Problem 24 - Chemistry The Molecular Science

Question

Calculate the binding energy in kJ per mole of \(\mathrm{P}\) for the formation of \({ }_{15}^{30} \mathrm{P}\) and \({ }_{15}^{31} \mathrm{P}\) $$ \begin{array}{l} 15{ }_{1}^{1} \mathrm{H}+15{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{15}^{30} \mathrm{P} \\ 15{ }_{1}^{1} \mathrm{H}+16{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{15}^{31} \mathrm{P} \end{array} $$ Which is the more stable isotope? The required masses (in \(\mathrm{g} / \mathrm{mol}\) ) are \({ }_{1}^{1} \mathrm{H}=1.00783 ;{ }_{0}^{1} \mathrm{n}=1.00867 ;{ }_{15}^{30} \mathrm{P}=\) \(29.97832 ;\) and \({ }_{15}^{31} \mathrm{P}=30.97376 .\)

Step-by-Step Solution

Answer

\({}^{31}_{15}P\) is the more stable isotope with higher binding energy.

Step 1: Define the reaction and process

To calculate the binding energy of isotopes, we need to look at the mass defect, which is the difference between the sum of the masses of individual nucleons and the actual mass of the nucleus formed. The relevant reactions are provided for each isotope, and we're given the masses involved: - Mass of proton \( ({}^1_1 \text{H}) = 1.00783 \, \text{g/mol} \)- Mass of neutron \( ({}^1_0 \text{n}) = 1.00867 \, \text{g/mol} \)- Mass of \( {}^{30}_{15} \text{P} = 29.97832 \, \text{g/mol} \)- Mass of \( {}^{31}_{15} \text{P} = 30.97376 \, \text{g/mol} \). Calculate the mass defect for each isotope, convert it to energy, and compare.

Step 2: Calculate Mass Defect for \( {}^{30}_{15} \text{P} \)

Mass defect for \( {}^{30}_{15} \text{P} \) is calculated by finding the difference between the total mass of individual nucleons and the mass of the isotope. The theoretical mass is calculated as follows: \[\text{Mass of nucleons} = 15 \times 1.00783 + 15 \times 1.00867 = 30.1395 \, \text{g/mol}\]Mass defect = Theoretical mass - Observed mass\[\Delta m = 30.1395 - 29.97832 = 0.16118 \, \text{g/mol}\]This mass defect will be used to calculate the binding energy.

Step 3: Calculate Mass Defect for \( {}^{31}_{15} \text{P} \)

For \( {}^{31}_{15} \text{P} \), calculate the theoretical mass of nucleons and compare with the actual mass:\[\text{Mass of nucleons} = 15 \times 1.00783 + 16 \times 1.00867 = 31.14817 \, \text{g/mol}\]Mass defect = Theoretical mass - Observed mass\[\Delta m = 31.14817 - 30.97376 = 0.17441 \, \text{g/mol}\]This mass defect will be used for the binding energy calculation.

Step 4: Convert Mass Defect to Energy for \( {}^{30}_{15} \text{P} \)

The energy equivalent of mass defect is given by Einstein's equation \( E = \Delta m \times c^2 \), where \( c \) is the speed of light (\( c = 3.00 \times 10^8 \text{m/s} \, \text{or} \, 3.00 \times 10^{10} \text{cm/s} \)). First, convert \( \Delta m \) to \( \text{kg/mol} \): \[0.16118 \, \text{g/mol} = 0.16118 \times 10^{-3} \text{kg/mol}\]Calculate energy in \( \text{kJ/mol} \):\[E = (0.16118 \times 10^{-3} \, \text{kg/mol}) \times (3.00 \times 10^8 \, \text{m/s})^2 = 1.448 \times 10^{13} \, \text{J/mol}\]Converting to kJ:\[E = 1.448 \times 10^{10} \, \text{kJ/mol}\]

Step 5: Convert Mass Defect to Energy for \( {}^{31}_{15} \text{P} \)

Similarly, convert the mass defect for \( {}^{31}_{15} \text{P} \) to energy:\[0.17441 \, \text{g/mol} = 0.17441 \times 10^{-3} \text{kg/mol}\]Calculate energy in \( \text{kJ/mol} \):\[E = (0.17441 \times 10^{-3} \, \text{kg/mol}) \times (3.00 \times 10^8 \, \text{m/s})^2 = 1.570 \times 10^{13} \, \text{J/mol} \]Converting to kJ:\[E = 1.570 \times 10^{10} \, \text{kJ/mol}\]

Step 6: Compare Binding Energies

Compare the calculated binding energies to determine the more stable isotope.- Binding energy for \( {}^{30}_{15} \text{P} = 1.448 \times 10^{10} \, \text{kJ/mol} \)- Binding energy for \( {}^{31}_{15} \text{P} = 1.570 \times 10^{10} \, \text{kJ/mol} \)The isotope with the higher binding energy per mole is more stable.

Key Concepts

Isotope Stability

Isotopes are different forms of the same element, having the same number of protons but different numbers of neutrons. Understanding the stability of isotopes helps us learn more about nuclear reactions and elemental behavior. The concept of binding energy comes into play here. The higher the binding energy, the more stable the isotope is.
In our example, we have two isotopes of phosphorus: \({ }_{15}^{30} \mathrm{P}\) and \({ }_{15}^{31} \mathrm{P}\). By calculating the binding energies, we found that \({ }_{15}^{31} \mathrm{P}\) has a higher binding energy. This means it's more stable compared to \({ }_{15}^{30} \mathrm{P}\), as it takes more energy to break its nucleus apart.
Isotope stability is crucial in fields like nuclear medicine and energy, where safe and controlled reactions are essential.

• Isotopes have the same number of protons, different neutrons.
• Binding energy indicates stability, with higher values meaning more stable isotopes.
• Stable isotopes have practical applications in various fields.

Mass Defect

Mass defect is the difference between the sum of the individual masses of protons and neutrons in a nucleus and the actual mass of that nucleus. This difference arises because a portion of the mass is converted into binding energy that holds the nucleus together.
To grasp mass defect, we look at the example of \({ }_{15}^{30} \mathrm{P}\). The theoretical mass (sum of individual nucleons' masses) is 30.1395 g/mol, while the actual mass is 29.97832 g/mol. This difference, 0.16118 g/mol, is the mass defect. Similarly, for \({ }_{15}^{31} \mathrm{P}\), the mass defect is 0.17441 g/mol.
Understanding mass defect helps us calculate how much energy is required to hold a nucleus together. This is key in processes like fission and fusion.

• Mass defect represents the lost mass converted into energy.
• It is crucial for calculating binding energy.
• Helps understand nuclear processes, like fission/fusion.

Nuclear Chemistry

Nuclear chemistry focuses on reactions involving changes in an atom's nucleus. It studies how particles like protons, neutrons, and electrons interact within the nucleus, altering the element itself.
The binding energy calculations for isotopes \({ }_{15}^{30} \mathrm{P}\) and \({ }_{15}^{31} \mathrm{P}\) are part of nuclear chemistry. By understanding binding energy, scientists can predict the resultant energy of nuclear reactions and the stability of isotopes involved.
Nuclear chemistry has many applications, from harnessing nuclear energy to diagnosing and treating diseases using radiolabeled compounds. It's an essential field that combines elements of physics and chemistry to explore the core of atoms.

• Involves the study of the atomic nucleus.
• Includes reactions that change the nucleus's structure.
• Crucial for energy production and medical applications.