Phosgene, COCl21g2, is a toxic gas used as an agent of warfare in...

Question

Phosgene, COCl21g2, is a toxic gas used as an agent of warfare in World War I.
(b) Using the table of bond dissociation energies (Table 9.3) and the value ΔH°f = 716.7 kJ>mol for C(g), estimate ΔH° f for COCl21g2 at 25 °C. Compare your answer to the actual ΔH° f given in Appendix B, and explain why your calculation is only an estimate.

Accepted Answer

welcome back everyone formaldehyde. Ch 20. Is a pungent colorless gas, famous for its use to preserve tissues or cells using bond association energies and entropy of formation equal to 7 16.7 kg per mole for carbon gas, estimate the value of entropy of formation for formaldehyde at 25 degrees Celsius explain why the value obtained is only an estimate And the literature of the entropy of formation for formaldehyde is 108.6 killed Jules. Permal were also given the average Bond Association energies for each type of bond that is present in the formation of formaldehyde. Our first step is to write out our reaction for how formaldehyde is formed. So we're going to recognize that its formation is going to consist of the atoms that make it up. So we would have carbon gas reacting with oxygen gas reacting with hydrogen gas to form c H 20 a k A R. Formaldehyde gas product. Now we want to make sure that this is balanced and to balance this out, we're going to have to place a coefficient of one half in front of oxygen gas. And now our next step is to draw out each of our molecules from the reaction. So we have just carbon gas as just a single carbon atom here. This is reacting with one half moles of our oxygen oxygen atoms which are double bonded to one another, where we have two lone pairs on each of our oxygen atoms because we recall that oxygen on our periodic table is located in group six A and should have six valence electrons to complete its octet. So it has 12 in Kovalev bonds 3456 where we have two sets of lone pairs that we drew in. Next. We're going to draw our third reactant, which is our hydrogen gas, where we just have two hydrogen atoms single bonded to one another. And then on our product side we have our formaldehyde molecule where we have carbon in the center surrounded by two hydrogen atoms, where it's then surrounded by an oxygen atom at the top where we would make our base connections. And then we would make a double bond with oxygen where oxygen still has two sets of lone pairs on itself. And this would complete our structure for for formaldehyde. Now we want to make note of the fact that we have a total of on our product side, three moles of our, we're sorry, we have one more of our hydrogen hydrogen single bonds on the reactant side, whereas on the product side we have carbon hydrogen single bonds. And we would say we have two moles of these types of bonds. We can also say that on the product side we have one mole of our oxygen double bond to carbon type of bond. And then we have on our reactant side only when half mole based on that coefficient of our oxygen oxygen double bond and as we stated, our carbon gas is just one atom here. Now that we have the the number of the types of bonds we have making up the reaction to form formaldehyde. We're going to now recognize that we can find the entropy change For our reaction which is ultimately going to tell us the entropy of formation of our only product being our formaldehyde product CH 20. And so we would recall that we can relate the entropy change of our reaction to the difference between the entropy of formation of our carbon gas added to the bond association energies of our reactant bonds. Where we would close off our brackets and then subtract this from where we would begin brackets by the bond association energy of our product bonds. And so now we're just going to plug in what we know from the prompt we would have that are entropy of our reaction Is equal to plugging in our entropy a formation of carbon that was given in the prompt as 716.7 kila jules, Permal for carbon gas. And we're adding this and sorry, this is our parentheses here. We're adding this to our bond association energy of our reactant. So beginning with our reactant bonds, we stated that we have one half moles which is multiplied by the bond association energy of our oxygen oxygen double bond given in the prompt as 498 kg joules per mole. And then added to this, we have our second type of bonds on our reactant side where we counted one mole multiplied by. And sorry, we don't need parentheses yet. We have one mole multiplied by the bond association energy of our hydrogen hydrogen single bonds given in the prompt as 436 killed jules per mole. Where we can close off our parentheses and end off our brackets for the bond dissociation energies of our products. And now we're going to begin subtraction where we're going to plug in the bond dissociation energies of our product bonds. And so we're going to have for our product bonds. two moles multiplied by the bond association energy of our carbon hydrogen single bonds given in the prompt as 410 kg joules per mole, which is then added to and will follow in the line beneath one mol multiplied by the bond association energy for our oxygen. Carbon single bonds given in the prompt as 732 kg joules per mole where we can now close up our parentheses and end off our brackets for the bond dissociation energies of our products. And now we're just going to simplify so that we have our entropy of our reaction equal to A value of 1,401 for the Bond Association energies of our reactant bonds. And we're going to be left with units of kilo jewels because we'll be able to rather will be left with units of killing joules per mole because we'll still have kilograms per mole after we cancel out our units of moles. We still have more here. So Now subtracting from this, we have our bond association energies of our product bonds which we will be able to simplify to a value of 1,552. Also just killing jewels are our final units here in the numerator. And so now we're going to take the difference here and we're going to find that our entropy of our reaction is equal to a value of negative 150 kg joules per mole. Now we understand that this is for the entropy of formation for formaldehyde. So that's C H 20. And That is why we can say that our only product formaldehyde has an entropy of formation equal to negative 150 kg joules per mole. And this estimate that we've just calculated is less than the given anthem PA formation for formaldehyde from the prompt which was given to us as a value of 108.6 kg per mole. This is because we want to recall that to calculate this Ethiopia formation for formaldehyde, we use these bond energies, the bond energies for the bonds that are involved in the reaction for formaldehyde. And these bond energies we want to recall our average values derived from different compounds. So because bond energies are average values obtained from several different compounds. That is why our entropy of formation didn't match up to the one given in the prompt that we just compared to. So we can confirm for our final answer that this value is an estimate obtained from average bond energies calculated from several different compounds. And this entire statement, as well as our calculated entropy of formation for our formaldehyde product, is going to be our final answer. To complete this example. I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

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