Bond that involves a die poll for the water molecule that has an O. H. Radical with a charge of minus 00.35 times electron unit, and hydrogen radical which is a positive end of the disciple um Which has a positive .35 e. f. charge. We're going to develop a model to determine the strength of the hydrogen bond, which is that red I'll make it red dash line Holding one Die poll to another within water. So this is an idea of what gives water its bonding strength. What we'll need is the electric potential energy for point charges. Um simple expression KQ big Q over our. So we're going to develop two expressions. Um and look at the interaction between the left water molecules with the right water molecule and we'll break it into two parts. There is the potential energy of the hydrogen. Okay, so we're thinking about the left molecule sitting in the potential of the right molecule and we're going to have to come up with two expressions. One for the bonding energy of the hydrogen on the left, to the molecule on the right and another one for the ohh part of the molecule um bonded to the auto molecule on the right. So if we we look at the picture, we have two charges, both the same amount Of .35 e. And we'll write down what he is and a little bit but it's the fundamental unit of charge. Um and then the hydrogen is interacting with the ohh, which is negative at a distance of what I've labelled as D. And it's interacting with the other hydrogen, which is a positive charge at a distance of D. Plus L. Again I have those distances labeled these, the length of the hydrogen bond and the L. Is the length of the bond uh within the disciple itself. Okay. And we'll have to do the same with the left end O. H. End of the day I poll. So again it's 0.35 E squared. And let's see that ohh is interacting with the other, ohh, so the charges have the same sign At a distance of one over d. plus L. And it's interacting with the opposite sign, which is a negative potential energy at a distance of D. Plus too well. And we're ready to evaluate that. I'll leave some intermediate steps but we have K. is nine times 10 to the right luytens times meter squared column squared. And he is the fundamental charge. Okay. And putting that all together. Um The total energy of the hydrogen bond is just the addition of those two. Um And so the molecule interacting with the other one. And yeah, we get a common factor out front and I'll put the nah note And no means 10 to the -9. I'll put the nano in the upper exponent. Get it out of the denominator. As you can see all those distances are in the denominator. So there's an intermediate step. Just to verify if you're stuck on a calculation and I won't put any more intermediate steps, but this is equal to -3.30 Times 10 to the -20 jewels. And that's per hydrogen bond. Um And this hydrogen bond has been attributed the strong bonding in liquid water which prevents it from being vaporized or boiled. Has been attributed to this hydrogen bond. So what we would like to do is think through how much energy you have to add To break one bond Is of course positive 3.3 Times 10 to the -20 jewels, the negative sign indicating that there is a bonding going on. Um But now we're going to try to compare this number to the latent heat of vaporization of water, which is so well known quantity. It's the amount of energy it takes at the phase change between water and water vapor. Uh and that's a very large number. 2.2, 6 Times 10 to the 6th, jules per kelvin. Sorry, jules per kilogram. Get my case mixed up. All right. So, um we know the energy it takes to break a bond. And now let's figure out the energy uh needed to break Bonds in one kg. Uh huh. And that should give us something comparable to the latent heat of vaporization. So, let's figure out how many bonds there are in one kg. So, we're going to have to take the molecular weight of water, Which is approximately 16 plus two. two for the hydrogen. Um 16 for the oxygen is approximately 18 grams per no. Okay. And we can divide one kg by this to figure out how many moles we have. And then multiply by Allah God rose number, convert grams to kilograms. So we'll use avocados number to figure out how many molecules there are. End um one kg mhm. So this is molecules per per mole. We lost her molecules per mole. Okay. And we work that out and it is approximately three 0.34 Maybe that's more accurate than I need times 10 to the 25th molecules. And so the energy to break The bonds in one kg is the energy per bond times the number of bonds. And I'm just being careful to show all my um little conversions there and we wind up with about what is that? Um 1.10, Times 10 to the six jules per kilogram. Okay, we see that that is about the same order of magnitude as the latent heat of vaporization, which is not surprising so that those hydrogen bonds are the ones that you're breaking as your turn, turning water into water vapor or boiling it. Um And probably what is off a little bit why it doesn't give an exact calculation Is the idea that there is only one bond per um molecule. Um that could be more, there could be less. So our model maybe a little too simplified, but it does give the right order of magnitude.