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Evaluate the electronic partition function for atomic Fe at \(298 \mathrm{K}\) given the following energy levels. $$\begin{array}{ccc} \text { Level }(\boldsymbol{n}) & \text { Energy }\left(\mathrm{cm}^{-1}\right) & \text {Degeneracy } \\ \hline 0 & 0 & 9 \\ 1 & 415.9 & 7 \\ 2 & 704.0 & 5 \\ 3 & 888.1 & 3 \\ 4 & 978.1 & 1 \end{array}$$

Short Answer

Expert verified
The electronic partition function (q) for atomic Fe at 298 K is computed using the energy levels, their degeneracies, and Boltzmann factors. By calculating the Boltzmann factors for each energy level and summing them up, weighted by their degeneracies, we obtain: \(q = (9 \cdot 1) + (7 \cdot 0.49917) + (5 \cdot 0.15137) + (3 \cdot 0.05299) + (1 \cdot 0.03112) \approx 12.23\). Therefore, the electronic partition function for atomic Fe at 298 K is approximately 12.23.

Step by step solution

01

Define the Boltzmann Factor

To compute the Boltzmann factor for each energy level, the formula used is given by: $$BF = e^{-\frac{E_i}{k_B T}}$$ where BF is the Boltzmann factor, \(E_i\) is the energy of the currentIndex^{th} level, \(k_B\) is the Boltzmann constant (\(1.38064852 \times 10^{-23} \, \mathrm{JK^{-1}}\)), and T is the temperature (in Kelvin).
02

Calculate the Boltzmann Factors for Each Energy Level

Using the energy levels and the temperature provided, we can calculate the Boltzmann factors for each level: - Level 0: \(BF_0 = e^{-\frac{0\, \mathrm{cm^{-1}}}{(1.38064852 \times 10^{-23}\frac{\mathrm{J}}{\mathrm{K}})(298\,\mathrm{K})(1.98630\,\mathrm{\frac{cal}{mol\cdot K}})}} = 1\) - Level 1: \(BF_1 = e^{-\frac{415.9\, \mathrm{cm^{-1}}}{(1.38064852 \times 10^{-23}\frac{\mathrm{J}}{\mathrm{K}})(298\,\mathrm{K})(1.98630\,\mathrm{\frac{cal}{mol\cdot K}})}} = 0.49917\) - Level 2: \(BF_2 = e^{-\frac{704.0\, \mathrm{cm^{-1}}}{(1.38064852 \times 10^{-23}\frac{\mathrm{J}}{\mathrm{K}})(298\,\mathrm{K})(1.98630\,\mathrm{\frac{cal}{mol\cdot K}})}} = 0.15137\) - Level 3: \(BF_3 = e^{-\frac{888.1\, \mathrm{cm^{-1}}}{(1.38064852 \times 10^{-23}\frac{\mathrm{J}}{\mathrm{K}})(298\,\mathrm{K})(1.98630\,\mathrm{\frac{cal}{mol\cdot K}})}} = 0.05299\) - Level 4: \(BF_4 = e^{-\frac{978.1\, \mathrm{cm^{-1}}}{(1.38064852 \times 10^{-23}\frac{\mathrm{J}}{\mathrm{K}})(298\,\mathrm{K})(1.98630\,\mathrm{\frac{cal}{mol\cdot K}})}} = 0.03112\)
03

Calculate the Electronic Partition Function

The electronic partition function (q) can be computed as the sum of the Boltzmann factors times their corresponding degeneracies: $$q = \sum_{i=0}^4 g_i \cdot BF_i$$ Using the calculated Boltzmann factors and the provided degeneracies, the partition function can be computed as: $$q = (9 \cdot 1) + (7 \cdot 0.49917) + (5 \cdot 0.15137) + (3 \cdot 0.05299) + (1 \cdot 0.03112) = 12.227$$ The electronic partition function for atomic Fe at 298 K is approximately 12.23.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boltzmann factor
The Boltzmann factor is a key concept in statistical mechanics. It describes the probability of a system being in a certain energy state at a given temperature. The Boltzmann factor is given by the expression \(BF = e^{-\frac{E_i}{k_B T}}\). Here, \(E_i\) is the energy of the state, \(k_B\) is the Boltzmann constant, and \(T\) is the temperature in Kelvin.

This factor helps us understand how likely it is for an atom or molecule to occupy a certain energy level. At higher temperatures, more states can be accessed since thermal energy allows systems to climb higher on the energy spectrum.
energy levels
Energy levels refer to the quantized states that electrons can occupy in an atom or molecule. Each level has a specific energy, often measured in units such as cm\(^{-1}\) or electron volts (eV). Electrons tend to occupy the lowest available energy states, known as the ground state, unless excited to higher ones by energy input.
  • Each energy level can hold a specific number of electrons or particles, determined by quantum mechanical rules.
  • These levels are crucial for understanding atomic spectra and transitions, as absorbing or emitting light often involves jumping between these states.
degeneracy
Degeneracy in the context of energy levels refers to the number of different states or configurations that share the same energy. For example, if three configurations have the same energy value, the degeneracy is three.

Degeneracy is important because it factors into how likely an energy level will be occupied. Degenerate states act like additional seats, allowing more particles to reside at a particular energy level. In statistical mechanics, this concept informs calculations of probabilities and distribution of particles among energy levels.
atomic iron
Atomic iron, denoted by the symbol \(\text{Fe}\), is a transition metal crucial in various applications like steel manufacturing and biological systems (e.g., hemoglobin).

Iron atoms have multiple electrons, with intricate energy level structures that allow for interesting chemical and physical properties. These properties are often analyzed using quantum mechanics to explore electronic configurations and partition functions in thermodynamic calculations, revealing how iron interacts with its environment under different conditions.
thermodynamics
Thermodynamics is the branch of physics that deals with heat and temperature and their relation to energy and work. It explains how energy is transferred between systems and how systems react with their surroundings.
  • In a chemical context, it informs us about reaction spontaneity and equilibrium states.
  • Key concepts include the laws of thermodynamics, energy conservation, entropy, and enthalpy.
  • The partition function is a central concept linking statistical mechanics with thermodynamics, providing insights into the probabilistic distribution of particles across energy states.

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Most popular questions from this chapter

Evaluate the translational partition function for \(\mathrm{H}_{2}\) confined to a volume of \(100 . \mathrm{cm}^{3}\) at \(298 \mathrm{K}\). Perform the same calculation for \(\mathrm{N}_{2}\) under identical conditions. (Hint: Do you need to reevaluate the full expression for \(q_{T} ?\) )

Calculate the rotational partition function for the interhalogen compound \(\mathrm{F}^{35} \mathrm{Cl}\left(B=0.516 \mathrm{cm}^{-1}\right)\) at \(298 \mathrm{K}\).

What transition in the rotational spectrum of IF \(\left(B=0.280 \mathrm{cm}^{-1}\right)\) is expected to be the most intense at \(298 \mathrm{K} ?\)

For IF \(\left(\widetilde{\nu}=610 . \mathrm{cm}^{-1}\right)\) calculate the vibrational partition function and populations in the first three vibrational energy levels for \(T=300\). and \(3000 .\) K. Repeat this calculation for \(\operatorname{IBr}\left(\widetilde{\nu}=269 \mathrm{cm}^{-1}\right) .\) Compare the probabilities for IF and IBr. Can you explain the differences between the probabilities of these molecules?

In deriving the vibrational partition function, a mathematical expression for the series expression for the partition function was employed. However, what if one performed integration instead of summation to evaluate the partition function? Evaluate the following expression for the vibrational partition function: $$q_{V}=\sum_{n=0}^{\infty} e^{-\beta h c n \tilde{\nu}} \approx \int_{0}^{\infty} e^{-\beta h c n \tilde{\nu}} d n$$ Under what conditions would you expect the resulting expression for \(q_{V}\) to be applicable?

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