Question

Isooctane \(\left(\mathrm{C}_{8} \mathrm{H}_{18} ; d=0.692 \mathrm{~g} / \mathrm{mL}\right)\) is used as the fuel in a test of a new automobile drive train. (a) How much energy (in \(\mathrm{kJ}\) ) is released by combustion of \(20.4 \mathrm{gal}\) of isooctane to gases \(\left(\Delta H_{\mathrm{rnn}}^{\circ}=-5.44 \times 10^{3} \mathrm{~kJ} / \mathrm{mol}\right) ?\) (b) The energy delivered to the automobile's wheels at \(65 \mathrm{mph}\) is \(5.5 \times 10^{4} \mathrm{~kJ} / \mathrm{h} .\) Assuming \(a l l\) the energy is transferred as work to the wheels, how far (in \(\mathrm{km}\) ) can the vehicle travel on the 20.4 gal of fuel? (c) If the actual range is \(455 \mathrm{mi}\), explain your answer to (b).

Short answer

Energy released is calculated using enthalpy change. Distance is determined from energy delivery and real-world factors account for the difference in range.

Step-by-step solution

Determine the mass of isooctane

First, convert the volume of isooctane from gallons to liters. Use the conversion factor: \[1 \text{ gallon} = 3.78541 \text{ liters} \]Then, calculate the mass using the formula: \[\text{mass} = \text{volume} \times \text{density}\]Given, \[ \text{Density of isooctane} = 0.692 \text{ g/mL} \]Convert density to \( \text{g/L} \): \[0.692 \text{ g/mL} = 692 \text{ g/L} \]Calculate the mass: \[ \text{mass} = 20.4 \text{ gal} \times 3.78541 \text{ L/gal} \times 692 \text{ g/L}\]

Convert mass to moles of isooctane

Use the molar mass of isooctane \((\text{C}_8 \text{H}_{18}) \):\[ \text{molar mass of C}_8 \text{H}_{18} = 8 \times 12.01 \text{ g/mol} + 18 \times 1.008 \text{ g/mol} = 114.23 \text{ g/mol} \]Calculate the moles of isooctane: \[\text{moles} = \frac{\text{mass}}{\text{molar mass}} \]

Calculate the energy released

Using the given enthalpy change of reaction, \[ \text{ΔH}_{\text{comb}}^{\text{°}} = -5.44 \times 10^3 \text{ kJ/mol}\]Calculate the energy released: \[ \text{Energy} = \text{moles} \times \text{ΔH}_{\text{comb}}^{\text{°}} \]

Calculate the distance the car can travel

Use the given energy delivered to the wheels, \[ 5.5 \times 10^4 \text{ kJ/h}\]Calculate the time the car can run on the released energy: \[ \text{time} = \frac{\text{Energy released}}{5.5 \times 10^4 \text{ kJ/h}}\]Determine the distance: \[ \text{Distance} = 65 \text{ mph} \times \text{time}\]Convert the distance to kilometers.Use the conversion factor: \[1 \text{ mile} = 1.60934 \text{ km}\]

Reconcile with actual range

Note that real-world efficiency often differs from the ideal theoretical calculations due to various factors such as energy losses, driving conditions, and mechanical inefficiencies. The difference between the calculated distance and the actual range (455 miles) highlights these practical inefficiencies.

Key concepts

enthalpy change

Understanding enthalpy change is essential when analyzing chemical reactions, especially combustion. Enthalpy change \(\Delta H\) represents the heat exchanged at constant pressure. For exothermic reactions, \(\Delta H\) is negative, indicating that energy is released. In this exercise, the combustion of isooctane has an enthalpy change of \(\Delta H_{\text{comb}}^{\text{°}} = -5.44 \times 10^3 \text{ kJ/mol}\). This signifies that for every mole of isooctane burned, 5440 kJ of energy is released. Recognizing how to use enthalpy change allows us to calculate the total energy produced in a reaction by multiplying the moles of the substance by the enthalpy change.
For example: \[\text{Energy} = \text{moles} \times \Delta H_{\text{comb}}^{\text{°}}\] This formula provides the total energy released during combustion, which we can then apply to real-world scenarios, such as determining how far a car can travel using this energy.

molar mass calculation

Calculating the molar mass of a compound is a crucial step in converting mass to moles. Molar mass is the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). To find the molar mass, sum the atomic masses of all atoms in the molecule.
For isooctane \(\text{C}_8 \text{H}_{18}\), the molar mass calculation is as follows:

• Carbon (C): 12.01 g/mol
• Hydrogen (H): 1.008 g/mol
• Formula: \[ \text{Molar mass of C}_8 \text{H}_{18} = 8 \times 12.01 \text{ g/mol} + 18 \times 1.008 \text{ g/mol} = 114.23 \text{ g/mol} \]

After determining the molar mass, we can convert the mass of isooctane to moles by using the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] This conversion helps us understand how many moles of isooctane are available for combustion, which is necessary for calculating the total energy released.

density conversion

Density conversion is often required to determine the mass from a given volume. Density is defined as mass per unit volume, usually expressed as g/mL or g/L. In our problem, the density of isooctane is converted from g/mL to g/L, making it easier to handle larger volumes.
For example:

• Given density of isooctane: 0.692 g/mL
• Convert to: \[0.692 \text{ g/mL} = 692 \text{ g/L}\]

To find the mass, multiply the volume by the density:

• Volume of isooctane in liters: 20.4 gal \(\times 3.78541 \text{ L/gal} = 77.21 \text{ L}\)
• Formula: \[ \text{mass} = 77.21 \text{ L} \times 692 \text{ g/L} = 53452.52 \text{ g} \]

This conversion allows us to proceed with further calculations, such as determining the number of moles of fuel available for combustion.

energy conversion efficiency

Energy conversion efficiency is a measure of how effectively energy is converted from one form to another. In a car, not all the energy released from fuel combustion is transferred to the wheels—some is lost as heat, sound, and other forms of energy. This discrepancy between theoretical and actual values can be significant.
For instance, if the theoretical energy released by burning 20.4 gal of isooctane is vastly different from the actual energy transferring to the wheels, it indicates inefficiencies in the energy conversion process. The ideal calculation of distance based on energy delivered to the wheels may not match the real-world data. Thus, when calculating the car's travel distance, acknowledging these inefficiencies explains why the theoretically calculated range differs from the actual range. Typical sources of energy loss in vehicles include:

• Heat loss due to engine inefficiency
• Frictional losses in transmission
• Aerodynamic drag and mechanical resistance

Recognizing these factors helps understand real-world performance versus theoretical calculations, enabling better energy planning and efficiency improvements.