Problem 59 - Chemistry The Central Science

Question

Complete and balance the nuclear equations for the following fission or fusion reactions: (a) \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{3} \mathrm{He}+\) (b) \({ }_{92}^{239} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{51}^{133} \mathrm{Sb}+{ }_{41}^{98} \mathrm{Nb}+{ }_{-0}^{1} \mathrm{n}\)

Step-by-Step Solution

Answer

(a) \({ }_{1}^{2} \mathrm{H} + { }_{1}^{2} \mathrm{H} \longrightarrow { }_{2}^{3}\mathrm{He} + { }_{0}^{1}\mathrm{n}\) (b) \({ }_{92}^{239}\mathrm{U} + { }_{0}^{1}\mathrm{n} \longrightarrow { }_{51}^{133}\mathrm{Sb} + { }_{41}^{98}\mathrm{Nb} + 9{ }_{0}^{1}\mathrm{n}\)

Step 1: Identify the reactions

First, let's identify the type of reactions given: (a) is a fusion reaction: two hydrogen isotopes combine to form helium-3. (b) is a fission reaction: uranium-239 is split into two smaller nuclei (antimony-133 and niobium-98) by neutron absorption.

Step 2: Balance the fusion reaction (a)

We must ensure that the atomic numbers and mass numbers are the same on both sides of the equation. On the left side of the equation for Reaction (a), we have: - Atomic number: 1 + 1 = 2 - Mass number: 2 + 2 = 4 On the right side, we have helium-3: - Atomic number: 2 - Mass number: 3 Since the mass number is not balanced, we need to add a missing particle to the right side of the equation: (a) \({ }_{1}^{2} \mathrm{H} + { }_{1}^{2} \mathrm{H} \longrightarrow { }_{2}^{3}\mathrm{He} + { }_{?}^{?}\mathrm{X}\) To balance the equation: - Atomic number: \(2 = 2 + ? \Rightarrow ? = 0\) - Mass number: \(4 = 3 + ? \Rightarrow ? = 1\) The missing particle has an atomic number of 0, and a mass number of 1. This particle is a neutron: (a) \({ }_{1}^{2} \mathrm{H} + { }_{1}^{2} \mathrm{H} \longrightarrow { }_{2}^{3}\mathrm{He} + { }_{0}^{1}\mathrm{n}\)

Step 3: Balance the fission reaction (b)

Now, we need to balance the equation for Reaction (b) in a similar way: On the left side of the equation for Reaction (b), we have: - Atomic number: 92 + 0 = 92 - Mass number: 239 + 1 = 240 On the right side, we have antimony-133 and niobium-98: - Atomic number: 51 + 41 = 92 - Mass number: 133 + 98 = 231 Since only the mass number is not balanced, we need to add a missing particle to the right side of the equation: (b) \({ }_{92}^{239}\mathrm{U} + { }_{0}^{1}\mathrm{n} \longrightarrow { }_{51}^{133}\mathrm{Sb} + { }_{41}^{98}\mathrm{Nb} + { }_{?}^{?}\mathrm{X}\) To balance the equation, we need a particle with an atomic number of \(0\) and a mass number of \(9\). As we can't find a single particle with a mass number of 9 and zero atomic number, we need multiple particles to balance this equation. We can use multiple neutrons. Therefore, to balance the equation: - Atomic number: \(0\) - Mass number: \(9\) Since a neutron has an atomic number of \(0\) and a mass number of \(1\), we need 9 neutrons: (b) \({ }_{92}^{239}\mathrm{U} + { }_{0}^{1}\mathrm{n} \longrightarrow { }_{51}^{133}\mathrm{Sb} + { }_{41}^{98}\mathrm{Nb} + 9{ }_{0}^{1}\mathrm{n}\)

Key Concepts

Fusion Reaction

A fusion reaction is a process where two light nuclei combine to form a heavier nucleus. This type of reaction powers stars, including our own sun. It's like nature's way of super-gluing tiny atomic Lego pieces together to form something new. They involve the fusion of light isotopes, such as hydrogen isotopes like deuterium ( _{1}^{2}H ) and another deuterium nucleus ( _{1}^{2}H ). When they fuse, they form helium-3 ( _{2}^{3}He ) along with the release of energy.
Fusion reactions are clean and powerful in terms of energy output, but replicating them on Earth presents technological challenges, largely due to the extreme temperatures and pressures required. Unlike fission, fusion doesn’t produce long-lived radioactive waste.
Understanding fusion is crucial because it offers the potential for a nearly limitless, carbon-free energy source. In our exercise, fusion helps us illustrate the step of balancing nuclear equations where light particles are the primary reactants.

Fission Reaction

In a fission reaction, a heavy nucleus splits into smaller nuclei, along with the release of energy. Picture a large Lego structure that suddenly falls apart into smaller sections. It usually happens when a nucleus absorbs a neutron, becoming unstable and dividing quickly. A common example of a fissionable material is uranium-239 ( _{92}^{239}U ).
In the exercise, the fission of uranium-239 leads to antimony-133 ( _{51}^{133}Sb ) and niobium-98 ( _{41}^{98}Nb ), releasing additional neutrons that may promote further fission in a chain reaction. This chain reaction is a key feature of nuclear reactors and atomic bombs.
Fission reactions are essential for understanding nuclear power, as they are the main process behind the energy produced in nuclear power plants. They allow us to balance energy needs with sustainability challenges as they do produce nuclear waste, requiring careful management and disposal.

Balancing Nuclear Equations

Balancing nuclear equations is crucial to ensure that the principles of conservation of mass and atomic numbers are obeyed. Just like a balanced chemical equation, a nuclear equation must account for all particles before and after a reaction.
In fusion reactions like the one in our exercise, we balance the equation by ensuring both atomic numbers and mass numbers are equal on each side. We had to identify the missing particle, in our case, a neutron ( _{0}^{1}n ), which balances out the equation for deuterium fusion into helium-3.
For fission reactions, the procedure is similar but often involves more nuclei and free neutrons. In the case of uranium fission, recognizing that missing particles (neutrons in this instance) must be counted ensures that the equation remains balanced. The by-products and leftover particles should equate to the initial elements in both atomic and mass numbers.
Mastering this skill is essential for anyone dealing with nuclear chemistry or physics. It helps illustrate how transformations occur in nuclear reactions, enabling the prediction of outcomes and balancing of nuclear reactions for practical applications.