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Diffusion of Gas and Graham’s Law .txt | Now, to find the mean free path or the average distance between a two collisions, I simply add up all the distances up and divide by the number of collisions. And that's my mean free path. Now, because of this, we can use Grams law to approximate diffusion and we'll see why in a second. So let's look at an illustration of diffusion. Suppose I have a cylindrical tube and I have two claps. Suppose I take this clap, I soak it into ammonia NH three and plug it up. |
Diffusion of Gas and Graham’s Law .txt | So let's look at an illustration of diffusion. Suppose I have a cylindrical tube and I have two claps. Suppose I take this clap, I soak it into ammonia NH three and plug it up. Suppose I take this cloth, I soak it up in hydrochloric acid and then plug it up as well. So it's plugged up on both sides and I have air molecules in the middle. So orange guys are air molecules, green guys armonia molecules and red guys are hydrochloric molecules. |
Diffusion of Gas and Graham’s Law .txt | Suppose I take this cloth, I soak it up in hydrochloric acid and then plug it up as well. So it's plugged up on both sides and I have air molecules in the middle. So orange guys are air molecules, green guys armonia molecules and red guys are hydrochloric molecules. So what will happen? Well, some of these guys will evaporate and some of these guys will evaporate and they will begin moving. But they won't move directly from this point to this point. |
Diffusion of Gas and Graham’s Law .txt | So what will happen? Well, some of these guys will evaporate and some of these guys will evaporate and they will begin moving. But they won't move directly from this point to this point. They will move via a crooked path because of something called the mean free path because there are air molecules present. And these air molecules will create collisions. And this green molecule, for example, will first collide with this guy, then move here, close the wall, then collide with this guy, and so on until it gets to some point here. |
Diffusion of Gas and Graham’s Law .txt | They will move via a crooked path because of something called the mean free path because there are air molecules present. And these air molecules will create collisions. And this green molecule, for example, will first collide with this guy, then move here, close the wall, then collide with this guy, and so on until it gets to some point here. Now, when the green guy reaches the red guy, something will happen. Well, the reaction is as follows. The green guy reacts with the red guy, or ammonia reacts with hydrochloric acid. |
Diffusion of Gas and Graham’s Law .txt | Now, when the green guy reaches the red guy, something will happen. Well, the reaction is as follows. The green guy reacts with the red guy, or ammonia reacts with hydrochloric acid. In a gas station to form a precipitate, it forms a solid. So when these guys meet, at whatever point they will meet, they will form a wall or a solid wall. The precipitate. |
Diffusion of Gas and Graham’s Law .txt | In a gas station to form a precipitate, it forms a solid. So when these guys meet, at whatever point they will meet, they will form a wall or a solid wall. The precipitate. And my question is, at which point will the wall lie? Will it be in the middle? Will it be on this side? |
Diffusion of Gas and Graham’s Law .txt | And my question is, at which point will the wall lie? Will it be in the middle? Will it be on this side? Or closer to hydrochloric acid? So, we can use Grounds law to approximate the position of this wall of solid. And the way we do it is the following. |
Diffusion of Gas and Graham’s Law .txt | Or closer to hydrochloric acid? So, we can use Grounds law to approximate the position of this wall of solid. And the way we do it is the following. So rate one over rate two equals square root of mass of two divided by the square root of mass of one equal. So what's the molecular weight or mass of my ammonia? Well, it's 14 plus three gives us 17 on the bottom. |
Diffusion of Gas and Graham’s Law .txt | So rate one over rate two equals square root of mass of two divided by the square root of mass of one equal. So what's the molecular weight or mass of my ammonia? Well, it's 14 plus three gives us 17 on the bottom. What about this guy? Well, one plus 35.5 gives us 36.5. So 36.5 on top. |
Diffusion of Gas and Graham’s Law .txt | What about this guy? Well, one plus 35.5 gives us 36.5. So 36.5 on top. We plug this into our calculator, and we get 1.5. So rate of molecule one is 1.5 times larger or faster than rate of two, because two is heavier. So it's not going to travel with the same velocity. |
Diffusion of Gas and Graham’s Law .txt | We plug this into our calculator, and we get 1.5. So rate of molecule one is 1.5 times larger or faster than rate of two, because two is heavier. So it's not going to travel with the same velocity. Velocity will be smaller. Remember, we're assuming constant temperature. So kinetic energies or average kinetic energies are equal. |
Naming of Alkenes.txt | So let's look at a few important rules that we have to use whenever we're naming alkanes. Rule number one, find the longest possible carbon chain containing all the double bonds. Rule number two, the lowest possible number value is given to the double bonds. Rule number three, if molecules contain more than one double bond, we give it a specific name. For example two double bonds, we name it a dying three double bonds, we name it a triangle. And rule number four, ring compounds containing double bonds are called cycloalkings. |
Naming of Alkenes.txt | Rule number three, if molecules contain more than one double bond, we give it a specific name. For example two double bonds, we name it a dying three double bonds, we name it a triangle. And rule number four, ring compounds containing double bonds are called cycloalkings. So here we have six examples. So let's look at example A, in which we're going to name our alkane. So our first and second step tells us that we have to find the longest possible carbon backbone and we have to assign our double bond the lowest possible number value. |
Naming of Alkenes.txt | So here we have six examples. So let's look at example A, in which we're going to name our alkane. So our first and second step tells us that we have to find the longest possible carbon backbone and we have to assign our double bond the lowest possible number value. So that means we have to begin on this end. So carbon number one, carbon number two, carbon number three, carbon number four. Notice in this case we have a four carbon backbone and our double bond has a one. |
Naming of Alkenes.txt | So that means we have to begin on this end. So carbon number one, carbon number two, carbon number three, carbon number four. Notice in this case we have a four carbon backbone and our double bond has a one. It gets assigned a number one because our double bond begins on carbon one. If we begin number eight, our backbone, from this end, our carbon double bond will get a three. And since we want the lowest possible number value according to step two or rule two, this is how we label it. |
Naming of Alkenes.txt | It gets assigned a number one because our double bond begins on carbon one. If we begin number eight, our backbone, from this end, our carbon double bond will get a three. And since we want the lowest possible number value according to step two or rule two, this is how we label it. So we name our alkene simply one. Butane so the in part simply means our double bond is found on the first position and we have a four carbon backbone. So butte means we have a four carbon backbone. |
Naming of Alkenes.txt | So we name our alkene simply one. Butane so the in part simply means our double bond is found on the first position and we have a four carbon backbone. So butte means we have a four carbon backbone. So let's go to example B. In example B, we have a symmetrical molecule, a symmetrical compound. And that simply means that it doesn't matter if we begin on this end or this end, we get the same alkyne, the same alkyne name. |
Naming of Alkenes.txt | So let's go to example B. In example B, we have a symmetrical molecule, a symmetrical compound. And that simply means that it doesn't matter if we begin on this end or this end, we get the same alkyne, the same alkyne name. So let's begin counting our carbons. Carbon one, carbon two, carbon three, and carbon four. Now now we have two double bonds. |
Naming of Alkenes.txt | So let's begin counting our carbons. Carbon one, carbon two, carbon three, and carbon four. Now now we have two double bonds. So according to rule number three, we have to name this compound a dyene. So our name becomes one three. Butene so the dying part means we have two double bonds, one on the first carbon and the second one on the third carbon. |
Naming of Alkenes.txt | So according to rule number three, we have to name this compound a dyene. So our name becomes one three. Butene so the dying part means we have two double bonds, one on the first carbon and the second one on the third carbon. Buta simply means we have a four carbon backbone. So let's go to example C.
So in example C, we have the following compound. So let's begin numbering. |
Naming of Alkenes.txt | Buta simply means we have a four carbon backbone. So let's go to example C.
So in example C, we have the following compound. So let's begin numbering. So remember, we want to find the longest possible carbon backbone that contains all the double bonds. So that means we either begin on this end and end on this end, or we begin on this end and go to this end. Since we want to find the lowest possible number values, we begin on this end. |
Naming of Alkenes.txt | So remember, we want to find the longest possible carbon backbone that contains all the double bonds. So that means we either begin on this end and end on this end, or we begin on this end and go to this end. Since we want to find the lowest possible number values, we begin on this end. So 123456 and seven. So we have a seven carbon backbone. Our first double bond begins on the first carbon. |
Naming of Alkenes.txt | So 123456 and seven. So we have a seven carbon backbone. Our first double bond begins on the first carbon. The second double bond begins on the third carbon. And also on the third carbon, we have this Ethyl group. So this group comes first, so three Ethyl. |
Naming of Alkenes.txt | The second double bond begins on the third carbon. And also on the third carbon, we have this Ethyl group. So this group comes first, so three Ethyl. And then we have one comma, three one comma, three Hecta, because we have a seven carbon backbone, dying because we have two double bonds. So once again, two double bonds, one on the first carbon, second one on the third carbon, we have an Ethyl group on the third carbon, and we have a seven carbon backbone hepta. So three Ethyl, one three Hepta, dying for this compound. |
Naming of Alkenes.txt | And then we have one comma, three one comma, three Hecta, because we have a seven carbon backbone, dying because we have two double bonds. So once again, two double bonds, one on the first carbon, second one on the third carbon, we have an Ethyl group on the third carbon, and we have a seven carbon backbone hepta. So three Ethyl, one three Hepta, dying for this compound. So compound D.
So now we have a ring, a compound, a ring structure. So let's begin labeling on this guy, on this carbon. So 123456. |
Naming of Alkenes.txt | So compound D.
So now we have a ring, a compound, a ring structure. So let's begin labeling on this guy, on this carbon. So 123456. So once again, I begin labeling or numbering on my double bond because I want my double bond to have the lowest possible number value. So this guy is known as one cyclone because it's a cyclic compound, and we have six, so hexene. So one cyclohexine, or simply cyclohexine. |
Naming of Alkenes.txt | So once again, I begin labeling or numbering on my double bond because I want my double bond to have the lowest possible number value. So this guy is known as one cyclone because it's a cyclic compound, and we have six, so hexene. So one cyclohexine, or simply cyclohexine. So let's go to part E, or compound E. So once again, I want to label my compound in a way such that my double bond gets the lowest possible number value, and this methyl group also gets the lowest possible number value. So I begin on this side. So one carbon, second carbon, third carbon, fourth carbon, fifth carbon, 6th carbon. |
Naming of Alkenes.txt | So let's go to part E, or compound E. So once again, I want to label my compound in a way such that my double bond gets the lowest possible number value, and this methyl group also gets the lowest possible number value. So I begin on this side. So one carbon, second carbon, third carbon, fourth carbon, fifth carbon, 6th carbon. So once again, I have a 6th carbon ring. And so I named this guy. So, notice that my methyl group is found on the third carbon. |
Naming of Alkenes.txt | So once again, I have a 6th carbon ring. And so I named this guy. So, notice that my methyl group is found on the third carbon. So I name it three methyl, one cycloxine. So three methyl simply means our methyl group is in the third carbon, and one cyclohexine means that my double bond is found on the first carbon, and we have a ring. So finally, we get to compound E. So in compound E, we have a cyclic six carbon backbone, and we have three double bonds. |
Naming of Alkenes.txt | So I name it three methyl, one cycloxine. So three methyl simply means our methyl group is in the third carbon, and one cyclohexine means that my double bond is found on the first carbon, and we have a ring. So finally, we get to compound E. So in compound E, we have a cyclic six carbon backbone, and we have three double bonds. So let's begin labeling or numbering. And actually doesn't matter where we begin numbering or labeling because this is a symmetrical compound. So 123456. |
Naming of Alkenes.txt | So let's begin labeling or numbering. And actually doesn't matter where we begin numbering or labeling because this is a symmetrical compound. So 123456. So we have one three five so one three and five cyclotex. One three five presents our double bonds. We have three double bonds, so we have a triangle, and we have a six member cyclic ring. |
Alkali and Alkaline Earth Metals .txt | Today we're going to compare and contrast two important groups or families found on our periodic table. We're going to look at alkali metals and alkaline earth metals. So let's begin with these guys. So the alkaline metals are found on group one or group one A on our periodic table. So that means these guys are metals or part of that a metal division. So that applies that they are soft, malleable and ductile. |
Alkali and Alkaline Earth Metals .txt | So the alkaline metals are found on group one or group one A on our periodic table. So that means these guys are metals or part of that a metal division. So that applies that they are soft, malleable and ductile. Duct till simply means they're stretchy or stretchable malleable simply means we can hammer them into thin sheets of metal and soft. Well, it simply means they're soft. These guys also display lust alike properties, which simply means they are shiny. |
Alkali and Alkaline Earth Metals .txt | Duct till simply means they're stretchy or stretchable malleable simply means we can hammer them into thin sheets of metal and soft. Well, it simply means they're soft. These guys also display lust alike properties, which simply means they are shiny. Now, metals are shiny, so that makes sense. And these guys, because they're metals, they conduct electricity. In other words, electrons are capable of moving from one point to another in alkali metals very easily. |
Alkali and Alkaline Earth Metals .txt | Now, metals are shiny, so that makes sense. And these guys, because they're metals, they conduct electricity. In other words, electrons are capable of moving from one point to another in alkali metals very easily. And that means because moving charge creates electricity, these guys create or conduct electricity very well. Now, they also form ions with a positive oxidation state. In other words, they're capable of losing electrons very easily, and therefore, they usually form plus one oxidation states. |
Alkali and Alkaline Earth Metals .txt | And that means because moving charge creates electricity, these guys create or conduct electricity very well. Now, they also form ions with a positive oxidation state. In other words, they're capable of losing electrons very easily, and therefore, they usually form plus one oxidation states. So a positive oxidation state. Now, these guys are highly reactive when you mix them with nonmetals and they form ionic compounds. For example, if we react these guys with an H, they will form something called metal Hydrides nah, lih, et cetera, in which the NA and the li both have a positive one oxidation state. |
Alkali and Alkaline Earth Metals .txt | So a positive oxidation state. Now, these guys are highly reactive when you mix them with nonmetals and they form ionic compounds. For example, if we react these guys with an H, they will form something called metal Hydrides nah, lih, et cetera, in which the NA and the li both have a positive one oxidation state. Now, if you mix the metals with water, they will react exothermically to produce a metal hydroxide and H two gas. Let's put this guy's in parentheses gas. So, for example, if you make two moles of sodium in a solid state with two moles of water, you will get two moles of metal hydroxide and 1 mol of H two or diatomic gas. |
Alkali and Alkaline Earth Metals .txt | Now, if you mix the metals with water, they will react exothermically to produce a metal hydroxide and H two gas. Let's put this guy's in parentheses gas. So, for example, if you make two moles of sodium in a solid state with two moles of water, you will get two moles of metal hydroxide and 1 mol of H two or diatomic gas. Now, let's look at the second type of group right next to our alkaline metals, known as alkaline earth metals. Now, these guys are obviously in group two or group two A on our periodic table. And these guys, just like the alkaline metals, are also part of the metal division. |
Alkali and Alkaline Earth Metals .txt | Now, let's look at the second type of group right next to our alkaline metals, known as alkaline earth metals. Now, these guys are obviously in group two or group two A on our periodic table. And these guys, just like the alkaline metals, are also part of the metal division. So they're metals. But unlike these guys, which are soft, these guys are harder and more dense. That means their molecules in a solid state are closer together per unit volume. |
Alkali and Alkaline Earth Metals .txt | So they're metals. But unlike these guys, which are soft, these guys are harder and more dense. That means their molecules in a solid state are closer together per unit volume. Now, since they're metals, they're also malleable duct till and they conduct electricity well. Now, these guys, unlike these guys, form plus two oxidation state. In other words, these guys lose not one electron but two electrons when they react with our nonmetals. |
Alkali and Alkaline Earth Metals .txt | Now, since they're metals, they're also malleable duct till and they conduct electricity well. Now, these guys, unlike these guys, form plus two oxidation state. In other words, these guys lose not one electron but two electrons when they react with our nonmetals. So that means they're more likely to lose electrons. Now, our alkaline earth metals are less reactive than the alkali metals, but still react with the nonmetals to form ionic compounds. For example, calcium reacts with hydrogen to form calcium Hydride CAH two. |
Rate Law .txt | And we said that rate law is a mathematical representation between the relationship of the concentration of reactants and our reaction rates. Now, we also said that rate law can only be determined using experimental results. And that's exactly right. Now, in this lecture, we're going to look at the following form reaction and try to determine our rate law using some experimental data. So let's begin. 1 mol of methyl acetate react with 1 mol of hydroxide to produce 1 mol of acetate ion and 1 mol of methanol. |
Rate Law .txt | Now, in this lecture, we're going to look at the following form reaction and try to determine our rate law using some experimental data. So let's begin. 1 mol of methyl acetate react with 1 mol of hydroxide to produce 1 mol of acetate ion and 1 mol of methanol. Now, let's conduct the following three experiments in which we measure in each experiment the concentration of methylacetate and hydroxide and we find the initial rate. Now, our goal is to see how our initial rate changes when we change our concentration of reactants. Now, the first experiment will serve as a control. |
Rate Law .txt | Now, let's conduct the following three experiments in which we measure in each experiment the concentration of methylacetate and hydroxide and we find the initial rate. Now, our goal is to see how our initial rate changes when we change our concentration of reactants. Now, the first experiment will serve as a control. We're going to basically compare our second and third experiment to our first experiment and see how our initial rate changes. So in the first experiment, we see that we have 0.5
molar of initial methyl acetate and 0.5 molar of our initial hydroxide. Now, when these two concentrations are 0.5
each, our initial rate is 0.2. |
Rate Law .txt | We're going to basically compare our second and third experiment to our first experiment and see how our initial rate changes. So in the first experiment, we see that we have 0.5
molar of initial methyl acetate and 0.5 molar of our initial hydroxide. Now, when these two concentrations are 0.5
each, our initial rate is 0.2. Next, our goal is to change one of these guys and see how our initial rate is influenced. So let's keep our initial hydroxide concentration the same and only change our initial concentration of methyl acetate. So let's double it. |
Rate Law .txt | Next, our goal is to change one of these guys and see how our initial rate is influenced. So let's keep our initial hydroxide concentration the same and only change our initial concentration of methyl acetate. So let's double it. So we double it to 0.1 molar, and this guy stays at 0.5
molar, and we see that our initial rate also doubles to 0.4. Now, that means that because we double this and our initial rate doubles, these guys are directly proportional. In other words, if you double this guy, you must double the initial rate. |
Rate Law .txt | So we double it to 0.1 molar, and this guy stays at 0.5
molar, and we see that our initial rate also doubles to 0.4. Now, that means that because we double this and our initial rate doubles, these guys are directly proportional. In other words, if you double this guy, you must double the initial rate. So let's conduct the same exact experiment. But now we keep our initial concentration of methyl acetate the same, and we double our concentration of our initial hydroxide. So let's stay at 0.1
molar and go from 0.5 molar of our hydroxide to 0.1 molar. |
Rate Law .txt | So let's conduct the same exact experiment. But now we keep our initial concentration of methyl acetate the same, and we double our concentration of our initial hydroxide. So let's stay at 0.1
molar and go from 0.5 molar of our hydroxide to 0.1 molar. We see that when we double our initial concentration of hydroxide, our initial rate also doubles. That means that our hydroxide is also proportional to our rate. So now with this result, we can find our rate law. |
Rate Law .txt | We see that when we double our initial concentration of hydroxide, our initial rate also doubles. That means that our hydroxide is also proportional to our rate. So now with this result, we can find our rate law. So, rate law is the following equation the rate of my reaction in the forward direction is equal to the rate constant of the forward reaction times the concentration of methylacetate times the concentration of hydroxide. Now, notice my exponents are each one. That means there is a direct relationship between our rate of reaction and our concentration. |
Rate Law .txt | So, rate law is the following equation the rate of my reaction in the forward direction is equal to the rate constant of the forward reaction times the concentration of methylacetate times the concentration of hydroxide. Now, notice my exponents are each one. That means there is a direct relationship between our rate of reaction and our concentration. In other words, if we double our concentration of methyl acetate while keeping this guy the same, we double our rate of reaction. Likewise, if we double this guy while keeping our methyl acetate the same, we also double our reaction rate. But if we double each guy, if this guy is multiplied by two and this guy is multiplied by two, that means this guy is quadrupled. |
Rate Law .txt | In other words, if we double our concentration of methyl acetate while keeping this guy the same, we double our rate of reaction. Likewise, if we double this guy while keeping our methyl acetate the same, we also double our reaction rate. But if we double each guy, if this guy is multiplied by two and this guy is multiplied by two, that means this guy is quadrupled. He's multiplied by four. So now I have this. I have this and I know my rate of reaction, but I don't know my rate constant. |
Rate Law .txt | He's multiplied by four. So now I have this. I have this and I know my rate of reaction, but I don't know my rate constant. Now, the rate constant can also be found using our experimental results. The way we do it is we choose any experiment we like and we plug in the data points and we find our KF. So let's use the first experiment. |
Rate Law .txt | Now, the rate constant can also be found using our experimental results. The way we do it is we choose any experiment we like and we plug in the data points and we find our KF. So let's use the first experiment. Let's plug in 0.5 for this guy, 0.5 for this guy, and 0.2
for our rate of reaction, the forward direction. So we plug these guys in and we solve for KF and we get 0.002 divided by zero. Five 0.5 times 0.05 gives us 0.08. |
Rate Law .txt | Let's plug in 0.5 for this guy, 0.5 for this guy, and 0.2
for our rate of reaction, the forward direction. So we plug these guys in and we solve for KF and we get 0.002 divided by zero. Five 0.5 times 0.05 gives us 0.08. So that means our KF, our rate constant for this reaction going this way is 0.8. So now we plug this into our reaction and we find the final rate law to be the rate of our four reaction is equal to 0.8 times the concentration of our methyl acetate to the first power. Times the concentration of our hydroxide to the first power. |
Rate Law .txt | So that means our KF, our rate constant for this reaction going this way is 0.8. So now we plug this into our reaction and we find the final rate law to be the rate of our four reaction is equal to 0.8 times the concentration of our methyl acetate to the first power. Times the concentration of our hydroxide to the first power. Now, let's think about it. So our initial concentration increases and that increase increases our initial rates. Why is that? |
Rate Law .txt | Now, let's think about it. So our initial concentration increases and that increase increases our initial rates. Why is that? Well, think about it. The only way that these guys react is if they collide. And if you increase the concentration of either guy, we have more collisions happening. |
Rate Law .txt | Well, think about it. The only way that these guys react is if they collide. And if you increase the concentration of either guy, we have more collisions happening. And if there are more collisions happening, that means our rate should increase. In other words, these guys will convert quicker to our products from the reactants. Now, likewise, if you decrease either concentration, our rate should decrease because we have less collisions occurring. |
Molarity Example .txt | Molarity is represented by the capital letter M, and it has the unit moles of solid over volume of solution. And this includes the volume of both the solvent and the solid. Now let's do an example using molar. The question tells us that we have 0.6 liters of a three molar solution. We need to find the number of liters that we need to add to go from three molar to a one molar solution. So we're diluting. |
Molarity Example .txt | The question tells us that we have 0.6 liters of a three molar solution. We need to find the number of liters that we need to add to go from three molar to a one molar solution. So we're diluting. Whenever we dilute, we want to keep the number of solid the same. We want to increase the number of solids. So this is our current situation. |
Molarity Example .txt | Whenever we dilute, we want to keep the number of solid the same. We want to increase the number of solids. So this is our current situation. We have a three molar solution where the red dots are the solid, blue dots are the solid. We want to go from a three molar to a one molar solution. So we need to ask ourselves, how many more blue dots do we need to add to go from a three molar to a one molar? |
Molarity Example .txt | We have a three molar solution where the red dots are the solid, blue dots are the solid. We want to go from a three molar to a one molar solution. So we need to ask ourselves, how many more blue dots do we need to add to go from a three molar to a one molar? So since the number of red dots stays the same, we need to find the constant. The constant here is the number of moles of solute. To find the number of moles of solute, we take 0.6 liters and we multiply it by three molar. |
Molarity Example .txt | So since the number of red dots stays the same, we need to find the constant. The constant here is the number of moles of solute. To find the number of moles of solute, we take 0.6 liters and we multiply it by three molar. 0.6
liters times three moles of solute over liters. The L's cross out 0.6 times three. We get 1.8 moles of solution. |
Molarity Example .txt | 0.6
liters times three moles of solute over liters. The L's cross out 0.6 times three. We get 1.8 moles of solution. Now, we found the number of red dots or the moles of red dots. Now, our goal is a one molar solution. So we set up an equation. |
Molarity Example .txt | Now, we found the number of red dots or the moles of red dots. Now, our goal is a one molar solution. So we set up an equation. One molar is equal to the thing that stays constant, one moles of solution. Over what the amount? We already have 0.6 liters plus the amount we need to add the mount of blue dots that we need to add to the system to get one molar solution. |
Sp2 Hybridization.txt | So we essentially took one S orbital, we took one P orbital, we combine them, and we formed two different sphypedized orbitals. Now we're going to look at SP two hybridized orbitals. So let's suppose we want to construct a BH three molecule. Now, in order to construct this molecule, we need three H atoms and one boron atom. Now, boron has five protons, so has five electrons. Two go into the one S, two go into the two S, and one goes into the two piece. |
Sp2 Hybridization.txt | Now, in order to construct this molecule, we need three H atoms and one boron atom. Now, boron has five protons, so has five electrons. Two go into the one S, two go into the two S, and one goes into the two piece. So we have three balanced electrons. Now, the H atom each has one electron. So the one electron goes into the one S orbital. |
Sp2 Hybridization.txt | So we have three balanced electrons. Now, the H atom each has one electron. So the one electron goes into the one S orbital. Now, before these atoms can combine to form, our BH three molecule hybridization of boron must take place. The question is, how many hybrid orbitals should boron form before this molecule can be created? The answer lies in this molecule itself. |
Sp2 Hybridization.txt | Now, before these atoms can combine to form, our BH three molecule hybridization of boron must take place. The question is, how many hybrid orbitals should boron form before this molecule can be created? The answer lies in this molecule itself. How many times does boron bond two H? Well, since there's one boron and three H atoms, that means there are three different orbitals. So that means we must develop three hybrid orbitals. |
Sp2 Hybridization.txt | How many times does boron bond two H? Well, since there's one boron and three H atoms, that means there are three different orbitals. So that means we must develop three hybrid orbitals. So that means we can no longer use SP hybridization, because SP hybridization produces only two hybrid orbitals. And we need three, as we see in this case here. So that means we're not combining S and P, but we're combining three orbitals. |
Sp2 Hybridization.txt | So that means we can no longer use SP hybridization, because SP hybridization produces only two hybrid orbitals. And we need three, as we see in this case here. So that means we're not combining S and P, but we're combining three orbitals. And these three orbitals are the two S orbital, the two PX orbital, and the two PY orbital. So we combine these atomic orbitals of the boron atom, and we form three identical SP, two orbitals or hybrid orbitals. And these guys look like this hybrid orbital here. |
Sp2 Hybridization.txt | And these three orbitals are the two S orbital, the two PX orbital, and the two PY orbital. So we combine these atomic orbitals of the boron atom, and we form three identical SP, two orbitals or hybrid orbitals. And these guys look like this hybrid orbital here. The only difference is they all lie in different directions. So they point in different directions. So that means because we're combining two P orbitals and one two S orbital, we're going to have 66% P character and 33% S character. |
Sp2 Hybridization.txt | The only difference is they all lie in different directions. So they point in different directions. So that means because we're combining two P orbitals and one two S orbital, we're going to have 66% P character and 33% S character. So now that we formed the three different hybrid orbitals, we are ready for these orbitals to interact with the one S orbitals of the H.
So we take three one S orbitals, we combine them with three SP two orbitals, and we form the following picture. So here we have our boron atom. The nucleus is in the middle here. |
Sp2 Hybridization.txt | So now that we formed the three different hybrid orbitals, we are ready for these orbitals to interact with the one S orbitals of the H.
So we take three one S orbitals, we combine them with three SP two orbitals, and we form the following picture. So here we have our boron atom. The nucleus is in the middle here. It's not shown. We have one SP hybridized orbital pointing into the page into the board, and it's bonding to one of the S's. One is coming out of the board, and that is bonding to a second one S orbital. |
Sp2 Hybridization.txt | It's not shown. We have one SP hybridized orbital pointing into the page into the board, and it's bonding to one of the S's. One is coming out of the board, and that is bonding to a second one S orbital. And a third is lying on the board, and it's bonding with a third one S orbital. And here we have our BH three molecule that also looks like this. So once again, the boron nucleus, one bond is coming out of the page. |
Sp2 Hybridization.txt | And a third is lying on the board, and it's bonding with a third one S orbital. And here we have our BH three molecule that also looks like this. So once again, the boron nucleus, one bond is coming out of the page. The second bond is going into the page, and the third is going this way. Now, if we grab the XYZ axis, we see that this molecule lies on the same plane. All the bonds lie on the same plane. |
Sp2 Hybridization.txt | The second bond is going into the page, and the third is going this way. Now, if we grab the XYZ axis, we see that this molecule lies on the same plane. All the bonds lie on the same plane. So that means since we have three angles, each angle must be 120 degrees. So, once again, let's review. So what is an SP two hybridized orbital? |
Sp2 Hybridization.txt | So that means since we have three angles, each angle must be 120 degrees. So, once again, let's review. So what is an SP two hybridized orbital? Well, an SP two hybridized orbital is simply the combination of three different atomic orbitals found within an atom. In this case, it was the Boron atom. When these three different atomic orbitals combined, they form identical hybrid orbitals. |
Polyprotic Acids .txt | Now, monoprotic acid is an acid that can donate a single H plus ion. Let's look at a few examples. Hydrochloric acid, acetic acid, nitric acid, hydrobromic acid, hydrochloric acid, chloro acid, and perchloric acid are all examples of monopolic acids because they can all donate a single H plus ion. Now let's look at the ionization of these acids in water. Let's choose the hypothetical monopolic acid. So let's choose Ha to be our acid. |
Polyprotic Acids .txt | Now let's look at the ionization of these acids in water. Let's choose the hypothetical monopolic acid. So let's choose Ha to be our acid. And Ha will react in water to produce hydronium and a conjugate base. Now, we've spoken about something called Ka, or the acid ionization concept. If you don't know what K is, check out the link above. |
Polyprotic Acids .txt | And Ha will react in water to produce hydronium and a conjugate base. Now, we've spoken about something called Ka, or the acid ionization concept. If you don't know what K is, check out the link above. But Ka is basically the product of the concentrations of the product. So the concentration of this guy times the concentration of this guy divided by the concentration of the reaction of this guy. And we said that if our Ka value is high, if it's bigger than one, that means our acid is a strong acid. |
Polyprotic Acids .txt | But Ka is basically the product of the concentrations of the product. So the concentration of this guy times the concentration of this guy divided by the concentration of the reaction of this guy. And we said that if our Ka value is high, if it's bigger than one, that means our acid is a strong acid. And if our Ka is low, if it's less than one, we're dealing with a weak acid. Now, by strong acid, I simply need an acid that's willing to dissociate, that's willing to go from this form to this form. So therefore, if our Ka is high, that means the concentration of these guys is high and the concentration of this guy is low. |
Polyprotic Acids .txt | And if our Ka is low, if it's less than one, we're dealing with a weak acid. Now, by strong acid, I simply need an acid that's willing to dissociate, that's willing to go from this form to this form. So therefore, if our Ka is high, that means the concentration of these guys is high and the concentration of this guy is low. And that makes sense. Now let's look at polyphonic acid. Polyportic acids are those acids that can donate more than one H plus ion. |
Polyprotic Acids .txt | And that makes sense. Now let's look at polyphonic acid. Polyportic acids are those acids that can donate more than one H plus ion. So let's look at a few examples. Sulfuric acid, hydronium ion, phosphoric acid and carbonic acid are all examples of polyphonic acids. And that's because they all have more than two or more than one H plus ion that they can donate. |
Polyprotic Acids .txt | So let's look at a few examples. Sulfuric acid, hydronium ion, phosphoric acid and carbonic acid are all examples of polyphonic acids. And that's because they all have more than two or more than one H plus ion that they can donate. This guy has three H plus ions that he can donate. So in the same way that we spoke about ionization reactions of monopolytic acids, we can also talk about ionization reactions of polyphotic acids. So let's look at a hypothetical example of a polyphotic acid, h two A. |
Polyprotic Acids .txt | This guy has three H plus ions that he can donate. So in the same way that we spoke about ionization reactions of monopolytic acids, we can also talk about ionization reactions of polyphotic acids. So let's look at a hypothetical example of a polyphotic acid, h two A. Now, for H two A, we're not going to have one reaction. We're going to have two reactions. And that's because we have two potential HS that can be lost to our environment. |
Polyprotic Acids .txt | Now, for H two A, we're not going to have one reaction. We're going to have two reactions. And that's because we have two potential HS that can be lost to our environment. So H two A will react in water to produce hydronium and our conjugate base. Now let's look at this conjugate base. This conjugate base can either go two ways. |
Polyprotic Acids .txt | So H two A will react in water to produce hydronium and our conjugate base. Now let's look at this conjugate base. This conjugate base can either go two ways. It can either act as a conjugate base, taking this H back and creating back our acid, or it can act as an acid itself and it can react with water to produce a second hydronium ion and the final conjugate base. Now, the same way we spoke about Ka or acidization constants for monoprotic acids, we can also talk about Ka's about polyproteic acids. Except now we're not going to have one Ka. |
Polyprotic Acids .txt | It can either act as a conjugate base, taking this H back and creating back our acid, or it can act as an acid itself and it can react with water to produce a second hydronium ion and the final conjugate base. Now, the same way we spoke about Ka or acidization constants for monoprotic acids, we can also talk about Ka's about polyproteic acids. Except now we're not going to have one Ka. We're going to have more than one Ka. Each Ka represents a single reaction. So, for this particular polyphonic acids, we have two reactions. |
Polyprotic Acids .txt | We're going to have more than one Ka. Each Ka represents a single reaction. So, for this particular polyphonic acids, we have two reactions. So we have two ka's. And normally, the first ka will have a higher value than the second ka. And that's because of the following fact. |
Polyprotic Acids .txt | So we have two ka's. And normally, the first ka will have a higher value than the second ka. And that's because of the following fact. Now, when we go from this acid to this base, we get a negative one charge. When we go from this acid to this base, we get a negative two charge. Which one of these is less stable? |
Polyprotic Acids .txt | Now, when we go from this acid to this base, we get a negative one charge. When we go from this acid to this base, we get a negative two charge. Which one of these is less stable? Well, this guy. And that's because the more charge we have, the less stability. And so this guy will not want to exist by itself. |
Polyprotic Acids .txt | Well, this guy. And that's because the more charge we have, the less stability. And so this guy will not want to exist by itself. This guy will want to exist in this form. And so our reaction for this guy is favored this way. And that's why our ka is much smaller for this reaction than this reaction. |
Polyprotic Acids .txt | This guy will want to exist in this form. And so our reaction for this guy is favored this way. And that's why our ka is much smaller for this reaction than this reaction. And in fact, this reaction is 10,000 times more likely than this reaction. So, in the same way that we spoke about titration curves of monopolic acids, we can also talk about titration curves of polyphonic acids. So here's our titration curve. |
Polyprotic Acids .txt | And in fact, this reaction is 10,000 times more likely than this reaction. So, in the same way that we spoke about titration curves of monopolic acids, we can also talk about titration curves of polyphonic acids. So here's our titration curve. While the Y axis is PH and the X axis is volume based at it. Now, this curve is for our hypothetical phypotic acid, h two A. Now, this guy will have not one equivalence point, but two equivalence points. |
Polyprotic Acids .txt | While the Y axis is PH and the X axis is volume based at it. Now, this curve is for our hypothetical phypotic acid, h two A. Now, this guy will have not one equivalence point, but two equivalence points. And let's look at what each represents. Let's look at the first one. What the first one represents is it's the point at which all of this guy has been neutralized to this guy. |
Polyprotic Acids .txt | And let's look at what each represents. Let's look at the first one. What the first one represents is it's the point at which all of this guy has been neutralized to this guy. So there's no more of this guy, and 100% of our solution is in this form at this point. Now, what the second point means is it's the point at which all of this guy has been neutralized into our final conjugate base. And at this point, as a second equivalence point, all of our solution is in this form. |
Polyprotic Acids .txt | So there's no more of this guy, and 100% of our solution is in this form at this point. Now, what the second point means is it's the point at which all of this guy has been neutralized into our final conjugate base. And at this point, as a second equivalence point, all of our solution is in this form. So, at this point, we notice we have a very high PH. And a high PH means we're in a basic environment. And basic means we don't have a lot of H plus ions. |
Polyprotic Acids .txt | So, at this point, we notice we have a very high PH. And a high PH means we're in a basic environment. And basic means we don't have a lot of H plus ions. So why is it that at this point, this guy really wants to associate into this guy? Well, let's examine why. Well, in our high PH or a basic environment, we don't have a lot of H plus, so this guy decreases. |
Polyprotic Acids .txt | So why is it that at this point, this guy really wants to associate into this guy? Well, let's examine why. Well, in our high PH or a basic environment, we don't have a lot of H plus, so this guy decreases. So to compensate that, according to Alicia clear principle, this guy will dissociate. So our equilibrium will be favored this way. And that's exactly why at a high PH, our reaction goes this way. |
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