Question

One fission reaction that was occurring in the Oklo fossil reactor was $$ { }_{0}^{1} \mathrm{n}+{ }_{92}^{235} \mathrm{U} \longrightarrow{ }_{56}^{140} \mathrm{Ba}+{ }_{36}^{93} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n} $$ How much energy in \(\mathrm{kJ} / \mathrm{mol}\) is released by the fission of uranium- 235 to form barium- 140 and krypton-93? The atomic masses are \({ }^{235} \mathrm{U}(235.0439 \mathrm{u}),{ }^{140} \mathrm{Ba}(139.9106 \mathrm{u}),{ }^{93} \mathrm{Kr}(92.9313 \mathrm{u})\), and \({ }_{0}^{1} \mathrm{n}(1.00867 \mathrm{u})\)

Short answer

The energy released is approximately \( 1.658 \times 10^{10} \mathrm{kJ/mol} \).

Step-by-step solution

Calculate Total Mass of Reactants

The reactants here are one uranium-235 nuclide and one neutron. The formula for the total mass of the reactants is: \[ M_{\text{reactants}} = M_{\mathrm{U}} + M_{\mathrm{n}} \] Substituting the given atomic masses: \[ M_{\text{reactants}} = 235.0439 \text{ u} + 1.00867 \text{ u} = 236.05257 \text{ u} \]

Calculate Total Mass of Products

The products in this fission reaction are barium-140, krypton-93 and three neutrons. Summing up their atomic masses gives: \[ M_{\text{products}} = M_{\mathrm{Ba}} + M_{\mathrm{Kr}} + 3 \times M_{\mathrm{n}} \] Substituting the values: \[ M_{\text{products}} = 139.9106 \text{ u} + 92.9313 \text{ u} + 3 \times 1.00867 \text{ u} \] \[ M_{\text{products}} = 139.9106 \text{ u} + 92.9313 \text{ u} + 3.02601 \text{ u} = 235.86791 \text{ u} \]

Calculate the Mass Defect

The mass defect is the difference between the total mass of the reactants and the total mass of the products. \[ \Delta m = M_{\text{reactants}} - M_{\text{products}} \] \[ \Delta m = 236.05257 \text{ u} - 235.86791 \text{ u} = 0.18466 \text{ u} \]

Convert Mass Defect to Energy in MeV

To convert the mass defect into energy, we use Einstein’s equation: \( E = \Delta m \times c^2 \) and the conversion factor that 1 amu (u) is equivalent to 931.5 MeV/c². \[ E = 0.18466 \text{ u} \times 931.5 \text{ MeV/u} \] \[ E \approx 171.91 \text{ MeV} \]

Convert Energy from MeV to Joules

1 MeV is equivalent to \( 1.60218 \times 10^{-13} \) Joules. Thus, the energy in Joules is calculated as: \[ E = 171.91 \times 1.60218 \times 10^{-13} = 2.754 \times 10^{-11} \text{ J} \]

Calculate Energy per Mole in kJ

To find the energy per mole, multiply the energy per fission reaction by Avogadro’s number \( 6.022 \times 10^{23} \text{ mol}^{-1} \) and convert to kJ. \[ E_{\text{mol}} = 2.754 \times 10^{-11} \text{ J} \times 6.022 \times 10^{23} = 1.658 \times 10^{13} \text{ J/mol} \] \[ E_{\text{mol}} = 1.658 \times 10^{10} \text{ kJ/mol} \]

Key concepts

Mass Defect

Mass defect plays a crucial role in understanding nuclear fission reactions. It refers to the difference between the mass of the reactants and the mass of the products in a nuclear reaction. In a fission reaction, like the splitting of uranium-235 into barium-140 and krypton-93 (along with neutrons), we observe a noticeable loss of mass. This missing mass is known as the mass defect.

Mathematically, the mass defect (\( \Delta m \)) is given by the equation:

• \[ \Delta m = M_{\text{reactants}} - M_{\text{products}} \]

When calculated, the mass defect for this reaction is 0.18466 atomic mass units (u). This phenomenon is significant because it directly translates into the energy released during the reaction according to Einstein’s famous equation (\( E = mc^2 \) ).

The understanding of mass defect helps us comprehend how nuclear reactions can release vast amounts of energy from relatively small masses, making it a foundation for the concept of nuclear power.

Energy Conversion

Energy conversion in nuclear fission is an extraordinary process where a tiny mass defect results in tremendous energy release. The anchored principle here is Einstein’s mass-energy equivalence equation, \( E = mc^2 \).

In this scenario, the mass defect produced during uranium-235 fission directly translates into energy. Using the conversion factor, where 1 atomic mass unit (u) is equivalent to approximately 931.5 million electron volts (MeV), it becomes apparent why nuclear reactions are potent energy sources.

The calculations show that a mass defect of 0.18466 u converts to about 171.91 MeV. This can further be converted into joules, as 1 MeV is equivalent to \( 1.60218 \times 10^{-13} \) joules. Hence, the total energy released per fission reaction is \( 2.754 \times 10^{-11} \) joules.

Converting this energy on a molar level by multiplying it by Avogadro’s number, results in approximately \( 1.658 \times 10^{10} \) kilojoules per mole. This immense energy explains the powerful capacity of nuclear reactors and the ability to generate large quantities of energy with minimal fuel.

Atomic Mass

Atomic mass, often measured in atomic mass units (u), is fundamental in calculations involving nuclear reactions. It represents the mass of an atom derived from the combined mass of its protons, neutrons, and electrons.

In nuclear fission problems, knowing the atomic masses of reactants and products is essential. For the fission reaction of uranium-235, atomic masses are pivotal in determining both the mass of the reactants and the products. For example, the mass of uranium-235 is 235.0439 u, and each of the resulting nuclei such as barium-140 and krypton-93 have masses of 139.9106 u and 92.9313 u, respectively.

Comparing the total mass of the products to that of the reactants allows us to identify the mass defect. This then enables us to calculate energy released in nuclear reactions.

Understanding atomic mass helps in converting between the mass of individual atoms and the mass of an entire mole of atoms, which is critical to predicting the yield of energy on a macroscopic scale.