Methanol (CH₃OH) has been suggested as a fuel to replace gasoline. Find ΔH°_rxn, and determine the mass of carbon dioxide emitted per kJ of heat produced. Use the information from the previous exercise to calculate the same quantity for octane, C₈H₁₈. How does methanol compare to octane with respect to global warming?

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Hi everyone for this problem, we're told that ethanol, a renewable fuel made from food sources, has been suggested to replace traditional fossil fuel gasoline. We need to find the standard entropy of reaction for the combustion of ethanol and calculate how much CO two is released. Particular jewel of energy produced. And then we're going to calculate the same for octane and compare the two values with respect to global warming. How does ethanol compare to octane? So we have two things we're going to be calculating here. So we have the combustion of ethanol and octane. So we'll go ahead and write that. And the balanced combustion reaction of ethanol is C two H five O H. This is a liquid plus 302. Gas is going to yield two CO 2 gas plus three H 20 quid. And in order for us to calculate the standard entropy of reaction, we're going to need the standard heats of formation for our products and react ints. So the standard heat of formations can be looked up and when we look them up we'll see that the standard heat of carbon dioxide gas is negative 393. killer jewels per mole. The standard heat of combustion or liquid water is going to be negative 285 eight killer jewels, Permal. And the standard heat of combustion of ethanol is going to be - killer jewels Permal. And for us to calculate this standard heat or the standard entropy change, it's going to be the some of our products minus the sum of our reactant. And we're going to use our balanced chemical equation and we're going to multiply our number of moles by our standard heats of formation. And so if we do the some of our products, we have three moles of H liquid. So we have three times the standard entropy change of water is negative 285.8 plus we have two moles of carbon dioxide and the standard heat of formation is negative 93.5. So that's the sum of our products. And we're going to minus the sum of our reactant. And for our reactant we have three moles of 02 gas and anything in its elemental state is going to be zero. So this is going to be three times zero. That's the standard heat of formation of anything in its elemental form. And we have one mole of ethanol. And the standard heat of formation of ethanol is negative 277. So once we go ahead and calculate that we'll get our standard entropy for our reaction for ethanol is going to be negative 1367.4 killer jewels per more. That is the first part of our answer. Find the standard entropy of our reaction. Next. It asks us to calculate how much carbon dioxide is released, particular jewel of energy produced. Okay, so we're going to need our mass, we're going to need our mass of carbon dioxide over killer jewels. Okay, so our molar mass of carbon dioxide is 44.01 g of carbon dioxide per one mm of carbon dioxide and looking at our balanced chemical equation, we're going to look at the multiple ratio of carbon dioxide to ethanol. So in in one more of ethanol we have two moles of carbon dioxide. And when we do that we can see which of our units cancel. So we have our moles of carbon dioxide cancel. So we're left with grams of carbon dioxide over a mole of ethanol. But we want our mass of carbon dioxide per kill a jewel. And now we can say We just calculated our standard entropy of reaction. So in one mole of ethanol we have the value we just calculated for our standard entropy of our reaction is 1367.4 kg jewels. And we'll keep that positive. Okay, so now we see that our moles of ethanol cancel. And the units that were left in is grams of carbon dioxide per kill a jewel, which is exactly what we want. And so when we go ahead and do that calculation, We get 0. grams of carbon dioxide per kill a jewel of heat. Okay, so that is or ethanol. And I'll make this smaller so we can have some space. Okay, so that is the second answer. So that is how much carbon dioxide is released per kilo jewel of energy produced. Now the question asked us to calculate the same for octane and then we're going to compare the two values. So we're going to do exactly what we did here for ethanol. But for octane, our balanced combustion reaction of octane is going to be two C 8 H 18. This is a liquid plus oxygen Gas, gives us 16 carbon dioxide gas plus 18. Make this a little smaller. We have a lot we're writing here And this is H 2 0 liquid. Now the same thing we're going to need our standard heats of formation and the only thing that we don't have is our standard heat of formation of octane. We have everything else from us calculating ethanol. So our standard heat of formation of octane is negative 0.4 killer jewels per. So now we can calculate our standard entropy of reaction and it's going to equal the sum of our products minus the sum of our reactant. So we're looking at our equation here and we're going to multiply our number of moles by the standard heats of formation. So we have starting with our products we have 18 moles of liquid water and the standard heat of formation for that is negative 200 and 85.8 kg joules per mole plus we have 16 moles of carbon dioxide which we said the standard heat of formation is negative 393.5 kg joules per mole. So that's the sum of our products minus the sum of our reactant. Since we have 25 moles of oxygen gas. And we said anything in its elemental state is going to have a standard heat of formation of zero. So that's 25 times zero. Plus we have two moles of octane. And we said the standard heat of formation of octane Is going to be negative .4 Killer Jules Permal. So that's the sum of our products minus the sum of our reactant. And when we do this calculation we get a standard entropy of our reaction for octane to equal negative 11, 23 0.6. Kill the jewels per mole. Okay, so this is our standard entropy of our reaction for octane. And we need to do the same thing, calculate how much carbon dioxide is released. Particular jewel of energy produced. So we want our massive carbon dioxide per killer jewel of heat. And so we need the molar mass will do that of carbon dioxide Which we said was 44.01 g of carbon dioxide her one mole of carbon dioxide. And we need to look at our multiple ratio of octane to carbon dioxide. So we have and two moles of octane. We have malls of carbon dioxide. And in one mole and one more of octane. We said that the standard heat of the standard entropy of our reaction that we just calculated was 11, 23.6 killer jewels of heat. So let's make sure our units cancel here. Our moles of carbon dioxide cancel our moles of octane cancel. And we're left with grams of carbon dioxide per kill a jewel of heat for octane. And we get a final answer of 0. g of carbon dioxide per kill a jewel of heat or octane. Okay, so let's just make this a little bit smaller. Okay, so the question asks us to with respect to the last part of the question says with respect to global warming, how does ethanol compared to octane? And here we see that ethanol produces more carbon dioxide or releases more carbon dioxide per kill a jewel of energy produced than octane does. That's all the parts to this problem. And that's the end of this problem. I hope this was helpful

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Key Concepts

Enthalpy of Reaction (ΔH°rxn)

Combustion and Carbon Dioxide Emission

Global Warming Potential (GWP)