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Rate Determining Step and Rate Law Part 1.txt | So since this is a bimolecular elementary reaction we can use the coefficients as the exponents. In other words, our rate of reaction is equal to KR rate constant times the concentration of hydrogen peroxide times the concentration of iodine. And each of the exponents is one because we have 1 mol of this guy and 1 mol of this guy react to produce these two intermediates. And this is indeed a balanced equation. So once again, the reason we're allowed to find a weight loss so quickly and without experimental results is because step one is an elementary bimolecular reaction with an overall reaction order of two. In other words, this has one coefficient and a second coefficient. |
Rate Determining Step and Rate Law Part 1.txt | And this is indeed a balanced equation. So once again, the reason we're allowed to find a weight loss so quickly and without experimental results is because step one is an elementary bimolecular reaction with an overall reaction order of two. In other words, this has one coefficient and a second coefficient. So one plus one, two. That's why. Now, once again, once you find this guy, you have to check to make sure this is in fact the rate law. |
Octet Rule.txt | So in this lecture, we're going to talk about the octave rule. But before we talk about the octet rule, let's recall what the electron configuration of an atom atom is. So let's look at helium, neon, and argon. So the electron configuration is simply the layout of the electrons found within that atom. So let's look at helium. Now, helium has two protons found in the nucleus and two electrons. |
Octet Rule.txt | So the electron configuration is simply the layout of the electrons found within that atom. So let's look at helium. Now, helium has two protons found in the nucleus and two electrons. These two electrons are going to be found in the one s orbital. So the one s orbital can have a maximum of two electrons. And because helium has two electrons, these two electrons will be found in the one s orbital. |
Octet Rule.txt | These two electrons are going to be found in the one s orbital. So the one s orbital can have a maximum of two electrons. And because helium has two electrons, these two electrons will be found in the one s orbital. So let's go to neon. Now, neon has ten protons down in the nucleus and ten electrons down in the orbital surrounding our nucleus. So our electron configuration for neon will be one f, two. |
Octet Rule.txt | So let's go to neon. Now, neon has ten protons down in the nucleus and ten electrons down in the orbital surrounding our nucleus. So our electron configuration for neon will be one f, two. So two electrons will be in the one s orbital, two electrons will be in a two s orbital, and six electrons will be in the two p orbital. Remember, there are actually three p orbitals. There's PX, PY, and PZ. |
Octet Rule.txt | So two electrons will be in the one s orbital, two electrons will be in a two s orbital, and six electrons will be in the two p orbital. Remember, there are actually three p orbitals. There's PX, PY, and PZ. And each orbital can hold a maximum of two electrons. So that means we're going to have a total of six electrons in our p orbitals. So, once again, our orbitals within neon are completely filled, just like they were in helium. |
Octet Rule.txt | And each orbital can hold a maximum of two electrons. So that means we're going to have a total of six electrons in our p orbitals. So, once again, our orbitals within neon are completely filled, just like they were in helium. So let's look at argon. Argon has 18 protons found in the nucleus, 18 electrons found in the surrounding orbital. So we're going to have two electrons in the one s orbital, two electrons in the two s orbital, six electrons in the three p orbitals. |
Octet Rule.txt | So let's look at argon. Argon has 18 protons found in the nucleus, 18 electrons found in the surrounding orbital. So we're going to have two electrons in the one s orbital, two electrons in the two s orbital, six electrons in the three p orbitals. So a total of six in the p.
Now we're going to have two electrons in the three s orbital, and three p will have six electrons. So, once again, every orbital within our gun is completely filled, just like it was in neon and helium. And in fact, these three atoms are noble gases. |
Octet Rule.txt | So a total of six in the p.
Now we're going to have two electrons in the three s orbital, and three p will have six electrons. So, once again, every orbital within our gun is completely filled, just like it was in neon and helium. And in fact, these three atoms are noble gases. Noble gases are atoms that have electron configurations that are completely filled. So they have a perfect electron configuration. All the orbitals that can possibly be filled are filled. |
Octet Rule.txt | Noble gases are atoms that have electron configurations that are completely filled. So they have a perfect electron configuration. All the orbitals that can possibly be filled are filled. And that's exactly why noble gases are very stable. They have very stable electron configurations. And in fact, any atom that is not a noble gas will try to attain. |
Octet Rule.txt | And that's exactly why noble gases are very stable. They have very stable electron configurations. And in fact, any atom that is not a noble gas will try to attain. So it will try to either lose or gain electrons such that its electron configuration matches that of one of the noble gases. And this is what we call the octave rule. So the octave rule is simply the process of filling electron shells or electron orbitals in a way such that an electron configuration that matches one of the noble gases is reached. |
Octet Rule.txt | So it will try to either lose or gain electrons such that its electron configuration matches that of one of the noble gases. And this is what we call the octave rule. So the octave rule is simply the process of filling electron shells or electron orbitals in a way such that an electron configuration that matches one of the noble gases is reached. So let's do an example. So, here we have a carbon atom, and we have a fluorine atom. So this carbon atom has six protons and six electrons. |
Octet Rule.txt | So let's do an example. So, here we have a carbon atom, and we have a fluorine atom. So this carbon atom has six protons and six electrons. So it has six protons within the nucleus along with the six neutrons. And we have six electrons surrounding our nucleus. So two electrons will be in the one s orbital, two electrons will be in the two s orbital, and only two electrons will be in the two p orbital. |
Octet Rule.txt | So it has six protons within the nucleus along with the six neutrons. And we have six electrons surrounding our nucleus. So two electrons will be in the one s orbital, two electrons will be in the two s orbital, and only two electrons will be in the two p orbital. So that means that there are four more electrons that can fit into our p orbitals because the p orbital can have a maximum of six electrons. So if we go up here, that means that we want, or carbon wants to gain four more electrons so that its electron configuration matches that of neon. So we basically want to attain this electron configuration. |
Octet Rule.txt | So that means that there are four more electrons that can fit into our p orbitals because the p orbital can have a maximum of six electrons. So if we go up here, that means that we want, or carbon wants to gain four more electrons so that its electron configuration matches that of neon. So we basically want to attain this electron configuration. Carbon wants to attain this electron configuration. So that means that it is very likely that it will gain four electrons to form our perfect electron configuration. So if this carbon had four electrons floating around this atom, that means it would gain those electrons, and these electrons would steal the remaining two p orbitals. |
Octet Rule.txt | Carbon wants to attain this electron configuration. So that means that it is very likely that it will gain four electrons to form our perfect electron configuration. So if this carbon had four electrons floating around this atom, that means it would gain those electrons, and these electrons would steal the remaining two p orbitals. And that means once it gains the four electrons, it will have the one f two, two f two, and two p six the perfect configuration that matches neon. So notice that a neutral atom of carbon has a charge of zero. That's because it has six protons and six electrons. |
Octet Rule.txt | And that means once it gains the four electrons, it will have the one f two, two f two, and two p six the perfect configuration that matches neon. So notice that a neutral atom of carbon has a charge of zero. That's because it has six protons and six electrons. Now, we have six protons. By the way, this six, this subscript means six protons. It's the atomic number which corresponds to the number of protons. |
Octet Rule.txt | Now, we have six protons. By the way, this six, this subscript means six protons. It's the atomic number which corresponds to the number of protons. So we have six protons, but now we have four plus six electrons. So we have a total of ten electrons, just like neon does. And that means our charge will be negative ten plus six. |
Octet Rule.txt | So we have six protons, but now we have four plus six electrons. So we have a total of ten electrons, just like neon does. And that means our charge will be negative ten plus six. So our charge will be four. So this is an anion. So now let's look at Florida. |
Octet Rule.txt | So our charge will be four. So this is an anion. So now let's look at Florida. Now, fluorine has nine protons and nine electrons. So once again, our nucleus will have nine protons, and our mutual fluorine atom will have nine electrons floating around the nucleus. So once again, let's apply our octave rule and let's see how many electrons this fluorine atom, this neutral fluorine atom needs to gain to obtain a noble gas configuration. |
Octet Rule.txt | Now, fluorine has nine protons and nine electrons. So once again, our nucleus will have nine protons, and our mutual fluorine atom will have nine electrons floating around the nucleus. So once again, let's apply our octave rule and let's see how many electrons this fluorine atom, this neutral fluorine atom needs to gain to obtain a noble gas configuration. So since nine is very close to ten, so this fluorine has nine electrons, which is only one away, one less than neon, that means fluorine will tend to gain an electron to form our neon noble gas configuration. So, once again, if we have an electron floating around this fluorine nucleus, it will tend to take that electron and place it into one of the two p orbitals so that once it places it there, it has a noble gas configuration. So, once again, this neutral fluorine atom gains one electron, puts it into the two p orbital, and it becomes an electron configuration that matches that of neon. |
Octet Rule.txt | So since nine is very close to ten, so this fluorine has nine electrons, which is only one away, one less than neon, that means fluorine will tend to gain an electron to form our neon noble gas configuration. So, once again, if we have an electron floating around this fluorine nucleus, it will tend to take that electron and place it into one of the two p orbitals so that once it places it there, it has a noble gas configuration. So, once again, this neutral fluorine atom gains one electron, puts it into the two p orbital, and it becomes an electron configuration that matches that of neon. So it has two electrons in the one s. It has two electrons in the two s and six electrons in the two p. Now, this guy is very happy because he's very stable, and that's because noble gases have very stable electron configuration. And that's basically what the Octave Rule is. It's basically a procedure, a process that you follow that will give you a noble gas electron configuration. |
Enthalpy, Heat of Reaction and Hess Law .txt | For example, I could take this water bottle, and I could measure the amount of water inside in terms of volume. I could measure the pressure. I could also measure the temperature. What I can say is, oh, this water bottle has x amount of enthalpy, okay? And that's because enthalpy is defined using a formula. So when somebody asks you what enthalpy is, you tell them it's a formula. |
Enthalpy, Heat of Reaction and Hess Law .txt | What I can say is, oh, this water bottle has x amount of enthalpy, okay? And that's because enthalpy is defined using a formula. So when somebody asks you what enthalpy is, you tell them it's a formula. And the formula is this the change in enthalpy is equal to change in U plus P times change in B. So let's see what this guy is. This guy is just a total internal energy of the system, okay? |
Enthalpy, Heat of Reaction and Hess Law .txt | And the formula is this the change in enthalpy is equal to change in U plus P times change in B. So let's see what this guy is. This guy is just a total internal energy of the system, okay? And if you look at this system here, which is a circle or a sphere, it's basically the collective energy of all the different molecules found within the system. So you sum their kinetic energies and you sum their potential energies, and that gives you, this guy here, the total infernal energy of the system. What this is is it's the work done to displace environment in creating the current state of the system, okay? |
Enthalpy, Heat of Reaction and Hess Law .txt | And if you look at this system here, which is a circle or a sphere, it's basically the collective energy of all the different molecules found within the system. So you sum their kinetic energies and you sum their potential energies, and that gives you, this guy here, the total infernal energy of the system. What this is is it's the work done to displace environment in creating the current state of the system, okay? And what that basically means is in order to get from nothing to this volume and pressure, these molecules have to do work on the environment and displacing them in forming this structure, this sphere or the circle. So what enthalpy is, it's the total energy of the system plus the work that the system has to do on the environment in creating that system. And that's what enthalpy is. |
Enthalpy, Heat of Reaction and Hess Law .txt | And what that basically means is in order to get from nothing to this volume and pressure, these molecules have to do work on the environment and displacing them in forming this structure, this sphere or the circle. So what enthalpy is, it's the total energy of the system plus the work that the system has to do on the environment in creating that system. And that's what enthalpy is. Now, enthalpy is a state function, which means that enthalpy is independent of the pathway and the reactions that lead to the system. In other words, enthalpy only depends on the current state of the system, okay? Now, enthalpy is also an external property, which also means that entropy depends on the amount of the system. |
Enthalpy, Heat of Reaction and Hess Law .txt | Now, enthalpy is a state function, which means that enthalpy is independent of the pathway and the reactions that lead to the system. In other words, enthalpy only depends on the current state of the system, okay? Now, enthalpy is also an external property, which also means that entropy depends on the amount of the system. So the more of the system that we have, the largest system, the more enthalpy we have. And that's because enthalpy is related to internal energy, and internal energy is related to number of moles, number of particles. So the more particles we have, the more enthalpy we have. |
Enthalpy, Heat of Reaction and Hess Law .txt | So the more of the system that we have, the largest system, the more enthalpy we have. And that's because enthalpy is related to internal energy, and internal energy is related to number of moles, number of particles. So the more particles we have, the more enthalpy we have. And that's why it's an external property. So because enthalpy is a man made concept, we can't talk about absolute values of enthalpy. They simply do not exist. |
Enthalpy, Heat of Reaction and Hess Law .txt | And that's why it's an external property. So because enthalpy is a man made concept, we can't talk about absolute values of enthalpy. They simply do not exist. What we do talk about are changes in enthalpy. And in order for changes in enthalpy to exist, we must assign zero values to Cherton atoms with the churchin's elements, okay? And in fact, under a standard state of 1 bar or 750 tor and a certain temperature such as 25 degrees Celsius, all elements such as diatomic, hydrogen diatomic, oxygen, carbon diatonic, fluorine and so on are assigned a value of zero entropy, zero joules per mole. |
Enthalpy, Heat of Reaction and Hess Law .txt | What we do talk about are changes in enthalpy. And in order for changes in enthalpy to exist, we must assign zero values to Cherton atoms with the churchin's elements, okay? And in fact, under a standard state of 1 bar or 750 tor and a certain temperature such as 25 degrees Celsius, all elements such as diatomic, hydrogen diatomic, oxygen, carbon diatonic, fluorine and so on are assigned a value of zero entropy, zero joules per mole. Okay? So when a reaction or a compound is created from its constituents or from its raw elements, there's a change in entropy. And this change in entropy is known as the standard entropy of formation. |
Enthalpy, Heat of Reaction and Hess Law .txt | Okay? So when a reaction or a compound is created from its constituents or from its raw elements, there's a change in entropy. And this change in entropy is known as the standard entropy of formation. And it's given this value where the small zero here called not is basically represents standard state or a 1 bar. Okay? And this f is formation. |
Enthalpy, Heat of Reaction and Hess Law .txt | And it's given this value where the small zero here called not is basically represents standard state or a 1 bar. Okay? And this f is formation. So change in enthalpy, okay? So standard enthalpy of formation and it basically means it's the change in enthalpy for a reaction that creates 1 mol of compound from its constituent elements. For example, let's take the creation or formation of water from its constituents, from hydrogen and from oxygen, okay? |
Enthalpy, Heat of Reaction and Hess Law .txt | So change in enthalpy, okay? So standard enthalpy of formation and it basically means it's the change in enthalpy for a reaction that creates 1 mol of compound from its constituent elements. For example, let's take the creation or formation of water from its constituents, from hydrogen and from oxygen, okay? These guys combine to form water and there is a decrease in enthalpy. And that means that the internal energy of the system decreases. There are less molecules within the system and so there is less Enthropy within the system. |
Enthalpy, Heat of Reaction and Hess Law .txt | These guys combine to form water and there is a decrease in enthalpy. And that means that the internal energy of the system decreases. There are less molecules within the system and so there is less Enthropy within the system. Now, different types of reactions exist. Some reactions are in a gas phase when other reactions are in a liquid and solid phases. Now, when you're dealing with reactions in a gas phase, you have to realize that gaseous systems are compressible. |
Enthalpy, Heat of Reaction and Hess Law .txt | Now, different types of reactions exist. Some reactions are in a gas phase when other reactions are in a liquid and solid phases. Now, when you're dealing with reactions in a gas phase, you have to realize that gaseous systems are compressible. You can expand them and compress them. And so pressure changes because volume changes, so pressure isn't constant. And if pressure changes, if volume changes, then PV work is done. |
Enthalpy, Heat of Reaction and Hess Law .txt | You can expand them and compress them. And so pressure changes because volume changes, so pressure isn't constant. And if pressure changes, if volume changes, then PV work is done. In other words, the system does work to expand by doing work in the surroundings, to move the surroundings away so that the system could expand. Same thing with compression. Therefore, when you're dealing with reactions of gases, this equation holds the change in enthalpy is equal to a change in internal energy plus p times change in volume. |
Enthalpy, Heat of Reaction and Hess Law .txt | In other words, the system does work to expand by doing work in the surroundings, to move the surroundings away so that the system could expand. Same thing with compression. Therefore, when you're dealing with reactions of gases, this equation holds the change in enthalpy is equal to a change in internal energy plus p times change in volume. But when you're dealing with reactions in liquid and solid phases, you have to realize that these type of systems are not compressible. If you can't compress them or expand them, that means no PD work is done. Okay? |
Enthalpy, Heat of Reaction and Hess Law .txt | But when you're dealing with reactions in liquid and solid phases, you have to realize that these type of systems are not compressible. If you can't compress them or expand them, that means no PD work is done. Okay? And that means this guy can go to zero. And now we could approximate the change in enthalpy to just simply change in internal energy, which is heat or the exchange of energy. Okay? |
Enthalpy, Heat of Reaction and Hess Law .txt | And that means this guy can go to zero. And now we could approximate the change in enthalpy to just simply change in internal energy, which is heat or the exchange of energy. Okay? So under lab conditions, normally this condition holds. We could approximate this, we could approximate change in Enthalpy as just simply being the change in internal energy. So what is the heat of reaction? |
Enthalpy, Heat of Reaction and Hess Law .txt | So under lab conditions, normally this condition holds. We could approximate this, we could approximate change in Enthalpy as just simply being the change in internal energy. So what is the heat of reaction? The heat of reaction is simply the change in internal energy or change in enthalpy of the reaction. And it's found using this formula. For example, let's take this equation here we have reactants and we have products. |
Enthalpy, Heat of Reaction and Hess Law .txt | The heat of reaction is simply the change in internal energy or change in enthalpy of the reaction. And it's found using this formula. For example, let's take this equation here we have reactants and we have products. To find the heat of reaction, we simply find a change in atrial energy or change in enthalpy of the product. We find the change in entropy or, I'm sorry, enthalpy of the reactants and we subtract this from this and that will give us the heat of reaction. So recall it, enthalpy is a state function and what that basically means that the pathway taken to get to the final product or to get to the final system does not affect the final change in enthalpy. |
Enthalpy, Heat of Reaction and Hess Law .txt | To find the heat of reaction, we simply find a change in atrial energy or change in enthalpy of the product. We find the change in entropy or, I'm sorry, enthalpy of the reactants and we subtract this from this and that will give us the heat of reaction. So recall it, enthalpy is a state function and what that basically means that the pathway taken to get to the final product or to get to the final system does not affect the final change in enthalpy. And this jumps directly into something called Hess's Law. Now, Hessa's Law basically states that the steps taken to get from the reactant to the products does not affect the final change in enthalpy, okay? And Hess's Law can be applied to reactions. |
Enthalpy, Heat of Reaction and Hess Law .txt | And this jumps directly into something called Hess's Law. Now, Hessa's Law basically states that the steps taken to get from the reactant to the products does not affect the final change in enthalpy, okay? And Hess's Law can be applied to reactions. Now, here we have an imaginary reaction. That's a two step reaction. And each step has its corresponding change in enthalpy. |
Enthalpy, Heat of Reaction and Hess Law .txt | Now, here we have an imaginary reaction. That's a two step reaction. And each step has its corresponding change in enthalpy. Now, what Hess Law basically states is that in order to find the final change in enthalpy, you simply add each corresponding enthalpy. Together we get the final result. So ten kilojoules per mole plus 20 kilojoules per mole gives you 30 kilojoules per mole. |
Enthalpy, Heat of Reaction and Hess Law .txt | Now, what Hess Law basically states is that in order to find the final change in enthalpy, you simply add each corresponding enthalpy. Together we get the final result. So ten kilojoules per mole plus 20 kilojoules per mole gives you 30 kilojoules per mole. And this is their final answer. Now, yield this side, you basically the same exact thing. You use basic algebra. |
Enthalpy, Heat of Reaction and Hess Law .txt | And this is their final answer. Now, yield this side, you basically the same exact thing. You use basic algebra. You add these guys up. You add these guys up, except instead of using equal sign like you would use in algebra, you use arrows. Okay? |
Enthalpy, Heat of Reaction and Hess Law .txt | You add these guys up. You add these guys up, except instead of using equal sign like you would use in algebra, you use arrows. Okay? So A plus B plus B plus C gives you a plus two b plus C arrows. C plus D. Now, one last thing left. You have to notice the molecules that appear on both sides of the equation. |
Enthalpy, Heat of Reaction and Hess Law .txt | So A plus B plus B plus C gives you a plus two b plus C arrows. C plus D. Now, one last thing left. You have to notice the molecules that appear on both sides of the equation. If they appear on both sides of the equation, you could simply cross them out, okay? So A appears only on this side, you leave it alone. Two B appears only on this side, you leave it alone. |
Enthalpy, Heat of Reaction and Hess Law .txt | If they appear on both sides of the equation, you could simply cross them out, okay? So A appears only on this side, you leave it alone. Two B appears only on this side, you leave it alone. C appears on this side, and on this side, you cross them out. And D appears only on this side. So you leave it alone. |
pH indicators .txt | In this lecture we're going to talk a bit more about titrations and equivalence points. Now, earlier in another video, we saw that we can actually calculate our equivalence point using a formula. Now, if you want to learn more about that, check out the link above. Now, actually calculating your equivalence point isn't very useful when you're conducting your experiment. When you're titrating, you want to actually visualize the equivalence point. And to visualize it, you use two things. |
pH indicators .txt | Now, actually calculating your equivalence point isn't very useful when you're conducting your experiment. When you're titrating, you want to actually visualize the equivalence point. And to visualize it, you use two things. The first thing you use is a PH meter. The PH meter will automatically tell you the PH of your solution. For example, suppose this is your PH meter and it's attached to your solution at any given time, you can look at it and it will tell you the PH of your buffer solution. |
pH indicators .txt | The first thing you use is a PH meter. The PH meter will automatically tell you the PH of your solution. For example, suppose this is your PH meter and it's attached to your solution at any given time, you can look at it and it will tell you the PH of your buffer solution. Now, the second thing you need to use is called an acid based indicator. Acid based indicators are compounds that change color in solution when they convert to their conjugate acid based form. Now, what you do is you take your indicator, you take a very small amount of it and you place it into your solution before you begin saturation. |
pH indicators .txt | Now, the second thing you need to use is called an acid based indicator. Acid based indicators are compounds that change color in solution when they convert to their conjugate acid based form. Now, what you do is you take your indicator, you take a very small amount of it and you place it into your solution before you begin saturation. And what the indicator does, it reacts with the H plus ions in the following way. When there is a lot of H plus ions associated in our solution, aka, when it's acidic, that means equilibrium will shift this way. Now, when there is very little H plus ions or our concentration of H plus ion is very low, equilibrium will shift this way. |
pH indicators .txt | And what the indicator does, it reacts with the H plus ions in the following way. When there is a lot of H plus ions associated in our solution, aka, when it's acidic, that means equilibrium will shift this way. Now, when there is very little H plus ions or our concentration of H plus ion is very low, equilibrium will shift this way. Meaning we're going to have more of the conjugate base of our conjugate acid. Note that this represents one color and this represents a different color. So in solution, when this guy predominates, our solution will be of this color. |
pH indicators .txt | Meaning we're going to have more of the conjugate base of our conjugate acid. Note that this represents one color and this represents a different color. So in solution, when this guy predominates, our solution will be of this color. And when this guy is now a solution and this guy predominates, the solution B will be of a different color. So once again, in acidic solutions, this guy dominates. And in basic solutions, this guy dominates. |
pH indicators .txt | And when this guy is now a solution and this guy predominates, the solution B will be of a different color. So once again, in acidic solutions, this guy dominates. And in basic solutions, this guy dominates. So, we can look at the ratio of the concentration of this guy to this guy. And when the ratio is ten or above, that means we're going to be of this guy, of this color, so of the active color, because there's much more of this than this in our solution. Now, the opposite is true as well. |
pH indicators .txt | So, we can look at the ratio of the concentration of this guy to this guy. And when the ratio is ten or above, that means we're going to be of this guy, of this color, so of the active color, because there's much more of this than this in our solution. Now, the opposite is true as well. When this guy will dominate, when there is more of this guy than this guy. Now, ratio is 0.1 or less. That means the base color will dominate. |
pH indicators .txt | When this guy will dominate, when there is more of this guy than this guy. Now, ratio is 0.1 or less. That means the base color will dominate. So this guy will dominate. Now, note that the end point is the point at which you observe a change in color. And the end point is not the same thing as the equivalence point. |
titration.txt | Now, whatever we're adding to our unknown sample, it's called a titren, and a titrate be their base or an acid. So let's illustrate Titration using a picture and then a graph. So suppose we have some known volume of acid, say, I don't know, hydrochloric acid in our flask, and we have a PH reader attached to the inside of our flask. So at any given time, we could measure our PH. Now suppose we have some known concentration of base and we begin adding this base drop by drop. Now, before we add the base, our PH is say, I don't know, two. |
titration.txt | So at any given time, we could measure our PH. Now suppose we have some known concentration of base and we begin adding this base drop by drop. Now, before we add the base, our PH is say, I don't know, two. Okay, that's our PH. Now, as I begin slowly adding drop by drop, what will happen to our PH? Well, it will increase. |
titration.txt | Okay, that's our PH. Now, as I begin slowly adding drop by drop, what will happen to our PH? Well, it will increase. Right, but how will it increase? Well, let's look at the graph of PH versus volume added, where PH is the Y axis, and volume added of our base in this case is the x axis. So according to this graph, we see that the relationship is based on a sigmoidal curve. |
titration.txt | Right, but how will it increase? Well, let's look at the graph of PH versus volume added, where PH is the Y axis, and volume added of our base in this case is the x axis. So according to this graph, we see that the relationship is based on a sigmoidal curve. And what that means is that initially, when you first add some known amount of concentration of base, the PH change will be very little. The PH won't change by that much. And in fact, from this much to this much volume added, the PH only rises by maybe 0.5. |
titration.txt | And what that means is that initially, when you first add some known amount of concentration of base, the PH change will be very little. The PH won't change by that much. And in fact, from this much to this much volume added, the PH only rises by maybe 0.5. But eventually we come to a point where any more volume added will increase the PH dramatically. And we will come to a point where the PH will increase by ten units. So why is that? |
titration.txt | But eventually we come to a point where any more volume added will increase the PH dramatically. And we will come to a point where the PH will increase by ten units. So why is that? Well, let's see. Now let's look at the reaction of sodium hydroxide with hydrochloric acid. Initially, we only have HCL in our mixture and HCL dissociates into H plus ion and the chloride ion. |
titration.txt | Well, let's see. Now let's look at the reaction of sodium hydroxide with hydrochloric acid. Initially, we only have HCL in our mixture and HCL dissociates into H plus ion and the chloride ion. Now, the H plus ion is what's responsible for creating the acidic solution for lowering our PH to a PH of two. So what will happen when we first dissociate or add NaOH to our mixture? Well, initially we only have a small amount of NaOH, and NaOH will dissociate into Ma plus and oh minus. |
titration.txt | Now, the H plus ion is what's responsible for creating the acidic solution for lowering our PH to a PH of two. So what will happen when we first dissociate or add NaOH to our mixture? Well, initially we only have a small amount of NaOH, and NaOH will dissociate into Ma plus and oh minus. But since we only have a small amount of NaOH, that means we only have a small amount of these guys. And these guys are the ones that associate with HMCL to form water and NaCl. So initially, only a small percentage of these guys will reassociate into this form. |
titration.txt | But since we only have a small amount of NaOH, that means we only have a small amount of these guys. And these guys are the ones that associate with HMCL to form water and NaCl. So initially, only a small percentage of these guys will reassociate into this form. And so we will see a decrease in the H plus ion, but a very small decrease. And that means we're only going to see a very small change or increase in PH initially. What will happen when the ratio of moles is one to one? |
titration.txt | And so we will see a decrease in the H plus ion, but a very small decrease. And that means we're only going to see a very small change or increase in PH initially. What will happen when the ratio of moles is one to one? So suppose we have some ratio of HCL, some unknown in our solution and suppose we add that same amount of moles, of NaOH. Well, that means we're going to have a ratio one to one to one to one. And all these guys are going to reassociate to form water and NaCl. |
titration.txt | So suppose we have some ratio of HCL, some unknown in our solution and suppose we add that same amount of moles, of NaOH. Well, that means we're going to have a ratio one to one to one to one. And all these guys are going to reassociate to form water and NaCl. And what will happen then? Well, then our PH will become seven, right? And in fact, at this point, when the ratio of moles added is the same as the ratio of this guy, this point is called the equivalence point. |
titration.txt | And what will happen then? Well, then our PH will become seven, right? And in fact, at this point, when the ratio of moles added is the same as the ratio of this guy, this point is called the equivalence point. And for NaOH and HCL, for a strong acid and strong base, this is a PH of seven. So we see that this point is the point at which we add enough Maoh that our ratio begins to equal out. Our ratio becomes closer and closer to one and one. |
titration.txt | And for NaOH and HCL, for a strong acid and strong base, this is a PH of seven. So we see that this point is the point at which we add enough Maoh that our ratio begins to equal out. Our ratio becomes closer and closer to one and one. And as we add more, as we get closer to a ratio of one to one, we get closer to our PH of seven. And that's why we see this large increase in PH. Now, if we begin to add more of our sodium hydroxide, our solution becomes basic. |
Introduction to PV work.txt | Recall what the conservation of energy tells us. According to the conservation of energy, energy cannot be destroyed and it cannot be created. Energy always exists, and energy of the Universe is constant. It does not change. But what energy can do is it can be transformed from one form to another. For example, from kinetic energy to another. |
Introduction to PV work.txt | It does not change. But what energy can do is it can be transformed from one form to another. For example, from kinetic energy to another. Another type of energy. Now there are two types of energy transfers. There's heat and work. |
Introduction to PV work.txt | Another type of energy. Now there are two types of energy transfers. There's heat and work. Heat is a natural Transfer of energy from an object with a Higher temperature to an object with a Lower temperature. And if you want to learn more about heat, check out the lecture on heat in the chemistry section. Here. |
Introduction to PV work.txt | Heat is a natural Transfer of energy from an object with a Higher temperature to an object with a Lower temperature. And if you want to learn more about heat, check out the lecture on heat in the chemistry section. Here. We're only going to talk about work. Now work is the transfer of energy due to forces. It could be individual forces acting on objects or the net forces acting on the entire object. |
Introduction to PV work.txt | We're only going to talk about work. Now work is the transfer of energy due to forces. It could be individual forces acting on objects or the net forces acting on the entire object. Now, work is a scalar, and that means it only has magnitude, no direction. And the units of work is like the units of energy. It's joules. |
Introduction to PV work.txt | Now, work is a scalar, and that means it only has magnitude, no direction. And the units of work is like the units of energy. It's joules. And we could also use electron volts when we're talking about very small microscopic objects. Now. Work does have negative and a positive. |
Introduction to PV work.txt | And we could also use electron volts when we're talking about very small microscopic objects. Now. Work does have negative and a positive. Negative work simply means our object loses energy, while positive work means our object gains energy. Now, whenever a nonfictional force is applied, work can be found using the following equation work is equal to the force applied on our object, times the distance our object moves. And since work is, in fact a scalar, a vector times a vector gives us a dot product. |
Introduction to PV work.txt | Negative work simply means our object loses energy, while positive work means our object gains energy. Now, whenever a nonfictional force is applied, work can be found using the following equation work is equal to the force applied on our object, times the distance our object moves. And since work is, in fact a scalar, a vector times a vector gives us a dot product. So that means we take the cosine of the angle. If you want to learn more about dot product and cross product, check out the lecture on that. So, basically, if we have some object, say, this block, and we apply a force with an angle theta to the horizontal of that object, and our object moves a distance d we can find the amount of work done on the object. |
Introduction to PV work.txt | So that means we take the cosine of the angle. If you want to learn more about dot product and cross product, check out the lecture on that. So, basically, if we have some object, say, this block, and we apply a force with an angle theta to the horizontal of that object, and our object moves a distance d we can find the amount of work done on the object. The amount of energy that object gain by using the following formula force times, distance travel times, cosine of the angle between them. Now, if we do not neglect friction, if we actually include friction, for example, when this box is traveling along this horizontal surface, there is friction in a real world situation. So if we do not neglect friction, we find our overall work by following the following formula the work done is equal to change in our potential energy of the object plus change in the kinetic energy of the object plus the change in the internal energy of our object due to friction. |
Introduction to PV work.txt | The amount of energy that object gain by using the following formula force times, distance travel times, cosine of the angle between them. Now, if we do not neglect friction, if we actually include friction, for example, when this box is traveling along this horizontal surface, there is friction in a real world situation. So if we do not neglect friction, we find our overall work by following the following formula the work done is equal to change in our potential energy of the object plus change in the kinetic energy of the object plus the change in the internal energy of our object due to friction. Now, for the most part, whenever friction acts on an object, it changes the object's internal energy. And internal energy is simply the energy of the individual molecules found in that object. If you sum up all the different types of energies on those individual molecules, you will get the internal energy of that system. |
Oxidation-Reduction Reactions Example .txt | So in this lecture we're going to look at three examples of redox reactions. And we're going to assign oxidation states to each atom in our molecule. So let's begin with the first one. So we begin with copper in its elemental state. And that means copper gets a charge of zero according to our rules. Let's look at nitric acid. |
Oxidation-Reduction Reactions Example .txt | So we begin with copper in its elemental state. And that means copper gets a charge of zero according to our rules. Let's look at nitric acid. Now, nitric acid, it contains an hmm and an O. That means we first assign to H, then we assign to O, then we assign to the N. Now our entire molecule must be neutral, so we must take that into consideration when assigning. So let's give our H a plus one according to our table. |
Oxidation-Reduction Reactions Example .txt | Now, nitric acid, it contains an hmm and an O. That means we first assign to H, then we assign to O, then we assign to the N. Now our entire molecule must be neutral, so we must take that into consideration when assigning. So let's give our H a plus one according to our table. And let's give an O a minus two according to our table. So our N should be a number, such that when you add that number to one and when you subtract negative six, we get here. Why negative six? |
Oxidation-Reduction Reactions Example .txt | And let's give an O a minus two according to our table. So our N should be a number, such that when you add that number to one and when you subtract negative six, we get here. Why negative six? Well, because there are three oxygen molecules. So three times negative two gives us a total of negative six. So negative six plus one gives us a five. |
Oxidation-Reduction Reactions Example .txt | Well, because there are three oxygen molecules. So three times negative two gives us a total of negative six. So negative six plus one gives us a five. So our N should be five. Let's check. One plus five. |
Oxidation-Reduction Reactions Example .txt | So our N should be five. Let's check. One plus five. Six minus six gives us a zero. So that works. Let's go to this side. |
Oxidation-Reduction Reactions Example .txt | Six minus six gives us a zero. So that works. Let's go to this side. So here we have copper nitrate. So from this part we know that N and three is a negative one, right, because the entire thing to balance a neutral charge, this must be a plus one. So this guy is minus one. |
Oxidation-Reduction Reactions Example .txt | So here we have copper nitrate. So from this part we know that N and three is a negative one, right, because the entire thing to balance a neutral charge, this must be a plus one. So this guy is minus one. So our three is minus one. So no three is minus one times two such molecules, this whole thing means it's negative two. So our copper must be plus two. |
Oxidation-Reduction Reactions Example .txt | So our three is minus one. So no three is minus one times two such molecules, this whole thing means it's negative two. So our copper must be plus two. Our copper is plus two. Now that means our oxygen is negative two, and our N must be plus five. So this whole thing must be minus two, including this two here. |
Oxidation-Reduction Reactions Example .txt | Our copper is plus two. Now that means our oxygen is negative two, and our N must be plus five. So this whole thing must be minus two, including this two here. So let's go to nitric oxide. So our O, we first assign our oxidation state to O, and our O gets a minus two. So that means our N must be plus two. |
Oxidation-Reduction Reactions Example .txt | So let's go to nitric oxide. So our O, we first assign our oxidation state to O, and our O gets a minus two. So that means our N must be plus two. Why is it plus two? Well, because this molecule is neutral and that means charge of zero. So to balance this negative two, this N must be a plus two. |
Oxidation-Reduction Reactions Example .txt | Why is it plus two? Well, because this molecule is neutral and that means charge of zero. So to balance this negative two, this N must be a plus two. And finally, let's look at water the oxygen. Well, we first assign to our H. So that gets a plus one. Plus one times two gives us a plus two. |
Oxidation-Reduction Reactions Example .txt | And finally, let's look at water the oxygen. Well, we first assign to our H. So that gets a plus one. Plus one times two gives us a plus two. So this guy must be a negative two to balance out the mutual charge. So now let's look at what gets oxidized and what gets reduced. So our copper goes from a zero charge to a plus two charge. |
Oxidation-Reduction Reactions Example .txt | So this guy must be a negative two to balance out the mutual charge. So now let's look at what gets oxidized and what gets reduced. So our copper goes from a zero charge to a plus two charge. And that means copper is oxidized because electrons are removed and these electrons transfer to some other atom. So let's look at what atom is reduced. So our N goes from a plus five to a plus two. |
Oxidation-Reduction Reactions Example .txt | And that means copper is oxidized because electrons are removed and these electrons transfer to some other atom. So let's look at what atom is reduced. So our N goes from a plus five to a plus two. That means all the electrons move from copper to N, creating that plus two charge. So our N is what's reduced. So our oxidizing agent is this guy nitric acid, the N specifically. |
Oxidation-Reduction Reactions Example .txt | That means all the electrons move from copper to N, creating that plus two charge. So our N is what's reduced. So our oxidizing agent is this guy nitric acid, the N specifically. And our reducing agent is copper. So let's write that. So this guy is oxidized and this guy is reduced. |
Oxidation-Reduction Reactions Example .txt | And our reducing agent is copper. So let's write that. So this guy is oxidized and this guy is reduced. And once again, our reducing agent, our oxidizing agent. So let's move on to example two. Now, in example two, we begin with the copper sulfide. |
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