Nuclear Chemistry and Energy Calculations

Nuclear Bombs vs. Nuclear Power Plants

Nuclear Bomb:
- Rapid, simultaneous explosion.
- Releases a large amount of energy quickly.
Nuclear Power Plant:
- Controlled energy release.
- Fission rate controlled by fuel rods.
- Fuel rods are inserted or removed to adjust the rate of fission.
- Example: Chernobyl - rods couldn't be lowered fast enough, leading to a runaway chain reaction.

Calculating Energy Change in Nuclear Reactions

Focus on products minus reactants; intermediate steps are not crucial.
Formula: $\Delta m = (mass \space of \space products) - (mass \space of \space reactants)$

Example Calculation: Uranium Fission

Reaction: Uranium-235 + neutron → Krypton-93 + Barium-140 + 3 neutrons
Change in mass calculation:
- $\Delta m = (mass \space of \space Kr<em>{93} + mass \space of \space Ba</em>{140} + 2 \times mass \space of \space neutron) - (mass \space of \space U_{235})$

Atomic Masses and Electron Considerations

Atomic masses include the mass of electrons.
In nuclear reactions, focus on nucleons (protons and neutrons).
No need to subtract electron masses in overall nuclear reactions because the electron masses cancel out.
- Subtracting electron masses for Krypton (36 electrons) and Barium (56 electrons) and Uranium (92 electrons) would negate each other.

Calculation Result and Energy Conversion

Example mass change: $\Delta m = -0.18478 \space amu$
Convert mass to energy using the conversion factor: $931.49 \space MeV/amu$
Energy change calculation:
- $E = \Delta m \times 931.49 \space MeV/amu$
- $E = -0.18478 \space amu \times 931.49 \space MeV/amu = -172.046 \space MeV$
Interpretation: Every Uranium-235 fission releases 172.046 MeV.

Contextualizing Energy Release

Convert MeV to kilojoules per mole to understand the energy released per gram of fuel.
Conversion factor: $9.65 \times 10^7 \space kJ/mol \space per \space MeV$
Total energy released for the fission of one mole of Uranium-235 is very large.
To find kilojoules per gram, divide kilojoules per mole by the molar mass of Uranium-235.

Nuclear Fission Reactors: Pros and Cons

Pros:
- Fission reactors work and are established.
Cons:
- Produce nuclear waste (radioactive material).
- Requires deep disposal sites.

Nuclear Fusion: A Better Alternative

Tokamak Reactor:
- A type of fusion reactor.
Nuclear Fusion Reaction:
- Deuterium + Tritium → Helium
Advantages of Fusion:
- Produces Helium, which is not radioactive (no nuclear waste).
Challenges:
- Requires very high temperatures (millions of Kelvin) to fuse deuterium and tritium.
- Difficult to confine materials at these temperatures; reactors are in the experimental stage.

Availability of Fusion Materials

Deuterium is a naturally occurring isotope of hydrogen (0.0115% natural abundance) and is readily available from seawater.
Tritium is an unstable isotope of hydrogen.
- Can be produced by bombarding lithium with a neutron, creating helium and tritium.
Helium produced can be used as a coolant for the tokamak reactor (can cool down to 4K).
Fusion reactions produce neutrons, which can be used to create more tritium from lithium.

Energy Change in Fusion Reactions

Calculating the energy change involves using the masses of deuterium, tritium, helium, and neutrons.
Fusion reaction releases $1.7 \times 10^9 \space kJ$ of energy per mole of tritium or deuterium.

Calculation of Deuterium and Tritium Mass for Individual Energy Consumption

Average US person consumes $334 \times 10^6 \space BTUs$ (British Thermal Units) per year.
$1 \space BTU = 1.055 \space kJ$
Calculation to find the mass of deuterium and tritium needed to power one person for a year:
- Assume reactors are 1% efficient.
Total energy needed per person: $334 \times 10^6 \space BTUs \times 1.055 \space kJ/BTU = 352 \times 10^6 \space kJ$
Moles of $H_2$ needed: $\frac{352 \times 10^6 \space kJ}{1.7 \times 10^9 \space kJ/mol} \times 100 = 20.7 \space moles$
Mass of Deuterium needed: $20.7 \space mol \times 2.014 \space g/mol = 42 \space grams$
Mass of Tritium needed: $20.7 \space mol \times 3.016 \space g/mol = 63 \space grams$

Astonishing Implications

Small amount of hydrogen (deuterium and tritium) can provide individual energy needs.
Contrast with methane: Far more methane is needed to generate the same energy.
The multiplication by 100 accounts for the 1% efficiency of the reactor (actual energy that would be used).

Steps in Nuclear Reactions and Energy Calculations

Understanding Reactants and Products:
- Focus on the products and reactants of the nuclear reaction.
- Intermediate steps are not crucial.
Change in Mass Calculation:
- Use the formula: $\Delta m = (mass \space of \space products) - (mass \space of \space reactants)$
Example Calculation for Uranium Fission:
- Consider the reaction: Uranium-235 + neutron → Krypton-93 + Barium-140 + 3 neutrons.
- Calculate the change in mass as follows:
  $\Delta m = (mass \space of \space Kr{93} + mass \space of \space Ba{140} + 2 \times mass \space of \space neutron) - (mass \space of \space U_{235})$
Atomic Mass Considerations:
- Remember that atomic masses include the mass of electrons.
- In nuclear reactions, focus on nucleons (protons and neutrons), as electron masses cancel out for the overall calculations.
Energy Conversion from Mass Change:
- Convert mass change to energy using the conversion factor: $931.49 \space MeV/amu$ .
- Calculate energy change with $E = \Delta m \times 931.49 \space MeV/amu$ .
  - For example: $E = -0.18478 \space amu \times 931.49 \space MeV/amu = -172.046 \space MeV$ .
  - Interpretation: Each fission of Uranium-235 releases 172.046 MeV of energy.
Contextualizing Energy Release:
- Convert to kilojoules per mole using the conversion factor: $9.65 \times 10^7 \space kJ/mol \space per \space MeV$ .
- Determine kilojoules released per gram of fuel by dividing total energy released by the molar mass of Uranium-235.
Pros and Cons of Nuclear Fission Reactors:
- Acknowledge both advantages (reactors are functional and established) and disadvantages (nuclear waste and disposal challenges).
Understanding Nuclear Fusion:
- Explore fusion reactions, data regarding Deuterium and Tritium, and energy calculations involved.
- Example calculations for energy needs and mass requirements for individuals can illustrate the effectiveness of fusion.
Implications of Energy Calculations:
- Recognize the significant implications of energy derived from small amounts of fusion fuel compared to traditional fuels like methane.