-13} is bigger than the 1.0x10^{-14}, right?--negative 13 is bigger than negative 14.2256

*That means that the equilibrium has actually shifted to the right; that means more product has been produced.*2267

*That means that the reaction shifted to the right when I increased the temperature.*2273

*Well, if it increases to the right when I increase the temperature, that means that temperature, or heat, must have been one of the reactants, because, if it's one of the reactants, and if I increase this temperature, the system is going to want to offset what I do to it (which is an increase) by decreasing the temperature.*2283

*Well, in order to decrease the temperature, it has to use up this extra heat that I have pumped into it.*2311

*In order to use up the heat, it has to shift the reaction that way; it has to produce more H and more OH*^{-}, which is exactly what it did, because now the equilibrium constant has gone up.2317

*When an equilibrium constant goes up, that means the reaction has moved to the right.*2329

*The numerator of the equilibrium constant has increased; so, in this case, because heat is on the left side of this equilibrium arrow, this is an endothermic reaction.*2333

*So notice, we used our discussion of equilibrium and Le Chatelier's Principle to actually find out, qualitatively, that this is an endothermic reaction.*2346

*In order to push this reaction forward--in order to produce more H*^{+} and more OH^{-}, I actually have to increase the temperature, which means that it is an endothermic reaction.2356

*I don't know what the ΔH is, but I do know that it is going to be positive for this reaction.*2368

*OK, B: what is the H*^{+} and OH^{-} concentration in a neutral solution at 100 degrees Celsius?2375

*Well, we just said that K*_{w} is equal to the concentration of OH^{-} times H^{+} at 100 degrees Celsius.2398

*We gave it already; it's 5.13x10*^{-13}.2407

*Well, in a neutral solution--go back a slide or two--neutral means that the OH*^{-} concentration is equal to the H^{+} concentration.2412

*So, if I call this X and call that X, I end up with: X squared is equal to 5.23x10*^{-13}.2425

*I get: X is equal to 7.16x10*^{-7} molarity.2435

*At 100 degrees Celsius, there is 7.16x10*^{-7} moles of H^{+} per liter of water, and 7.16x10^{-7} moles of OH^{-} per liter of water.2444

*That is all that is going on here; molarity is moles per liter.*2461

*OK, let's see what else we can do.*2466

*Now, we are going to close this discussion off by introducing the pH scale.*2474

*Now, I have to (I probably shouldn't say this, but I'm going to go ahead and do it, because I tell all my students this)...working with concentrations is perfectly valid--concentration is something we always work with; concentration of H*^{+}, concentration of OH^{-}, concentration of Cl^{-}...2479

*With acids and bases, because these concentrations tend to be really, really small, as you have noticed (you know, 10*^{-7}, 10^{-6}...), chemists have come up with a way to have numbers that actually are a little more easy to deal with--normal numbers like 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.2499

*So, they have created this thing called a pH scale.*2516

*Here is what it means: the p of anything is equal to negative log of the concentration of that anything.*2519

*So, in other words, if I talk about pH, that is equal to the negative log of the hydrogen ion concentration.*2532

*If I talk about pOH, that is equal to the negative log of the hydroxide ion concentration.*2540

*If I say pK*_{a} (so this p function means whatever it is...p of whatever--just take the negative log of that whatever), pK_{a} is the negative log of K_{a}.2546

*It is a way of taking numbers that are really odd, like 6.6x10*^{-3}, and turning them into actual numbers that you can deal with: 3.16--something like that.2562

*Personally, I don't care for it myself; I like dealing strictly with what we are dealing with.*2571

*We are dealing with concentrations; the equilibrium constant is expressed in concentrations; so, I think it is perfectly valid to deal with concentrations.*2578

*But again, this is so deeply entrenched in chemistry that literally nobody talks about concentration anymore when they speak about acid-base.*2588

*They speak about pH and pOH and pK*_{a}.2596

*That is the only reason; but they are actually the same thing.*2599

*It is not like we are introducing a new concept; we are just introducing a way of changing the numbers of concentrations that tend to be really small, and turning them into numbers that make more sense, that look better--not make more sense, just look better, aesthetically.*2605

*OK, so one more example: if I said, "What is the pCl*^{-}?" well, that is going to be the -log of the Cl^{-} concentration, if we happen to be talking about Cl^{-}.2618

*It is just a p function, and it is the negative log of the concentration we are talking about.*2630

*OK, let's do an example.*2636

*What is the pH and pOH of a solution which is 1.4x10*^{-3} Molar hydroxide?2642

*OK, well, they want the pH, and they want the pOH.*2668

*Well, I know the pH is the negative log of the hydrogen ion concentration; pOH is the negative log of the hydroxide ion concentration.*2672

*Well, they give me the hydroxide ion concentration, so why don't I just deal with that one first.*2681

*pOH is equal to negative log of the OH*^{-} concentration, which is negative log of 1.4x10^{-3}.2685

*I just stick in this in the calculator; I take the logarithm of it; I change the sign; and I end up with 2.85.*2699

*So notice, 2.85 is a much more attractive number than 1.4x10*^{-3}; at least, that is what many people think.2709

*I, personally, don't think so, but that is fine; I'm a chemist; we deal with pHs; we'll deal with pHs.*2716

*But, don't think it is something different; it is not.*2722

*If you want to, you are more than welcome to deal strictly in concentrations, and if you ever have to find a pH, just at the last minute take the negative log of it: no harm, no foul.*2724

*Now, we want the pH; well, the pH--we know that the H concentration, times the OH*^{-} concentration, is equal to 1.0x10^{-14}.2735

*The H*^{+} concentration equals 1.00x10^{-14} (well, let me just stick with 1.0; let's just do 2 significant figures here), divided by 1.4x10^{-3}.2748

*That is going to equal 7.2x10*^{-12}, and then, when I take the pH of that (that is going to be the negative log of the H concentration), -log of 7.2x10^{-12} equals 11.14.2767

*11.14 pH; 2.85 pOH: notice something--what is 11.14 plus 2.85? Yes, you guessed it--it's about 14.*2791

*Now, let's explain why.*2806

*We said that K*_{w} is equal to the hydrogen ion concentration, times the hydroxide ion concentration.2812

*Let's take the negative log of both sides.*2822

*-log of K*_{w} is equal to -log of this whole thing: H^{+}, OH^{-}...2824

*Well, -log of K*_{w} is pK_{w}; that is it--equals -log of H^{+}, right?--plus the negative log of OH^{-}.2836

*You end up with pH plus pOH; well, pK*_{w} is the negative log of 1.0x10^{-14}, equals pH plus pOH.2860

*Because K*_{w} is 1.0x10^{-14}, the negative log of that is 14; that is our final relationship.2884

*So, for any aqueous solution at 25 degrees Celsius, pH plus the pOH of the solution equals 14.*2895

*This is just a restatement of the fact that the H*^{+} concentration times the OH^{-} concentration equals 1.0x10^{-14}.2914

*These are the same thing, except one deals with these numbers (these scientific notation numbers, the small ones); one deals with numbers that are a little bit more tractable.*2928

*In any aqueous solution, the product--the **product*--of the hydrogen ion and hydroxide concentrations is 10^{-14}.2937

*The pH plus the pOH of that solution is equal to 14.*2948

*OK, so we have introduced acid and base; actually, we have introduced acid mostly; we have talked a little bit about base, mostly in the context of conjugate base.*2955

*We have talked about the dissociation of acids and what happens in water.*2964

*We have introduced the idea of an equilibrium constant for that acid dissociation.*2970

*We have introduced the idea of water acting as both acid and base, an amphoteric substance.*2976

*And we have introduced the very, very important property that the product of the hydrogen ion and the hydroxide ion concentrations is 10*^{-14}; or, if we want to use the p scale, we can talk about the pH and the pOH equaling 14 for any aqueous solution at 25 degrees Celsius.2981

*Next time, we will continue our discussion of acids and bases, and we will actually get into some of the more equilibrium-type problems that are involved.*3000

*Thank you for joining us here at Educator.com and AP Chemistry.*3008

*We'll see you next time; goodbye.*3010

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

*In our last discussion, we introduced the basic notions of acids and bases; actually, we didn't talk about bases so much--we will actually talk about those mostly in the next lesson.*0004

*We introduced the notion of an acid, and we said an acid is something that has hydrogen ions that it can give up.*0014

*Often, what happens is: you will take a particular acid, like hydrogen chloride, drop it in water, and that will dissociate, and it will separate into hydrogen ions and chloride ions.*0020

*Or, it may not separate at all, which is part of the topic we are going to be talking about today: weak acids.*0030

*We talked a little bit about the pH scales, the relationship between the hydrogen ion concentration and the hydroxide ion concentration in water.*0035

*We said that each one of them is 10*^{-7}, to come up with a total of 10^{-14}.0045

*Right now, in a minute, I am actually going to write down some of the basic things we discussed, and then we are just going to launch right into a series of problems on how to handle the pHs of weak base solutions.*0050

*A weak base--I'm sorry, a weak acid--is something that doesn't dissociate completely; so let's just jump in and get started and see what we can work out.*0064

*OK, so, last time, we said that, remember, we have water, which acts as both an acid and a base.*0073

*H*_{2}O--it dissociates into H^{+} plus OH^{-}.0080

*We said that the equilibrium constant for this--because this is liquid water, this is aqueous, this is aqueous--so our K*_{a} for this, which we ended up calling K_{w} for water, was just the hydrogen ion concentration, times the hydroxide ion concentration.0087

*It always equals 1.0x10*^{-14} (let me make that 14 a little bit more clear), at 25 degrees Celsius.0106

*So, the K*_{eq}, the K_{a}...these things change at temperature, but normally, we are talking about 25 degrees Celsius.0119

*This means that any aqueous solution, no matter what is in it, the hydrogen ion concentration times the hydroxide ion concentration is always going to equal 1.0x10*^{-14}.0125

*That is very, very convenient, because, no matter what one of these is, we can always find the other--and that is the whole idea.*0137

*Today, we are going to talk about weak acids; and weak acids are ones that actually don't dissociate completely.*0145

*Let's talk about a strong acid first: hydrogen chloride--when we drop this is water, the dissociation reaction for this is H*^{+} + Cl^{-}; it's a strong acid--strong means it completely dissociates in water, just like sodium chloride (salt) or anything else that is just completely soluble.0153

*Well, as it turns out, a weak acid doesn't do that; a weak acid behaves like this: like we said last time, HA--it will react a little bit with water, and it will actually establish a little bit of an equilibrium.*0174

*What you will end up getting is something like H*_{3}O^{+}, plus the conjugate base of this weak acid.0187

*Another way of writing this, which I personally prefer, is just the standard dissociation, without mentioning the water: H*^{+} + A^{-}.0195

*Now, we said there is a K*_{a} associated with this; and the K_{a} is equal to the H^{+} concentration.0205

*And again, sometimes I will use the square brackets; sometimes I won't, simply to save myself a little bit of time.*0210

*But, these are all concentrations in moles per liter.*0216

*A*^{-} over HA: now, most of the time, for the weak acids, these K_{a}s are going to be very, very small.0219

*They are going to be things on the order of -4, -5, -8, -10, even -12 and -15.*0227

*What that means: because a K*_{a} is small, a small equilibrium constant means that the reaction has not moved very far forward.0235

*That means most of it is here; so, if I took, let's say, some hydrofluoric acid (which is not really hydrofluoric acid--it's hydrogen fluoride, and I dropped it in water to create a solution of hydrofluoric acid), it doesn't actually dissociate too much.*0244

*Yes, it does produce a little bit of this and this; it does come apart a little bit; but it is a weak acid, which means that most of the hydrofluoric acid will be floating around as HA, as the entire species, in aqueous solution.*0260

*You still can't see the H*_{a} in there; it is still a homogeneous solution; but it is solvated, so it isn't a hydrogen...0274

*This is why, when we talk about acids, acids are always that particular compound, found in water.*0281

*We just automatically presume that we are talking about an aqueous solution.*0289

*As it turns out, hydrogen fluoride, when you make it, is not an acid--it's just hydrogen fluoride.*0293

*Things become acids when you drop them in water; but, because we always find them in water, because we are always talking about aqueous chemistry, we just often refer to them as acids.*0298

*A weak acid is one that doesn't dissociate very much.*0307

*Now, we need to find a way to actually find the pH of these solutions; and that is pretty much what we are going to be doing today, in this lesson; we are just going to be doing several problems.*0311

*Some of them are exactly like the ones before--maybe slightly different; but the idea is to develop a sense of turning what we have discussed (as far as the basic definitions of pH acid)--now we want to quantify it--now we want to use some of the math behind it.*0319

*What you are going to discover is: these problems that we are going to do are exactly like the equilibrium problems that we did in the previous chapter.*0336

*This whole ICE chart--the Initial, the Change, the Equilibrium concentration--that is going to happen over and over and over and over again.*0343

*You are going to be doing a ton of these; and I am specifically concentrating on doing a number of these problems, because it is in the process of doing these problems that we actually get a sense of what is going on with these acid-base chemistries.*0350

*There is going to be a lot of commentary, a lot of talk.*0362

*This is where most of the deep chemistry is happening; and, if you understand this stuff, the AP exam will be an absolute piece of cake, I promise you.*0365

*This is where most of the problem takes place, and we want to make sure we really understand it, because it's beautiful chemistry; it's applicable chemistry; and...well, let's just do it.*0375

*OK, so we are going to start off, actually, with a strong acid, just to get a sense.*0385

*The first example is going to be: Calculate the pH of a 0.15 Molar nitric acid solution (nitric acid is HNO*_{3}), and the pH of a 1.5x10^{-10} Molar HCl solution.0389

*OK, so let's do the first part: let's calculate the pH of a .15 Molar nitric acid solution.*0420

*All right, when we do these acid-base problems (strong acid, weak acid...later on when we work with weak base, strong base; and later on when we work with buffers), it is always the same thing.*0426

*We want to develop a nice, systematic approach, because, as the species in the solution get more complicated, it's all that much more important to keep track of what is going on.*0437

*We are going to use these problems to develop a sense of the chemistry; we are going to let the chemistry decide how the math works.*0446

*You need to understand the chemistry; if you don't understand the chemistry, the math is irrelevant.*0451

*It will mean nothing to you; but, if you understand the chemistry, the math is a piece of cake.*0456

*In fact, it is pure instinct; there is nothing here, mathematically, that you don't know or can't follow--there is nothing mysterious happening.*0460

*We just want to make sure you understand what is happening.*0467

*OK, so first of all, we notice that it is a strong acid; a strong acid means full dissociation--that means this is going to happen.*0470

*That means HNO*_{3} is going to completely dissociate into H^{+} ion plus NO_{3}^{-}; that means H^{+} is going to be floating around freely, NO_{3} is going to be floating around freely, and there is going to be nothing that is left.0479

*That is why there is an arrow pointing one way--this is not an equilibrium.*0491

*Strong acid--full dissociation.*0494

*So now, what we ask ourselves--**always* the first question we ask ourselves: What is the major species--list the major species in the solution.0497

*Well, the major species in solution are...this will decide what the chemistry is, because there is always going to be something that is going to dominate the chemistry in the solution.*0504

*Well, major species: we have...of course, we have H*_{2}O--that is our solvent; we have H^{+} floating around; and we have NO_{3}^{-} floating around.0517

*That is it; well, they are asking for the pH--what is pH?--remember, we said that pH is just a different way of dealing with the hydrogen ion concentration.*0527

*It is the negative log of the hydrogen ion concentration, in moles per liter.*0537

*Well, look at the major species we have in solution: we have H*_{2}O; we have H^{+} and NO_{3}.0543

*There are two sources of hydrogen ion: one of the sources is the dissociation of the nitric acid itself; well, the other source is the dissociation of water (remember, water also comes apart and produces a little bit of this).*0548

*But, because this is such a strong acid that all of these hydrogen ions...actually, what it is going to end up doing is: you know that this is (under normal conditions) just 10*^{-7} molarity--the hydrogen ion concentration.0564

*It is 7 orders of magnitude smaller than this; it is virtually...you could completely neglect it; as a matter of fact, because this dissociates (remember Le Chatelier's Principle?), all of the hydrogen ion that is floating around from the dissociation of this strong acid actually pushes this equilibrium this way.*0579

*So, there is even less hydrogen ion coming from water itself.*0598

*For all practical purposes, we can ignore that; the dominant species here is this; our problem is already done.*0602

*It says we have .15 Molar HNO*_{3}; well, strong acid...0.15 moles per liter HNO_{3} produces .15 Molar H^{+} and .15 Molar NO_{3}^{-}, because it's 1:1, 1:1; that is the ratio.0611

*This is easy; the pH is just minus the logarithm of 0.15; that is it, because the concentration of H*^{+} in a .15 Molar HNO_{3} solution is .15 moles per liter of H^{+}.0626

*That is the free thing floating around; and we get 0.82.*0642

*That is it--nice and simple: strong acid--all you have to do is use the molarity of that acid.*0648

*Just take the negative log of it; you automatically have the pH, because you have full dissociation.*0655

*Now, let's do the second part, B: Now, we have a 1.5x10 to the negative (11 or 10?--what did we say here)...1.5x10*^{-10} Molar HCl solution.0660

*OK, again, let's see what the major species are in solution.*0676

*Well, again, we have HCl; it's a strong acid; it breaks up into H*^{+} + Cl^{-}; well, 1.5x10^{-10} Molar HCl is full dissociation.0680

*That implies that it is 1.5x10*^{-10} Molar H^{+}.0696

*Well, think about this for a minute (and again, this is why you don't just want to jump into the math--you are going to get the answer wrong if you just all of a sudden take the negative log of this, like we did the first one): look at this number.*0701

*1.5x10*^{-10}: that is a really tiny number; compared to the auto-ionization of water, water is 1.0x10^{-7}; 1.0x10^{-7} is three orders of magnitude bigger than 1.5x10^{-10}.0712

*So, virtually, for all practical purposes, this hydrogen ion concentration that comes from this hydrochloric acid, which is a strong acid, is completely negligible.*0730

*The reason it is negligible is because the concentration of the H is so little.*0741

*Yes, it's a strong acid, so our instinct tells us, "Well, if it's a strong acid, then it must be the dominant force in the solution"; it's not.*0746

*There is so little of it that it doesn't have any effect; so for all practical purposes, the H*^{+} concentration here is what it would be in water at 25 degrees Celsius--which is 10^{-7}.0753

*Well, the negative log of 10*^{-7} is 7.0768

*So, our pH is 7 in this particular case.*0773

*We took a look at the chemistry; we took a look at the numbers involved in the chemistry--the concentrations; that helps us decide what is going to happen next--what math to use.*0777

*Here, .15 is a pretty high concentration; 1.5x10*^{-10}: that is three orders of magnitude less than the normal concentration of just standard water--the hydrogen ion concentration in standard water.0787

*We can ignore it, despite the fact that it is a strong acid.*0801

*Don't let names fool you; let the chemistry tell you what is happening--pull back from the problem and think about what is going on.*0804

*Science and chemistry--well, all of science, not just chemistry...there is a lot happening; you have to be able to keep track of multiple things.*0811

*It is not just about solving the problem; solving the problem is easy if you know what is happening.*0820

*If you don't know what is happening, solving the problem will always be a nightmare; you will always be a victim--you will always be sort of a slave to algorithmic procedures.*0826

*And once something--once a problem deviates from that algorithmic procedure, you are going to be lost.*0835

*You don't want to be lost; you want to be able to think--always think; that is the idea.*0840

*OK, so now, let's go on to weak bases; and this is where the mathematics and the situation actually gets interesting.*0844

*We are going to present--we are going to do the same thing: we are going to present a nice, systematic approach for all of these problems.*0852

*Weak acid, weak base; strong acid, strong base; buffer solution; the same steps are the habit we want to get into.*0857

*OK, so let's do Example 2: Calculate the pH of a 1.5 Molar hydrofluoric acid solution.*0865

*So again, we could say hydrogen fluoride solution; it's the same thing--it just means you have taken hydrogen fluoride, and you have dropped it into some water, and you have created a solution.*0888

*Hydrogen fluoride is your solute; water is your solvent; now, it becomes an acid.*0897

*Even though we speak about HF being hydrofluoric acid, it's only an acid when it has been dropped in water; it's very important to remember that.*0902

*OK, the K*_{a} for hydrofluoric acid (because it is a weak acid, it has a K_{a}; remember, strong acids don't have K_{a}--the equilibrium lies far to the right, so the denominator is virtually 0) is 7.2x10^{-4}.0911

*That is a kind of a small number: 7.2x10*^{-4}; well, what does a small K_{a} mean?0927

*A small K*_{a}--it means that there is very little dissociation--that is the whole idea.0935

*You need to know what these things mean; they are not just numbers and parentheses and concentrations; it means something, physically.*0941

*It means there is very little dissociation.*0948

*Very little dissociation--that means that HF...the dissociation of HF into H*^{+} + F^{-}...that means mostly, if we looked at this, it's going to be mostly hydrogen fluoride, not hydrogen ion and fluoride ion.0952

*That is what the small K*_{a} means.0966

*OK, in other words, it is mostly here; the equilibrium is mostly here; in solution, it's still just HF floating around, not H*^{+} and F^{-} floating around freely; so, we will say mostly here.0969

*Well, the next step is...let's list our major species to see what is going on.*0981

*Let me actually list the major species here: our major species in this solution are going to be...again, because there is very little dissociation, it's a weak acid, you are going to find mostly HF, and you are going to find H*_{2}O.0986

*Both of these are weak acids: HF, as we said, has a K*_{a} of 7.2x10^{-4}; well, H_{2}O has a K_{a}, which we call K_{w}; it is 1.0x10^{-14}.1001

*Well, between 10*^{-4} and 10^{-14}, this is 10 orders of magnitude bigger; it's huge--this is virtually nonexistent compared to this.1018

*Because this is so much bigger, we can ignore any contribution that H*_{2}O does in its dissociation.1027

*Its contribution of hydrogen ion is completely negligible--not even measurable, to be honest with you.*1035

*This is going to dominate the equilibrium in the solution.*1040

*So now, we want to see: at equilibrium, when everything has stopped and settled down, what is the pH of that solution?*1043

*Well, here is how we do it: We write down the equilibrium expression, HF goes to H*^{+} + F^{-}; we have I, we have C, and we have E (our initial concentration, change in concentration, and equilibrium concentration).1050

*Our initial concentration was 1.5 Molar; there is, before the system has come to equilibrium, 0 and 0; so again, imagine, when you are doing these problems--you are taking something; you are dropping it in water; and it is before anything happens.*1065

*Before it starts to come apart, that is the initial conditions.*1084

*Now, HF is going to dissociate a little bit, so that means a little bit of it is going to disappear; and for every amount that disappears, that is how much H and F show up, right?*1088

*1:1, 1:1...when 1 molecule of this breaks up, it produces 1 ion of H*^{+} and 1 ion of F^{-}.1100

*That is why we have -x, +x, +x, when they put +'s here.*1108

*And now, our equilibrium concentration is just adding these: 1.5-x; when everything has come to a stop, that is how much we have of the HF; that is how much we have of the H; and that is how much we have of the F*^{-}.1113

*Now, we can write our expression.*1126

*K*_{a}, which is equal to 7.2x10^{-4}, is equal to...well, it's equal to the H^{+} concentration, times the F^{-} concentration, over the HF concentration, at equilibrium.1128

*That equals x, times x, divided by 1.5-x.*1145

*Now, let me rewrite this over here; so let's...part of one equation: 7.2x10*^{-4}; now, what we want to do...again, we want the pH.1155

*We want to take hydrogen ion concentration and take the negative log of it; so we are looking for x.*1166

*We want to solve this equation for x.*1172

*Now notice, we are going to end up with a quadratic equation here; there are a couple of ways that we can handle this.*1175

*We can go ahead and (because you have graphing calculators, presumably--most of you) you can go ahead and use your graphing calculator to find the roots of this quadratic equation.*1180

*Or, you could solve the quadratic equation--turn this into a quadratic--multiply through and turn it into a quadratic--use the -b, plus or minus radical b*^{2} minus 4ac over 2a to solve for x.1189

*Or, we can actually simplify the procedure to make this a little better for us.*1201

*And again, the simplification is based on standing back and taking a look at the chemistry.*1205

*This is a weak acid; there is going to be very little dissociation here.*1210

*Even though there is an equilibrium, there is going to be very little, actually, of the H*^{+} concentration--so little, in fact, that compared to 1.5 Molar--compared to this, this is going to be so tiny that the 1.5 probably won't even notice that it is gone.1214

*So, we can simplify this expression by writing it as: 7.2x10*^{-4} equals x^{2} over 1.5.1230

*Now, when we find the value of x, we need to check to see if our assumption, if our approximation, is valid; and we will show you how to do that in just a minute.*1245

*Let's go ahead and just solve this first, and then we'll check to see if it's valid.*1253

*When we multiply through, we get x*^{2} = 1.08x10^{-3}, and we get x=0.033 moles per liter.1257

*Well, this is the hydrogen ion concentration, because that was x in our ICE chart.*1270

*Now, we need to check if our approximation--our eliminating x from the denominator here--was valid.*1275

*Here is how you do it: most K*_{a}s that are listed in tables have an error of about plus or minus 5%.1280

*Now, this value, .033, 0.033--if it is 5% or less than the original concentration that you subtracted from...1.5 times 100; you want to calculate a percentage; if this is 5% or less of this number, then our approximation is valid, meaning the difference is not really going to be noticeable--that your numbers are actually pretty good.*1289

*If it is above about 5 or 6 percent, that means you can't make this approximation--you have to solve the quadratic equation.*1322

*So again, we are taking a look at the chemistry, and we are letting the K*_{a} tell us that we have a weak acid--very little dissociation.1329

*So, this x, compared to 1.5, is probably going to be very, very tiny--so much so that we are probably not even going to notice a difference: 1.5, 1.49--it's going to be so small.*1337

*Well, if it falls into the 5% rule, then it's valid; if it doesn't, then it's not valid; so the first thing to do is go ahead and simplify, and check to see if this is the case.*1349

*In this particular case, we get 2.2%; well, 2.2% is definitely less than 5%, which means our approximation is valid, and we can use this value of x; it's perfectly good.*1359

*So, now that we have that, we have our H*^{+} concentration of .033, our pH equals -log of 0.033, is equal to 1.48; there you go.1372

*Now, I personally (just a little aside) don't care for pH; I think once you have found the hydrogen ion concentration, you are talking about concentration.*1387

*Taking a number and fiddling with it so that it has a more attractive number, like 1.48--I personally don't think 1.48 is more attractive or easier to deal with than .033, or 3.3x10*^{-2}.1396

*It's a perfectly valid number; here, we know we're talking about concentration; pH--it's a little weird, because again, the lower the pH, the more acidic.*1409

*If you remember, you have a pH scale that runs from 0 to 14; 7 is neutral--that is the pH of plain water.*1419

*Below 7, we are talking about an acidic solution; above 7, you are talking about a basic solution.*1427

*The lower the number (6, 5, 4, 3, 2, 1)--that is more acidic; what that means is that there is a greater concentration of hydrogen ion.*1434

*It's a little strange: a lower number means more powerful; but again, this has just become part and parcel of the standard chemical practice in the industry, across the board.*1444

*But again, as long as you understand what is happening, that is what is important.*1456

*OK, so let's do another one; this time, let's calculate...let's see; this is going to be Example 3.*1461

*Let's calculate the pH of a 0.15 (this time, 0.15) Molar solution of HClO, hydrogen hypochloride--otherwise known as hypochlorous acid.*1472

*The K*_{a} for this equals 3.5x10^{-8}.1501

*The first you want to notice is: look at the K*_{a}; this is a very, very small number--that means we are talking about a very weak acid, very little dissociation.1507

*When you take hydrogen hypochloride and drop it into water, it's going to stay hydrogen hypochloride; it's not going to come apart into hydrogen ion and hypochloride ion too much.*1515

*A little bit, it will; but not too much.*1526

*The fact that we have a number that we can measure means there is some dissociation; but again, this says that it is a weak dissociation.*1529

*OK, so now let's see which equilibrium is actually going to dominate in this.*1536

*We have two sources of hydrogen ion: we have the HClO, which breaks up into H*^{+} + ClO^{-}, potentially; and the K_{a} for this, as we just said, is 3.5x10^{-8}.1542

*Then, we have H*_{2}O; that is the major species in the water; that is H^{+} + OH^{-}, but again, the K_{w} for this is 1.0x10 to the (oops, not 14) negative 14.1558

*Well, 10 to the -8--even though it is a small number, it is a lot bigger than 10 to the -14; 6 orders of magnitude bigger.*1577

*The dominant reaction here, the dominant equilibrium, is this one.*1584

*This is virtually not even noticeable.*1589

*So, let's go ahead and do the problem.*1596

*Since the HClO equilibrium is going to dominate, we are going to write HClO, and I just am one of those people that likes to write everything--so, H*^{+} + ClO^{-} (I know that it's a little...I'm just writing it over here, but I just like to make sure that everything is clear).1598

*This is initial, change, equilibrium (and let's erase some of these stray lines here); so, our initial concentration of HClO, before the system has come to equilibrium--when we just drop it in before it has come to equilibrium--this is 0.15 Molar.*1619

*There is no hydrogen ion, and there is no hypochloride ion.*1635

*Well, a certain amount of it dissociates; and it produces, for every mole that (we're not going to have that; we want this to be clean; +x) dissociates, it produces a mole of H*^{+}; it produces a mole of ClO^{-}.1639

*Equilibrium concentration is 0.15-x; this is +x; this is +x; and now, we can go ahead and write our equilibrium.*1658

*Our K*_{a} is our hydrogen ion concentration, times our hypochloride ion concentration, over our hydrogen hypochloride.1671

*When we do this, in terms of numbers, we get: 3.5x10*^{-8}, which is our K_{a}; it is equal to x times x, over 0.15-x.1681

*And again, we want to simplify, so this is going to be approximately equal to x squared, over 0.15.*1697

*Then, we'll check to see if that approximation is actually valid.*1705

*When we solve this, we get x*^{2} is equal to 5.25x10^{-9}, and we get that the x, which is equal to the hydrogen ion concentration, is equal to 7.24x10^{-5}.1709

*Now, let's check the validity; so I'll write "check the validity of our approximation"--in other words, what we did here.*1728

*We'll take 7.24x10*^{-5}, and we'll divide it by the initial concentration, 0.15; so, when we do that and multiply by 100, it equals (let me drop it down a little bit here) 0.048%.1735

*Wow, look at that; that means only 0.048% is actually dissociated; that is a very, very, very weak acid--virtually none of it has come apart.*1753

*This is clearly below 5%; our approximation is good; so we can use this number, and when we take the negative log of 7.24x10*^{-5}, we end up with (let's see what we ended up with here) 4.14.1765

*4.14--and this our pH, and this is our hydrogen ion concentration.*1788

*Again, my preference is for the actual concentration; pH is just a number.*1794

*OK, that is good; so hopefully, you are getting a sense of what it is that we are doing; we are looking at K*_{a}; we're choosing the major species; we're deciding which of those major species is going to dominate in the solution.1800

*We know what chemistry is going to dominate; usually there is one species, one equilibrium, that dominates--everything else can be ignored.*1812

*And then, we just write out our ICE chart: Initial concentration, the Change equilibrium, and then we set it equal to the equilibrium constant, and we just deal with the math at that point; it's a simple algebra problem.*1820

*OK, now let's do a problem where we have a mixture of weak acids.*1831

*Same exact thing: it's going to be more species in water, but one of them is going to dominate; so let's do this.*1834

*Example 4: Calculate the pH of a solution that contains 1.2 moles per liter of hydrogen cyanide, or hydrocyanic acid, whose K*_{a} is 6.2x10^{-10}--very, very weak--and 4.0 Molar HNO_{2} (hydrogen nitrite or nitrous acid), whose K_{a} is equal to 4.0x10^{-4}.1843

*Also, calculate the concentration of the cyanide ion at equilibrium.*1902

*OK, so we want to find the pH; so we have a mixture--we have this certain amount of water; we drop in some 4-Molar hydrogen nitrite, or nitrous acid; we drop in 1.2-Molar hydrocyanic acid; we want to know what the pH of the solution is, and we want to know what the concentration of the free cyanide ion is in that solution.*1925

*OK, let's list our major species with their particular equilibriums.*1949

*Well, let me go back to blue here; so, major species in water.*1955

*That is what we are doing; we always want to do the major species to decide what chemistry is going to dominate.*1963

*Well, HCn is a weak acid; that means it is mostly HCn--it hasn't dissociated.*1967

*A strong acid dissociates--weak acids, not very much.*1973

*We also have HNO*_{2} floating around in that, and we have H_{2}O.1977

*Well, the HCn equilibrium (H*^{+} + Cn^{-})--we said that the K_{a} is equal to 6.2x10^{-10}.1983

*The (oops, wow, look at that; that is a crazy line; OK) HNO*_{2} equilibrium (NO_{2}^{-})--the K_{a} for this one is equal to 4.0x10^{-4}.1996

*And of course, we have the (oh, here we go again; wow)...last but not least, we have the H*_{2}O equilibrium, which is also another source of hydrogen ion, plus OH^{-}; the K_{w} equals 1.0x10^{-14}.2022

*So now, let's compare: 10*^{-10}; 10^{-4}; 10^{-14}.2042

*We can virtually ignore the 10*^{-14}--it's too tiny.2047

*Between 10*^{-4} and 10^{-10}, this one--the HNO_{2}; this is going to dominate the chemistry in this solution.2050

*Because that is going to dominate, we can ignore any contribution of hydrogen ion from hydrogen cyanide, and we can ignore any contribution of H*^{+} from water.2058

*And again, we are doing this because there are three sources of hydrogen ion: some can come from the HCn; some can come from HNO*_{2}; some can come from H_{2}O.2069

*But, the K*_{a} of the HCn and the H_{2}O are so tiny that they are negligible, so this is what controls the chemistry of the solution; that is the chemistry we concentrate on.2078

*Therefore, let's go with HNO*_{2}: H^{+} + NO_{2}^{-}; now, you are going to do the same thing that you did before.2089

*You are going to make a little ICE chart: Initial, Change, Equilibrium; you are going to do your simplification; you are going to check it; everything is going to be fine; you are going to end up with a pH of...*2103

*I hope you actually run through and do this, based on the previous two examples.*2113

*So, for this one, you are going to end up with a pH of 1.40.*2118

*Now, the question is (the second part): How do we find the cyanide ion concentration?*2126

*OK, let's write down the cyanide equilibrium; we need to find the cyanide ion concentration, so we need to work with the cyanide equilibrium.*2132

*HCn goes to H*^{+} + Cn^{-}.2141

*K*_{a} equals 6.2x10^{-10}.2147

*Now, let's stop and think about what this means.*2152

*This is saying that, in a given solution, where you have cyanide ion, hydrogen ion, and HCn (hydrogen cyanide) floating around in solution--these three species floating around in some concentration each--the equilibrium concentration (in other words, the concentration of this, times the concentration of that, divided by the concentration of this) equals 6.2x10*^{-10}.2156

*Remember, that is what an equilibrium expression is: equilibrium positions can change, but the constant stays the same--the relationship among these three--that is what equals this.*2184

*It doesn't matter where these H*^{+} come from; this is talking about: at any given moment, if you have cyanide, hydrogen ion, and hydrocyanic acid in solution, the concentration of this times that, divided by the concentration of this, always equals that.2196

*This gives us a way to find it; it doesn't matter where these H*^{+} come from--they can come from HCn, or they can come from any other source, *any* other source.2215

*In this case, the primary source of the hydrogen ion is the nitrous acid.*2226

*It is the dominant species in the water; it is the one that is going to give up its hydrogen ion and suppress the others.*2232

*So, at equilibrium, there is a certain concentration of hydrogen ion, and that concentration of hydrogen ion was the 1.4--well, the pH was 1.4; the 1.4 came from the negative log of the hydrogen ion concentration.*2239

*Now, let's go ahead and work this out.*2257

*This says...so we know that the K*_{a} is equal to H^{+} concentration, times Cn^{-} concentration, over HCn concentration.2263

*That equals 6.2x10*^{-8}; this is the equilibrium expression.2278

*Well, let's see: let's go ahead...so at equilibrium, we have...not 6.2x10*^{-8}; we have 6.2x10^{-10}; yes, that is 10.2285

*That means 6.2x10*^{-10}; now, we go back on our ICE chart, and we read off this, this, and this.2300

*We ended up with: our equilibrium concentration is 0.04.*2310

*That was the hydrogen ion concentration that gave us the pH of 1.4.*2321

*We have the Cn*^{-} concentration, and, at equilibrium, the HCn concentration was 1.2-x.2325

*OK, again, we are just dealing with a basic math problem, but as it turns out, we have a little bit of a problem; we have this x here...smaller than normal...suppresses the HCn dissociation...let's see here.*2335

*Let me see, how do I want to go ahead and explain this to make it most reasonable?*2358

*We have that; we have that; we have the cyanide concentration; OK.*2364

*Let me erase something here.*2369

*OK, so let me rewrite this here: HCn, H*^{+} + Cn^{-}.2378

*Now, at equilibrium, we started off with 1.2 Molar of the HCn, right?*2389

*This was 1.2 Molar.*2404

*Now, we have the H*^{+} concentration, which is 0.04; this is the equilibrium concentration.2407

*This is what we want--this Cn*^{-} concentration.2412

*Now, it's true that, of the HCn, some of it will dissociate; so the fact of the matter is, the equilibrium concentration is going to be 1.2-x, but again, remember what we said: the dominant equilibrium here was the nitrous acid.*2417

*It is so dominant that it is actually going to suppress this dissociation.*2437

*Anything that HCn would actually produce, any H*^{+}, is virtually nonexistent, because not only was it small to begin with, but because there is so much of the H^{+} from the nitrous acid, it's actually going to push this even more that way, so there is going to be even less.2442

*So, for all practical purposes, we don't even have to worry about any HCn dissociating.*2458

*Therefore, our final equilibrium will end up actually being: 6.2x10*^{-10} is equal to 0.04 (which was the hydrogen ion concentration we found from the HNO_{2} dissociation), times the Cn^{-} concentration, over the HCn concentration (which was 1.2 Molar).2464

*And then, when we solve for this, we end up getting a Cn*^{-} concentration (and this time, I will use the square brackets): 1.86x10^{-8} Molar.2490

*As you can see, 1.86x10*^{-8} Molar is so tiny--that means virtually...there is virtually no cyanide ion floating around.2504

*That is based on this equilibrium.*2515

*So again, if you have a mixture of acids, one of those acids will dominate the equilibrium.*2518

*If you are asked for the concentration of a species from the other acid, the one that is not dominant, you can still use...you just write out its equilibrium, and it just means that--in this particular case, we wanted cyanide; well, the equilibrium expression for HCn says that cyanide concentration, times H*^{+}, divided by this, equals this.2524

*Well, we have had this from what we just calculated from the dominant species; we have this, which is the original concentration.*2546

*And then, all we have to do is solve for this, because we already have this.*2555

*1, 2, 3, 4; we have 1, 2, 3 of them; we solve for the fourth.*2559

*There you go; I hope that made sense.*2564

*OK, let's see: we have taken a look at some weak acid problems; we have listed the major species; we take a look at the major species to decide which one is dominant.*2569

*And usually, in this particular case (for weak acids), we just see which one has the higher K*_{a} value.2582

*Whichever one has the higher K*_{a} value, that equilibrium, that acid, will dominate the chemistry of the solution.2588

*We write its equation; we list its initial concentration, the change in the concentration, and the equilibrium concentration; and then, we put the equilibrium concentrations, in terms of x, into the equilibrium expression, and we solve for x.*2595

*More often than not, they are going to ask for calculating the hydrogen ion concentration or, actually, the pH.*2610

*More often than not, they will just straight-out ask for the pH.*2616

*So, thank you for joining us here at Educator.com and AP Chemistry for weak acids.*2619

*Our next topic we are going to be discussing is going to be percent dissociation; we are going to spend a fair amount of time on weak bases.*2624

*Take care; goodbye.*2631

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

*Today, we are going to continue our discussion of acids and bases--a profoundly important topic.*0004

*We are going to be talking a little bit about percent dissociation, sort of a continuation of our weak acid topic from last lesson.*0009

*But mostly, we are going to be talking about strong and weak bases; and, as you are going to see, how you handle the calculations is exactly like we did for weak acids.*0017

*Anyway, let's just jump in and talk about percent dissociation, and move forward and see what we can do.*0027

*So again, the idea is to do a fair number of problems to get you comfortable with what is going on--comfortable with the chemistry--how to handle this intuitively and turn that intuition into mathematics.*0033

*OK, so weak acids: remember, we said that a weak acid is some species, like for example, let's say, hydrogen fluoride (which was an example that we did in the last lesson): that dissociates a little bit into H*^{+} and F^{-}.0045

*Now, not too much; again, that is the whole idea behind a weak acid.*0062

*Well, there is this notion called percent dissociation; we want to know how much of the original HF actually came apart--how much of it dissociated.*0065

*Well, a percent is always the same thing; a percent is always the part over the whole, times 100.*0075

*So, by definition, our percent dissociation is equal to the amount dissociated (and the amount dissociated is always this amount, or this amount, because again, if .2 moles dissociates of this, well, .2 moles is produced of that and .2 moles is produced of that...so it's always the amount dissociated), divided by the initial amount (or the initial concentration--either way--the initial concentration); it's the part over the whole, times 100.*0080

*That is it; a percent dissociation--you are just measuring: "To what extent has this weak acid actually come apart? 10%? 5%? 3%? 2%?"*0121

*Strong acids--100% dissociation; percent dissociation for an acid, a strong acid, is 100%.*0132

*You are going to find, in general, for weak acids, anywhere from about .5% up to maybe about 5 or 6% for a weak acid.*0138

*Like, for example, your vinegar solution: you are talking about acetic acid, which is maybe 2% dissociated; it's very little.*0146

*It is actually kind of interesting--with such little dissociation, and yet, it has that really, really strong, strong acidic quality to it.*0156

*OK, as a quick example, let's think about the last example that we did when we talked about the hydrogen fluoride.*0164

*Remember, we found that the hydrogen ion concentration for the fluoride was 0.033 Molar, in the last lesson.*0173

*Well, the initial hydrogen fluoride concentration that we started with (so initial is a little 0 down at the bottom)--it was 1.5 Molar.*0183

*Therefore, our percent dissociation is equal to 0.033 over 1.5, times 100%; you ended up with 2.2 percent.*0193

*That means, when we stuck that 1.5 Molar hydrogen fluoride solution...well, when we have that 1.5 Molar hydrogen fluoride solution...only 2.2% of the hydrogen fluoride has dissociated into H*^{+} and F^{-}.0207

*That means the other 97.8% is still hydrogen fluoride, floating around in solution--pure hydrogen fluoride, not hydrofluoric acid.*0225

*Acid is when it is this way; but again, we just sort of have become accustomed to saying "hydrofluoric acid," but it's important to distinguish.*0234

*When it is together like this, yes, it is an acid, because it is in water--but it is not dissociated; it is actually still together.*0242

*Only 2.2% of the original amount has dissociated; that is it--it's all this is.*0249

*OK, let's list a general trend, actually--a general trend with percent dissociation.*0256

*The more dilute the solution, the greater the percent dissociation.*0269

*It is a nice little thing just to sort of keep in the back of your mind: in other words, if I have, let's say, a 1.0 Molar solution of HF, it's going to be some percentage.*0287

*Well, if I have a .5 Molar solution of HF, which is more dilute (right?--lower the concentration), the percent dissociation is actually going to be higher.*0299

*If I have a 0.1 Molar HF solution, the percent dissociation is going to be even higher.*0309

*That doesn't mean that the pH is going to be lower; that doesn't mean that the concentration of the hydrogen ion is going to be higher.*0317

*As you have lower and lower concentrations, you are actually going to end up with less concentration of H*^{+}; it just means that there is a greater dissociation, that more of the original acid has actually completely come apart into free H^{+} and free conjugate base ion.0324

*That is what that means; be very, very certain to distinguish between the two.*0341

*Lower concentration, more dilute solution, gives a greater percent dissociation.*0346

*The percent dissociation is not a measure of acidity; pH is a measure of acidity (in other words, how much hydrogen ion is actually floating around freely).*0356

*Remember, we said: when we talk about an acid, we are talking about free H*^{+} floating around in solution; more H^{+}, more acidic, more damage.0366

*That is it; so let's use this idea of percent dissociation to actually calculate a K*_{a} (an equilibrium constant).0375

*Our first example is going to be: In a 0.100 Molar lactic acid solution, the lactic acid is 3.8% dissociated.*0384

*OK, and for those of you who are biology majors and biochemists, lactic acid is the product that is produced by your cells under anaerobic conditions (under anaerobic respiration).*0423

*When you start to get really, really fatigued, that is the lactic acid building up in your muscles, until your body can get enough oxygen going to actually start burning the sugar aerobically (meaning with oxygen), as opposed to anaerobically (without oxygen).*0435

*So, dissociated...Calculate the K*_{a} for this acid.0449

*Calculate the equilibrium constant for this acid.*0456

*OK, so let's go ahead and write down an equilibrium expression.*0466

*I'm just going to write HL for lactic acid; actually, you know what--I think I'll write HLac, is in equilibrium with H*^{+} + lactate ion.0471

*Well, they tell me that the initial concentration is 0.100 Molar, and we have none of that; the change was -x, so it becomes x and x.*0484

*That means the equilibrium concentration is .100-x, x, and x.*0496

*Well, the K*_{a} for this is equal to...well, sure enough, the same equilibrium expression: the hydrogen ion concentration, times the lactate ion concentration, times the concentration of lactic acid (or hydrogen lactate).0502

*Now, we are looking, this time, for K*_{a}; that is what we want to find, which means that we need this number, this number, and this number in order to multiply and divide--which means we need x.0516

*Once we have x, we have this and this, and we can subtract from .1 to get this; we multiply and divide, and we get our K*_{a}.0527

*How can we find x?--well, they tell us that the percent dissociation is 3.8 (is it 3.8?)...yes, 3.8%.*0535

*What does that mean?--that means that this number, the amount that is dissociated (which is the same as this number, x), divided by the original .100, times 100, equals 3.8.*0545

*So, we have a way to find x; and now, we just go ahead and solve it.*0561

*We get: x is equal to 0.0038; there you go--nice and simple!*0565

*Now, we have our values; now, the hydrogen ion concentration (which is this one) is going to be 0.0038; the lactate ion concentration is 0.0038 (that is what this is, right?--they are the same; for every mole of this broken up, it produces a mole of this and produces a mole of that); and our HLac concentration is going to equal the original .100, minus 0.0038.*0574

*When we put all of these values in, we get: K*_{a} is equal to 0.0038, times 0.0038, divided by 0.0962 (that is this one....962...).0613

*And, when we run this number, we get 1.5x10*^{-4}.0639

*So again, if you understand the chemistry, you should be able to do the math.*0645

*We have this; we set up the ICE chart; we knew that we needed to find x.*0651

*But, they gave us x already--they gave us a way to find x; they told us the percent dissociation.*0656

*That means the amount that was dissociated; well, the amount dissociated is the amount produced of the H*^{+}, the amount produced of the lactate, divided by the initial amount, which was .1.0661

*So, this over that or this over that, times 100, is 3.8; we have x; we plug it in; and we get our K*_{a}.0671

*This is a common problem; you may actually run across it in your AP exam.*0679

*OK, now let's go on to discuss bases.*0684

*All right, now, the base is the opposite of an acid.*0688

*An acid is something that has hydrogens that it wants to give away; a strong acid--it gives up all of its Hs (in other words, it comes apart completely); a weak acid is one that sort of doesn't really give up its Hs too easily--it actually holds onto them.*0693

*A base is the opposite; a base is something that actually...strong base: let's actually define a strong base.*0707

*A strong base--it fully dissociates to produce OH*^{-}--so, for example, sodium hydroxide.0716

*Sodium hydroxide: when you take solid sodium hydroxide (it's kind of like hard rocks) and drop it in water, it dissolves; it dissociates.*0731

*It produces free sodium ion and free hydroxide ion.*0741

*It is a basic solution; it is a base, because when it dissociates, it produces OH*^{-}.0745

*Remember, water is also a base; water breaks up into hydrogen ion and OH*^{-}; it's amphoteric--it's an acid and a base.0751

*This is just a base.*0758

*Potassium hydroxide, another strong base: notice, I wrote the arrow in one direction--full dissociation.*0760

*K*^{+}, OH^{-}: if you had a solution of sodium or potassium hydroxide, you wouldn't find any NaOH, any KOH, in solution; it would all be this, this, this, this--free ions in solution.0766

*Let's do a quick example.*0783

*Handle it the exact same way, except now, in reverse.*0787

*Calculate the pH of a 4.0x10*^{-2} Molar NaOH solution.0791

*Well, let's see: NaOH is a strong base; any of your alkali metals (sodium, potassium, all of those in the first group)--with a hydroxide, they are all strong bases; they all fully dissociate.*0808

*"Strong base" means full dissociation.*0822

*OK, so the major species in water: again, nothing new--we handle it the same way.*0829

*Take a look at what it is: it's a strong base; there is going to be full dissociation; now, let's see what is floating around in water to decide what is going to dominate the equilibrium.*0834

*The major species in water floating around: you have sodium ion; you have hydroxide ion; and you have H*_{2}O.0844

*Well, we know that H*_{2}O is also a source of hydroxide ion, right?--because H_{2}O dissociates into H^{+} + OH^{-}.0853

*But, this is 10*^{-7}; let's write the K_{a} for this--the K_{w} is 10^{-14} (1x10^{-14}; I just ignored the 1).0863

*Well, this is a strong base, and this is 4.0x10*^{-2} Molar; that means 4.0x10^{-2} Molar sodium hydroxide has fully dissociated, and has produced 4.0x10^{-2} moles of hydroxide.0876

*Well, 10*^{-2}; 10^{-14}; a huge difference--that is 12 orders of magnitude--so this is virtually...you can ignore it.0894

*The species that is going to dominate the chemistry is this; the sodium is not going to do anything at all--it just sits there like a spectator ion.*0905

*So, our pOH (or actually, let me)...our OH*^{-} concentration is equal to 4.0x10^{-2}, because we have full dissociation.0913

*In other words, NaOH goes to Na*^{+} + OH^{-}, in case you are not sure what I am talking about.0930

*It starts off with 4.0x10*^{-2} Molar--full dissociation.0941

*This is 00 minus 4.0x10*^{-2}; all of it dissociates--this ends up being 0.0946

*This ends up being 4.0x10*^{-2}, 4.0x10^{-2}.0954

*That is what we mean by the OH*^{-} concentration, is that; because all of it is gone away--there is no more of that left in solution.0961

*It is that; now, let's calculate the pOH; remember, p is just a function--it means the negative log of something.*0970

*Negative log of 4.0x10*^{-2}; we end up with...actually, you know what, I did this a little differently.0976

*I did it this way: OH*^{-} equals that; well, what do we know about the hydroxide ion concentration and the hydrogen ion concentration in any aqueous solution?0988

*They multiply to 10*^{-14}.1000

*So, OH*^{-} times H^{+} equals 10^{-14}.1002

*We are looking for pH, not pOH; so the H*^{+} concentration is 10^{-14}, divided by 4.0x10^{-2}, equals 2.5x10^{-13}.1013

*And then, we will take the pH, equals the negative log of that number (2.5x10*^{-13}), and we get 12.6--basic solution.1030

*Remember what we said: if the pH is above 7 and below 14, or just above, you get a basic solution.*1042

*This confirms the fact that this is a basic solution--the pH.*1049

*pH is the standard by which we decide.*1053

*OK, now let's talk about weak bases.*1059

*Weak base--all right, the general reaction for a weak acid, we said, was this: we said, if we had an acid, plus water, we'll go to hydronium ion (which is the same as H*^{+}), plus the conjugate base of the weak acid.1065

*Now, the weak base general reaction is this (the general reaction; I'll just use B for base): it is: the base, plus water, goes to BH*^{+} + OH^{-}.1095

*In other words, let me rewrite this another way: B, plus (let me write H*_{2}O as HOH)--what is happening is that the base actually takes a hydrogen ion away from water.1120

*But, it doesn't take it as a hydrogen atom; it takes just the hydrogen ion.*1135

*Hydrogen leaves its electron with the hydroxide, which is why you end up with BH*^{+} + OH^{-}.1139

*I want to show you why this happens.*1148

*Most bases, weak bases--they have a lone pair of electrons, and we'll do an example in a minute.*1151

*H is here; this is usually not something that you are going to do until Organic Chemistry, but I want you to see what is happening, because I want it to make sense to you.*1157

*These electrons--they reach out and they actually grab the H; they rip it away from the water molecule.*1165

*When these electrons move in, these electrons move out, OK?*1173

*H cannot be attached to two simultaneously (well, it can for hydrogen bonding; but for our purposes); this base wants this, so when it takes this, it goes this way, and it kicks these electrons; they move onto the hydroxide, and what you end up with is this.*1180

*Because this is only coming as a proton, minus its electron--it's coming as an H*^{+}--now this whole species has a plus charge, and you have an OH^{-}.1201

*This general reaction is the reaction of a weak base with water; this is very important reaction--it always happens like this.*1211

*We said that a base is something that produces hydroxide ion; well, here you go.*1221

*A base doesn't necessarily have to have hydroxide ion in it; so, for example, you know that sodium hydroxide produces hydroxide by dissociation.*1226

*Well, an example like ammonia--well, let's actually use a specific example.*1236

*NH*_{3}, which has a lone pair, plus HOH (I'm going to write it as HOH, water) goes to NH_{4}^{+} + OH^{-}.1245

*OH*^{-} is still produced in solution, but the OH^{-} doesn't come from the base itself; it is the base that reacts with water.1259

*Water is now acting as the acid; it is giving up its H; this ammonia is acting as the base--it pulls the H off.*1268

*Now, it's NH*_{4}^{+} and it's OH^{-}; so it's producing OH^{-} in a roundabout way--not by dissociation--it's doing it by breaking up the water.1274

*This is...and there is, of course, a K*_{b} associated with this.1286

*It is an equilibrium constant expression for the reaction of a base (or something, some species) that will extract a hydrogen ion from water to produce the conjugate acid of this base and hydroxide ion.*1294

*The K*_{b} is exactly what you think it is; this is liquid; this is aq; this is aq; and this is aq; it is equal to NH_{4}^{+} times OH^{-}, over NH_{3}.1311

*That is it; this is the generic expression for a weak base; it is the base, plus the water, plus the BH*^{+}, plus OH^{-}.1332

*It is taking the hydrogen from water, leaving the hydroxide ion.*1342

*In the process, it produces the hydroxide ion.*1346

*It produces the hydroxide ion; the solution becomes basic; that is why it is called a base.*1351

*It is also called a base because it actually takes the hydrogen ion from something (in this case, water).*1356

*It is acting like the base; water is acting like the acid.*1363

*In the weak acid equilibrium, this is the acid; this is the base; now, these are the lone pair of electrons that are taking this hydrogen and producing that as a conjugate.*1367

*Back and forth: it's just a competition between two things for the hydrogen ion.*1379

*Here, the base...it just depends on what the equilibrium is; that is all acid-base chemistry is--it's a competition between two different species for the hydrogen ion in between.*1385

*It is a tennis game between the two: which one is stronger?--the stronger one will take the hydrogen ion.*1393

*OK, so again, a base--whether weak or strong--produces OH*^{-}.1401

*This is the take-home lesson: a weak base, strong base...they all produce OH*^{-}.1418

*A strong base does it by dissociation; a weak base does it by hydrogen abstraction from water.*1423

*OK, so now, let's do a couple of examples here.*1430

*Oh, let me just give a couple of versions: Strong base--we already mentioned one, potassium hydroxide; it dissociates into potassium ion plus OH*^{-}; here is your OH^{-}.1436

*A weak base--I'm going to use something called pyridine, and it is a molecule that looks like this, and it reacts with water.*1450

*Let me actually write it as HOH; you will find it very, very convenient to write water as HOH--I certainly prefer to do so, although a lot of people censure me for it--I don't know why.*1462

*It becomes N + OH*^{-}; again, here is your OH^{-}; it happens in a roundabout way.1475

*Here it comes apart; here it takes hydrogen from the water to produce the hydroxide.*1488

*Let's do an example.*1494

*All of it will make sense as we sort of do more examples: OK.*1498

*Let's see: what can we do?*1503

*Oh, calculate the pH of a 13.0 Molar NH*_{3} solution.1508

*The K*_{b} is equal to 1.8x10^{-5}.1526

*OK, so here, because we are dealing with a weak base, we are producing hydroxide ion concentration.*1533

*The x-value that we found is actually going to be hydroxide ion concentration; we are going to have to use the relationship of hydroxide, times hydrogen ion concentration, equals 10*^{-14} to find the hydrogen ion concentration, and then take the negative log of that.1540

*If we just take the negative log of the hydroxide ion concentration we find for a weak base problem, we are not going to get the pH; we are going to get the pOH.*1555

*Now, you can do that; that is fine, as long as you take the pOH and subtract it from 14, and that will give you the pH; because again, we have chosen hydrogen ion concentration as the standard--pH as the standard.*1563

*Well, we can calculate the p of anything.*1576

*OK, so let's see what we have: well, let's take a look at the major species.*1580

*Again, same procedure: major species: we have NH*_{3} (weak base--it's a weak base, which means that it is not dissociated--it stays mostly NH_{3}), and its K_{a} is 1.8x10^{-5}.1586

*The other species is water, because water is also a source of hydroxide.*1607

*But, its K*_{a} (which is K_{w}) is equal to 1.0x10^{-14}; -5; -14; I think I'll go with the -5.1612

*We can ignore this one; this is going to dominate the equilibrium--this is going to dominate the solution.*1622

*The hydroxide ion in this solution is going to come mostly from the chemistry of ammonia.*1629

*So, let's write it out: NH*_{3} + HOH (same equation over and over again: base plus water goes to conjugate acid plus hydroxide; "conjugate acid": conjugate acid just means stick an H on top of it--attach an H to it and put a plus sign on it) + OH^{-}; the Initial; the Change; the Equilibrium.1635

*I know you are going to get sick of these ICE charts.*1658

*Let me see where are we (where are we, where are we, where are we); oh, there we are: 13 Molar; this is before anything happens--initial means before the system comes to equilibrium.*1662

*Before this takes place, water doesn't matter; there is no ammonium, and there is no hydroxide yet.*1672

*A certain amount of ammonia disappears; OK, now this is kind of interesting.*1680

*NH*_{3} is not coming apart like an acid is, but we put a -x because NH_{3}, as ammonia, is disappearing.1686

*What is forming is ammonium; does that make sense?*1696

*Before, when we had H, like HF, and it dissociated, we knew it was -x; and it produced x amount of H and x amount of F*^{-}.1700

*But here, it is still -x, even though this NH*_{3} is actually taking something from the water and becoming more.1715

*It is becoming NH*_{3} to NH_{4}^{+}.1724

*But what is happening is that NH*_{3} is disappearing as a species; NH_{4}^{+} is showing up; OH^{-} is showing up.1726

*That is the idea; you have to get your mind around the physical reality of it.*1735

*This doesn't matter; x is showing up; hydroxide is showing up; our equilibrium concentration is going to be 13...actually, this is 13.0, if I am not mistaken.*1740

*Yes, I think it's three significant figures; sorry about that: 13.0, 13.0-x; that doesn't matter; let's make sure this is clear; that doesn't matter; that doesn't matter; that doesn't matter; this is +x, +x; handle it the same exact way.*1753

*So, we have: K*_{b} is equal to the ammonium ion concentration, times the hydroxide ion concentration, divided by the ammonia concentration.1773

*Let's put some numbers on this: 1.8x10*^{-5} is equal to x, times x, divided by 13.0-x.1788

*Well, again, because we are talking about a weak base (1.8x10*^{-5}), x is probably going to be pretty small compared to the 13.0.1800

*Let's just ignore it for the time being, and we will check the validity of our approximation in a minute; and just leave it as 13.0.*1812

*x squared is equal to 2.34x10*^{-4}, and x is equal to 0.0153 Molar, which equals the hydroxide ion concentration.1820

*Well, the pOH equals the negative log of the hydroxide ion concentration, equals negative log of 0.0153, is equal to 1.82; and pH + pOH is equal to 14; pH plus 1.82 is equal to 14, so our big fat pH is 12.18 (bigger than 7--a lot bigger than 7).*1838

*This is a basic solution; it's confirmed.*1876

*There you go; that is it--handle it the exact same way as an acid, except it's different; it's a base--it's a weak base.*1882

*Instead of a K*_{a}, we have a K_{b}; nothing is different; everything is exactly the same.1891

*Initial, Change, Equilibrium; decide what the major species are; decide which species is going to dominate the solution--which is going to produce, in this particular case, the most hydroxide ion.*1896

*It is the weak base; it's weak, but it is still stronger than the water.*1907

*OK, now let's do the percent; I'll put in quotes "dissociation."*1913

*Well, percent dissociation works for acid, because an acid is going from HA; it is actually dissociating into H*^{+} + A^{-}.1920

*A base, like NH*_{3}, is actually becoming BH^{+} + OH^{-}...so we say "percent dissociation," but what is more appropriate for a base is percent association (n other words, percent of the association of B with the H to produce H^{+}).1929

*They still call it percent dissociation; but as long as you know that now, you are talking about a base; it is a reverse process.*1953

*So, percent "dissociation"--I actually call it percent association, because I like things to make sense--I don't like things to just drop out of the sky.*1959

*OK, so we said that it is the OH*^{-}; it's the amount of thing produced--over the initial concentration of what we had.1969

*We have 0.0153 Molar, over 13.0 Molar, times 100, equals 0.12%.*1983

*That means, of the NH*_{3}, only .12% of the ammonia actually ripped off a hydrogen from the water to become ammonium ion and produce that much hydroxide.1998

*That is it; that is all this means.*2016

*OK, so let's see what we have here: let's try one more, and I think we'll wrap it up for weak base discussion.*2020

*Example 3: Calculate the pH of a 0.10 Molar (this time) methylamine, which is CH*_{3}NH_{2}, solution; so a methylamine is just like an ammonia, except it has, instead...I've taken out one of the hydrogens and put a CH_{3} on there.2031

*And the K*_{a} for methylamine is 4.38x10^{-4}.2070

*I would like you to see what this actually looks like, because I'm a big fan of structures.*2078

*NH*_{3} looks like this, as you know, and it has a lone pair of electrons; well, methylamine--it has its 2 hydrogens (not a problem), but now, it has a CH_{3} attached to it; so, it's the same thing; it's like an ammonia; it has a lone pair there; but instead of an H, it just has that.2082

*It behaves exactly the same way; it's a base; it pulls a hydrogen off the water to produce hydroxide.*2104

*So, let's write our major species: well, we have the CH*_{3}NH_{2}, and we have our H_{2}O.2110

*Well, 4.38x10*^{-4} versus 1.0x10^{-14}: yes, I think we can ignore the 1.0x10^{-14} as a source of hydroxide.2123

*Most of the hydroxide in this solution is going to come from this--a weak base, but still a stronger base than that.*2134

*So, let's write our CH*_{3}NH_{2} + HOH (or you can write H_{2}O, not a problem--same equation--plus water) goes to CH_{3}NH_{3}^{+} (just add a hydrogen and stick a plus charge on it) + OH^{-}.2140

*We have an initial; we have a change; we have an equilibrium.*2159

*0.10; nothing; 00 (before anything happens); as the system comes to equilibrium, this species disappears; this species appears; and this species appears.*2163

*At equilibrium, we are left with .100-x; this doesn't matter; that doesn't matter; this is +x; this is +x; now, we have: 4.38x10*^{-4} is equal to this times that, divided by that.2176

*x squared over 0.10-x approximately equals x squared over 0.10.*2196

*Now, when we do this, we end up with the following (when we do this approximation): we end up with x equal to 6.6x10*^{-3}.2206

*Let's check the validity of this; let's see if our approximation here, from going from here to here, is valid.*2219

*Well, 6.6x10*^{-3} over 0.10, times 100: guess what--it actually equals 6.6%.2224

*6.6% is too high; it's close to the 5, but it really is too high.*2237

*That means that you have a choice; you can either...you have to go back; you can't use this approximation, in other words.*2243

*You have to actually solve this whole equation as it is.*2249

*You can't eliminate the x from here; you have to solve this equation; you have to do it--either solve it as a quadratic equation (which is not a problem--it's easy enough to do; it's just numbers; you have a calculator--you can do it), or I'm going to show you a method called the method of successive approximations, which is a really, really great technique, if you don't want to use the quadratic.*2255

*It is an older technique; it still works--there are a lot of computer programs that are actually based on this method of successive approximations.*2276

*Here is how it works: OK, so now let's go back, and we said that we have 6.6x10*^{-3}, right?2283

*So, when we did this approximation of 4.38x10*^{-4} equals x^{2} over 0.10-x, approximately equal to x^{2} over 0.10; we got a value of...our first value...we got 6.6x10^{-3} for x.2295

*Let me write x=this.*2325

*x equals 6.6x10*^{-3}; OK, we checked this 6.6x10^{-3}; we divided by the .10, and we got 6.6%; that is too high.2329

*Instead of going back and solving this, here is what you do: you take this first value, and you put it back in for x, and you solve this equation again.*2342

*In other words, you take .10; you subtract 6.6x10*^{-3}; and you solve the equation x^{2} over the number that you get here, when you subtract this from this.2351

*You solve this, and you get a second value for x.*2365

*Now notice, I didn't put it here and here; that doesn't make sense.*2369

*What I am doing is: I'm going to use this first value to get closer with a second value.*2374

*I'm going to use the second value to get closer with a third value.*2380

*When any two successive values actually match each other, that means I have hit my point.*2384

*For those of you that are familiar with something called the Newton-Raphson method of solving for the roots of an equation, this is somewhat similar to that.*2389

*You are basically just sort of converging on a value, and when two successive values are equal, that means you have hit your point; you are not going to go any further.*2397

*So here, we were slightly off; I use this to put it back into this one, leaving this alone; I solve for x, and a second value I get is x=6.39x10*^{-3}.2407

*This is a pretty significant difference; this is fairly significant here.*2422

*And now, I check this one; well, I actually don't really--I just stick it back into this again; I take .10-6.39x10*^{-3} in the denominator; I leave the x^{2} on top; and I solve this equation again.2427

*My third value that I get when I solve for x: I get x=6.40x10*^{-3}.2443

*I stick this back in there; I do it again; I get a fourth value, x=6.40x10*^{-3}.2453

*Two values, one after the other, match; I can stop there.*2464

*x=6.40x10*^{-3}.2468

*x happens to be my hydroxide ion concentration of 6.4x10*^{-3}.2473

*Now, let me see: how else did I handle that?*2481

*And the pOH, which is the negative log of the 6.4x10*^{-3}, gives me 2.19, and then pH is equal to 14-pOH; I end up with 11.8.2483

*That is my basic solution.*2502

*So again, when you are presented with a value, and you try to approximate it by leaving this x off, and it turns out that that x value is higher than 5% of the original value that you took it from; well, you can take the value that you got and stick it in there.*2507

*Just subtract it from that value in the denominator, run this calculation again for x, and just keep doing that over and over again.*2523

*Whatever value you get, put it back in; when you get two values that are the same, that is when you stop--that is your value; you have converged on it.*2531

*You have gone from...if this is the real value, let's say you started over here; you are going bounce here; you are going to bounce here; you are going to bounce here; you are going to converge on it.*2538

*It really is nothing more than a Newton-Raphson method.*2547

*Or, you can just solve the quadratic equation.*2552

*It's up to you; personal taste--I think it's sort of nice to do whatever you feel comfortable with, because it certainly makes the act of problem-solving much more relaxing and much more enjoyable whenever you are doing something that you like to do.*2554

*OK, so I'll go ahead, and we will stop there for our discussion of weak bases.*2567

*Next time, when we meet, we are going to talk about polyprotic acids and the acidic and the acid-base properties of regular salts.*2573

*So, thank you for joining us here at Educator.com, and we'll see you next time; goodbye.*2581

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

*Today, we are going to continue our discussion of acid-base equilibria, and we are going to talk about polyprotic acids.*0004

*If you remember, in the first lesson that we discussed acids in, we said that polyprotic acids are simply a fancy name for acids that have more than one hydrogen that they could possibly give up.*0010

*So, for example, sulfuric acid (H*_{2}SO_{4}), carbonic acid (H_{2}CO_{3})...probably the most important of the polyprotic acids is H_{3}PO_{4}, phosphoric acid.0021

*It is how the inside of a cell actually maintains its pH; it is the buffer solution inside of the cell.*0033

*And the carbonic acid, the H*_{2}CO_{3}, is how the blood maintains its pH; it is a buffer solution outside of the cell.0040

*We are going to discuss polyprotic acids today and talk about they behave; and it is not altogether different than anything that you have seen already, as far as weak acids and strong acids.*0048

*We'll just jump right in, do a quick couple of definitions, and we'll start with the examples.*0058

*OK, so once again, the really important thing about all acids (polyprotic or otherwise--well, polyprotic especially) is that acids dissociate one hydrogen ion at a time.*0065

*So, in the case of, say, carbonic acid (let me actually write that out first: so acids dissociate one hydrogen ion at a time), which is H*_{2}CO_{3}, it's in equilibrium, so it dissociates into H^{+} + HCO_{3}^{-}, and there is a K_{a} associated with this, an acid dissociation constant, 4.3x10^{-7}.0076

*And notice, we have a little 1 here, so K*_{a1}, because it gives up one ion at a time.0112

*Now, this HCO*_{3}^{-} has another H that it can give up, and it also behaves like an acid; so it now shows up on this side, and it is in equilibrium with the proton that it gave up: CO_{3}^{2-}.0118

*And this is K*_{a2}; this is the second acid dissociation constant, which is equal to 5.6x10^{-11}.0135

*One thing you want to notice here, which is a general trend for all polyprotic acids, is that the first K*_{a} is going to be a certain number; the second K_{a} is one that is very, very small: 10^{-11}...very tiny--certainly much smaller than this.0143

*So basically, when it comes to polyprotic acids, it's really the first dissociation that counts the most.*0160

*The only exception to this case (which we will see in a little bit when we discuss sulfuric) is sulfuric acid, and the reason being that sulfuric acid is strong in its first dissociation--it behaves just like hydrochloric acid or hydrobromic acid--but in the second dissociation, it's actually a weak acid, so it's slightly different depending on the particular molarity of the solution that we are trying to find the pH of.*0168

*But, we will get to that in a minute.*0191

*Another example would be phosphoric acid, like we mentioned, and it has three dissociations, because it has three protons to give up.*0193

*H*_{3}PO_{4} is in equilibrium with H^{+} + H_{2}PO_{4}^{-}, and the first acid dissociation constant is 7.5x10^{-3}.0200

*And then, this H*_{2}PO_{4}^{-} dissociates into H^{+} + HPO_{4}^{2-}; so make sure you keep track of the charges.0214

*It's very, very important, especially with these polyprotic acids; they all sort of look alike--the H*_{2}PO_{4}, the HPO_{4}, the PO_{4}, H_{3}PO_{4}; it gets very, very confusing, so definitely work very carefully, very, very slowly.0228

*Chemistry is very symbol-heavy, and it's easy to get lost in the symbols and make silly mistakes, simply for that--as opposed to conceptual mistakes.*0241

*The second dissociation constant, the K*_{a2}, is 6.2x10^{-8}.0251

*And then, this species, HPO*_{4}, the hydrogen phosphate (which is 2-), becomes H^{+} + PO_{4}^{3-}.0258

*Now, the phosphate is finally alone, and the third dissociation constant is equal to 4.8x10*^{-13}; so clearly, 10^{-3}; 10^{-8}; 10^{-13}: you see the pattern.0270

*At is dissociates more, it becomes harder and harder to actually pull off that second and third hydrogen.*0285

*Let's just go ahead and do an example, and I think everything will make sense.*0291

*Example 1: We want to calculate the pH of a 4.0 Molar H*_{3}PO_{4} solution, as well as the equilibrium concentrations of H_{3}PO_{4}, H_{2}PO_{4}^{-}, HPO_{4}^{2-}, and PO_{4}^{3-}.0297

*So, not only do we want the hydrogen ion concentration (in other words, the pH, when we actually take the negative log of it), but we want to find the concentration of all of the species in solution, once you take that phosphoric acid and drop it into water to create a 4 Molar solution.*0349

*What is the concentration of each of those species, in addition to the pH?*0365

*OK, well, let's just go ahead and start.*0368

*The first thing we do is the same thing we always do: we check to see the major species in the water before anything happens.*0371

*That is the whole idea; we want to see what is in there before the situation comes to equilibrium, before the system comes to equilibrium.*0377

*We want to check out the chemistry; we want to make some decisions about what is going to dominate--what chemistry is going to dominate.*0384

*There is always going to be one that dominates.*0390

*OK, so major species: well, phosphoric acid is a weak acid; you notice, K*_{a1} is 7.5x10^{-3}; that is kind of small, so it's a weak acid.0392

*A weak acid means that it's not going to dissociate very much.*0405

*So, most of it is going to be in this form: H*_{3}PO_{4}.0409

*That H*_{3}PO_{4} is just going to be floating around in solution, not very dissociated.0413

*The major species in the solution: well, you have an H*_{3}PO_{4}, and your other major species is just the water.0418

*Well, both of these contribute hydrogen ions; well, the K*_{a} of, or the K_{w} of, water (which happens to be the K_{a} of water) is 1x10^{-14}.0425

*Well, the K*_{a} of this one is 7.5x10^{-3}, 11 orders of magnitude bigger; so we can ignore water's contribution to the hydrogen ion concentration; this is the dominant species.0435

*Because that is the dominant species, that is the equilibrium that we are going to work with.*0447

*The same thing we have always done: major species; decide which is dominant, which is going to control the chemistry; that is the one you concentrate on--ignore everything else.*0451

*So, let's go ahead and write our equilibrium expression and do our ICE chart.*0459

*Let's see...yes, that is fine.*0465

*OK, so we have: H*_{3}PO_{4} that dissociates into H^{+} + H_{2}PO_{4}^{-}.0470

*We have our initial concentration, our change, and our equilibrium.*0482

*Our initial concentration is 4.00 molarity, and this is 00, so this is before anything has had a chance to come to equilibrium.*0486

*A certain amount of this H*_{3}PO_{4} is going to dissociate, and for each amount that dissociates, 1:1, 1:1 ratio--that is the amount that shows up in solution of the other species.0495

*Therefore, our equilibrium concentration is 4.00-x; this is x; this is x; and now that we have our equilibrium concentrations in terms of x, and we know what the K*_{a} is, we set it equal to each other.0505

*7.5x10*^{-3} is the K_{a}; it is the x, the hydrogen ion concentration, times x, the dihydrogen phosphate concentration, divided by the phosphoric acid concentration, 4.00-x.0521

*We do our normal approximation, which we can check the validity of; so this is equal to x*^{2} over 4.00, just to make our math a little bit easier.0540

*When we do this, we get x*^{2} (actually, you know what, I'm just going to skip this line altogether; and I'm just going to go ahead)...and you multiply by the 4; you take the square root; you end up with x is equal to 0.173.0548

*That is equal to the hydrogen ion concentration; therefore, the pH is the negative log of the hydrogen ion concentration--you get a pH of 0.76.*0563

*That is our first part: 0.76 is our pH; let's check the validity of what we have; by checking the validity, that means...*0574

*So let's do it over here; let's say "check validity of our approximation"--I mean, we know it's good, but it's good to just sort of check it.*0583

*0.173, which is x, divided by 4.0, times 100; you end up with 4.3%.*0592

*It's close to 5, but it's still under the 5% rule, so we're still actually pretty good; it's not a problem.*0600

*A perfectly good, valid approximation: we don't need to do the quadratic equation; this is a good concentration; this is the pH.*0606

*Now let's see where we stand: we found the pH, but we also wanted to find the concentrations of all the other species in solution.*0613

*Let's see what we have; we have the H*^{+} concentration; that is equal to x, which is 0.173 Molar.0620

*Well, the H concentration also happens to be the H*_{2}PO_{4}^{-} concentration.0629

*The H*_{2}PO_{4}^{-} concentration is the same thing, 0.173 Molar.0634

*Well, since we have x, we also have the phosphoric acid concentration (H*_{3}PO_{4}), which is equal to 4.00-0173, and I'll just go ahead and...no, we'll just do 3.8 Molar (I'll just go ahead and round it up a little bit; sorry about that).0641

*So, that is going to be 3.8 Molar; so now, the only thing that we are actually missing is: we are missing the HPO*_{4}^{-} concentration and the PO_{4}^{3-} (this is 2-; see, again, I am making the same mistakes; I have symbols and charges all over the place).0667

*So now, we need the concentration of the hydrogen phosphate, and we need the concentration of the phosphate ion.*0688

*Well, like we said, an equilibrium is established; it doesn't matter where these species come from (the phosphate, the hydrogen...); the K*_{a} that we have for a given species, for a given equilibrium; if it involves the species that we want, that is good enough.0693

*The equilibrium that we do have, that involves the HPO*_{4}^{2-}, is the following.0713

*We have the H*_{2}PO_{4} concentration; well, that is in equilibrium with H^{+}, plus the HPO_{4}^{2-}, which is the species that we are looking for.0721

*Well, we have this concentration, and we have this concentration; that is this and this; and we also have the second K*_{a}, the second dissociation constant.0734

*That is what this is here: this is the second dissociation of phosphoric acid, and the K*_{a} for this one, K_{a2}, is equal to 6.2x10^{-8}.0742

*So now, we can just basically rearrange the equilibrium expression and solve for HPO*_{4}^{2-}.0757

*And then, for the PO*_{4}^{3-}, we do the same thing with K_{a3}, the third dissociation constant.0764

*Let's go ahead and write everything out, so that we see it.*0769

*K*_{a2}, based on the equation that was just written (the dissociation of the H_{2}PO_{4}^{-}), becomes the concentration of H^{+}, times the concentration of HPO_{4}^{2-}, over the concentration of H_{2}PO_{4}^{-} (oh, this is crazy--all kinds of charges and symbols going on).0774

*Therefore, I rearrange this to solve for this species, because I have this, I have this, and I have this; it's a simple algebra problem.*0797

*We have HPO*_{4}^{2-}, the concentration--moles per liter--is equal to the K_{a2} times the H_{2}PO_{4}^{-} concentration, divided by the H^{+} concentration.0806

*We get: that is equal to (I'm going to move this over here) 6.2x10*^{-8}, which is our K_{a}, times 0.173, which was our concentration of H_{2}PO_{4}; it also happens to be the concentration of our H^{+}.0825

*So, these cancel, of course, leaving us with 6.2x10*^{-8} molarity for...that is the HPO_{4}^{2-} concentration--that is one of the last things that we needed.0849

*Notice, it's very, very, very small.*0866

*Well, now we want to know the PO*_{4}^{3-} concentration; that is the final concentration--the final species whose concentration we want.0870

*OK, well, now that we have found the HPO*_{4}^{2-} concentration (it's this), we have an equilibrium.0880

*The third equilibrium is (the third dissociation of phosphoric acid--excuse me): HPO*_{4}^{2-} (oh, wow, you see what I mean--all of these charges, all of these symbols floating around; you're bound to make mistakes--you have to go very carefully) is in equilibrium--it dissociates into H^{+} + PO_{4}^{3-}.0886

*We can write a K*_{a} for this: the third dissociation constant is equal to the H^{+} concentration, times the PO_{4}^{3-} concentration, over the HPO_{4}^{2-} concentration (which we just found).0912

*We rearrange this, and you get: the PO*_{4}^{3-} concentration is equal to the K_{a3}, times the HPO_{4}^{2-} concentration, divided by the H^{+} concentration.0929

*So, the K*_{a3} is going to be (let me see: what is the K_{a3}?) 4.8x10^{-13}, times the HPO_{4} concentration, which we just found, which is 6.2x10^{-8}, over the hydrogen ion concentration, which we found in the first step: .173.0953

*Well, we get a very, very tiny number: we get 1.7x10*^{-19} molarity; that equals the PO_{4}^{3-} concentration.0976

*There you have it.*0990

*So, a polyprotic acid: you handle it in the exact same way--most polyprotic acids are going to be weak acids.*0993

*The only exception is sulfuric; sulfuric is strong in its first dissociation, weak in its second dissociation; we are going to handle a problem in just a minute.*0998

*But aside from that, you handle it in exactly the same way.*1007

*And then, if you happen to need the concentrations of the other species--a further dissociation (so, phosphoric acid, dihydrogen phosphate, hydrogen phosphate, phosphate), you use the concentrations that you found, and then you use the next dissociation, and the next dissociation, with the appropriate K*_{a}, to find the concentrations of all the species that you want.1010

*About the only difficult problem here, as you saw, was just keeping track of the symbols and not losing your way in the symbology.*1035

*That is going to be the hardest problem, which, as far as I am concerned, that is the problem you always want--you don't want conceptual problems; you want mechanical problems; those are easy to deal with.*1042

*OK, so let's do another example; this time, we are going to deal with sulfuric acid, and we are going to do a couple of variations of it.*1050

*So, let's just start Example 2: We want to calculate the pH of a 1.2 Molar H*_{2}SO_{4}.1057

*OK, well, so sulfuric acid: let's write down its dissociations.*1074

*Its first dissociation is: H*_{2}SO_{4} dissociates into H^{+} + HSO_{4}^{-} (the hydrogen sulfate ion, also called the bisulfate ion).1079

*This K*_{a} is large: we said it's a strong acid in its first dissociation; it's large--it doesn't even have a number.1091

*You remember, strong acids don't have K*_{a}s; they are huge.1097

*The second dissociation is: HSO*_{4}^{-} dissociates into H^{+}...1101

*Actually, I'm not even going to give the equilibrium sign here; this is going to be a one-sided arrow pointing in one direction--full dissociation; there is no H*_{2}SO_{4} left--very, very important to remember that for strong acids.1107

*For strong acids, there is an arrow pointing to the right, and that is it; it's not in equilibrium.*1119

*We have HSO*_{4} going to H^{+} + SO_{4}^{2-}, and the K_{a} (it's small, but it's actually still pretty large, relatively speaking): 1.2x10^{-2}.1125

*So, a little larger than the other weak acids, but it still behaves as a weak acid; there is an equilibrium here; it is still less than 1.*1138

*OK, we want to calculate the pH; well, let's do what we always do--let's see what the major species are.*1145

*Be careful here: the major species in solution--you have taken sulfuric acid; you have dropped it into water; sulfuric acid is strong in its first dissociation.*1152

*Therefore, a solution of sulfuric acid is entirely composed of free hydrogen ion, free hydrogen sulfate ion, and (the other species in water is) H*_{2}O.1163

*There you go; now, again, we are calculating a pH here; once again, strong dissociation; it's a strong acid in its first dissociation; that means in species, before any equilibrium is reached, it's this and this and this.*1179

*"Strong acid" means full dissociation--it breaks up completely: there is none of this left.*1197

*It is all that in solution: this floating around, this floating around, this floating around; what is the hydrogen ion concentration?*1202

*Well, the question is: Since it's a strong acid in its first dissociation, can we treat it just like any other strong acid and just take the molarity and just take the negative log of it?*1211

*Can we just do -log of 1.2?*1220

*Because 1.2 moles per liter produced 1.2 moles per liter of H and 1.2 moles per liter of that, and this is actually pretty weak, it's probably not going to contribute too much.*1223

*However, we need to make sure; so let's ask our question: Can we just treat this like any other strong acid problem?*1235

*The answer is maybe; so, we have to check.*1263

*The reason we have to check is, we need to see if this actually does contribute anything.*1269

*It is small, but it's not so small (like on the order of 10*^{-5} or 10^{-6}, 10^{-7}; it's 10^{-2}.1274

*When you get to the 10*^{-2} range, you're going to have to be careful; and we'll see in a minute what a general rule of thumb is.1284

*A general rule of thumb is: when you have second-dissociation constants, such as 10*^{-2}, if you are below about 1 molarity, you can't really ignore it; you have to include it.1291

*So, not only do you have to take the 1.2 Molar, but some of this HSO*_{4}, because it also dissociates, is going to produce a little bit more H^{+}; so you might have to include that; that is what we are going to check.1302

*We are going to check to see if the second dissociation is significant--produces enough H*^{+} so that we have to include it into the 1.2.1317

*Let's write: first of all, maybe, so we have to check; so now, mind you, our final H*^{+} concentration that we are going to take the negative log of to find the pH is going to equal, in this case, the 1.2+x.1327

*This x is going to be any hydrogen ion that comes from this dissociation.*1345

*It's strong in this first dissociation, so I know that, before anything happens, I know that at least 1.2 moles per liter of hydrogen ion is floating around in solution.*1352

*However, I need to know if the second dissociation also produces enough hydrogen ion; so I have to add it to this.*1360

*That is the thing; we are going to compare x and 1.2 to see what the real difference is; if it's small, we can ignore it, and just treat it as a regular strong acid problem; if it's not small, we have to include it before we take the final negative log of that final hydrogen ion concentration.*1369

*I hope that makes sense.*1384

*The ICE chart is going to look slightly different; OK.*1386

*Well, let's go ahead and do our ICE chart, then: so again, we know that we have the 1.2of that, so we don't have to worry about this.*1389

*We have to worry about this equilibrium: how much H*^{+} is this HSO_{4} going to produce under these circumstances?1397

*So, the equilibrium that we want to look at is the HSO*_{4}^{-}: H^{+} + SO_{4}^{2-}.1405

*I hope that makes sense: major species, ICE charts...you do ICE charts with weak acids; you don't do them with strong acids.*1414

*I have at least 1.2 moles per liter of H*^{+} floating around, because it's a strong acid in its first dissociation.1421

*I need to check the second dissociation to see if it produces a significant amount of H*^{+} to add to the 1.2.1428

*So, Initial, Change, Equilibrium: our initial concentration of HSO*_{4} is 1.2 Molar, because it is produced in the same way that this is produced.1436

*When H*_{2}SO_{4} dissociates, it produces 1.2 Molar of this, 1.2 Molar of that.1448

*Now, what is our initial H*^{+} concentration?1453

*Here is where we have to be careful, and where the ICE chart is different: our initial H*^{+} concentration is also 1.2, because we have free H^{+} floating around and free HSO_{4} floating around; that is what these two are.1455

*But, there is no SO*_{4}^{2-} yet, before anything else happens.1471

*In solution, before the system comes to equilibrium, this and this are the same from the first dissociation.*1475

*Now, the change is going to -x here; this is going to be +x here; +x here; we add vertically down; we get 1.2-x for the HSO*_{4}; we get 1.2+x...like I said, 1.2+x is going to be our final hydrogen...and we get +x.1482

*Now, we form our equilibrium; let's see, our K*_{a} is (oops, here we go with the stray lines again; they show up in the most interesting places; OK--let me just do it off to the side here) 1.2x10^{-2}, equals 1.2+x, times x, over 1.2-x.1504

*So now, we have to solve this equation.*1542

*OK, here is what we are going to do: we are going to presume that x is small in order to do our approximation; then, we are going to check our approximation to see if it's valid.*1544

*So, when we take x small, watch what happens to this approximation.*1556

*OK, it's approximately equal to...that means this x disappears--not this x, this x.*1561

*So, it becomes 1.2 times x, over 1.2.*1568

*Well, these cancel, and I am left with: x is equal to 1.2x10*^{-2}, which is the same as 0.012.1579

*I hope you'll forgive me; I actually prefer to work in decimals, as opposed to scientific notation.*1590

*I like scientific notation, but I just like the way that regular numbers look.*1594

*OK, now, we said that x is .012; now, we have to go up here: our final hydrogen ion concentration is going to be 1.2 (the original amount floating around), plus the x, plus the 0.012 that came from the second dissociation.*1599

*That equals 1.2.*1627

*It equals 1.212, but it's 1.2 to the correct number of significant figures.*1632

*So again, significant figures become a little bit of an issue; I personally don't care about significant figures all that much; a lot of people sort of criticize me for that, but you know what, it doesn't really matter; I'm more interested in concepts than I am in actual significant figures.*1640

*I think they are fine, and I think it's good to sort of use them, and in this context, yes, it sort of helps, because you notice: 1.2 plus .012--this is actually pretty small compared to this.*1652

*Since, working with significant figures, you still end up with 1.2, this says that you are actually justified in ignoring the second dissociation.*1665

*In other words, it doesn't really contribute all that much, in terms of hydrogen ion; you can just take the negative log of the 1.2.*1674

*So, here we have the 1.2; so the pH is equal to -log of 1.2, and you get a (let me see what...yes) -0.079 for a pH.*1682

*Yes, a pH can be negative; that means it's a really, really strong acid.*1699

*Now, if you ignore the significant figures, like I personally tend to do, I'll just go ahead and put that number down, too.*1703

*If you ignore the significant figures, and take the final hydrogen ion concentration to equal 1.212--if I include that .012, there is a little bit of extra hydrogen ion in there--well, you know what--your pH is going to be slightly different.*1711

*This is .084; 0.079, 0.084...you know, it's a small difference; if you are doing analytical work, it's important--for normal work, it's not altogether that important.*1737

*But you are going to end up with a pH of .08; that is really what it comes down to.*1749

*So, as it turns out here, the rule of thumb ends up being the following.*1754

*We notice that, for a solution which is about 1.2 Molar, roughly 1 Molar, the contribution of the hydrogen sulfate dissociation can be ignored.*1765

*You can treat it like a strong acid problem.*1776

*Below about 1.0 Molar, 1 Molar, .9 Molar...you can't ignore it; the second dissociation does produce a significant amount of hydrogen ion that has to be included in the concentration that came from the first dissociation.*1779

*You have to add those two; so, the first dissociation and then a little bit of the second dissociation...that is when you have your final hydrogen ion concentration.*1796

*For solutions of H*_{2}SO_{4} more dilute than approximately 1 Molar, the full expression for the equilibrium constant must be used.1805

*In other words, we can't approximate; we have to do the 1+x, times x, over 1-x; we have to solve the quadratic, which means the quadratic expression must be solved.*1842

*OK, so now, we'll do a quick example; I'm actually just going to set it up for you (I'm not actually going to solve it), just so you see what it looks like.*1864

*The Example 3 is: Calculate the pH of a 0.010 Molar H*_{2}SO_{4}.1872

*This is pretty dilute: .01 Molar--a lot less than 1.*1889

*So, we can't ignore the second dissociation; OK.*1896

*Let's take a look at what we have; we have our major species, and again, we have the H*^{+} from the first dissociation; we have the HSO_{4} from the first dissociation; and we have water.1900

*These two are going to dominate the equilibrium; we know what this is already--this is just .01 Molar, because it's fully dissociated (strong).*1918

*The final hydrogen ion concentration is going to be 0.010+x, and x is the hydrogen ion concentration we get from the dissociation of the HSO*_{4}^{-}.1927

*That is the equilibrium we want to look at, HSO*_{4}^{-} goes to H^{+} + SO_{4}^{2-}; we have Initial; we have Change; we have Equilibrium.1943

*The initial concentration is 0.010, right?--and 0.010, strong, dissociates into this and this (first dissociation).*1955

*This is 0; the change is -x, +x, x; we get 0.010-x over here; we get 0.010+x over here; we get x over here, and we write K*_{a2}, which is 1.2x10^{-2}, is equal to 0.010+x, times x (oops, let me just take out this parentheses here), and then, that is going to be over 0.010-x.1965

*That is it; we have to solve this equation.*2003

*Multiply through; get your x*^{2}; get your x term; get your...it's going to be ax^{2}+bx+c=0.2007

*b and c and a--they can be negative, if they need to be.*2019

*And then, you just plug it into the quadratic equation, or use your graphing utility (your TI-83, 84 calculator, your graphing calculator) to find the roots of this equation.*2022

*Now, you are going to get two roots; check to see what makes sense--one of the roots is completely not going to make sense.*2031

*The other root will make sense, and that is the whole idea.*2037

*So, your final hydrogen ion concentration (oops, let's make this a little bit...): when you get x, you are going to get x equal to some number; we'll just put a little box there.*2041

*Your final hydrogen ion concentration is going to equal the 0.010, plus this boxed number, and your pH is going to equal the negative log of your final hydrogen ion concentration.*2054

*And that is it: so, for polyprotic acids, treat them like any other equilibrium; usually, most polyprotic acids are weak acids; it is going to be the first dissociation that dominates.*2071

*You can pretty much ignore the second, third, fourth...I don't think...I don't even know if there is a...well, there is, but...*2081

*Weak acid--you can pretty much ignore the second and third dissociations.*2086

*For sulfuric acid, maybe; maybe not.*2092

*If you are talking about 1 Molar or above, you can ignore the second dissociation; if you are talking about more dilute than 1 Molar (.8, .7, .6...below that), the second dissociation is going to contribute a significant amount of hydrogen ion.*2094

*You have to take the amount that comes from the original concentration, the first dissociation; find the amount that comes from the second; add them; and then take the negative log of that.*2110

*I hope that helps; again, you see: it's pretty much all the same--slightly different when you are dealing with multiple dissociations in the case of sulfuric acid.*2119

*Thank you for joining us here at Educator.com.*2128

*We will see you next time to discuss the acid-base properties of salts.*2130

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

*Today, we are going to talk about salts and their acid-base properties.*0004

*Let's just jump on in, start with a definition, and start doing problems, because I think that is probably the best approach.*0008

*A salt is a generic term for an ionic compound.*0015

*Sodium phosphate; silver chloride; lead iodide; sodium chloride; you name it--potassium permanganate--these are all salts, because they are ionic compounds.*0036

*It is just a positive ion and a negative ion; we just call them all salts.*0050

*Salt, sodium chloride, is specific--just happens to take that name; but when we talk about salts, we are talking about any ionic compound.*0054

*Now, as you know (or as you should know), when you take an ionic compound and you drop it in water, it is either going to dissolve, or it's not.*0062

*If it dissolves, it breaks up into its free ions, and those ions are just floating around freely in the water; so let's write that down, actually.*0069

*When salts dissolve, their ions float around as free ions.*0081

*OK: so, for example, if we had something like sodium chloride (which is a solid, as you know--it's salt), if we drop it in water (an arrow with a little water on it--that means you have dissolved it in water), you end up creating Na*^{+} + Cl^{-}.0107

*Now, I am not usually going to write the aq, the subscripts, but that is what that means--it just means "dissolved in water"--aq.*0119

*You have sodium ions floating around, and you have chloride ions floating around.*0125

*Well, often, when this happens, the anion of the salt that you drop into water and dissolve--it is actually going to be the conjugate base of a weak acid.*0129

*Let's say that again: Often (let's write it down), the anion--the negatively-charged ion, this one--will be the conjugate base of a weak acid.*0141

*Example: sodium fluoride--if we take sodium fluoride, and if we dissolve it in water, we produce sodium ion, and we produce fluoride ion; so there is F*^{-} floating around, and there is Na^{+} floating around.0166

*Now, you have seen this F*^{-} before; it is the conjugate base of the weak acid hydrofluoric acid, which means--a conjugate base just means you have taken an acid and you have taken off the H.0182

*The thing that you are left with--that is the conjugate base.*0195

*So, over here, I am going to write; you have seen it as (oops, let me do this in red to separate) this: HF is in equilibrium with H*^{+} + F^{-}.0197

*There is a certain K*_{a} associated with this; in fact, it's 7.2x10^{-4}.0215

*The K*_{a} equals 7.2x10^{-4}.0221

*So notice, this time, it doesn't show up as the acid; you didn't take the acid and drop it in solution--you dropped a salt whose negative ion, the anion, happens to be the conjugate base of a weak acid.*0226

*When this happens--when the anion of a salt that you dissolve in water happens to be the conjugate base of a weak acid--it actually reacts with water as a base.*0240

*Because remember--notice--as an acid, this is a weak acid; a weak acid means the equilibrium is on the left; that means it wants to be this way--it doesn't want to be this way.*0249

*That means, if you put some free F*^{-} anywhere near some H^{+}, or anywhere near a source of H^{+}, it is going to take that H^{+}, and it is going to move over in this direction until this equilibrium is established.0261

*So, weak acid--strong conjugate base; a strong base means that it has a tendency, a very high affinity, for hydrogen ions--for protons--for hydrogen ions; it wants to take them.*0275

*When you take a salt and dissolve it in water, and all of a sudden create this free base, which happens to be the conjugate base of a weak acid (hydrofluoric acid), here is what it does: it actually behaves as a base now.*0286

*It does this reaction: it reacts with the water that is floating around in solution, and it actually takes the hydrogen ion from the water; it is acting as a base--it takes hydrogen ion.*0301

*This time, water is acting as the base; it becomes HF + OH*^{-}.0313

*Notice what we have done: in this process, this base takes the H, creates hydrofluoric acid, and in the process it creates hydroxide ion.*0321

*When this happens, you create a basic solution; so if you took a neutral, normal water, which is pH 7, and if you drop in some sodium fluoride, well, the fluoride ion will pull hydrogens off of the water, creating hydroxide ion.*0330

*The pH of that solution is going to go up; it is going to become a basic solution--that is what a basic solution is: it's when the concentration of hydroxide ion is higher than usual.*0345

*So, any time you have a salt where the anion happens to be the conjugate base of a strong acid...conjugate base of a weak acid; forgive me--where it happens to be the conjugate base of a weak acid, it is going to react as a base in this reaction.*0356

*People call this a hydrolysis reaction; I don't really like that name--just know that now, this conjugate base is going to act as a base.*0374

*You have seen this reaction before; this is the reaction of a base with water.*0382

*There is a K*_{b} associated with this.0387

*There is a K*_{b} associated with this: it is HF, OH^{-}, over F^{-} (because water is liquid water).0390

*Well, how do we know what the K*_{b} is?0403

*Here is the best part about it: the K*_{b}, the relationship--we know the K_{a} already, but what is the relationship between the K_{a} and the K_{b}, since now the F^{-} is actually behaving as a base, and not this way, as part of the weak acid equilibrium?0407

*Now, it is involved in a basic equilibrium, where it is actually pulling off a hydrogen ion.*0422

*Here is the relationship--all K*_{a}s and K_{b}s are related by this equation: K_{a}, times the K_{b}, is equal to K_{w}.0429

*So in this case, if you want to do a problem with this, and you need to treat this as a K*_{b} problem, base problem, because this reaction is that of a base with water, all you do is: you take the K_{b} is equal to K_{w} over K_{a} for that species.0438

*This is 10*^{-14}; in this particular case, it's 7.2x10^{-4}, and you find the K_{b}, and you run this problem as a base problem.0459

*We have already done it: weak acids, weak bases--that is all that is going on here.*0470

*So, let's actually do a problem, and it will make sense.*0474

*Let's see...OK; so, Example 1: Calculate the pH of a 0.35 Molar sodium fluoride solution.*0479

*The K*_{a} is equal to (sorry, I had better say which K_{a} of what)...the K_{a} of hydrofluoric acid (the actual acid, the weak acid, where the base is coming from) is 7.2x10^{-4}.0507

*OK, what do we always do first?--we check the major species in solution to decide what chemistry is going to dominate.*0529

*Major species: well, you have sodium fluoride: sodium fluoride is completely soluble--remember the solubility chart from earlier in the year?*0535

*If you haven't memorized it, not a problem--just check it out; sodium fluoride, alkali metal, halogen--completely soluble.*0545

*That means what you have floating around in solution is sodium ion; you have fluoride ion; and you have H*_{2}O.0552

*Well, we notice sodium ion doesn't do anything; it doesn't affect anything.*0560

*However, the anion happens to be the conjugate base of a strong acid, HF.*0565

*I keep saying "strong acid"; what is wrong with me?--weak acid, weak acid, weak acid.*0571

*So, F*^{-} is the conjugate base of a weak acid, hydrofluoric acid.0577

*Well, it's the conjugate base of a weak acid; so what it is going to do--it is actually going to react as a base with the water, the following reaction.*0582

*It is going to be: F*^{-} + HOH goes to HF + OH^{-}.0591

*That is the reaction; this reaction is the reaction of a base with water.*0608

*There is a K*_{b} associated with it.0614

*We want to find the pH of this; well, now that we have our reaction, that we know what reaction is going to take place, we do our ICE chart.*0616

*Well, what is the initial concentration of the F*^{-}?--well, since we have full dissociation, it's 0.35.0626

*Water doesn't matter; there is no HF formed yet--this is before anything happens, and there is no OH*^{-} before anything happens.0634

*A certain amount of F*^{-} is going to disappear; as a species, it's going to react with H to become HF.0642

*It is going to disappear; water doesn't matter; that means HF is going to show up, and OH*^{-} is going to show up.0649

*There is absolutely nothing under the sun when it comes to these; we have done these several times--we know how it works, but now, because this is a base reaction, we need the K*_{b}.0659

*K*_{b} is equal to K_{w} over K_{a}, equals 1.0x10^{-14}, divided by 7.2x10^{-4}, and we get a K_{b} of 1.4x10^{-11}.0670

*So now, we do 1.4x10*^{-11} is equal to x, times x, divided by 0.35-x.0696

*Well, look how small this is; you know what, x is going to be pretty small, so chances are, we can do the approximation.*0709

*It equals x*^{2} over 0.35.0716

*When we solve for x, we get x=2.2x10*^{-6}; but notice, this was not hydrogen ion concentration; we created a base in this reaction: x is equal to the OH^{-} concentration, which implies that the pOH is equal to the negative log of this.0722

*The pOH ends up being 3.2 (was that correct?--yes), and (wait, is that...yes, it is) then the pH is equal to 14 minus the pOH, which equals 10.8.*0742

*Sure enough, a pH of 10.8 means that it is a basic solution.*0773

*We had a salt; the anion was the conjugate base of a weak acid; therefore, it is going to react as a base with the water in a standard base reaction.*0778

*That is what bases do: bases take hydrogen from water to create hydroxide ion.*0794

*The equilibrium expression for that: we said it's a K*_{b}, not a K_{a}; it's not an acid dissociation--it is a base association, if you will; it's a base constant.0799

*Well, we have the K*_{a}; normally, hydrofluoric acid is listed as a K_{a}, because it behaves mostly as an acid.0810

*Therefore, the K*_{a} is listed; but the relationship between K_{a} and K_{b} is K_{a} times K_{b} equals 10^{-14}, which is K_{w}.0818

*We solve for the K*_{b}; we treat it like any other weak base problems; and we solve it; we get the pOH, in this case, because it's a base--because we want the pH, we take 14 minus that, 10.8.0826

*I hope that makes sense.*0839

*OK, let's do Example 2: oh, actually, before that, let me actually...so let's stop there, and now let me go back to blue ink...*0841

*Now, if you have a salt (now we are going to talk about the cation; we mentioned the anion; now we are going to talk about the cation) where the cation is the conjugate acid of a weak base, then this cation will act as an acid and create hydrogen ion--create an acidic solution.*0858

*So, you have to watch the salt; you take a look at the salt, dissolve it; you take a look now--you not only look at the anion--you look at the cation as well.*0919

*For the first one, we said if the anion happens to be the conjugate base of a weak acid; now, if the cation happens to be the conjugate acid of a weak base.*0928

*It is going to act as an acid, and it is going to produce an acidic solution.*0944

*Let's do an example, and I think it will make sense.*0949

*And again, it is the chemistry that you want to understand: you want to take a look at the species and decide how it is acting.*0954

*That is really what is going on; in this previous problem, we saw that, when we dissolved the sodium fluoride, we have F*^{-} floating around freely.0962

*Well, what is F*^{-} going to do?--F^{-} happens to be the conjugate base of a weak acid, so it is going to actually behave as a base; it is going start taking Hs away from water.0969

*If you write down the reaction, everything should fall out.*0980

*OK, calculate the pH of a 0.10 Molar ammonium chloride solution; the K*_{b} of NH_{3} equals 1.8x10^{-5}.0985

*OK, our major species: well, we know that anything that involves ammonium and chloride is going to be fully soluble; therefore, what is floating around in solution is ammonium ion, chloride ion, and H*_{2}O.1013

*Chloride--let's look at the anion first--chloride is the conjugate base of a strong acid, HCl.*1031

*So, Cl is not going to take any H from anything; it is just going to float around freely--we can ignore it.*1040

*However, NH*_{4}^{+}, ammonium, is the conjugate acid of a weak base.1047

*How do we find the conjugate acid?--just stick a hydrogen ion onto it...of the weak base ammonia.*1056

*Ammonia is the weak base; its conjugate acid is that, right?--because it comes from ammonia.*1062

*When you put ammonia in water, it takes a hydrogen from water; it becomes ammonium, and it creates hydroxide.*1074

*However, in this case, we didn't just drop ammonia into solution; we dropped the actual ammonium ion into solution as a salt.*1084

*We dropped it as a salt; now, there is just free ammonium ion floating around--all ammonium ion.*1093

*Well, ammonium ion is the conjugate acid of a weak base, which is ammonia.*1099

*Therefore, it is going to now behave as an acid.*1104

*Its equilibrium is going to be this: it is going to actually dissociate into H*^{+} + Cl^{-}...no, plus NH_{3}.1107

*There is a K*_{a} associated with this, because this is an acid dissociation reaction: it is something that has a hydrogen ion to give up.1122

*It gives it up, and it creates its other side; so now, this K*_{a} is equal to K_{w} over the K_{b}.1130

*K*_{b} is the K_{b} that we have for ammonia, the conjugate base of that.1138

*So now, let's solve the problem.*1144

*NH*_{4}^{+} is the dominant species; it is the conjugate acid of a weak base--it is going to behave as an acid, so we write it: behaving as an acid.1147

*We are going to create H*^{+}; we are going to create an acidic solution, because we dropped it in water that was neutral.1164

*We have Initial, we have Change, and we have Equilibrium; well, the initial concentration of NH*_{4} is 0.10.1171

*There is no H to start with; there is no ammonia to start with; this is -x; this is x; this is x; +x; this is 0.10-x; this is +x; this is +x.*1180

*Now, it's behaving as an acid, so we need a K*_{a}.1194

*Well, a K*_{a} is equal to K_{w} over K_{b}.1199

*That is equal to 1.0x10*^{-14}, divided by 1.8x10^{-5}; the K_{a} for this reaction is equal to 5.6x10^{-10}.1204

*That tells me, this 5.6x10*^{-10}...it's a small number; it is a weak acid.1222

*But, it is still an acid--it is going to produce some H*^{+}...it's weak, but it still behaving as a weak acid--it's going to give up its H^{+} into floating around freely in the water; that is the acidic solution.1228

*The pH of the solution is going to drop.*1239

*So now, let's take 5.6x10*^{-10} is equal to x, times x, over 0.10-x, which is approximately equal to x squared over 0.10; we end up with x=7.4x10^{-6}, which does equal the hydrogen ion concentration, because here, we are directly forming hydrogen ion, and therefore, the pH of this is the negative log of that, which ends up being 5.1.1241

*Sure enough, 5.1 is an acidic solution.*1279

*So there you have it: recap: if you have a salt floating around in solution (if you have a salt and it dissolves), if the anion is the conjugate base of a weak acid, it will create a basic solution.*1284

*You treat it as a base equilibrium that will react with water, in other words--pull off the hydrogen ion.*1301

*If the cation is the conjugate acid of a weak base, it will behave as an acid--you write the acid equilibrium and use the K*_{a} to solve it as a weak acid problem, like you have done before.1307

*It will create an acidic solution.*1318

*A second type of cation--a second type of species (you know, it might be nice if I actually wrote my words properly here--what do you think?) that creates an acidic solution--is a highly-charged metal ion.*1321

*A good example is aluminum 3+; 3+ is a pretty high charge; you are also going to find solutions like, for example, chromium 6+, manganese 5+, 4+...things that have a really high charge after about 1 or 2.*1362

*Here is what is going on: if you take a salt like aluminum chloride, I'm going to actually write out the chemistry of what happens, because I want you to see what happens--how the species forms in solution, and then how we treat that species in solution, just like any other weak acid.*1376

*If you take aluminum chloride, and you dissolve it in water--well, you know that aluminum chloride is completely soluble, so you are going to end up with aluminum 3+, plus 3 chloride ions--it completely dissociates into its free ions.*1395

*Chloride doesn't do anything; it's the conjugate base of a strong acid, so it doesn't behave as a base--it just floats around.*1408

*However, aluminum (because it is so highly charged, and because it is in water)--the water molecules, the actual molecules themselves, surround the aluminum; and in the case of aluminum, you get something that looks like this.*1415

*I am actually going to draw the structure, so you see it; I'm going to write it out first.*1431

*Aluminum (so I'm going to bring it over here)--the aluminum 3+ actually associates with 6 water molecules (and I'm going to draw the lone pairs on them) to create this species called a complex ion.*1435

*Al(H*_{2}O)_{6}, and the whole charge on the species is 3+; this is usually how we write it.1456

*Any time you have a metal that is surrounded by, in this case, water (we will just deal with water for the moment), and this actually looks like something--you have an aluminum ion, and you have an H*_{2}O, an H_{2}O...I'm going to draw this 3-dimensional structure.1465

*I really shouldn't, because you really don't have to know this for the time being; we may talk about it a little bit later, towards the end of the course, but I think it's important; it's nice--there is nothing here that you shouldn't be able to sort of visualize.*1487

*There are these 6 water molecules: OH*_{2}, OH_{2}, OH_{2}, OH_{2}...and it is actually the oxygen, believe it or not, that is attached to the aluminum here.1499

*In a normal bond, it is just surrounding it; so you have these 6 water molecules, and they are arranged; four of them are arranged in a plane; so aluminum is in the middle; one here, here; one is here; and back here.*1512

*These dashed lines mean it is going back; it's facing away from you.*1525

*These wedges mean it's coming towards you; these straight lines mean there is one on top and one on bottom.*1528

*What you have is the following: what you have is aluminum, in the center; you have a water molecule here, a water molecule here, a water molecule back here, a water molecule back here; one up here, and one down here.*1535

*There are six of them around it; and this whole species is carrying a 3+ charge; well, of course it's carrying a 3+ charge--aluminum is 3+, and water is a neutral molecule, so when they aggregate (when they sort of grab onto aluminum, if you will--6 of them), you create this thing called a complex ion.*1548

*Well, here is what happens with this complex ion: now I'm going to take this complex ion; as it turns out, it's so highly charged that it actually ends up pulling a lot of the electrons towards itself, and actually creates an acidic hydrogen.*1569

*One of these hydrogens, believe it or not, actually comes off.*1584

*And here is the chemistry of it (and this is what is important: the structure you don't have to understand; the chemistry is what is going on): in solution, this is produced.*1588

*When you drop aluminum chloride into water, aluminum ion is formed; water is attracted to the aluminum ion, and six water molecules arrange themselves in a pattern around the aluminum to create this complex ion.*1598

*This complex ion has an acidic hydrogen; it gives it up.*1612

*Here is how it gives it up: Al(H*_{2}O)_{6}^{3+} (I'm going to leave off the brackets; I think it's just extra symbolism that is not necessary)--it dissociates into H^{+} + Al; now there is just a hydroxide attached to it, and there is 5 waters.1615

*One of the waters has lost its hydrogen; now it's 2+.*1639

*Notice, this is just a standard dissociation of an acid.*1644

*You have this species that has this H that it can give up; it gives it up; it's right there.*1649

*The rest of it--it doesn't even matter what it is; all that matters is this.*1656

*You have a species that has a hydrogen ion; it gives up that hydrogen ion to create some conjugate base; this is what is important.*1660

*You treat this like any other weak acid; there is a K*_{a} associated with this--we measured it, and the K_{a} of this happens to be 1.4x10^{-5}.1667

*That is it; this is just HA dissociating into H*^{+} + A^{-}; this A^{-}--yes, it happens to be a very, very complex-looking thing (we call it a complex ion), but you treat it the exact same way.1679

*Don't let the makeup of the thing that you are discussing confuse you; it's the chemistry that matters--the chemistry is just: some species gives up a hydrogen ion, and then ends up as something else.*1694

*The hydrogen ion is what is important; the equilibrium is handled the exact same way.*1708

*So now, let's do a problem.*1713

*Example 3: This is what is important in science--you need to understand what is going on underneath.*1717

*The individual identities of the species--they don't really matter, as far as what is going on; they matter for the individual case that you happen to be dealing with, as a researcher, as a lab scientist, as a doctor, whatever it is, but the chemistry is all the same.*1726

*It is still just some species, some acid, that has a hydrogen to give up, and it gives it up.*1742

*The mathematics is handled exactly the same way; the species, the identity, is entirely irrelevant.*1747

*It is entirely irrelevant; this is what we want you to do.*1752

*Our ideal is to get you to think abstractly, to think big-picture; if you can handle the big picture, you will know what is going on with the little picture; the little details are just incidental--they change from problem to problem.*1756

*But, the big picture doesn't change; that is what is important.*1769

*In science, what you want to concentrate on is what doesn't change.*1772

*The things that change...well, they are incidental.*1777

*That is when you know something is important--if something is not changing, that is what you want to concentrate on.*1780

*It is the chemistry that is important; OK.*1785

*Sorry about that lecture.*1788

*So, Example 3--we have: Calculate the pH of a 0.010 Molar AlCl*_{3} solution.1790

*OK, so the first thing we do: major species. *1808

*Well, here is what we know: when we have some aluminum salt that is dropped in water, the aluminum is going to float around freely as ion; that is the first thing that is going to happen--the aluminum chloride is going to dissociate.*1813

*Aluminum is...the water molecules are going to aggregate around aluminum, 6 of them are, and they are going to form the species Al(H*_{2}O)_{6}^{3+}.1827

*Now, it isn't important that you know the name of it, but this is called hexaaqua-aluminum (3).*1841

*You will actually do the naming towards the end of the course, when you talk about coordination compounds, but this is the species that actually forms in solution.*1846

*If we took a picture of the solution, that is the species that we find.*1854

*We find every aluminum ion surrounded by six water molecules, and that whole thing is carrying a 3+ charge.*1858

*The only other species in there is water.*1864

*Well, it is true--water does contribute some hydrogen ion--but it is K*_{a} of 10^{-14}; this one has a K_{a} of 1.4x10^{-5}.1867

*10 to the negative 5 is a lot bigger than 10 to the negative 14, so water can be ignored.*1881

*This is the dominant species in the water that will control the pH of the solution.*1885

*We know what this does: it behaves as an acid.*1893

*Al(H*_{2}O)_{6}^{3+} dissociates into H^{+} + AlOH(H_{2}O)_{5}^{2+}--just an acid dissociation; that is it.1896

*This is being created; let's do Initial; let's do Change; let's do Equilibrium.*1919

*What is the initial concentration?--well, all of the aluminum is dissolved; that means all of this is formed; 0.010--there is nothing formed yet; there is nothing formed yet.*1924

*A certain amount is going to dissociate--that is how much is going to show up of the other species.*1935

*0.010-x, +x, +x; we have the K*_{a}; we have the equilibrium expression; so we just put it in.1940

*1.4x10*^{-5} (I hope you're not getting sick of these problems; I know it's just over and over--it's the same thing; that is nice--we like patterns) equals x times x, divided by 0.010-x.1951

*We can approximate this with x squared over 0.010; now, you might think to yourself, "Well, wait a minute; 0.010 is pretty small, and x...we are talking about 10 to the negative 5 here...maybe."*1972

*As it turns out, when you check the validity, the error ends up being about 3.7%; we are still below 5, so we are good; this is a perfectly good approximation.*1987

*So, x is equal to 3.7x10*^{-4}, which equals the hydrogen ion concentration; that implies that the pH is equal to 3.43.1998

*3.43--OK, I think I want to write this a little bit slower, so that all these wacky lines don't show up--equals 3.43.*2013

*How is that?*2026

*So notice, it's handled exactly the same way; the identity of the species is irrelevant--it's behaving as an acid; it's giving up a hydrogen ion.*2027

*It's a weak acid, 1.4x10*^{-5}; there is an equilibrium; we have to use an ICE chart.2036

*We are done; that is nice; OK.*2043

*Now, our final little situation here: what if we have this situation?*2048

*What if we have the situation: NH*_{4}F--what if we have ammonium fluoride?--it is a perfectly valid salt.2052

*Ammonium chloride...we can form ammonium fluoride.*2070

*Salt--you drop it into water; what happens?--well, it's going to dissolve, if it's completely soluble; it's going to dissolve into NH*_{4}^{+} and F^{-}.2074

*Well now, I have a little bit of a problem: we have an anion which is the conjugate base of a weak acid, hydrofluoric acid; so it is going to behave as a base, and it is going to pull hydrogen off of water to produce hydroxide ion.*2083

*So, it is going to create a basic solution; but, we have ammonium ion also floating around in solution.*2099

*It is the conjugate acid of a weak base, ammonia, and it is going to behave as an acid, as a weak acid, itself.*2106

*It is going to give up its hydrogen ion to create an acidic solution.*2115

*The F*^{-} is going to go ahead and create basic solution; this is going to create an acidic solution; well, what is the final solution going to be--acidic or basic? How do we decide?2118

*Well, when you have a situation where both of the ions are species that react, and one produces base; one produces water, it gets very, very complicated--that is the short answer.*2128

*The equilibrium gets complicated, and you actually will be dealing with stuff like this, if you go on to study analytical chemistry.*2141

*If you are a chemistry major, usually in your third year, you will take an analytical chemistry course, and there are ways to handle this mathematically.*2147

*Pretty complex; for our purposes, we just want to be able to sort of give a qualitative answer.*2154

*We want to be able to say is the solution acidic or basic, without specifying what the pH is.*2159

*That is actually very, very easy to do.*2164

*It comes down to this: you do a quick test--if (I'm actually going to write this a little further to the left--excuse me) the K*_{a} for the acidic ion (meaning this one) is bigger than the K_{b} for the basic ion, which is this one, well, the solution is acidic.2168

*In other words, if the K*_{a} for this is bigger than this, that means that the equilibrium for this is farther to the right; it produces more hydrogen ion than this produces hydroxide ion.2207

*Therefore, there will be more hydrogen ion in solution; therefore, the solution will be acidic.*2219

*That is what this is saying: so, you need to find the K*_{a} of this; you find the K_{b} of this; remember, K_{b} is 10 to the 14 over K_{a} of the acid, and this K_{b} is 10 to the 14 over K_{a}.2223

*I'm sorry; K*_{a} is 10^{-14}/K_{b} for the conjugate base, what we did earlier.2238

*You compare the two; the bigger one--that will dominate.*2245

*If this is bigger, it will be acidic; if this is bigger, it will be basic.*2249

*So, if the K*_{a} is less than the K_{b} (the K_{a} for the acidic ion, less than the K_{b} for the basic ion), then your solution is going to be basic.2254

*And of course, the last possibility (always three possibilities when it comes to ordering: less than, greater than, or equal to): If the K*_{a} is equal to the K_{b}, well, you know exactly what that is; that is going to get a neutral solution.2265

*And now, let's do our final example: Example 4: Will a solution of aluminum sulfate (AlSO*_{4}, Al_{2}(SO_{4})_{3}) be acidic or basic?2283

*Well, Al*_{2}(SO_{4})_{3} dissociates into Al_{2}, aluminum 3^{+}, plus 3 SO_{4}^{2-}, so yes; we have an anion--negative ion--which is the conjugate base of a weak acid.2315

*The weak acid, in this case--just add one H.*2339

*OK, it's HSO*_{4}^{-}; not H_{2}SO_{4}; it's HSO_{4}^{-}--add one H.2343

*And aluminum happens to be that thing that forms that species, Al, the hexaaqua-aluminum (3); so this is going to create a basic solution; this is going to create an acidic solution; we need to compare the two.*2350

*So, we want to know the K*_{a} of this; the K_{a}--we already know that one; that is going to be 1.4x10^{-5}.2369

*And we want to know the K*_{b} of this; the K_{b} of this is 10 to the -14, over the K_{a} of this, which is 1.2x10^{-6}.2383

*I hope you guys saw what I did; I found out the species that are floating in solution; this--the conjugate acid of that is the HSO*_{4}.2401

*This actually just forms this species; we have the K*_{a} of this; we find the K_{b}; we don't take the H_{2}SO_{4}; we add one H to it, OK?--one H at a time.2411

*It dissociates one at a time; it associates one at a time.*2422

*We get a K*_{b}, is equal to 1.3x10^{-13}; well, this is hugely bigger than this, which means that our solution will be acidic.2427

*This aluminum, this aluminum species, will dominate the acidity of the solution; we will get an acidic solution; and that is how you handle it.*2444

*So, thank you for joining us here at Educator.com to discuss the acid-base properties of salts.*2454

*In our next lesson, I'm going to close off with just a brief discussion of some oxides, and then we will go ahead and move on to further aspects of acid-base equilibria.*2459

*We will talk about buffer solutions and titration curves.*2470

*Take care; see you next time; goodbye.*2472

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

*Today, we are going to close out with a couple of comments--close out the acid-base discussion with a couple of comments on some oxides--the covalent oxides and the non-covalent oxides.*0004

*We are just going to briefly touch on them--not too much, because I really want to get into the common ion effect and buffer solutions.*0017

*It is going to be a very, very, very important application of acid-base chemistry, probably the most important application of acid-base chemistry.*0023

*But, I didn't want to leave out this tail-end of sort of straight, normal, run-of-the-mill, routine acid-base stuff, and I wanted to talk about some of the oxides, because the chemistry does come up, and it's a little strange how certain oxides produce acidic and basic solutions.*0032

*If you remember, last time we talked about salts; and when they dissolve (the ones that are soluble), either the anion happens to be the conjugate base of a weak acid, or the cation happens to be the conjugate acid of a weak base.*0050

*Well, these things react with water, and they actually produce acidic or basic solutions.*0064

*Well, kind of the same thing happens with oxides; so let's take a quick look, and we'll just write down what they are.*0069

*We won't get too much into the chemistry of it, but I am going to demonstrate, actually, how it happens, because I want you to be able to see that this doesn't just fall out of the sky--that it is just a movement of electrons and things actually do make sense.*0077

*It isn't just about memorizing reactions; it's about understanding why things react the way that they react.*0090

*OK, so let's write down...covalent oxides (and you will see what we mean in just a minute) will produce acidic solutions in water.*0095

*In other words, if I take a covalent oxide, if I drop it in water, it is actually going to form an acidic solution.*0120

*Here are some examples: SO*_{3}: SO_{3} is a covalent oxide--"covalent" because you are talking about a nonmetal-nonmetal bond.0125

*Sulfur and oxygen are both nonmetals; it's a covalent bond.*0134

*Plus H*_{2}O: it ends up forming H_{2}SO_{4}, and when H_{2}SO_{4} forms, it's a strong acid in its first dissociation, so it dissociates into H^{+} + HSO_{4}^{-}.0138

*That is why it is an acidic solution: you drop SO*_{3} into water; it forms H_{2}SO_{4}; H_{2}SO_{4} dissociates, producing hydrogen ion; therefore, you have an acidic solution.0153

*SO*_{2}: if you bubble that into water, you end up with H_{2}SO_{3}, which is sulfurous acid, which is a weak acid.0164

*That is why we have a double arrow here, showing equilibrium; but again, the dissociation is the same: HSO*_{3}^{-}.0175

*CO*_{2}: if you bubble CO_{2} into water at high pressure, you end up with carbonic acid.0182

*Carbonic acid is a weak acid, and there is an equilibrium with that and the hydrogen carbonate ion.*0193

*So, in each case, you see that H*^{+} is being produced down the line, downstream of the actual reaction that is taking place.0199

*And then, let's just do one final one; let's do NO*_{2} + H_{2}O: it produces (it's actually going to be 2 NO_{2}) nitric acid, and it also produces nitrous acid.0207

*You will find both species in solution; nitric acid is strong, so it dissociates into H*^{+} + nitrate; and here, we are going to have an equilibrium with H^{+} + nitrite.0221

*There you go: covalent oxide--"covalent," meaning that the thing bonded to the oxygen--they are both nonmetals, basically; the thing bonded to oxygen is a nonmetal, so sulfur...any nonmetal.*0235

*These oxides, when dissolved in water, react with water to form acidic solutions.*0249

*Now, ionic oxides, on the other hand: ionic oxides produce basic solutions in water.*0255

*And you know, ionic means metal-nonmetal; so ionic oxide is basically a metal with oxygen; so, an example would be calcium oxide.*0275

*With calcium oxide, when it reacts with water, what you end up producing is calcium hydroxide.*0284

*Potassium oxide (K*_{2}O) plus water: you end up producing potassium hydroxide.0294

*Potassium hydroxide is a strong hydroxide, and it dissociates into 2 hydroxides, plus 2 potassium ions.*0304

*Calcium hydroxide is moderately soluble; actually, it's not very soluble at all, but it is slightly soluble; therefore, it dissociates into 2 OH*^{-}, plus calcium 2+.0313

*We write it as an equilibrium, because in fact, most of it is actually over here.*0326

*This is mostly a solid; this is potassium hydroxide; this is an aqueous solution; so this just means that it is not very soluble, but enough of it does dissolve to produce a hydroxide ion, and hydroxide ion--that is why we have a basic solution.*0330

*So again, an oxide--a covalent oxide--produces acidic solutions when reacting with water; an ionic oxide (basically, oxygen plus a metal, so any ionic bond between oxygen and a metal)--when those react with water, they produce basic solutions.*0349

*OK, now I'm going to quickly discuss how these are actually formed--just going to give some quick examples.*0370

*This is definitely not something that you have to know; however, I want you to see it, because I think it is important to see it; it is important to get a sense of the chemistry and to feel comfortable with it.*0375

*Those of you who go on to study organic chemistry: this is going to be a huge part of what you do--this sort of what we call arrow-pushing...movement of electrons.*0385

*Let's do the CO*_{2} example: the CO_{2}, plus H_{2}O, goes to H_{2}CO_{3}, which dissociates into H^{+} + HCO_{3}^{-}.0394

*Here is what actually happens: CO*_{2} is a linear molecule; each oxygen is double-bonded to the carbon.0409

*Water, as you know, or as you should know, maybe (and if you don't know, that is not a problem--we are actually going to get to it, because you remember: in this particular AP Chemistry course, we actually skipped over the bonding--I wanted to get to this stuff--equilibrium and acids and base--and I will actually return to the bonding)--a water molecule consists of oxygen single-bonded to hydrogens, and it has a couple of lone pairs.*0415

*Well, as it turns out, these lone pairs--what they do is: they attack the carbon, and they push the electrons onto one of the oxygens--one pair of electrons that are part of the double bond here.*0443

*What you end up getting is something that looks like this: O (now that oxygen has an extra electron), and then you have the oxygen, this and this, and 1 lone pair.*0457

*Well, because you have three things bonded to an oxygen, that is actually an extra bond; so now, this is carrying a positive charge.*0469

*You remember, charge has to balance on both sides of an arrow; here, this is neutral--everything is neutral here--net.*0476

*Here, this is minus; this has to be plus; that is how this works.*0483

*You will learn more about that later; I am just trying to get you to see what, exactly, goes on.*0487

*What happens next is: there is a transfer of a proton, and I am going to represent it like this: these negative charges actually take this hydrogen, and they push the electrons onto this oxygen, and what you get is OH, OH; there, you have your H*_{2}CO_{3}.0492

*Now, this ends up going into equilibrium with...that is how it happens.*0514

*Water attacks the carbon dioxide; there is a shift; there is a transfer of a proton over to oxygen; and what you end up with is carbonic acid.*0522

*Carbonic acid dissociates to produce H*^{+}.0529

*I just wanted you to see that things don't just fall out of the sky; it actually makes sense what happens--this is reasonable.*0533

*OK, now let's do a calcium oxide example.*0541

*We have calcium oxide, plus H*_{2}O, going to calcium hydroxide.0545

*Here is how it happens: calcium...this is an ionic bond--this is 2+; this is O*_{2}^{-}; well, O_{2}^{-}...here is what happens: H, again, with the oxygen; these electrons over here...this is fully...there are 8 electrons around here.0554

*They actually take one of these; they push the electrons onto here; what you end up with is, now, calcium 2+ plus an OH*^{-} plus another OH^{-}.0570

*This O*_{2}^{-} takes an H^{+} and becomes OH^{-}, right?--2 minus, 1 plus, is minus 1; it's an OH.0583

*That leaves an OH, so that is the other OH*^{-}.0589

*Now, these form your calcium hydroxide, which is mostly insoluble--it's actually kind of a solid--it's pretty solid; but it is moderately (you know what, I'm not going to have these stray lines running around all over here, so let me do it this way--let me go slow)...*0594

*Calcium hydroxide is in equilibrium with calcium 2+, plus 2 hydroxide ion, and there is your hydroxide ion to form a basic solution.*0615

*I just wanted you to see it: covalent oxides form acidic solutions; ionic oxides form basic solutions--standard chemistry.*0627

*It will show up on the AP Chemistry exam, in terms of actual reactions; so they might say something like "aluminum oxide," "iron oxide--does it form an acidic or a basic solution?"*0639

*Well, iron oxide is an ionic oxide; therefore, it forms a basic solution.*0649

*If you have sulfur trioxide, does that form an acidic or a basic solution when dropped in water?*0655

*Sulfur trioxide is a covalent oxide: when you put it into water, it forms an acidic solution.*0661

*Qualitatively, it does show up on the AP exam; but I wanted you to see why--what actually happens chemically.*0667

*I want you to get a sense that these things--it's not magic; it's just atoms that are slamming into each other, and electrons are moving around; that is all that is happening--very intuitive stuff, really.*0674

*OK, now we are going to move on to a profoundly, profoundly important concept called the common ion effect.*0685

*The common ion effect, in its application, is used to talk about buffer solutions; so let's talk about the common ion effect--let's get a little bit of the math and chemistry under our belts.*0693

*And then, we will talk at length about buffer solutions; there are going to be, actually, two or three lessons strictly devoted to buffer solutions, because they are profoundly important.*0705

*Your blood is a buffer solution; it is buffered by the carbonic acid buffer system, the carbonate buffer system.*0715

*It is used to maintain the pH of the blood somewhere between about 7.2 and 7.4.*0719

*The phosphate buffer system inside the cell--that maintains the pH at a certain value inside the cell.*0727

*If pH goes up or down, terrible things start to happen; there is actually a very, very narrow range, as far as healthy body function, when it comes to blood pH and intracellular pH.*0735

*OK, so let's start with a definition.*0748

*Definition: When an equilibrium already exists in a solution, the shift that occurs when you add an ion (a little metathesis here) already involved in the equilibrium is called the common ion effect.*0752

*It is just an application of Le Chatelier's Principle, which you will see in just a moment.*0829

*Let's say this again: When an equilibrium already exists in a solution, the shift that occurs when you add an ion already involved in the equilibrium is called the common ion effect.*0833

*So, for example, if we have an equilibrium that exists in hydrofluoric acid solution, H*^{+} + F^{-}, we know that hydrofluoric acid is a weak acid--there is a partial equilibrium.0844

*It is not completely dissociated, but enough of it is dissociated to produce F*^{-} and H^{+}.0858

*Now, the question is: What happens to the above equilibrium when some KF is added (when some potassium fluoride is added)?*0864

*Well, you know that potassium fluoride is a soluble salt, which means that potassium fluoride will completely dissociate into free K*^{+} + F^{-}.0883

*OK, you have this equilibrium right here (let's do it in red); you have an equilibrium that has been established, and all of a sudden, to the solution, you decide to add some potassium fluoride.*0895

*Well, the potassium fluoride dissolves; now, there is fluoride ion floating around, but there is already fluoride ion floating around.*0904

*Here is what we mean by an ion that is already involved in the equilibrium.*0910

*There is already fluoride involved in this equilibrium; now, you have added a salt that produces more of this ion; which way is it going to shift this equilibrium?*0915

*Well, Le Chatelier's Principle says: now that we have added more of this F*^{-} (which is over here), the system is going to shift in the direction that opposes the change.0924

*You have added F*^{-} over to the right, the product side...well, you have just added F^{-}; now, the equilibrium is going to shift in a direction which reduces the concentration of F^{-}, which means it's going to shift that way.0936

*It is going to use up F*^{-}; it is going to produce HF.0949

*So literally, what is happening, though--what is actually happening is: this F*^{-} is going to end up suppressing the actual dissociation of this HF.0953

*So, you can either establish this equilibrium, and then add the F*^{-}, and then push the equilibrium back that way; or you can add them simultaneously, and the existence of this F^{-} floating around will keep this from actually moving forward.0962

*It will keep the HF from dissociating.*0977

*There are two ways of looking at it, because again: you can establish the equilibrium, then add the common ion; or, you can just throw everything in and then let everything come to equilibrium on its own.*0979

*But, what is happening is Le Chatelier's Principle: you have an ion involved in an equilibrium, and then you have the presence of that same ion, and it is going to affect this equilibrium.*0988

*That is all that is happening--a common ion effect.*0998

*Two things: HF and KF--they share a common ion, the F*^{-}; this is the common ion effect.1002

*OK, let's just do an example, and I think it will start to make sense.*1009

*Example 1: Calculate the hydrogen ion concentration of a 1.2 Molar HF solution (something we have done before); then, calculate the hydrogen ion concentration of a solution that is 1.2 Molar hydrogen fluoride and 1.2 Molar sodium fluoride.*1016

*So, calculate the hydrogen ion concentration (or the pH, if you want to take the negative log of it)--calculate the H*^{+} concentration--of a 1.2 Molar HF solution; then, calculate the H of a solution that is