Question

The fundamental vibrational frequencies for \(^{1} \mathrm{H}_{2}\) and \(^{2} \mathrm{D}_{2}\) are 4401 and \(3115 \mathrm{cm}^{-1},\) respectively, and \(D_{e}\) for both molecules is \(7.677 \times 10^{-19} \mathrm{J}\). Using this information, calculate the bond energy of both molecules.

Short answer

The bond energy for both \(^{1}H_{2}\) and \(^{2}D_{2}\) is 4.796 eV.

Step-by-step solution

Relating vibrational frequency to reduced mass and force constant

The vibrational frequency (\(\nu\)) of a diatomic molecule is related to the reduced mass (\(\mu\)) and force constant (\(k\)) as follows: \[\nu = \(\dfrac{1}{2\pi}\)\sqrt{\dfrac{k}{\mu}}\] We need to find the force constant for both molecules and then relate it to their bond energy.

Calculate reduced mass for \(^{1}H_{2}\) and \(^{2}D_{2}\)

The reduced mass (\(\mu\)) of a diatomic molecule can be calculated using the atomic masses of the constituent atoms: \[\mu = \dfrac{m_{1}m_{2}}{m_{1} + m_{2}}\] For \(^{1}H_{2}\), the atomic mass of Hydrogen is 1 amu. Therefore, the reduced mass for \(^{1}H_{2}\) is: \[\mu_{H_{2}} = \dfrac{1 \times 1}{1+1} = 0.5 \, amu\] For \(^{2}D_{2}\), the atomic mass of Deuterium is 2 amu. Therefore, the reduced mass for \(^{2}D_{2}\) is: \[\mu_{D_{2}} = \dfrac{2 \times 2}{2 + 2} = 1 \, amu\]

Calculate force constant for \(^{1}H_{2}\) and \(^{2}D_{2}\)

Now, we will use the vibrational frequencies and the reduced masses to calculate the force constants for both molecules. For \(^{1}H_{2}\): \[k_{H_{2}} = 4\pi^{2}(\mu_{H_{2}})(\nu_{H_{2}})^{2}\] Plugging in the values, we get: \[k_{H_{2}} = 4\pi^{2}(0.5)(4401)^{2} = 3.864 \times 10^{12}\, \textrm{dynes/cm}\] For \(^{2}D_{2}\): \[k_{D_{2}} = 4\pi^{2}(\mu_{D_{2}})(\nu_{D_{2}})^{2}\] Plugging in the values, we get: \[k_{D_{2}} = 4\pi^{2}(1)(3115)^{2} = 1.218 \times 10^{12}\, \textrm{dynes/cm}\]

Calculate bond energy using dissociation energy

Since the dissociation energy (\(D_{e}\)) is the same for both molecules, we can equate their bond energy (\(E_{b}\)) with their dissociation energy: \[E_{b} = D_{e}\] We are given that the dissociation energy is \(7.677 \times 10^{-19} \, \textrm{J}\) for both molecules. Therefore, to find the bond energy, we just need to convert this value to the desired unit. For example, we can convert from Joules to electron volts (eV) using the conversion factor, \(1 \, \textrm{eV} = 1.602 \times 10^{-19} \, \textrm{J}\). So, the bond energy for both molecules, in electron volts, is: \[E_{b} = \dfrac{7.677 \times 10^{-19} \, \textrm{J}}{1.602 \times 10^{-19} \, \textrm{J/eV}} = 4.796 \, \textrm{eV}\] #Solution# Step 1: Relate vibrational frequency to reduced mass and force constant. Step 2: Calculate reduced mass for \(^{1}H_{2}\) and \(^{2}D_{2}\). Step 3: Calculate force constant for \(^{1}H_{2}\) and \(^{2}D_{2}\). Step 4: Calculate bond energy using dissociation energy. The bond energy for both \(^{1}H_{2}\) and \(^{2}D_{2}\) is 4.796 eV.

Key concepts

Vibrational Frequency

Vibrational frequency is a fundamental concept in understanding how molecules behave. It refers to the frequency at which the atoms in a molecule move relative to each other. This movement can be thought of as a vibration, similar to a vibrating string.
The vibrational frequency of a diatomic molecule like hydrogen ( H_2 or D_2 ) depends on the reduced mass and the force constant of the bond. In the given problem, the frequencies provided are 4401 cm⁻¹ for H_2 and 3115 cm⁻¹ for D_2 .
This difference in frequency is due to the fact that deuterium, a hydrogen isotope, is heavier than protium. The heavier the atomic masses, the lower the vibrational frequency.

Reduced Mass

Reduced mass (\mu) simplifies the analysis of two-body systems by combining the masses of the two atoms into a single effective mass. It is calculated using the formula:

• \(\mu = \dfrac{m_1m_2}{m_1 + m_2}\)

For the hydrogen molecule (H_2), with each hydrogen atom having a mass of approximately 1 amu, the reduced mass is 0.5 amu.
For deuterium (D_2), with each deuterium atom having a mass of approximately 2 amu, the reduced mass is 1 amu.
The reduced mass is crucial for calculating the vibrational frequency, as it allows us to treat the two atoms as if they were a single particle.

Force Constant

The force constant ({k}) is a measure of the bond strength between two atoms. It appears in the formula that relates it to vibrational frequency and reduced mass:

• \(u = \frac{1}{2\pi}\sqrt{\frac{k}{\mu}}\)

This equation shows that the force constant is directly related to how strongly the two atoms are bound together. Stronger bonds have higher force constants.
For H_2, using the given vibrational frequency and reduced mass, the force constant is calculated as \(3.864 \times 10^{12} \text{dynes/cm}\).
For D_2, it is calculated as \(1.218 \times 10^{12} \text{dynes/cm}\).
These values tell us that the hydrogen molecules have different bond strengths, influenced by the heavier isotopes.

Dissociation Energy

Dissociation energy (D_e) is the energy required to break the bond between two atoms in a molecule.
In the exercise, it is provided as \(7.677 \times 10^{-19}\text{J}\) for both H_2 and D_2. This implies that both molecules would need a similar amount of energy to separate into individual atoms.
To convert from joules to electron volts (eV), which is a more convenient unit for bond energy, you use the conversion factor \(1 \text{eV} = 1.602 \times 10^{-19}\text{J}\).
So, the bond energy for both molecules is calculated as 4.796 eV. This conversion is essential for comparative analysis and practical applications where eV is more commonly used.