Problem 24 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

The radiation absorbed by \(^{12} \mathrm{C}^{16} \mathrm{O}\) during a vibrational transition occurs at \(2168 \mathrm{cm}^{-1}\). (Section 10.5 ) (a) Caiculate the ground state vibrational energy in \(\mathrm{kJmol}^{-1}\) (b) What is the ratio of the number of molecules in the first vibrational excited state \((\mathrm{v}=1)\) compared with the ground state \((\mathrm{v}=0)\) at \(298 \mathrm{K} ?\) (c) Calculate the force constant of the bond, assuming \(^{12} \mathrm{C}^{16} \mathrm{O}\) behaves as a simple harmonic oscillator. (d) Estimate the change in the position of the peak if \(^{12} \mathrm{C}\) was replaced by \(^{13} \mathrm{C}\)

Step-by-Step Solution

Answer

(a) The ground state energy is approximately 25.9 kJ/mol. (b) The ratio of molecules in the excited to ground state is about 0.0025. (c) The force constant is approximately 1866 N/m. (d) The peak shifts to about 2116 cm⁻¹ with \(^{13}C\).

Step 1: Calculate the Energy of Transition

The given wavenumber for the transition is \(2168 \ cm^{-1}\). The energy for the vibrational transition can be calculated using the formula \(E = h \ c \ u\), where \(h\) is Planck's constant \(6.626 \times 10^{-34} \ J \cdot s\), \(c\) is the speed of light \(3.00 \times 10^8 \ m/s\), and \(u\) is the wavenumber in \(cm^{-1}\). First convert \(u\) to \(m^{-1}\) by multiplying by \(100\), then compute \(E\). Afterward, convert the energy from joules to \(kJ/mol\) by converting to per molecule using Avogadro's number \(6.022 \times 10^{23} \ molecules/mol\).

Step 2: Boltzmann Factor and Population Ratio

To find the ratio of molecules in the first excited state \(v=1\) to the ground state \(v=0\), use the Boltzmann distribution: \(\frac{N_1}{N_0} = e^{-\Delta E/(kT)}\), where \(\Delta E\) is the energy difference from part (a), \(k\) is Boltzmann's constant \(1.38 \times 10^{-23} \ J/K\), and \(T\) is the temperature in Kelvin \(298\ K\). Calculate this ratio using the energy in Joules.

Step 3: Determine the Force Constant

For a simple harmonic oscillator, the force constant \(k\) is related to the vibration wavenumber \(u\) by the formula \(u = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}}\), where \(\mu\) is the reduced mass of \(^{12} \text{C}^{16} \text{O}\). To calculate \(\mu\), use \(\mu = \frac{m_{12C} \times m_{16O}}{m_{12C} + m_{16O}}\), with the molar masses \(m_{12C} \approx 12 \ g/mol\) and \(m_{16O} \approx 16 \ g/mol\) converted to kg. Rearrange to solve for the force constant \(k\).

Step 4: Isotopic Effect and Peak Position Change

To estimate the shift in position when \(^{12}C\) is replaced by \(^{13}C\), calculate the reduced mass for \(^{13}\text{C}^{16}\text{O}\) and find the new frequency \(u'\). Compare this new frequency with the original frequency using the formula \( u' = u \sqrt{\frac{\mu}{\mu'}} \), where \(\mu'\) is the reduced mass of \(^{13}\text{C}^{16}\text{O}\). This gives the shift in wavenumber due to isotopic substitution.

Key Concepts

Simple Harmonic Oscillator

The concept of a simple harmonic oscillator is central to understanding vibrational spectroscopy. When we consider molecules like \(^{12} \text{C}^{16} \text{O}\), they can be visualized as two masses connected by a spring. This spring represents the chemical bond. In this model, the molecule can vibrate around its equilibrium position similar to how a spring expands and contracts.

The energy of these vibrations depends on several factors:

• Masses of the atoms involved
• Strength of the bond, known as the force constant \(k\)

In a simple harmonic oscillator, the wavenumber \(u\) of the vibration can be related to the force constant and the reduced mass \(\mu\) of the molecule by the formula:\[u = \frac{1}{2\pi}\sqrt{\frac{k}{\mu}} \]By knowing the wavenumber, you can infer the strength of the bond and the characteristics of the vibration, making it a powerful tool in spectroscopy.

Boltzmann Distribution

The Boltzmann distribution is a statistical principle that helps us understand the distribution of energy levels within a molecular system at thermal equilibrium. When molecules vibrate, they can occupy different energy levels. These levels are often denoted as \(v=0\) for the ground state and \(v=1\) for the first excited state.

The ratio of the population of molecules in these states at a temperature \(T\) can be calculated using the Boltzmann factor:\[\frac{N_1}{N_0} = e^{-\Delta E/(kT)}\]Here, \(\Delta E\) is the energy difference between the levels, \(k\) is Boltzmann's constant, and \(T\) is the temperature in Kelvin. By applying this equation, you can determine how many molecules are likely to be in an excited state at a given temperature, which is crucial for techniques such as infrared spectroscopy.

Understanding this distribution shows how temperature affects molecular vibrations and energy levels.

Force Constant

The force constant \(k\) represents the strength of the bond in a molecule and is a crucial parameter in vibrational spectroscopy. In the analogy of a simple harmonic oscillator, it can be thought of as the stiffness of the spring connecting two atoms.

A higher force constant indicates a stronger bond, which leads to higher vibrational frequencies. To determine the force constant, scientists use the vibrational wavenumber \(u\) and the reduced mass \(\mu\) of the molecule:\[u = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}}\]By rearranging this equation, you can solve for \(k\):\[k = (2\pi u)^2 \mu\]Where \(\mu\) is the reduced mass, computed from individual atomic masses. This equation illustrates the direct connection between the microscopic property of a bond's stiffness and an observable vibrational frequency, providing insights into molecular interactions and structures.

Isotopic Mass Effect

The isotopic mass effect in vibrational spectroscopy examines how changing an isotope in a molecule affects its vibrational characteristics. An isotope is an atom with the same number of protons but a different number of neutrons, resulting in different masses.

Replacing one isotope with another changes the reduced mass \(\mu\) of the molecule. This change, in turn, affects the vibrational wavenumber \(u\). The relationship between the original and new wavenumber is:
\[u' = u \sqrt{\frac{\mu}{\mu'}}\]Here, \(u'\) is the new wavenumber, \(\mu\) is the original reduced mass, and \(\mu'\) is the reduced mass after substitution. This formula helps predict how isotopic substitution shifts vibrational frequencies, often observed as a change in infrared spectra.

Studying the isotopic mass effect is valuable in identifying molecular structures and dynamics, especially in distinguishing between different isotopic species in a sample.