Problem 36 - Chemistry: The Central Science

Question

Complete and balance the following nuclear equations by supplying the missing particle: (a) \({ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \mathrm{He} \longrightarrow\) ? \(+{ }_{1}^{1} \mathrm{H}\) (b) \({ }_{19}^{40} \mathrm{~K}+{ }_{-1}^{0} \mathrm{c}\) (orbital electron) \(\longrightarrow\) ? (c) \(?+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{1}^{1} \mathrm{H}\) (d) \({ }_{26}^{58} \mathrm{Fe}+2{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{27}^{60} \mathrm{Co}+\) ? (e) \({ }_{42}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{54}^{135} \mathrm{Xe}+2{ }_{0}^{1} \mathrm{n}+\) ?

Step-by-Step Solution

Answer

(a) \({ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{8}^{17} \mathrm{O} +{ }_{1}^{1} \mathrm{H}\) (b) \({ }_{19}^{40} \mathrm{~K}+{ }_{-1}^{0} \mathrm{c} \longrightarrow{ }_{18}^{40} \mathrm{Ar}\) (c) \({ }_{13}^{27} \mathrm{Al} +{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{1}^{1} \mathrm{H}\) (d) \({ }_{26}^{58} \mathrm{Fe}+2{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{27}^{60} \mathrm{Co}+{ }_{-1}^{0} \mathrm{c}\) (orbital electron) (e) \({ }_{42}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{54}^{135} \mathrm{Xe}+2{ }_{0}^{1} \mathrm{n}+{ }_{-12}^{98} \mathrm{Mo}\)

Step 1: Analyze the given nuclear equation

We are given the following nuclear equation to balance: \({ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \mathrm{He} \longrightarrow\) ? \(+{ }_{1}^{1} \mathrm{H}\)

Step 2: Apply conservation of mass and charge

We must find the particle that ensures both mass and charge are conserved. Mass conservation: \(14 + 4 = 1 + X\) → \(X = 17\) Charge conservation: \(7 + 2 = 1 + Y\) → \(Y = 8\)

Step 3: Write the balanced nuclear equation

Now that we have the mass and charge of the missing particle, we can write the balanced nuclear equation: \({ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{8}^{17} \mathrm{O} +{ }_{1}^{1} \mathrm{H}\) (b) Balancing potassium with an orbital electron

Step 1: Analyze the given nuclear equation

We are given the following nuclear equation to balance: \({ }_{19}^{40} \mathrm{~K}+{ }_{-1}^{0} \mathrm{c}\) (orbital electron) \(\longrightarrow\) ?

Step 2: Apply conservation of mass and charge

We must find the particle that ensures both mass and charge are conserved. Mass conservation: \(40 + 0 = X\) → \(X = 40\) Charge conservation: \(19 + (-1) = Y\) → \(Y = 18\)

Step 3: Write the balanced nuclear equation

Now that we have the mass and charge of the missing particle, we can write the balanced nuclear equation: \({ }_{19}^{40} \mathrm{~K}+{ }_{-1}^{0} \mathrm{c} \longrightarrow{ }_{18}^{40} \mathrm{Ar}\) (c) Balancing helium with silicon and hydrogen

Step 1: Analyze the given nuclear equation

We are given the following nuclear equation to balance: \(?+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{1}^{1} \mathrm{H}\)

Step 2: Apply conservation of mass and charge

We must find the particle that ensures both mass and charge are conserved. Mass conservation: \(4 + X = 30 + 1\) → \(X = 27\) Charge conservation: \(2 + Y = 14 + 1\) → \(Y = 13\)

Step 3: Write the balanced nuclear equation

Now that we have the mass and charge of the missing particle, we can write the balanced nuclear equation: \({ }_{13}^{27} \mathrm{Al} +{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{1}^{1} \mathrm{H}\) (d) Balancing iron and neutrons with cobalt

Step 1: Analyze the given nuclear equation

We are given the following nuclear equation to balance: \({ }_{26}^{58} \mathrm{Fe}+2{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{27}^{60} \mathrm{Co}+\) ?

Step 2: Apply conservation of mass and charge

We must find the particle that ensures both mass and charge are conserved. Mass conservation: \(58 + 2\times 1 = 60 + X\) → \(X = 0\) Charge conservation: \(26 + 2\times 0 = 27 + Y\) → \(Y = -1\)

Step 3: Write the balanced nuclear equation

Now that we have the mass and charge of the missing particle, we can write the balanced nuclear equation: \({ }_{26}^{58} \mathrm{Fe}+2{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{27}^{60} \mathrm{Co}+{ }_{-1}^{0} \mathrm{c}\) (orbital electron) (e) Balancing uranium and neutron with xenon and other particles

Step 1: Analyze the given nuclear equation

We are given the following nuclear equation to balance: \({ }_{42}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{54}^{135} \mathrm{Xe}+2{ }_{0}^{1} \mathrm{n}+\) ?

Step 2: Apply conservation of mass and charge

We must find the particle that ensures both mass and charge are conserved. Mass conservation: \(235 + 1 = 135 + 2\times 1 + X\) → \(X = 98\) Charge conservation: \(42 + 0 = 54 + 2\times 0 + Y\) → \(Y = -12\)

Step 3: Write the balanced nuclear equation

Now that we have the mass and charge of the missing particle, we can write the balanced nuclear equation: \({ }_{42}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{54}^{135} \mathrm{Xe}+2{ }_{0}^{1} \mathrm{n}+{ }_{-12}^{98} \mathrm{Mo}\)

Key Concepts

Mass conservation

In nuclear equations, mass conservation is a critical principle used to balance the equations. This means that the total mass number (protons + neutrons) on the left side of the equation must equal the total mass number on the right side. For example, consider the reaction involving nitrogen and helium where nitrogen-14 combines with helium-4 to form oxygen-17 and hydrogen-1. To ensure mass conservation, we must calculate:

• The initial side mass numbers: 14 from nitrogen and 4 from helium, totaling 18.
• The resulting side mass numbers: Initially, we have 1 from hydrogen so the missing particle must have a mass number of 17 to ensure the total is 18.

Balancing mass is essential for accuracy in predicting reaction products, ensuring neither mass nor particles are incorrectly "added" or "destroyed." In our example, the participants yielded the mass-equivalent oxygen particle that's correctly aligned via mass conservation.

Charge conservation

Charge conservation is another fundamental principle used to balance nuclear equations, ensuring the total electrical charge is equal on both sides of the equation. In atomic nuclei, the charge is determined entirely by the number of protons, because neutrons are neutral. Let's illustrate this with the reaction involving potassium and an orbital electron:

• The initial charge on the left side: Potassium has 19 protons (thus charge +19), and an electron brings minus one (-1).
• The overall charge: This gives us 18 as the net charge initially.

To achieve balance, the resulting element must have the same charge. It turns out being argon, which has 18 protons, aligning perfectly with our conservation requirement. Therefore, charge conservation ensures no discrepancy or loss/gain of charge without corresponding particles.

Balancing chemical equations

Balancing chemical equations in nuclear reactions goes beyond just simple number matching. It involves ensuring both mass and charge conservation principles, thus preserving all original characteristics of particles. Each side of the equation must mirror the same physical properties such as:

• Equality in total mass number.
• Net charge balance.

For instance, in a reaction where uranium combines with a neutron to yield xenon and another element, consider:

• Initial masses include 235 from uranium and 1 from the neutron, totaling 236.
• Reactant charge remains at 92 (uranium charge), which helps identify other resultant particles (e.g., xenon holdings, neutron adjustments, and a missing particle).

Balancing involves reconciling these totals, developing reactions that reflect real-world behavior and predict outcomes. With practice, it becomes intuitive, allowing for quick assessments of nuclear stability and reaction predictability in nuclear chemistry.