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Lewis Acids and Bases .txt | And we form SP three hybridized bonds. So 1234 SP three hybridized bonds. And now this carbon and this H share a pair of electrons. So let's look at the energy diagram of our interaction of our two lowest acids and bases. So, once again, our one S is lowering energy than the two p. That's because this is closer to our nucleus than this orbital is. And so this guy will be found lower on the energy level. |
Lewis Acids and Bases .txt | So let's look at the energy diagram of our interaction of our two lowest acids and bases. So, once again, our one S is lowering energy than the two p. That's because this is closer to our nucleus than this orbital is. And so this guy will be found lower on the energy level. So the y axis in the energy, this will be lower than our two P orbital. So the two electrons will come directly from the one S from the Hydride. And when they interact, they will form a bonding molecular orbital and an antibounding molecular orbital. |
Lewis Acids and Bases .txt | So the y axis in the energy, this will be lower than our two P orbital. So the two electrons will come directly from the one S from the Hydride. And when they interact, they will form a bonding molecular orbital and an antibounding molecular orbital. The two electrons will go into the bonding molecular orbital, forming our SP 30 hybridized molecular orbitals. So, though, now let's define what a broncidlaric acid and a broncillary base is. A bronzedidlori acid is a molecule that donates an H ion, while a broncillary base is a molecule that accepts an H ion. |
Lewis Acids and Bases .txt | The two electrons will go into the bonding molecular orbital, forming our SP 30 hybridized molecular orbitals. So, though, now let's define what a broncidlaric acid and a broncillary base is. A bronzedidlori acid is a molecule that donates an H ion, while a broncillary base is a molecule that accepts an H ion. So acid strength of a broncillary acid increases with increasing S character. So why is that? Well, to examine that, let's recall one simple concept. |
Lewis Acids and Bases .txt | So acid strength of a broncillary acid increases with increasing S character. So why is that? Well, to examine that, let's recall one simple concept. So, here we have a protons. We have a nucleus and protons inside that nucleus. And this is our one S orbital, our two S orbital and the two P orbital. |
Lewis Acids and Bases .txt | So, here we have a protons. We have a nucleus and protons inside that nucleus. And this is our one S orbital, our two S orbital and the two P orbital. So recall that as our electron gets closer and closer to our protons in the nucleus, the energy level of the entire atom decreases, so it becomes more stable. So the closer our electron is to our nucleus, the more stable that atom is. And let's look what happens when an acid, a brothed lauric acid, reacts. |
Lewis Acids and Bases .txt | So recall that as our electron gets closer and closer to our protons in the nucleus, the energy level of the entire atom decreases, so it becomes more stable. So the closer our electron is to our nucleus, the more stable that atom is. And let's look what happens when an acid, a brothed lauric acid, reacts. So, when a brothed Laric acid, reacts, it creates a conjugate base. So a brothelary base, and it creates an hion. Now, if this has a lot of S character, that means the electron pair here will be found in that S character. |
Lewis Acids and Bases .txt | So, when a brothed Laric acid, reacts, it creates a conjugate base. So a brothelary base, and it creates an hion. Now, if this has a lot of S character, that means the electron pair here will be found in that S character. So, the more S character we have, the closer our electrons are to our nucleus and the more stable our conjugate bases. And if we have a stable conjugate base, that means our acid will be more likely to form this conjugate base, and therefore, our acid will be more likely to dissociate. And that means our asset will be a stronger acid. |
Cahn-Ingold-Prelog Priority System .txt | Now, the system is used for either one of two things. It's either used in determining the absolute configuration of your enantomers to either R or S absolute configuration, or it's used to help you rank groups attached to double bonds. So in this lecture, we're going to focus primarily on this usage here. So whenever you're using this system, four important rules must be followed that will help you to get or find the highest priority groups. So let's look at rule number one. So, atom with higher atomic number receives higher priority. |
Cahn-Ingold-Prelog Priority System .txt | So whenever you're using this system, four important rules must be followed that will help you to get or find the highest priority groups. So let's look at rule number one. So, atom with higher atomic number receives higher priority. So let's see what that means via this example. So here we have a carbon carbon double bond. So let's examine this carbon carbon number one. |
Cahn-Ingold-Prelog Priority System .txt | So let's see what that means via this example. So here we have a carbon carbon double bond. So let's examine this carbon carbon number one. So carbon number one is attached to two groups. It's either attached to the H or it's attached to the carbon here. So which atom has a higher atomic number? |
Cahn-Ingold-Prelog Priority System .txt | So carbon number one is attached to two groups. It's either attached to the H or it's attached to the carbon here. So which atom has a higher atomic number? Well, clearly the carbon has a higher atomic number, and that means this group attached this carbon has a higher priority than this group here. Likewise, let's examine the second carbon. The second carbon of the double bond is also attached to a carbon, and it's also attached to an H.
Which one of these two groups has a higher atomic number? |
Cahn-Ingold-Prelog Priority System .txt | Well, clearly the carbon has a higher atomic number, and that means this group attached this carbon has a higher priority than this group here. Likewise, let's examine the second carbon. The second carbon of the double bond is also attached to a carbon, and it's also attached to an H.
Which one of these two groups has a higher atomic number? Well, clearly, the carbon has a higher atomic number, and so the carbon wins. And let's label it with an asterisk. So notice that in this compound, our groups with the higher priorities are on the same side of the double bond. |
Cahn-Ingold-Prelog Priority System .txt | Well, clearly, the carbon has a higher atomic number, and so the carbon wins. And let's label it with an asterisk. So notice that in this compound, our groups with the higher priorities are on the same side of the double bond. And that means this must be a Z isomer. So let's look at rule number two for isotopes. The isotope with a higher atomic weight wins. |
Cahn-Ingold-Prelog Priority System .txt | And that means this must be a Z isomer. So let's look at rule number two for isotopes. The isotope with a higher atomic weight wins. Remember, two compounds or two atoms are isotopes if they have the same number of protons and electrons, but different number of neutrons. So they differ in atomic weight. So let's look at the following example. |
Cahn-Ingold-Prelog Priority System .txt | Remember, two compounds or two atoms are isotopes if they have the same number of protons and electrons, but different number of neutrons. So they differ in atomic weight. So let's look at the following example. Let's suppose we have a carbon carbon double bond. So let's examine this carbon here. This carbon is attached to an H group and also to a D group. |
Cahn-Ingold-Prelog Priority System .txt | Let's suppose we have a carbon carbon double bond. So let's examine this carbon here. This carbon is attached to an H group and also to a D group. D is simply deteriorate. It's the isotope of H. So since D has a higher atomic weight, d must have a higher priority than H.
So this group has a higher priority than this group. Likewise, for the second carbon in a double bond, we have the following two groups. |
Cahn-Ingold-Prelog Priority System .txt | D is simply deteriorate. It's the isotope of H. So since D has a higher atomic weight, d must have a higher priority than H.
So this group has a higher priority than this group. Likewise, for the second carbon in a double bond, we have the following two groups. Once again, we have the H and we have the D.
So our D, the deteriorium, has a higher atomic weight, so it has a higher priority. And now we have an allochene where our two higher priority groups are on opposite sides of the double bond. And that means this must be an Eisomer. |
Cahn-Ingold-Prelog Priority System .txt | Once again, we have the H and we have the D.
So our D, the deteriorium, has a higher atomic weight, so it has a higher priority. And now we have an allochene where our two higher priority groups are on opposite sides of the double bond. And that means this must be an Eisomer. So let's look at rule number three. If we have the same atom, if we have a tie between our atoms, we move to the next atom. So let's see exactly what that means. |
Cahn-Ingold-Prelog Priority System .txt | So let's look at rule number three. If we have the same atom, if we have a tie between our atoms, we move to the next atom. So let's see exactly what that means. Let's look at the following example. We have a carbon carbon double bond. So let's begin with this carbon, the first carbon in the double bond. |
Cahn-Ingold-Prelog Priority System .txt | Let's look at the following example. We have a carbon carbon double bond. So let's begin with this carbon, the first carbon in the double bond. So this carbon is attached to a carbon of this side and a carbon on that side. So far, we have a tie. We can't determine the atomic number, so we move on to the next atom. |
Cahn-Ingold-Prelog Priority System .txt | So this carbon is attached to a carbon of this side and a carbon on that side. So far, we have a tie. We can't determine the atomic number, so we move on to the next atom. So there is no next atom here. But here we have a following carbon atom. So that means this side wins. |
Cahn-Ingold-Prelog Priority System .txt | So there is no next atom here. But here we have a following carbon atom. So that means this side wins. It has a higher atomic number and it also has a higher atomic weight. So this side, this group, has a higher priority than the lower group. Likewise, for this carbon, we have the same exact situation. |
Cahn-Ingold-Prelog Priority System .txt | It has a higher atomic number and it also has a higher atomic weight. So this side, this group, has a higher priority than the lower group. Likewise, for this carbon, we have the same exact situation. So, once again, as the first example, we have the Z Isomer, because our two higher priority groups are the same size. Finally, let's look at rule number four. A double bond to a carbon is considered as two single bonds. |
Cahn-Ingold-Prelog Priority System .txt | So, once again, as the first example, we have the Z Isomer, because our two higher priority groups are the same size. Finally, let's look at rule number four. A double bond to a carbon is considered as two single bonds. So let's see exactly what that means. Let's suppose we have our double bond here. So a carbon and carbon double bond. |
Cahn-Ingold-Prelog Priority System .txt | So let's see exactly what that means. Let's suppose we have our double bond here. So a carbon and carbon double bond. We want to examine the groups attached to our first carbon. So here we have a carbon in a carbon. So so far, no one wins. |
Cahn-Ingold-Prelog Priority System .txt | We want to examine the groups attached to our first carbon. So here we have a carbon in a carbon. So so far, no one wins. And a carbon and a carbon. So no one wins. But we have a double bond here. |
Cahn-Ingold-Prelog Priority System .txt | And a carbon and a carbon. So no one wins. But we have a double bond here. And this double bond, according to rule number four, is actually, or actually looks something like this. So every double bond is considered as having two single bonds. So we erase this double bond and we replace one carbon and a second carbon. |
Cahn-Ingold-Prelog Priority System .txt | And this double bond, according to rule number four, is actually, or actually looks something like this. So every double bond is considered as having two single bonds. So we erase this double bond and we replace one carbon and a second carbon. So two more single covalent bonds. So now this side, this group has a higher atomic weight. And so this group wins, and this group must be the higher priority group. |
Cahn-Ingold-Prelog Priority System .txt | So two more single covalent bonds. So now this side, this group has a higher atomic weight. And so this group wins, and this group must be the higher priority group. Now, on this side of the carbon, we have two identical methyl groups. So we have carbon and carbon. So no one wins here. |
Stability of Alkenes.txt | So let's suppose we have a certain alky, let's say hexine, for example, and let's write out all different types of isomers of hexine. Now, if we compare the stability of one isomer of hexine to another isomer of exe, we'll see a difference in stability. In other words, some isomers are more stable than other isomers. So in general, why is that? Why is it that some isomers of alkans are more stable than other isomers of that same alkane? So we're going to address that question in this lecture. |
Stability of Alkenes.txt | So in general, why is that? Why is it that some isomers of alkans are more stable than other isomers of that same alkane? So we're going to address that question in this lecture. So let's begin by defining change in Enthalpy affirmation. So loosely speaking, enthalpy affirmation is the energy difference between the final product and its constituent elements. In other words, if this value is negative, that means our final products are stable or more stable than the constituent elements. |
Stability of Alkenes.txt | So let's begin by defining change in Enthalpy affirmation. So loosely speaking, enthalpy affirmation is the energy difference between the final product and its constituent elements. In other words, if this value is negative, that means our final products are stable or more stable than the constituent elements. And in fact, the more negative this value is, the more stable our product is. So let's list a few isomers of hexine. So here's list of five different isomers of hexine and each corresponding change in enthalpy of formation. |
Stability of Alkenes.txt | And in fact, the more negative this value is, the more stable our product is. So let's list a few isomers of hexine. So here's list of five different isomers of hexine and each corresponding change in enthalpy of formation. So for example, for this isomer of hexane, we have negative ten kilo cals per mole of change in enthalpy of formation. And we see that as we go from one to five, our values become more and more negative. And five is the most stable isomer and one is the least stabilizomer. |
Stability of Alkenes.txt | So for example, for this isomer of hexane, we have negative ten kilo cals per mole of change in enthalpy of formation. And we see that as we go from one to five, our values become more and more negative. And five is the most stable isomer and one is the least stabilizomer. So stability increases as we go down from one to five. So why is that? Why is it that number one is less stable than number five? |
Stability of Alkenes.txt | So stability increases as we go down from one to five. So why is that? Why is it that number one is less stable than number five? So to examine this, let's recall an important detail. Remember, when electrons are found in the s orbital, those electrons are more stable than if the electrons were found in the p orbital. And generally speaking, s character is more stable than p character because of that same concept. |
Stability of Alkenes.txt | So to examine this, let's recall an important detail. Remember, when electrons are found in the s orbital, those electrons are more stable than if the electrons were found in the p orbital. And generally speaking, s character is more stable than p character because of that same concept. So let's examine the different bonds, let's compare one and five and examine different bonds that exist within that molecule. And let's see if we can find where the difference in stability comes from. So let's look at one. |
Stability of Alkenes.txt | So let's examine the different bonds, let's compare one and five and examine different bonds that exist within that molecule. And let's see if we can find where the difference in stability comes from. So let's look at one. So here we have the first isomer. So notice we have a double bond. And within this double bond, the Sigma bond contains SP two, SP two character. |
Stability of Alkenes.txt | So here we have the first isomer. So notice we have a double bond. And within this double bond, the Sigma bond contains SP two, SP two character. Remember, SP two simply means there's 33% S character and 66% P character. SP three means there's only 25% S character and 75% P character. So SP two bonds are more stable than SP three bonds because SP two bonds contain more S character. |
Stability of Alkenes.txt | Remember, SP two simply means there's 33% S character and 66% P character. SP three means there's only 25% S character and 75% P character. So SP two bonds are more stable than SP three bonds because SP two bonds contain more S character. So this Sigma bond within the double bond contains SP two SP two character. This bond, this covalent bond has SP two SP three character. And each of these three covalent bonds has SP three SP three character each. |
Stability of Alkenes.txt | So this Sigma bond within the double bond contains SP two SP two character. This bond, this covalent bond has SP two SP three character. And each of these three covalent bonds has SP three SP three character each. So we have three SP three SP three bonds. We have one SP two SP two bond, and we have one SP two SP three bonds. So now let's compare the bonds within compound pi within itemer number five. |
Stability of Alkenes.txt | So we have three SP three SP three bonds. We have one SP two SP two bond, and we have one SP two SP three bonds. So now let's compare the bonds within compound pi within itemer number five. So once again, we have that same exact SP two SP two bond like we have here and we have 1234 SP three SP two bonds. So notice we have more stable bonds within this compound than this compound because SP two is more stable than SP three. Here, we only have one SP two SP three bond and the rest are SP three SP threes, while here, all four bonds are SP three SP two bonds. |
Stability of Alkenes.txt | So once again, we have that same exact SP two SP two bond like we have here and we have 1234 SP three SP two bonds. So notice we have more stable bonds within this compound than this compound because SP two is more stable than SP three. Here, we only have one SP two SP three bond and the rest are SP three SP threes, while here, all four bonds are SP three SP two bonds. So in other words, electrons are more stable in the s orbitals. Thus, the more S character in a bond, the more stable that bond. And alkines are more stable when they have bonds with a lot of S character. |
Chemical Equilibrium and Equilibirium Constant .txt | Today we're going to talk about the concept of chemical equilibrium or dynamic equilibrium. Now, let's look at the following hypothetical reversible first order elementary reaction in which x converts to Y in a single step. Elementary simply means that we can use the coefficients of X and Y to produce the rate law. So let's begin at time equals zero. The concentration of our reactants is very high and the concentration of our product is very low. It's actually zero. |
Chemical Equilibrium and Equilibirium Constant .txt | So let's begin at time equals zero. The concentration of our reactants is very high and the concentration of our product is very low. It's actually zero. So at times zero before our reaction even occurred. We don't have any of our products yet the products haven't formed, so our Y is zero. Now, our X, the concentration of X of our reactants is at its maximum because none of this guy has converted to Y yet. |
Chemical Equilibrium and Equilibirium Constant .txt | So at times zero before our reaction even occurred. We don't have any of our products yet the products haven't formed, so our Y is zero. Now, our X, the concentration of X of our reactants is at its maximum because none of this guy has converted to Y yet. However, as the reaction begins to proceed, the concentration of X begins to decrease, while the concentration of Y begins to increase because some of this X is converting or becoming Y. Eventually a point is reached in which the concentration of the reactants of X and the concentration of the products Y does not change. And this is known as chemical equilibrium. |
Chemical Equilibrium and Equilibirium Constant .txt | However, as the reaction begins to proceed, the concentration of X begins to decrease, while the concentration of Y begins to increase because some of this X is converting or becoming Y. Eventually a point is reached in which the concentration of the reactants of X and the concentration of the products Y does not change. And this is known as chemical equilibrium. This is the point of greatest entropy. In other words, our entropy of our system is at its highest at equilibrium. Now, let's look at the rates of the reactions, the four reaction and the reverse reaction at each point. |
Chemical Equilibrium and Equilibirium Constant .txt | This is the point of greatest entropy. In other words, our entropy of our system is at its highest at equilibrium. Now, let's look at the rates of the reactions, the four reaction and the reverse reaction at each point. Let's go back here. Now, at this point, what's the rate of our forward reaction of X to Y? Well, our rate is determined by our rate constant as well as the concentration of X. |
Chemical Equilibrium and Equilibirium Constant .txt | Let's go back here. Now, at this point, what's the rate of our forward reaction of X to Y? Well, our rate is determined by our rate constant as well as the concentration of X. And because our constant K stays the same at the same temperature going this way, that means our rate is strictly dependent on our concentration. And because we said our concentration is at its maximum at the beginning at time equals zero, that means our rate going this way forward of X converting to Y is at its highest. What about the reverse rate of going from Y to X? |
Chemical Equilibrium and Equilibirium Constant .txt | And because our constant K stays the same at the same temperature going this way, that means our rate is strictly dependent on our concentration. And because we said our concentration is at its maximum at the beginning at time equals zero, that means our rate going this way forward of X converting to Y is at its highest. What about the reverse rate of going from Y to X? Well, at the beginning we said we don't have any of the Y, our Y or concentration of our product is zero. And that means since this guy is zero for the reverse reaction, that means our rate going backwards is zero. And that makes sense because if none of the Y has formed yet, that means none of the Y can convert back to X. |
Chemical Equilibrium and Equilibirium Constant .txt | Well, at the beginning we said we don't have any of the Y, our Y or concentration of our product is zero. And that means since this guy is zero for the reverse reaction, that means our rate going backwards is zero. And that makes sense because if none of the Y has formed yet, that means none of the Y can convert back to X. So let's go to this point somewhere in between our initial and our final. Well, somewhere here, our rate begins to decrease going this way. In other words, because our concentration of X begins to slowly diminish, this guy begins to get smaller. |
Chemical Equilibrium and Equilibirium Constant .txt | So let's go to this point somewhere in between our initial and our final. Well, somewhere here, our rate begins to decrease going this way. In other words, because our concentration of X begins to slowly diminish, this guy begins to get smaller. And so, since this guy remains the same throughout the experiment going this way, that means our rate also begins to diminish going from X to Y. How about going from Y to X? Well, going from Y to X, our concentration of that product begins to increase. |
Chemical Equilibrium and Equilibirium Constant .txt | And so, since this guy remains the same throughout the experiment going this way, that means our rate also begins to diminish going from X to Y. How about going from Y to X? Well, going from Y to X, our concentration of that product begins to increase. And that means as the reaction is going this way, we're getting more of y. That means our Y, our rate law for going from the products to reactants begins to increase the rate of reaction, rate of reverse reaction. That's because the concentration of Y begins to increase. |
Chemical Equilibrium and Equilibirium Constant .txt | And that means as the reaction is going this way, we're getting more of y. That means our Y, our rate law for going from the products to reactants begins to increase the rate of reaction, rate of reverse reaction. That's because the concentration of Y begins to increase. Eventually, the concentration of this guy and this guy at equilibrium will be exactly constant. They won't be the same, although they could be the same, but they will be constant. The concentration of x will not change and the concentration of Y will also not change. |
Chemical Equilibrium and Equilibirium Constant .txt | Eventually, the concentration of this guy and this guy at equilibrium will be exactly constant. They won't be the same, although they could be the same, but they will be constant. The concentration of x will not change and the concentration of Y will also not change. And that means our rate of the reverse reaction this way and the rate of the four reaction known this way will be the same. Now, once again, it's important to understand that the concentration of x and Y in equilibrium will not be the same. They could be the same, but it's not necessary for them to be the same. |
Chemical Equilibrium and Equilibirium Constant .txt | And that means our rate of the reverse reaction this way and the rate of the four reaction known this way will be the same. Now, once again, it's important to understand that the concentration of x and Y in equilibrium will not be the same. They could be the same, but it's not necessary for them to be the same. The reason that their rates are equal is because the K constant going this way and the K constant for the reverse reaction are different values. And so the rates are equal even though the concentrations might not be equal because the K's balance the rates out. Now, once again, it's important to understand that entropy at this point is at its highest because entropy defines the most probable state and equilibrium is the most probable state. |
Chemical Equilibrium and Equilibirium Constant .txt | The reason that their rates are equal is because the K constant going this way and the K constant for the reverse reaction are different values. And so the rates are equal even though the concentrations might not be equal because the K's balance the rates out. Now, once again, it's important to understand that entropy at this point is at its highest because entropy defines the most probable state and equilibrium is the most probable state. Now, it's also important to understand that at this point it's dynamic equilibrium. In other words, the reactions are still occurring. X is still being converted to Y and y is still being converted to x. |
Chemical Equilibrium and Equilibirium Constant .txt | Now, it's also important to understand that at this point it's dynamic equilibrium. In other words, the reactions are still occurring. X is still being converted to Y and y is still being converted to x. It's just because the two rates are equal, the concentrations remain the same. So even though it seems like the reactions have stopped occurring, x is not going to Y and Y is not going to x. That's not true. |
Chemical Equilibrium and Equilibirium Constant .txt | It's just because the two rates are equal, the concentrations remain the same. So even though it seems like the reactions have stopped occurring, x is not going to Y and Y is not going to x. That's not true. Reactions forward and backward are still occurring. It's just they're occurring at the same rate. And that's exactly why the concentrations are remaining the same. |
Chemical Equilibrium and Equilibirium Constant .txt | Reactions forward and backward are still occurring. It's just they're occurring at the same rate. And that's exactly why the concentrations are remaining the same. Now suppose, for example, if I added more x, if I add more x, I would change my concentration of x. And that means for a few seconds or for some time, my concentration of x would change and this would shift equilibrium this way, causing x to be converted to Y at a faster rate than y converted to x. And that's called leisurely air principle and we'll talk about that in another lecture. |
Chemical Equilibrium and Equilibirium Constant .txt | Now suppose, for example, if I added more x, if I add more x, I would change my concentration of x. And that means for a few seconds or for some time, my concentration of x would change and this would shift equilibrium this way, causing x to be converted to Y at a faster rate than y converted to x. And that's called leisurely air principle and we'll talk about that in another lecture. So now let's suppose we have the following elementary reaction in which our reactants x and y convert to our product z and W. Now, A-B-C and D are the coefficients that represent the moles of each respective atom. Now let's suppose our reaction is a dynamic equilibrium. And what that basically means once again is that the rates at which x and Y are converting to z and W is the same as the rate at which z and w is converting to x and Y. |
Chemical Equilibrium and Equilibirium Constant .txt | So now let's suppose we have the following elementary reaction in which our reactants x and y convert to our product z and W. Now, A-B-C and D are the coefficients that represent the moles of each respective atom. Now let's suppose our reaction is a dynamic equilibrium. And what that basically means once again is that the rates at which x and Y are converting to z and W is the same as the rate at which z and w is converting to x and Y. Now, once again, we're making an assumption that this is elementary reaction. So that means we can write the rate law for each reaction going this way and going that way by simply using the coefficient ADC and d.
So let's do exactly that. So the rate of our four reaction is equal to the constant for going this way. |
Chemical Equilibrium and Equilibirium Constant .txt | Now, once again, we're making an assumption that this is elementary reaction. So that means we can write the rate law for each reaction going this way and going that way by simply using the coefficient ADC and d.
So let's do exactly that. So the rate of our four reaction is equal to the constant for going this way. K one times the concentration of A to the A power times the concentration of y to the b power, the coefficients. And this equals the rate of the reverse reaction, the backwards reaction z and w converting back to x and y. And this equals k minus one, which is a different constant than this constant. |
Chemical Equilibrium and Equilibirium Constant .txt | K one times the concentration of A to the A power times the concentration of y to the b power, the coefficients. And this equals the rate of the reverse reaction, the backwards reaction z and w converting back to x and y. And this equals k minus one, which is a different constant than this constant. Remember, the two constants going this way and this way are different times the concentration of z to the c coefficient or exponent times the concentration of w to the d power. Now this guy is equal to this guy because we're a dynamic equilibrium. And the only reason we are able to write the rate laws like this using the coefficients is because this is an elementary reaction. |
Chemical Equilibrium and Equilibirium Constant .txt | Remember, the two constants going this way and this way are different times the concentration of z to the c coefficient or exponent times the concentration of w to the d power. Now this guy is equal to this guy because we're a dynamic equilibrium. And the only reason we are able to write the rate laws like this using the coefficients is because this is an elementary reaction. So now let's bring all the constant to one side and everything else the concentrations to this side. So we get the following. K one divided by k minus one equals this guy divided by this guy. |
Chemical Equilibrium and Equilibirium Constant .txt | So now let's bring all the constant to one side and everything else the concentrations to this side. So we get the following. K one divided by k minus one equals this guy divided by this guy. Now notice that these two guys are constants. They're the same or they don't change at the same temperature. And that means we can represent this guy as another constant, namely K. Now this K is known as the equilibrium constant. |
Chemical Equilibrium and Equilibirium Constant .txt | Now notice that these two guys are constants. They're the same or they don't change at the same temperature. And that means we can represent this guy as another constant, namely K. Now this K is known as the equilibrium constant. And the relationship between our K, the equilibrium constant. And the chemical equation above is known as the law of mass action. So what's the meaning of K? |
Chemical Equilibrium and Equilibirium Constant .txt | And the relationship between our K, the equilibrium constant. And the chemical equation above is known as the law of mass action. So what's the meaning of K? Well, K is simply the ratio of the concentration of products and the concentration of reactants at equilibrium when the Navy equilibrium has been established. And what K does is it tells us how far the reaction proceeded at equilibrium. In other words, we can have three situations. |
Chemical Equilibrium and Equilibirium Constant .txt | Well, K is simply the ratio of the concentration of products and the concentration of reactants at equilibrium when the Navy equilibrium has been established. And what K does is it tells us how far the reaction proceeded at equilibrium. In other words, we can have three situations. K can either be greater than one and if K is greater than one, that means at equilibrium we have more products than reactants. And that means our reaction is a product favored, it's spontaneous going this way. Now, if K equals one, that means at equilibrium our concentration of products is the same as the concentration of reactants. |
Chemical Equilibrium and Equilibirium Constant .txt | K can either be greater than one and if K is greater than one, that means at equilibrium we have more products than reactants. And that means our reaction is a product favored, it's spontaneous going this way. Now, if K equals one, that means at equilibrium our concentration of products is the same as the concentration of reactants. Now, if K is less than one, that means this denominator is larger than our enumerator. And that means we have more concentration of reactants of these guys at equilibrium than of our products than these guys. And that means our reaction is not product favorite, it's not spontaneous, in fact it's reactant favorite. |
Chemical Equilibrium and Equilibirium Constant .txt | Now, if K is less than one, that means this denominator is larger than our enumerator. And that means we have more concentration of reactants of these guys at equilibrium than of our products than these guys. And that means our reaction is not product favorite, it's not spontaneous, in fact it's reactant favorite. This reaction is spontaneous, but this reaction isn't if our K is less than one. And that's the meaning of K. Now, a few more important things that I want to mention about equilibrium constants. Now an equilibrium constant is unitless. |
Chemical Equilibrium and Equilibirium Constant .txt | This reaction is spontaneous, but this reaction isn't if our K is less than one. And that's the meaning of K. Now, a few more important things that I want to mention about equilibrium constants. Now an equilibrium constant is unitless. And that's because we're dividing concentration by concentration. So our units at the end will cancel out. Now, our equilibrium constant depends strictly on temperature. |
Chemical Equilibrium and Equilibirium Constant .txt | And that's because we're dividing concentration by concentration. So our units at the end will cancel out. Now, our equilibrium constant depends strictly on temperature. And that's because our constant is actually a rate constant divided by a rate constant. So it's the ratio of the rate constant going this way to the rate constant going in the reverse direction. And because these guys are dependent only on temperature, these guys also depend upon temperature. |
Chemical Equilibrium and Equilibirium Constant .txt | And that's because our constant is actually a rate constant divided by a rate constant. So it's the ratio of the rate constant going this way to the rate constant going in the reverse direction. And because these guys are dependent only on temperature, these guys also depend upon temperature. It does not depend on the concentration. Now note there is a big difference between equilibrium constant and chemical equilibrium. Although the two things are related, they're two different separate ideas. |
Chemical Equilibrium and Equilibirium Constant .txt | It does not depend on the concentration. Now note there is a big difference between equilibrium constant and chemical equilibrium. Although the two things are related, they're two different separate ideas. Once again, equilibrium constant is a ratio of products to reactants, and it depends on temperature. While chemical equilibrium refers to a condition, a system. And if we add, for example, more reactants to our system now our chemical equilibrium is shifted to the right, more reactants will be produced. |
Chemical Equilibrium and Equilibirium Constant .txt | Once again, equilibrium constant is a ratio of products to reactants, and it depends on temperature. While chemical equilibrium refers to a condition, a system. And if we add, for example, more reactants to our system now our chemical equilibrium is shifted to the right, more reactants will be produced. And that's because of Washacliere's principle. We'll discuss that in a bit. But remember to have this distinction between equilibrium constant and chemical equilibrium. |
Chemical Equilibrium and Equilibirium Constant .txt | And that's because of Washacliere's principle. We'll discuss that in a bit. But remember to have this distinction between equilibrium constant and chemical equilibrium. There are different things. Last thing I want to mention is about this expression, this chemical equilibrium expression. Now, notice we included every single reactant product. |
Chemical Equilibrium and Equilibirium Constant .txt | There are different things. Last thing I want to mention is about this expression, this chemical equilibrium expression. Now, notice we included every single reactant product. And that's because we assume that X-Y-Z and W were either in the aqueous state or the gas state. Now only aqueous or gas molecules are included or expressed in our final expression. Solid molecules and liquid molecules are not included in our expression. |
Neuron Cells Part II .txt | So let's say our outside is 0.003 molar, and on the inside is 0.135
molar. So how would we find the cell voltage due to the potassium ions? Will we use the Nurse equation? What this equation says is our cell voltage at any given concentration is equal to our standard cell voltage. But this guy is zero. We just said that the cell voltage of this reaction and this reaction are equal but opposite. |
Neuron Cells Part II .txt | What this equation says is our cell voltage at any given concentration is equal to our standard cell voltage. But this guy is zero. We just said that the cell voltage of this reaction and this reaction are equal but opposite. So when you add them, this guy goes to zero. That means our cell voltage is just simply this whole guy. Where gas constant T is our temperature, n is the Mozzo electrons, epic Faradays constant. |
Neuron Cells Part II .txt | So when you add them, this guy goes to zero. That means our cell voltage is just simply this whole guy. Where gas constant T is our temperature, n is the Mozzo electrons, epic Faradays constant. And Q is our expression. Now, let's look at Q first. What is Q? |
Neuron Cells Part II .txt | And Q is our expression. Now, let's look at Q first. What is Q? Well, Q is the concentration of products divided by the concentration of reactants, right? And our products is this guy, it's 0.3 molar, our 0.3 molar. Sorry. |
Neuron Cells Part II .txt | Well, Q is the concentration of products divided by the concentration of reactants, right? And our products is this guy, it's 0.3 molar, our 0.3 molar. Sorry. And this guy is zero point 13 five molar. So our Q is 0.3
over zero point 13 five. The M cancel out. |
Neuron Cells Part II .txt | And this guy is zero point 13 five molar. So our Q is 0.3
over zero point 13 five. The M cancel out. Now, our T is our temperature of our body. It's not 25 degrees Celsius, it's 37 degrees Celsius. So we have 37 to 273, and we get 310. |
Neuron Cells Part II .txt | Now, our T is our temperature of our body. It's not 25 degrees Celsius, it's 37 degrees Celsius. So we have 37 to 273, and we get 310. So it's 310 right here. This is our gas constant. It's just a constant 8.3, 114. |
Neuron Cells Part II .txt | So it's 310 right here. This is our gas constant. It's just a constant 8.3, 114. Our fahrenheit is constant. So what is N is the moles of electrons produced per potassium or a mole of potassium. So notice that our mole here is one. |
Neuron Cells Part II .txt | Our fahrenheit is constant. So what is N is the moles of electrons produced per potassium or a mole of potassium. So notice that our mole here is one. It's a ratio of one to one. That means we have 1 mol of electron. So number one goes for N.
We plug these guys into the calculator. |
Neuron Cells Part II .txt | It's a ratio of one to one. That means we have 1 mol of electron. So number one goes for N.
We plug these guys into the calculator. Notice that natural log of a number smaller than one gives you a negative number. So the negative is becoming positive. And this is our final cell voltage, zero point 102 volts. |
Neuron Cells Part II .txt | Notice that natural log of a number smaller than one gives you a negative number. So the negative is becoming positive. And this is our final cell voltage, zero point 102 volts. So then we do the same exact thing for calcium, for sodium, and for chloride. Add all the guys up and we should get our final resting electrical potential to sell. Now, I want to talk more about the meaning of this number. |
Neuron Cells Part II .txt | So then we do the same exact thing for calcium, for sodium, and for chloride. Add all the guys up and we should get our final resting electrical potential to sell. Now, I want to talk more about the meaning of this number. What is meant by this number? Remember, we have the electrochemical gradient of our cell. And this is the gradient due to the concentration of ions and due to charge. |
Neuron Cells Part II .txt | What is meant by this number? Remember, we have the electrochemical gradient of our cell. And this is the gradient due to the concentration of ions and due to charge. So it's the chemical gradient and electrical gradient or voltage gradient. And these guys are opposite of each other. In other words, notice that our potassium ion, there is a larger concentration on the inside than outside. |
Neuron Cells Part II .txt | So it's the chemical gradient and electrical gradient or voltage gradient. And these guys are opposite of each other. In other words, notice that our potassium ion, there is a larger concentration on the inside than outside. And that means these guys will tend to move down their chemical gradients, right? Because there are more of these guys on the outside. So equilibrium will want to establish and these guys will want to move to the outside down their chemical gradient. |
Neuron Cells Part II .txt | And that means these guys will tend to move down their chemical gradients, right? Because there are more of these guys on the outside. So equilibrium will want to establish and these guys will want to move to the outside down their chemical gradient. Now, electrical gradient is the opposite of that. Because electrons travel this way, electrons will want to travel to the place where there is more positive charge. That means it's opposite. |
Neuron Cells Part II .txt | Now, electrical gradient is the opposite of that. Because electrons travel this way, electrons will want to travel to the place where there is more positive charge. That means it's opposite. Now, what this number means is that when our electrical gradient and our chemical gradient equal to this number, when they're both this number, that means equilibrium will be established between the potassium ions and the same number of potassium ions will be going in as they will be coming out, right? So the rates will equal, and that's what this number means. So actually, our electrical potential should be negative of this because they're opposite of each other. |
Diffusion of Gas and Graham’s Law .txt | So we already spoke about a process called effusion which is the movement of gas molecules from a high pressure to a low pressure via a very small hole. And we said that we can find the rates at which gas molecules effuse via something called Grams law which is given right down here. And Grams law states that rate rate of gas molecule one over rate of gas molecule two is equal to the square root of mass of two divided by the square root of mass of one. And what this says is that the lighter the molecule, the faster its rate or the higher its rate. Now we're going to talk about something called diffusion. Diffusion is the movement or the flux of one type of gas molecule into another gas or an empty space. |
Diffusion of Gas and Graham’s Law .txt | And what this says is that the lighter the molecule, the faster its rate or the higher its rate. Now we're going to talk about something called diffusion. Diffusion is the movement or the flux of one type of gas molecule into another gas or an empty space. And now we can use grammar's law to approximate the rates of diffusion. Now, the reason that we approximate is because of something called mean free path. Now, the average or the mean free path of a gas molecule is the distance it travels between any two collisions. |
Diffusion of Gas and Graham’s Law .txt | And now we can use grammar's law to approximate the rates of diffusion. Now, the reason that we approximate is because of something called mean free path. Now, the average or the mean free path of a gas molecule is the distance it travels between any two collisions. So let's look at this system here. In this system we have a square and we have a bunch of red molecules. We have one green molecule. |
Diffusion of Gas and Graham’s Law .txt | So let's look at this system here. In this system we have a square and we have a bunch of red molecules. We have one green molecule. Suppose this green molecule wants to go from this corner to this corner. Now, the best way would be to go directly across from this point to this point. But notice that we have a bunch of red molecules in the way. |
Diffusion of Gas and Graham’s Law .txt | Suppose this green molecule wants to go from this corner to this corner. Now, the best way would be to go directly across from this point to this point. But notice that we have a bunch of red molecules in the way. So this guy will have to push its way to the other side. And by pushing, I mean colliding. So it's going to travel some distance A, then distance B, then CDEF, and finally G.
So each time it collides and it bounces off. |
Diffusion of Gas and Graham’s Law .txt | So this guy will have to push its way to the other side. And by pushing, I mean colliding. So it's going to travel some distance A, then distance B, then CDEF, and finally G.
So each time it collides and it bounces off. 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. |
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