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Some automobiles and buses have been equipped to burn propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) .\) Compare the amounts of energy that can be obtained per gram of \(\mathrm{C}_{3} \mathrm{H}_{8}(g)\) and per gram of gasoline, assuming that gasoline is pure octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l) .\) See Example \(6.11 .\) ) Look up the boiling point of propane. What disadvantages are there to using propane instead of gasoline as a fuel?

Short Answer

Expert verified
In comparing propane (C3H8) and gasoline (assumed to be pure octane, C8H18), propane produces about \(2.46\, kJ/g\) more energy than octane per gram (\(-50.34\, kJ/g\) for propane and \(-47.88\, kJ/g\) for octane). However, propane has a low boiling point of \(-42^\circ C\) and must be stored in pressurized tanks, leading to increased weight and complex fuel systems. Additionally, propane has lower energy density than gasoline, resulting in larger fuel tanks and lower range before refueling. Infrastructures for propane fueling stations are not as widespread as gasoline, posing difficulties in finding refueling options. Furthermore, propane handling and maintenance require special training and may pose increased safety hazards.

Step by step solution

01

Write the balanced combustion reactions for propane and octane

For this step, we'll write the balanced combustion reactions for both propane and octane. Combustion reactions involve reacting with oxygen to produce carbon dioxide and water. For propane: C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l) For octane: C8H18(l) + 12.5O2(g) -> 8CO2(g) + 9H2O(l)
02

Calculate the energy produced per gram for propane and octane

To compare the amounts of energy produced per gram of propane and octane, we'll need their enthalpy of combustion values. The enthalpy of combustion for propane is \(-\,2220.0\, kJ/mol\), and for octane, it is \(-\,5471.0\, kJ/mol\). The molar mass of propane is \(44.10\, g/mol\) and that for octane is \(114.22\, g/mol\). Now, we will calculate the amount of energy produced per gram by dividing the enthalpy of combustion by the molar mass. For propane: Energy per gram = \(\frac{-2220.0\, kJ/mol}{44.10\, g/mol} = -50.34\, kJ/g\) For octane: Energy per gram = \(\frac{-5471.0\, kJ/mol}{114.22\, g/mol} = -47.88\, kJ/g\)
03

Compare the energy produced per gram of propane and octane

From the calculations in step 2, we can compare the energy produced per gram of propane and octane: Propane: \(-50.34\, kJ/g\) Octane: \(-47.88\, kJ/g\) Propane produces around \(2.46\, kJ/g\) more energy than octane per gram.
04

Look up the boiling point of propane

The boiling point of propane is \(-42^\circ C\).
05

Identify disadvantages of using propane instead of gasoline as a fuel

Some disadvantages of using propane as a fuel instead of gasoline are: 1. Due to its lower boiling point, propane must be stored in pressurized tanks, which leads to increased weight and more complex fuel systems. 2. Propane has lower energy density than gasoline, which means that vehicles running on propane will need larger fuel tanks and have a lower range before refueling compared to vehicles running on gasoline. 3. The infrastructure for propane fueling stations is not as widespread as gasoline, making it difficult to find refueling options, especially in remote areas. 4. Propane handling and maintenance require special training, and potential safety hazards to the vehicle owner and the mechanic increase. In conclusion, while propane can produce more energy per gram compared to gasoline, the disadvantages of using propane as a fuel are the low boiling point, complex storage system, lower energy density, and less developed infrastructure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy of Combustion
The concept of enthalpy of combustion is a vital aspect in understanding how much energy can be derived from a fuel. It refers to the heat released when one mole of a substance is completely burned in oxygen. When we look at propane, its enthalpy of combustion is \(-2220.0\, kJ/mol\), while octane, representing gasoline, has a value of \(-5471.0\, kJ/mol\). These values are essential to calculating how much energy you get per gram of fuel. A fascinating takeaway here is that despite propane having a lower total enthalpy of combustion than octane, when you break down the energy per gram, propane shines as the more efficient option. It's essential to realize that the molar masses of the two fuels play a significant role in these computations, leading to propane producing more energy per gram.
Energy Density
Energy density is an important term when discussing fuels like propane and gasoline. It essentially describes how much energy can be packed into a given amount of the substance, often calculated on a per-gram basis. In the exercise, we learned that propane releases \(-50.34\, kJ/g\) compared to octane's\(-47.88\, kJ/g\). This means propane has a slightly higher energy density than gasoline, offering more energy for less weight. However, energy density alone doesn't provide the whole picture of fuel efficiency. The way fuel is stored, handled, and transported can influence practical usability, and here propane faces challenges due to its physical characteristics.
Boiling Point
The boiling point of a substance tells us the temperature at which it transforms from liquid to gas. For propane, this is extremely low at \(-42^\circ C\). This low boiling point means that propane needs to be stored under pressure in liquid form.
Therefore, it requires specialized equipment like pressurized tanks. This can lead to increased costs and complexity in vehicle design as compared to fuels like gasoline, which have more manageable storage requirements. The necessity for pressurized storage is one of the significant hurdles in using propane as a widespread alternative to gasoline.
Propane vs Gasoline
Propane and gasoline have both similarities and differences that affect their use as fuels. The analysis begins with understanding that propane can technically provide more energy for the same weight of fuel. However, the practical use of propane is limited.
  • The infrastructure for gasoline refueling is highly developed and ubiquitous, whereas propane stations are rare, especially in rural areas.
  • Gasoline does not require the complex pressurized tank systems that propane needs, making gasoline vehicles simpler to maintain.
  • Gasoline offers higher energy density per volume, allowing for smaller fuel tanks and longer refueling intervals.
While propane presents certain advantages in energy output per gram, the logistics of its storage, combined with the ubiquity of gasoline, currently makes gasoline a more practical choice for most vehicles.

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Most popular questions from this chapter

In a coffee-cup calorimeter, 150.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) HCl is added to 50.0 \(\mathrm{mL}\) of 1.00 \(\mathrm{M} \mathrm{NaOH}\) to make 200.0 \(\mathrm{g}\) solution at an initial temperature of \(48.2^{\circ} \mathrm{C}\) . If the enthalpy of neutralization for the reaction between a strong acid and a strong base is \(-56 \mathrm{kJ} / \mathrm{mol}\) , calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is 4.184 \(\mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) and assume no heat loss to the surroundings.

One of the components of polluted air is NO. It is formed in the high- temperature environment of internal combustion engines by the following reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) \quad \Delta H=180 \mathrm{kJ} $$ Why are high temperatures needed to convert \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) to NO?

Given the following data $$ \begin{array}{ll}{\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-23 \mathrm{kJ}} \\ {3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-39 \mathrm{kJ}} \\ {\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=18 \mathrm{kJ}}\end{array} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

The overall reaction in a commercial heat pack can be represented as $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \quad \Delta H=-1652 \mathrm{kJ} $$ a. How much heat is released when 4.00 moles of iron are reacted with excess \(\mathrm{O}_{2} ?\) b. How much heat is released when 1.00 mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is produced? c. How much heat is released when 1.00 \(\mathrm{g}\) iron is reacted with excess \(\mathrm{O}_{2} ?\) d. How much heat is released when 10.0 \(\mathrm{g}\) Fe and 2.00 \(\mathrm{g} \mathrm{O}_{2}\) are reacted?

If the internal energy of a thermodynamic system is increased by \(300 . \mathrm{J}\) while 75 \(\mathrm{J}\) of expansion work is done, how much heat was transferred and in which direction, to or from the system?

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