The automobile fuel called E85 consists of 85% ethanol and 15% gasoline. E85 can be used in the so-called flex-fuel vehicles (FFVs), which can use gasoline, ethanol, or a mix as fuels. Assume that gasoline consists of a mixture of octanes (different isomers of C 8 H 18 ), that the average heat of combustion of C 8 H 18 (l) is 5400 kJ/mol, and that gasoline has an average density of 0.70 g/mL. The density of ethanol is 0.79 g/mL. (a) By using the information given as well as data in Appendix C, compare the energy produced by combustion of 1.0 L of gasoline and of 1.0 L of ethanol.

Hi everyone for this problem, we're told to assume that the average heat produced by the combustion of diesel is 44.80 kg jules program, compare the energy produced by five liters of diesel and five liters of methanol. If the heat of combustion of methanol is negative, 726 kg joules per mole. The density of diesel and methanol is 0.85 g per milliliter and 0.79 g per milliliter respectively. So for this problem, we're comparing two values. We're comparing diesel and methanol and we're starting off with the volume. So we're told that we have five liters of each and we need to compare the Energy produced by both. Okay, so our goal is to go from 5l of both to energy produced by both. So let's go ahead and get started. So let's start off with methanol. Okay, so methanol is c H 30. H. And we need to know this because we're going to need to know the molar mass of methanol. So let's start off with what they tell us. They tell us we have five leaders of methanol and we want to go from leaders of methanol to kill a jewels produced energy produced. So let's go ahead and we're going to need the density of methanol and they tell us that density is 0.79 g per mil leader. So we need to first go from leaders of methanol to mill leaders. So we say that in one middle leader, We have 10 to the negative three leaders. Okay, so now we were in units of mill leaders and we can use the density that was given in the problem, they tell us that The density is .79 g per mil leader and that's our Unit conversion here. So in one millimeter There is 0.79 g of methanol. Okay, so now we're in grams of methanol and remember we want to go to energy. And what we can do now is use our molar mass of methanol to go from grams to moles. So in one mole of methanol, our molar mass of methanol Is 32.05 g. Okay, so let's just make sure our units are canceling properly here. So our leaders of methanol cancel. Mill leaders cancel grams of methanol cancel. Now we're in molds of methanol and we want to go to kill a jules. And what we can use here is we're told in the problem that methanol produces 726 kg jewels per mole. Okay, so that is the conversion that we can use to go from moles to kill a jewels. So in one mole of methanol we have 726. Kill a jewels of energy produced. So now our molds of methanol cancel and we're left with killer jewels. So let's go ahead and calculate that number. So we have eight 0.95 times 10 to the four killer jewels of energy produced by this five liters of methanol. And the question is asking us to compare the energy produced between methanol and diesel. So now we need to do the same thing. But for diesel and then we can compare the two. Okay, so for diesel, Our formula is C12 h 23. So we're told that we have five leaders and we want to do the same thing. We want to go from leaders of diesel to energy of diesel produced. So we need to convert this to mill leaders in one millimeter. We have 10 to the negative three leaders and were also given the density of diesel. So now we can use that here And one militia leader of diesel, we have 0.85 grams of diesel. This is our density. Okay, and For diesel, the problem tells us that the combustion of diesel is 44.80 killer jewels per gram. So we can use that here because if you take a look at our units are leaders cancel our middle leaders canceled. And now we're in grams for methanol, we were in moles and we used the combustion of methanol to solve. But here for diesel we have grams. So we're going to use this combustion of diesel at 44.80 kg joules per gram to figure out our energy produced. So when we plug that unit conversion in for every one g Of diesel, we have 44.80 kila jewels of energy produced. So if you see here are grams cancel and we're left with kila jewels of heat. And so this gives us 1.90 times 10 to the 5th killer jewels. Okay, so these are our two energies, it says to compare the the energy produced by both, and these are our respective energies produced by both. That's the end of this problem. I hope this was helpful.

Ch.5 - Thermochemistry

Brown - Chemistry: The Central Science 14th Edition

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Ch.5 - Thermochemistry

Problem 98a

Chapter 5, Problem 98a

The automobile fuel called E85 consists of 85% ethanol and 15% gasoline. E85 can be used in the so-called flex-fuel vehicles (FFVs), which can use gasoline, ethanol, or a mix as fuels. Assume that gasoline consists of a mixture of octanes (different isomers of C₈H₁₈), that the average heat of combustion of C₈H₁₈(l) is 5400 kJ/mol, and that gasoline has an average density of 0.70 g/mL. The density of ethanol is 0.79 g/mL. (a) By using the information given as well as data in Appendix C, compare the energy produced by combustion of 1.0 L of gasoline and of 1.0 L of ethanol.

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Calculate the mass of 1.0 L of gasoline using its density. Use the formula: mass = density × volume. Here, the density of gasoline is 0.70 g/mL, and the volume is 1000 mL (since 1.0 L = 1000 mL).

Calculate the number of moles of octane (C₈H₁₈) in 1.0 L of gasoline. Use the molar mass of octane, which is approximately 114 g/mol. The number of moles can be calculated using the formula: moles = mass / molar mass.

Calculate the total energy released from the combustion of the octane in 1.0 L of gasoline. Use the heat of combustion of octane, which is 5400 kJ/mol. The total energy can be calculated using the formula: total energy = moles × heat of combustion.

Calculate the mass of 1.0 L of ethanol using its density. Use the formula: mass = density × volume. Here, the density of ethanol is 0.79 g/mL, and the volume is 1000 mL.

Refer to Appendix C for the heat of combustion of ethanol, and calculate the total energy released from the combustion of ethanol in 1.0 L. Use the molar mass of ethanol, which is approximately 46 g/mol, to find the number of moles, and then calculate the total energy using the formula: total energy = moles × heat of combustion.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Heat of Combustion

The heat of combustion is the amount of energy released when a substance undergoes complete combustion with oxygen. It is typically expressed in kilojoules per mole (kJ/mol). For gasoline, the average heat of combustion is 5400 kJ/mol, which indicates how much energy can be obtained from burning a specific amount of gasoline. Understanding this concept is crucial for comparing the energy output of different fuels.

Density and Volume Relationships

Density is defined as mass per unit volume and is a critical factor in determining how much of a substance can fit into a given volume. In this context, the densities of gasoline (0.70 g/mL) and ethanol (0.79 g/mL) will help calculate the mass of each fuel in 1.0 L. This relationship is essential for converting volume measurements into mass, which is necessary for calculating the total energy produced from combustion.

Combustion Reaction Stoichiometry

Combustion reaction stoichiometry involves the balanced chemical equations that describe the combustion of fuels, such as ethanol and gasoline. For each fuel, the stoichiometry allows us to determine the moles of reactants and products involved in the reaction. This is important for calculating the total energy produced from a specific volume of fuel, as it links the amount of fuel burned to the energy released during combustion.

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Textbook Question

It is interesting to compare the 'fuel value' of a hydrocarbon in a hypothetical world where oxygen is not the combustion agent. The enthalpy of formation of CF₄(g) is -679.9 kJ/mol. Which of the following two reactions is the more exothermic?

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)

CH₄(g) + 4 F₂(g) → CF₄(g) + 4 HF(g)

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Textbook Question

At the end of 2012, global population was about 7.0 billion people. What mass of glucose in kg would be needed to provide 1500 Cal/person/day of nourishment to the global population for one year? Assume that glucose is metabolized entirely to CO₂(𝑔) and H₂O(𝑙) according to the following thermochemical equation: C₆H₁₂O₆(s) + 6 O₂(𝑔) → 6 CO₂(𝑔) + 6 H₂O(𝑙) ΔH° = -2803 kJ

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Textbook Question

The air bags that provide protection in automobiles in the event of an accident expand because of a rapid chemical reaction. From the viewpoint of the chemical reactants as the system, what do you expect for the signs of q and w in this process?

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Textbook Question

Consider a system consisting of the following apparatus, in which gas is confined in one flask and there is a vacuum in the other flask. The flasks are separated by a valve. Assume that the flasks are perfectly insulated and will not allow the flow of heat into or out of the flasks to the surroundings. When the valve is opened, gas flows from the filled flask to the evacuated one. (a) Is work performed during the expansion of the gas? (b) Why or why not?

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Textbook Question

A sample of gas is contained in a cylinder-and-piston arrangement. There is an external pressure of 100 kPa. The gas undergoes the change in state shown in the drawing. (b) Now assume that the cylinder and piston are made up of a thermal conductor such as a metal. During the state change, the cylinder gets colder to the touch. What is the sign of q for the state change in this case? Describe the difference in the state of the system at the end of the process in the two cases. What can you say about the relative values of E?

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