Ethyl alcohol, C2 H5 OH, has been suggested as an...

Ethyl alcohol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},$ has been suggested as an alternative to gasoline as a fuel. In Example 18.5 we calculated $\Delta G^{\circ}$ for combustion of $1 \mathrm{~mol}$ of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} ;$ in Example 18.6 we calculated $\Delta G^{\circ}$ for combustion of $1 \mathrm{~mol}$ of octane. Let's assume that gasoline has the same properties as octane (one of its constituents). The density of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ is $0.7893 \mathrm{~g} / \mathrm{mL} ;$ the density of octane, $\mathrm{C}_{8} \mathrm{H}_{18}$, is $0.7025 \mathrm{~g} / \mathrm{mL}$. Calculate the maximum work that could be obtained by burning 1 gallon $(3.78$ liters) each of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ and $\mathrm{C}_{8} \mathrm{H}_{18} .$ On a volume basis, which is the better fuel?

Key Concepts

Energy Density

Energy density refers to the amount of energy stored in a given system or region of space per unit volume (volumetric energy density) or per unit mass (gravimetric energy density). This concept is crucial when assessing fuels because it provides a measure of how much usable energy can be delivered from a given amount of fuel, directly impacting the overall efficiency and performance of fuel-dependent systems.

Fuel Density

Fuel density is the mass per unit volume of a fuel and is a critical parameter when comparing fuels on a volumetric basis. It allows for the conversion between volume and mass, enabling the calculation of the total energy available in a given volume of fuel by linking it to the thermodynamic properties like ?G per mole.

Combustion Reactions

Combustion reactions involve the rapid oxidation of a fuel, releasing energy in the form of heat and light. These reactions are highly exothermic and are fundamental to the operation of engines and power plants, serving as the basis for comparing the energy output of different fuels under controlled conditions.

Gibbs Free Energy

Gibbs free energy (?G) is a thermodynamic function that indicates the maximum usable work that can be obtained from a chemical reaction at constant temperature and pressure. It is essential for evaluating the efficiency and spontaneity of energy-releasing reactions such as combustion, where a negative ?G signifies that the reaction can proceed spontaneously while doing useful work.

Transcript

00:01 Okay, so this is another huge question about the anthropology.

00:08 So this question, center of this question is the calculation of the entropy.

00:14 So first, let's look at the question itself.

00:18 Question a is right the balance aquarium for the combustion of the ethanol, right? then the product should be carbon dioxide and water.

00:28 So let's start writing this reaction.

00:35 First, we have the ethanol and it combusts as a liquid phase.

00:45 It combust to form carbon dioxide and water.

00:58 So the way we use to balance the equation, oh sorry, there's one thing missing right here.

01:07 It's oxygen, right? we need oxygen.

01:14 The way that we use to balance is to look at the fuse itself.

01:20 So in the fuels here, right, we say there are two carbons.

01:27 So as a product, there should be two carbon dioxide.

01:32 And you say there are six hydrogen in the mass anode as a reactant.

01:37 So there should be three water molecules as a product because each water has two hydrogen.

01:44 And now we can try to balance the oxygen.

01:47 You can see as a product, the total 2 times 2 plus 3 times 1 equals to 7 oxygens, right? so 7 oxygen atoms as in the product.

02:08 And in the ethanol, there's 1 oxygen.

02:11 So you need extra 7 minus 1, 6 oxygens, right, as reactant.

02:18 So here, oxygen should be three, right, because each oxygen gas molecule has two oxygen atoms.

02:28 And this equation is then balanced.

02:33 And also it asks us to calculate the entropy of the combustion of one more ethanol, using the formation, energy of the ethanol and carbon dioxide and water.

02:50 So, now we need to do the calculation, right? so the first thing, what we need to do is write down the formation energy of the different chemicals and their equation.

03:09 The first is the carbon dioxide.

03:14 That's right down as one, right? the formation energy of the carbon dioxide is already given.

03:19 It's not given here, but it's given in.

03:24 The table from the book, which equals to minus 3903 .5 kilo per more.

03:43 And what we need to do is write down its corresponding chemical equations, which is carbon right, plus oxygen and forms the carbon dioxide.

03:57 And the second piece in formation is the entropy of the formation of the water, right? and it's equal to minus 285 .8 kilojou per more.

04:22 And the same, we need to write down its corresponding chemical equations, the hydrogen react with oxygen, or form liquid phase water.

04:37 And the third one is the formation energy of the methanol or ethanol.

04:47 Yeah, it's ethanol, right, c2h5 -oh.

04:50 It says you can find it in a table, right, which is appendix.

04:56 Here i would just give you the number.

04:58 I checked on the appendix.

05:01 It's actually minus 277 .6 kilojou per more.

05:11 And same, we need to write down its corresponding equations.

05:16 So in order to write down, first we need to write down the product.

05:20 It's a mass node, right? it's a liquid phase.

05:27 And you say, so the reactant should be carbon, hydrogen, right? and then since it has oxidite inside, right, so she also has oxygen as the reactants.

05:45 And now we balance it.

05:48 There are two carbons, right, and six hydrogen, so that would be three hydrogen gas, and one half of the oxygen.

05:59 So we need to use this three quirin, right, to form the combustion of the mass known.

06:08 Equation.

06:10 We can indicate this equation as equation 4.

06:20 So through the observation, you will find the aquarium 4 actually equals to 2 equation 1, right? because there are two carbon dioxide in the formation of the combustion with the ethanol and plus the sweet water formation, right? so three water equine 2 and minus 1 equation 3.

06:56 And then same for the anselp rate, be equal to 2 times the formation energy of the carbon dioxide, plus 3 times the formation energy of the water, minus one formation energy of the methanol.

07:31 And let's write down.

07:33 It's, let's plug in the number two times minus 393 .5 plus three times minus 285.

07:50 A minus 285.

07:51 A minus, minus, right, 200.

07:57 177 .6.

07:59 And you can use your calculator.

08:01 You'll find the answer is minus 1 ,366 .8 kilchow per more...