Small molecules with C'C double bonds, called monomers,
can join ...

Question

Small molecules with C'C double bonds, called monomers,
can join with one another to form long chain molecules
called polymers. Thus, acrylonitrile, H2C'CHCN, polymerizes
under appropriate conditions to give polyacrylonitrile,
a common starting material for producing the carbon fibers
used in composites.
(b) Use the bond dissociation energies in Table 7.1 to calculate
ΔH per H2C'CHCN unit for the conversion of
acrylonitrile to polyacrylonitrile. Is the reaction endothermic
or exothermic?

Accepted Answer

welcome back everyone in this example, we need to calculate the approximate molar entropy change for the formation of Kevlar, which is a polyamide. And what poly means is that we have are a mite groups, which we should recall is a type of group in organic molecules where we have a carbon chain which we represent as are bonded to our carbon Neil, so that's C double bonded to oxygen and then that is bonded to nitrogen, which is bonded to a hydrogen atom and H. So that's our aiming right here and as a whole, this is our group. And so now that we understand that part, we're going to recognize that we're given the following reaction where we have our two monomer units being this first structure and then this second structure where we have boxed in where the reaction occurs. And so according to the diagram, we can say that the bond between carbon and oxygen here is breaking. So this is breaking. So we can say it breaks as well as the bond between nitrogen hydrogen breaks. And these bonds break to form our one mole of water, where we have a second occurrence where the nitrogen hydrogen bond here breaks and the oxygen hydrogen bond here breaks to form our second mole of water, giving us two moles of water on our product side. Total. And when these bonds break, that leaves a bond between carbon and nitrogen to form. So we would have a bond now between carbon and nitrogen which forms and that is why carbon and nitrogen are bonded here together on our product side. For our Kevlar polymer, we know that this is a polymer because the way that our product is written, we have just a single focus, which is why we have these brackets here on this single unit of our polymer structure of our particularly poly amid structure where we can count for our types of bonds here that are formed. We have two moles of our carbon nitrogen single bond that is formed because we have one here and actually let's use the color green so that it's clear we have two moles of our carbon nitrogen bond where we said we have a bond between carbon nitrogen formed here and recall that a corner in our structures here represents carbon atoms. So we have a bond between carbon and nitrogen here giving us two moles of our carbon nitrogen bond for bonds formed because we know that this is our product side. And then we also recognize that we have a bond that forms between our oxygen and hydrogen giving us our two moles of our oxygen hydrogen single bond because we know that that makes up our water molecule here. Now going back to our bonds broken, which is our product side, we can count a total of two moles of our carbon oxygen single bond, which breaks because it's breaking in this first circled position and also breaking between these two atoms here. So this also breaks giving us two moles of our carbon oxygen single bond breaking. And then our second type of bond we see that breaks is our two moles of our nitrogen hydrogen single bonds, which breaks in this first circled location and also breaks over here in this second circled location where the hydrogen atoms are going to bond with the oxygen's that break off from the carbon oxygen bond. And so these are our bonds broken, which is our react inside. And so now that we understand this image, we're going to recall that to calculate our molar entropy change, which we should recall is represented by the symbol delta H. We would recall that delta H is calculated by taking the difference between the bonds broken. So specifically the bond energies which will say as B dot e of our bonds broken or reactant minus the bond energies of our bonds formed which is our products. And so plugging in from our values of bond energies for each type of bond. In our textbooks, we're going to begin with our reactant side, which is our bonds broken, where we have in brackets. First, our two moles of our carbon oxygen single bond multiplied by the bond energy of a carbon oxygen. Single bond of 358 kg joules per mole. We have our second type of bond broken, which is our two moles of our nitrogen hydrogen single bonds that's bringing multiplied by the bond energy of a nitrogen hydrogen single bonds, which we see in our textbooks has a value of 391 kg joules per mole. And so now taking the closing off our brackets here and taking the difference from our bond energies of our bonds formed, which is our products. We'll begin our brackets where we have our first type of bond formed, which is our two moles of our carbon nitrogen single bond that forms multiplied by the bond energy of a carbon nitrogen single bond, which we see in our textbooks, has a value of 293 kg joules per mole, adding to this to our second type of bond formed. We have two molds of our oxygen hydrogen single bond that forms multiplied by the bond energy of an oxygen hydrogen single bonds being from our textbooks, 463 Killah jewels per mole. And so we can close off our brackets for bonds formed and what we would get from our product and some here or sorry. The product and differences here. So let's do for our reactant is first We would have a value of 1,498 from the sum of our from the summoned products from our bonds broken which is our react inside. This is subtracted from our product side where we would get a value of 1,512. And we have units of kill jules per more here. So that's for both sides. Killer jewels Permal. And let's make this clear. So that's sorry, 1498 kg joules per mole. So taking the difference between these two values will result in a molar entropy change equal to the value of negative 14. Killer joules per mole. Which we will interpret as per reactant. So we'll extend this and say per reactant, Per mole of our reactant that is being lost as our bonds are broken on our reactant side. And so we can say that this applies to both of our monomer units of our reactant because we have a 1-1 molar ratio of each of these react ints. And so for our final answer, we can confirm that our molar entropy change delta H. Is equal to 14 kg joules per mole of each reactant that we have for the formation of Kevlar. And this is our final answer to complete this example. I hope everything I reviewed was clear. If you have any questions, leave them down below and I will see everyone in the next practice video

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