Problem 57 - Chemistry: Principles and Reactions

Question

Consider the fission reaction $${ }_{0}^{1} n+{ }_{92}^{235} \mathrm{U} \longrightarrow{ }_{3}{\underline{\phantom{xx}}}_{3}^{89} \mathrm{Rb}+{ }_{55}^{144} \mathrm{Ce}+3_{-1}^{0} e+3{ }_{0}^{1} n$$ How many liters of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), the primary component of gasoline, must be burned to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) to produce as much energy as the fission of one gram of U-235 fuel? Octane has a density of \(0.703 \mathrm{~g} / \mathrm{mL} ;\) its heat of formation is \(-249.9 \mathrm{~kJ} / \mathrm{mol}\).

Step-by-Step Solution

Answer

Around \(4.516 \times 10^{-13} \, \mathrm{L}\) of octane must be burned to produce the same amount of energy as the fission of 1 gram of U-235 fuel.

Step 1: Calculate the energy produced from 1 gram of U-235

To find the energy produced in the fission of 1 gram of U-235, we will use the famous equation that relates energy (E) with mass (m) and the speed of light (c): \(E=mc^2\). Since we are given the mass of U-235 (1 gram), we can easily convert it into energy using this equation. We need to convert the mass to kg before using the equation: 1 gram = 0.001 kg Speed of light, c = \(3 \times 10^8 \: \mathrm{m/s}\) Now, calculate the energy produced: \(E = (0.001 \, \text{kg}) \times (3 \times10^8 \: \mathrm{m/s})^2\) \(E = 9 \times 10^{13} \: \mathrm{J}\)

Step 2: Calculate the energy produced by burning 1 mole of octane

We are given the heat of formation of octane, which is \(-249.9 \, \mathrm{kJ/mol}\). This means that 249.9 kJ of energy is released when 1 mole of octane is burned.

Step 3: Calculate the amount of octane required to produce the same energy

Now, we need to find how many moles of octane must be burned to produce \(9 \times 10^{13} \: \mathrm{J}\) of energy. We can use the following proportion: \(\frac{\text{moles of octane}}{249.9 \times 10^3 \, \mathrm{J}} = \frac{1 \, \text{g of U-235}}{9 \times 10^{13} \, \mathrm{J}}\) Solving for moles of octane: \(\text{moles of octane} = \frac{249.9 \times 10^3 \, \mathrm{J}}{9 \times 10^{13} \, \mathrm{J}}\) \(\text{moles of octane} \approx 2.777 \times 10^{-12}\)

Step 4: Convert moles of octane to liters

We need to convert the moles of octane required to liters, as given in the problem. Use the density of octane and its molecular weight: \(\text{Density of octane} = 0.703\, \mathrm{g/mL}\) \(\text{Molecular weight of octane} (#_\textrm{C} = 8, #_\textrm{H} = 18)\) = \(8 (12.01 \, \text{g/mol}) + 18 (1.008 \, \text{g/mol}) \approx 114.23 \, \mathrm{g/mol}\) First, find the mass of octane in grams: \(\text{Mass of octane} = (\text{moles of octane}) \times (\text{molecular weight})\) \(\text{Mass of octane} = (2.777 \times 10^{-12} \, \mathrm{mol}) \times (114.23 \, \mathrm{g/mol})\) \(\text{Mass of octane} \approx 3.173 \times 10^{-10} \, \mathrm{g}\) Now, convert the mass of octane to volume using the density: \(\text{Volume of octane} = \frac{\text{mass}}{\text{density}}\) \(\text{Volume of octane} = \frac{3.173 \times 10^{-10} \, \mathrm{g}}{0.703 \, \mathrm{g/mL}}\) \(\text{Volume of octane} \approx 4.516 \times 10^{-10} \, \mathrm{mL}\) Convert mL to L: \(\text{Volume of octane} = 4.516 \times 10^{-13} \, \mathrm{L}\) Therefore, around \(4.516 \times 10^{-13} \, \mathrm{L}\) of octane must be burned to produce the same amount of energy as the fission of 1 gram of U-235 fuel.

Key Concepts

Nuclear Chemistry

Nuclear chemistry is an intriguing area of chemistry focusing on the reactions, structures, and processes of atomic nuclei. It covers various topics from radioactivity and decay processes to nuclear fission and fusion—the reactions that power the sun and can be harnessed in nuclear reactors.

One of the most well-known and historically significant nuclear reactions is nuclear fission, where a heavy nucleus such as uranium-235 splits into two smaller nuclei when struck by a neutron. This process releases a tremendous amount of energy due to the conversion of mass into energy, a principle described by Albert Einstein's famous equation, E=mc². The released energy can be harnessed for power generation, as occurs in nuclear power plants. However, understanding and controlling these reactions are crucial, given the potential for both significant power production and the danger of radioactive contamination.

Energy-Mass Equivalence

Energy-mass equivalence is a fundamental concept rooted in Albert Einstein's theory of relativity. According to Einstein, mass and energy are two forms of the same thing and can be converted into each other. This is expressed in the iconic equation, E=mc², where E stands for energy, m for mass, and c represents the speed of light. In this context, nuclear fission energy calculation leverages this equation to determine how much energy is released when a nucleus splits.

When 1 gram of uranium-235 undergoes fission, the tiny amount of mass lost is converted into a substantial amount of energy. The ability to calculate this energy allows scientists and engineers to understand how much fuel is needed in a nuclear power plant or how much energy could be released in a nuclear explosion. The precision of these calculations is critical for advancements in nuclear energy and the development of sustainable and safe nuclear power sources.

Thermochemistry

Thermochemistry, a branch of physical chemistry, is the study of the heat involved in chemical reactions. It relates to the energy changes occurring during a reaction, particularly the energy exchanged with the surroundings. This field is critical for comparing the energy outputs of different types of reactions, such as the combustion of octane in gasoline versus the fission of uranium-235.

Heat of formation, or enthalpy of formation, is a thermochemical concept used to evaluate the energy released or absorbed during the formation of a compound from its elements in their standard states. It is a crucial variable when calculating energy content in fuels like octane. For example, the heat of formation of octane can be used to determine how much octane must be burned to release enough energy to equate that produced by the fission of uranium-235. Such comparisons are not only essential for pedagogical reasons but also provide practical insights into the energy density of various fuels and the feasibility of their use in different energy systems.