Calculate ΔHrxn for the combustion of octane (C8H18), a component of gasoline, by using average bond energies and then calculate it using enthalpies of formation from Appendix IIB. What is the percent difference between your results? Which result would you expect to be more accurate?

Video transcript

Hey everyone in this example we have propane a three carbon al cane with the formula C three H eight. We need to calculate the heat of combustion for our propane by using average bond energies and then using its standard heat of formation. Next we want to calculate what percent the two values will differ by and explain which value is expected to be more accurate. So we're going to first go ahead and find our entropy of our reaction for our propane. And in order to do so we should recall that this is going to be equal to the entropy of our bonds for our reactant, minus the entropy of the bonds of our products. So what that will look like is we're going to write our reaction out where we would recall that in a combustion reaction, we have our molecule which should be reacting with oxygen. So to to produce carbon dioxide and water. And to balance this equation out, we're going to place the coefficient of three in front of our C 02, A five in front of our 02 and then a coefficient of four in front of our water. And so to clarify our entropy of our reaction for propane, we would say that that's going to equal the Difference between first the bond, the bond entropy totals from our reactant. So for our eight carbon hydrogen bonds that make up our propane, that's then going to be added to our two Carbon Carbon bonds from our propane molecule Which is then also added to our reactant, oxygen which contains a total of five oxygen, oxygen double bonds. And then this quantity is going to be subtracted from the bond n therapies of our products, which begins with Our total of six of our carbon double bonded to oxygen bond types. Then added to a total of eight of our oxygen hydrogen single bonds from our water molecule. And so this would complete the some of the essential piece of our products here in our reaction. And so what we can do is now look up the bond entropy values for each of these bond types we've outlined in our formula above. And so what we would have is that our entropy of our reaction is equal to a value of eight multiplied by the bond entropy of a carbon hydrogen single bond in our textbooks which we would find equal to about 413 kg joules per mole. This is then going to be added to two times the bond entropy of a carbon carbon single bond equal to 3 47 kg joules per mole. And then this quantity is added to five times our bond entropy value for a oxygen oxygen double bond, which we would find in our textbooks equal to about 4 95 kg joules per mole. And so our brackets for our bond entropy is of reactant would end here and now we will take the difference where our bond entropy of our products begin with six times the bond entropy value of a carbon double bonded to an oxygen bond with a value of 99 kg jewels Permal and then added to eight times the bond entropy of an oxygen hydrogen bond, which should equal a value in our textbooks of about 4 67 kayla jewels per mole. And so this completes the some of the bond entropy of our products. And when we add all of these quantities up, we're going to get that the entropy of our reaction is equal to a value of the bond entropy of our reactant, which added to about a value of 6473 kg jewels per mole. Subtracted from the ban entropy of our products, which added up to a value of 8530 kg joules per mole. And so this gives us a final entropy of our reaction equal to a difference of negative 2057 kg joules per mole. And so this is going to be our first answer for finding the heat of combustion. Using our bond n therapies for our reactant and products. But now the problem wants us to use heat of formation to find our bond entropy of our reaction. Or sorry, entropy of our reaction. And so we would rewrite our reaction where we have propane reacting in a combustion with oxygen to produce carbon dioxide and water. We still have our same coefficients. So we would still have four in front of water, three in front of carbon dioxide and five in front of our oxygen. And we want to make note of our entropy of formations for each of our molecules. So for our entropy of formation of propane We would look that up in our textbooks and see that it's equal to a value of -103.85. For our entropy a formation of our second reactant which is our oxygen gas, we would see that that's equal to a And there'll be a formation value in our textbooks of zero. Then for our entropy a formation of our carbon dioxide which is our product, We would have a value equal to negative 3 93.5. And then for Ethiopia formation for our second product water, we would get a value in our textbooks of negative 2 85.8 and all of our units are still kill a jules Permal And so to make things less confusing, we're going to separate our two methods of solution and so now we can go ahead and say therefore now that we've noted down all of our entropy of formation of all of our e agents, we can say that our entropy of our reaction is going to equal the entropy of formation of our products minus the sum of the entropy of formation of our reactant. And so what we would have is that our anthill beat of our reaction is equal to the three moles of our carbon dioxide molecule, which has an entropy of formation equal to a value of we said negative 3 93.5 kg joules per mole. And this is then added to the entropy of formation of our second product, which is water. So we had four moles of the entropy of formation of our water which we said has a value of. And I'm just going to make this four clear here. But our value for our entropy of formation of our full morals of water, we set has a value of negative 2 85.8 kg joules per mole. And I'm just going to scoot this over so that we have enough room here. And so then we would want to subtract our entropy of formation of our reactant which begins with our first reactant, where we said we have one mole of the entropy of formation of our propane, which is equal to a value of negative one oh 3.85 kg joules per mole. And then this is added to the or rather because this is our reactant, we would say subtracted from the entropy of formation of our five moles of our oxygen gas, which we said has an entropy of formation. So we would just say -5 Delta HF of zero kg joules Permal for our oxygen gas. And so this difference here would give us a value for our entropy of our reaction of our combustion reaction equal to negative 2219.8 kg joules per mole. And so for our final question answer to this question we need to figure out by what percent These two values of our entropy of our reaction for our combustion of propane differs. And we need to explain which value is more accurate. So we're going to continue down here with our solution where we should recall that our percent difference is going to equal a quotient. Where in our numerator we're going to take the value of our difference between first hour entropy of formation for our combustion reaction. Found from Yeah from found from the entropy of formation solution which was equal to a value of negative 2219.8 kg joules per mole. And we're subtracting this from our solution which we found for our entropy of our reaction from using our bond n therapies which gave us a result in the first part of our solution of negative 2057 kg joules per mole. And we would finish off our parentheses here and then in our denominator. We're gonna plug in that value from our first or sorry our second part of our solution using entropy of formations to find our reaction entropy as again negative 2219. kg joules per mole. So what we're going to do is then multiply this whole quotient By 100% to give us a percentage here. And this just gave us a value Of 7.34% as our difference here. And so overall, all of our final answers are boxed in blue for each part of our question here. So our first method of our solution was finding our entropy of our reaction using bond energies. The second part of our solution here that we boxed in which gave us negative 2219.8 as our entropy of our reaction was our method of using heat of combustion or NLP a formation and then this is our third answer here, which gave us our percent difference between these two methods. And so overall, our final answer is going to be using heat of combustion is the most accurate method to find our entropy of our reaction because our heat of combustion is specific forgiven compounds compared to our bond energies which are more general. And so this would be our fourth final answer to complete this example. So I hope that everything we went through is clear. But if you have any questions, please leave them down below. Otherwise, I'll see everyone in the next practice video