Use the data in Table 9.3 to calculate an approximate ∆H° in kilojoules for the synthesis of hydrazine from ammonia: 2 NH3(g) + Cl2(g) → N2H4(g) + 2 HCl(g)

Video transcript

welcome back everyone. We need to estimate the entropy of reaction and kill jewels for the formation of methane tile CH three S. H. From methanol CH three oh H. Below were given our equation where we have methanol reacting with hydrogen sulfide gas to produce methane gas and gaseous water. And when we check out the number of atoms on both sides of our reaction, we have a balanced reaction. So we don't need to add any coefficients. But what we do need to check for is the number of types of bonds we have in each of our reactant and product molecules. And that is important because we're given below a chart with the average bond association energies of each type of bond present in our molecules of our reaction. So we need to figure out the types of bonds or the number of the types of bonds that we have. So beginning with our structure for our methanol, we have CH 30. H. Where we have one carbon atom surrounded by three hydrogen atoms which is bonded to an oxygen where the oxygen is bonded to a hydrogen atom and we would just connect all of our atoms together. And this is our structure for methanol. We also want to recognize that oxygen should have two lone pairs on itself and it's reacting with our H two S. Gas. So hydrogen sulfide gas is just one atom of sulfur bonded to two atoms of hydrogen where we have single bonds to hydrogen and sulfur will have two sets of lone pairs on itself because it does have six valence electrons being in group six A of the periodic table. Now on our product side, we have our methane tile product where we have carbon bonded to three hydrogen atoms bonded to sulfur and then brought it to hydrogen where we would connect everything with single bonds. And again our sulfur atom has two sets of lone pairs on itself. And then we have our water product which is just our water molecule with two lone pairs on the oxygen bonded to two hydrogen atoms with a single bond. And so we can count that we have a total and we'll write this below actually, we have a total of three of our carbon hydrogen single bonds on the reactant side. We're on the product side. We can also confirm that we have three of these carbon hydrogen single bonds. Now moving on to our carbon oxygen single bond, we just have one of our carbon oxygen single bond on the reacting side and one on the product side as well. Now we have also on the product side, one of our carbon sulfur single bonds. And then considering our oxygen hydrogen single bonds on the reactant side, we have one of our oxygen hydrogen single bonds. And on the product side we can see that we have two of our oxygen hydrogen single bonds in our molecule of water and then our hydrogen sulfur single bonds. We would count one or sorry, we would count two of our hydrogen silver single bond on the reactant side Where on the product side, we would only have just one hydrogen sulfur single bond, which is just in this part of our molecule here for methane tile. And so now that we have each number of the types of bonds noted down, we're going to next recall that we can relate the entropy of our reaction to the difference between the the bond association energy of our reactant bonds subtracted from the bond association energy of our product bonds. And so plugging in our knowns from the given chart we would have that are entropy of our reaction is going to equal one. So one mole multiplied by our bond association energy of our carbon oxygen single bond Which is then added to three moles of our carbon hydrogen bond association energy, sorry, three moles times the And let's write this accordingly. So plus three moles times the bond dissociation energy of our carbon hydrogen single bonds on the reactant side. And then added to this, we have as we stated, one mole times the bond association energy of our oxygen hydrogen single bond. This is going to end in brackets here where we're going to now subtract from the bond dissociation energy of our products product bonds where we would begin our bracket and have our one mole of our carbon sulfur single bond bond association energy which is then added to and will follow in the line below three moles multiplied by the bond association energy of our carbon hydrogen single bonds, which is added to one mole, multiplied by the bond association energy of our single hydrogen sulfur bond, which is then added to two moles of the sorry, two moles multiplied by the bond association energy of our oxygen hydrogen single bond. And this would complete our bracket for the bond association energy of our product bonds. So now we want to plug in the known values given from the chart and we would have that the entropy of our reaction is equal to. We have one mole and we'll just simplify this to just say one. Actually, no. Let's write one mole multiplied by the bond association energy for a carbon oxygen single bond given in the prompt as 350 joules per mole. This is then added to three moles times the bond association energy of our carbon hydrogen single bonds Given as kg joules Permal. Then added to this, we have one mole multiplied by the bond association energy of an oxygen hydrogen single bond given us 460 kg joules per mole. And then added to this, we have two moles, multiplied by the bond association energy of our hydrogen sulfur single bond given as 340 kg joules per mole. Closing off our brackets. We subtract this and begin our brackets for the bond association energy of our product bonds. Where we have one mole, sorry, Beginning with one mole multiplied by the bond association energy of our carbon sulfur single bond being to 60 kg per mole in the prompt added to this, we have our three moles times the bond association energy of our carbon hydrogen single bonds given in the prompt as 410 kg joules per mole added to this, we have one mole multiplied by the bond association energy of our hydrogen sulfur single bond given in the prompt as kg joules per mole. And then added to this, we have finally two moles of our bond association energy for oxygen, hydrogen single bonds being 460 kg per mole. And then we can end off our brackets for the bond association energy of our products. And so simplifying both sides of our brackets. We're going to have that our entropy of our reaction is equal to a value of 2720. And we would be left with units of killing jewels. After we cancel out all of our units of moles for the Bond Association energy of our reactant reactant bonds. So we have 2720 kill a jules left subtracted from our bond dissociation energies of our product bonds. We're going to simplify in our brackets to a value of killed jules. After we cancel out our units of moles for the products product bonds. And so now taking the difference between these two values, we would find that our entropy of our reaction Is equal to a value of negative 30 when we take that difference there and we have units of killing jewels. And this would be our final answer for the entropy change of our reaction and kill a jewels for the formation of methane tile being CH three S. H. From methanol CH 30. H. So it's highlighted in yellow is our final answer. I hope everything I reviewed was clear. If you have any questions, leave them down below and I'll see everyone in the next video.