Problem 50 - Chemistry An Atoms First Approach

Question

The easiest fusion reaction to initiate is $$\frac{2}{1} \mathrm{H}+\frac{3}{1} \mathrm{H} \longrightarrow_{2}^{4} \mathrm{He}+\frac{1}{0} \mathrm{n}$$ Calculate the energy released per \(\frac{4}{2} \mathrm{He}\) nucleus produced and per mole of \(^{4}_{2}\) He produced. The atomic masses are \(\frac{2}{1} \mathrm{H}\) \(2.01410 \mathrm{u} ; \frac{3}{1} \mathrm{H}, 3.01605 \mathrm{u} ;\) and \(\frac{4}{2} \mathrm{He}, 4.00260 \mathrm{u} .\) The masses of the electron and neutron are \(5.4858 \times 10^{-4} \mathrm{u}\) and \(1.00866 \mathrm{u}\) respectively.

Step-by-Step Solution

Answer

mass loss = 5.03015 u - 5.01126 u mass loss = 0.01889 u #tag_title#Calculate the energy released per Helium-4 nucleus#tag_content# To calculate the energy released, we will use the mass-energy equivalence formula: \(E = mc^2\), where \(E\) is the energy, \(m\) is the mass, and \(c\) is the speed of light in a vacuum. We will convert the mass loss from atomic units (u) to kilograms (kg) using the conversion factor: 1 u = 1.66054 × 10^{-27} kg. The speed of light is approximately 3 × 10^8 m/s. Energy = (0.01889 u) (1.66054 × 10^{-27} kg/u) (3 × 10^8 m/s)^2 Energy ≈ 1.6904 × 10^{-12} J #tag_title#Calculate the energy released per mole of Helium-4#tag_content# Now that we have the energy released per Helium-4 nucleus, we can calculate the energy released per mole of Helium-4. Since there are \(6.022 \times 10^{23}\) particles in one mole, we simply multiply the energy per nucleus by Avogadro's number to obtain the energy per mole. Energy per mole = (1.6904 × 10^{-12} J/nucleus) (6.022 × 10^23 nuclei/mole) Energy per mole ≈ 1.017 × 10^12 J/mole Thus, the energy released per Helium-4 nucleus produced is approximately \(1.6904 \times 10^{-12}\) joules, and the energy released per mole of Helium-4 produced is approximately \(1.017 \times 10^{12}\) joules.

Step 1: Calculate the mass loss in the reaction

First, we need to find the mass change between the initial reactants and the final products. The mass change can be calculated as follows: mass loss = (mass of reactants) - (mass of products) The mass of reactants includes one deuterium atom and one tritium atom. The mass of products includes one helium-4 atom and one neutron. mass loss = (mass of deuterium + mass of tritium) - (mass of helium-4 + mass of neutron) Using the provided atomic masses, we can plug in the values: mass loss = (2.01410 u + 3.01605 u) - (4.00260 u + 1.00866 u)

Key Concepts

Energy Release Calculation

In nuclear fusion, calculating the energy released during a reaction is essential for understanding its power output. To determine this, we first calculate the mass lost in the fusion process. This loss results from the conversion of the reactants' mass into energy, in line with Einstein's theory of relativity.

For the reaction \(^{2}_1\mathrm{H} + ^{3}_1\mathrm{H} \rightarrow ^{4}_2\mathrm{He} + ^{1}_0\mathrm{n}\), the energy release comes from the mass difference (mass defect) between the initial reactants and the final products.

• Mass of Reactants: Deuterium (\(^{2}_1\mathrm{H}\)) and Tritium (\(^{3}_1\mathrm{H}\))
• Mass of Products: Helium-4 (\(^{4}_2\mathrm{He}\)) and a neutron (\(^{1}_0\mathrm{n}\))

The energy released can be calculated with the mass-energy equivalence formula, that relates mass loss to energy.

Mass-Energy Equivalence

Einstein's famous equation \(E=mc^2\) forms the foundation for understanding how mass is converted into energy in nuclear reactions. In this formula, \(E\) is the energy produced, \(m\) is the mass defect that occurs during the reaction, and \(c\) is the speed of light in a vacuum (\(c = 3 \, \times\ 10^8 \, \mathrm{m/s}\)).

This equation implies that the conversion of even a small amount of mass can yield a significant amount of energy, due to the speed of light being such a large number:

• Fusion reactions show this principle clearly—where the small atomic mass units lost are transformed into substantial energy outputs.
• The energy released in nuclear fusion is significantly more per unit mass than chemical reactions, such as burning coal or gasoline.

Understanding mass-energy equivalence is crucial to grasp why nuclear fusion is a potent source of energy.

Atomic Mass Units

The atomic mass unit (u) is a standard unit of mass that quantifies the mass of atomic and subatomic particles. It is defined as one-twelfth of the mass of a carbon-12 atom. This unit is crucial when calculating mass differences in nuclear reactions.

For example, in our reaction:

• Deuterium (\(^{2}_1\mathrm{H}\)): 2.01410 u
• Tritium (\(^{3}_1\mathrm{H}\)): 3.01605 u
• Helium-4 (\(^{4}_2\mathrm{He}\)): 4.00260 u
• Neutron (\(^{1}_0\mathrm{n}\)): 1.00866 u

The precision of atomic mass values allows for accurate measurements of the mass defect, essential for calculating energy release in nuclear reactions.

The atomic mass unit simplifies the complex calculations of nuclear chemistry, translating minute mass differences into significant energy outputs.

Nuclear Reactions

Nuclear reactions involve changes in an atom's nucleus and typically require or release large amounts of energy. Here, two light nuclei combine to form a heavier nucleus, releasing energy in the process.

Fusion reactions, such as the deuterium-tritium reaction discussed here, are a prime example. These reactions aim to replicate the processes occurring in stars, where immense pressure and temperature allow for continuous nuclear fusion.

Key characteristics of nuclear reactions include:

• Involvement of nuclear forces, which are much stronger than chemical forces acting between atoms.
• Energy changes that are significantly larger than in typical chemical reactions.
• Products of fusion reactions are often stable or emit less radioactive material, compared to other types of nuclear reactions, such as fission.

Understanding nuclear fusion reactions is fundamental to advancing nuclear energy technology and potentially solving global energy challenges.