Consider the following energy levels, each capable of holding two objects: \(E=2 \mathrm{~kJ}\) ______ \(E=1 \mathrm{~kJ}\) ______ \(E=0 \quad \mathrm{XX}\) Draw all the possible arrangements of the two identical particles (represented by X) in the three energy levels. What total energy is most likely, that is, occurs the greatest number of times? Assume that the particles are indistinguishable from each other.

Short Answer

Expert verified
The unique arrangements of the two indistinguishable particles in the three energy levels are as follows: 1. \(XX\), _, _. (Total Energy = 0 kJ) 2. _, \(XX\), _. (Total Energy = 2 kJ) 3. _, X, X. (Total Energy = 3 kJ) 4. _, _, \(XX\). (Total Energy = 4 kJ) The most likely total energy is 4 kJ, as it occurs twice in the unique arrangements.

Step by step solution

01

List initial arrangements

Before distributing the particles to the other energy levels, let's list the initial arrangement where both particles are in the E=0 kJ energy level: \(XX\), _, _.
02

Distribute particles to energy levels

Now, we will distribute the particles into different energy levels: 1. Move one particle from E=0 kJ to E=1 kJ level: _, \(XX\), _. 2. Move the other particle from E=0 kJ to E=1 kJ level: _, _, \(XX\). 3. Move both particles from E=0 kJ to E=1 kJ and E=2 kJ levels: _, X, X. 4. Move one particle from E=0 kJ to E=2 kJ and the other from E=1 kJ to E=2 kJ: _, _, \(XX\). 5. Move both particles from E=0 kJ to E=2 kJ level: _, _, \(XX\).
03

Identify unique arrangements

From the list of arrangements in Step 2, we must eliminate repetitions by identifying unique arrangements: 1. \(XX\), _, _. (Total Energy = 0 kJ) 2. _, \(XX\), _. (Total Energy = 2 kJ) 3. _, X, X. (Total Energy = 3 kJ) 4. _, _, \(XX\). (Total Energy = 4 kJ)
04

Identify most likely total energy

Now, we will determine the total energy that occurs the most number of times: 1. The total energy of 0 kJ occurs 1 time. 2. The total energy of 2 kJ occurs 1 time. 3. The total energy of 3 kJ occurs 1 time. 4. The total energy of 4 kJ occurs 2 times (arrangements 4 and 5). The most likely total energy is 4 kJ, as it occurs twice in the unique arrangements.

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Key Concepts

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

Quantum Mechanics
Quantum mechanics is a fundamental theory in physics that explains the nature of matter and energy on the atomic and subatomic scale. At the heart of quantum mechanics is the concept of wave-particle duality, where particles such as electrons can exhibit both wave-like and particle-like properties.

In the microscopic world, quantum mechanics shows that the energy of particles is quantized, meaning it comes in discrete amounts, often referred to as 'quanta'. An electron in an atom, for example, can only exist at specific energy levels and can transition between these levels by absorbing or emitting a photon with energy equal to the difference between the levels.

Another peculiar aspect of quantum mechanics is the principle of uncertainty, which states that certain pairs of physical properties, like position and momentum, cannot be precisely known simultaneously. This uncertainty creates a probabilistic nature in the outcomes of quantum events, leading to a non-deterministic model of the physical universe that differs from classical physics.
Particle Distribution in Energy Levels
In chemistry, particle distribution in energy levels can be understood through the context of molecular orbitals or electronic structures in atoms. Particles, such as electrons, occupy discrete energy levels that are determined by quantum mechanics. The Pauli exclusion principle further dictates that no two fermions (particles like electrons with half-integer spin) can occupy the same quantum state simultaneously.

When considering the distribution of particles across energy levels, the Aufbau principle guides the order in which electrons fill these levels, typically filling the lowest energy states first. However, in a given system, not all distributions are equally probable. The likelihood of a particular energy state being occupied is related to its energy, temperature, and the statistics governing the particles either Fermi-Dirac for fermions or Bose-Einstein for bosons (particles with integer spin).

In the exercise provided, identifying different arrangements of particles within given energy levels and calculating their corresponding total energy illustrates how multiple configurations can exist within the bounds of quantum mechanical rules.
Indistinguishability in Quantum Systems
Indistinguishability plays a crucial role in quantum mechanics, affecting the statistical distribution of identical particles and their allowable states. In quantum systems, indistinguishable particles are identical in every way—they cannot be distinguished even in principle. This means that swapping two indistinguishable particles does not result in a new state.

This feature of indistinguishable particles is exemplified by their quantum statistics: bosons follow Bose-Einstein statistics and can occupy the same quantum state, while fermions follow the Pauli exclusion principle and Fermi-Dirac statistics, prohibiting them from sharing a state. This premise has profound implications for how particles distribute themselves among energy levels, as observed in phenomena such as the Bose-Einstein condensate and the behavior of electrons in atoms.

In the exercise context, treating the identical particles 'X' as indistinguishable means that some of the arrangements we can draw are essentially the same state and cannot be counted separately. This concept allows us to determine the most likely total energy outcome when considering all unique arrangements of a system, as done in the step-by-step solution for the exercise.

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Most popular questions from this chapter

Consider the following system at equilibrium at \(25^{\circ} \mathrm{C}\) : $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \quad \Delta G^{\circ}=-92.50 \mathrm{~kJ}$$ What will happen to the ratio of partial pressure of \(\mathrm{PCl}_{5}\) to partial pressure of \(\mathrm{PCl}_{3}\) if the temperature is raised? Explain completely.

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from \(99.90 \%\) to \(99.99 \%\) purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix 4, predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature dependent. Predict the sign of \(\Delta S_{\text {sarr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) at \(298 \mathrm{~K}\). Using these values and data in Appendix 4, calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C}\). The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the preceding reaction at \(50 .{ }^{\circ} \mathrm{C}\). f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\). The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\). g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\). Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\). The boiling point for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(29.0 \mathrm{~kJ} / \mathrm{mol}\).

a. Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{\mathrm{f}} / k_{\mathrm{r}}\), where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{2} / R T}\right)\). b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."

Consider the following reaction at \(800 . \mathrm{K}\) : $$\mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g)$$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{~atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{~atm}, P_{\mathrm{NF}_{3}}=0.48 \mathrm{~atm} .\) Calculate \(\Delta G^{\circ}\) for the reaction at \(800 . \mathrm{K}\).

Using data from Appendix 4, calculate \(\Delta G\) for the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{~S}_{\text {mombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ for the following conditions at \(25^{\circ} \mathrm{C}\) : $$\begin{array}{l}P_{\mathrm{H}_{2} \mathrm{~S}}=1.0 \times 10^{-4} \mathrm{~atm} \\\P_{\mathrm{SO}_{2}}=1.0 \times 10^{-2} \mathrm{~atm} \\ P_{\mathrm{H}_{2} \mathrm{O}}=3.0 \times 10^{-2} \mathrm{~atm}\end{array}$$

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