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Chapter 17: Problem 33

Consider the following energy levels, each capable of holding two particles: $$\begin{array}{l}{E=2 \mathrm{kJ}} \\ {E=1 \mathrm{kJ}} \\ {E=0 \quad \underline{X X}}\end{array}$$ 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
We have six possible arrangements of the two identical particles in the three energy levels: \(XX \ \_ \ \_\), \(\_ \ XX \ \_\), \(\_ \ \_ \ XX\), \(X \ X \ \_\), \(X \ \_ \ X\), and \(\_ \ X \ X\). The corresponding total energies are 0 kJ, 2 kJ, 4 kJ, 1 kJ, 2 kJ, and 3 kJ. The most likely total energies are 1 kJ and 2 kJ, as they occur two times each.

Step by step solution

01

Understanding the problem statement

We are given three energy levels, each capable of holding two particles, and we have two distinguishable particles represented by X. The goal is to find all the possible arrangements of the two particles in the three energy levels.
02

Enumerate all possible arrangements

Let's start with the possible arrangements for placing the two particles in the energy levels. Remember, each energy level can hold up to two particles, and the particles are indistinguishable. 1. Both particles in the 0 kJ energy level: \(XX \ \_ \ \_\) 2. Both particles in the 1 kJ energy level: \(\_ \ XX \ \_\) 3. Both particles in the 2 kJ energy level \(\_ \ \_ \ XX\) 4. One particle in 0 kJ and another in 1 kJ energy level: \(X \ X \ \_\) 5. One particle in 0 kJ and another in 2 kJ energy level: \(X \ \_ \ X\) 6. One particle in 1 kJ and another in 2 kJ energy level: \(\_ \ X \ X\)
03

Calculate the total energy for each arrangement

Now let's calculate the total energy for each arrangement: 1. Both particles in the 0 kJ energy level: Total energy = 0 kJ 2. Both particles in the 1 kJ energy level: Total energy = 1 kJ + 1 kJ = 2 kJ 3. Both particles in the 2 kJ energy level Total energy = 2 kJ + 2 kJ = 4 kJ 4. One particle in 0 kJ and another in 1 kJ energy level: Total energy = 0 kJ + 1 kJ = 1 kJ 5. One particle in 0 kJ and another in 2 kJ energy level: Total energy = 0 kJ + 2 kJ = 2 kJ 6. One particle in 1 kJ and another in 2 kJ energy level: Total energy = 1 kJ + 2 kJ = 3 kJ
04

Find the total energy that occurs the most

Now that we have the total energy for each arrangement, let's count how many times each energy value appears: - 0 kJ occurs 1 time - 1 kJ occurs 2 times - 2 kJ occurs 2 times - 3 kJ occurs 1 time - 4 kJ occurs 1 time Since the total energy of 1 kJ and 2 kJ occur the greatest number of times (2 times each), they are the most likely total energy values.

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Most popular questions from this chapter

Consider the system $$\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)$$ at \(25^{\circ} \mathrm{C}\) a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{J} / \mathrm{mol},\) calculate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if 1.00 mole of \(\mathrm{A}(g)\) at 1.00 atm and 1.00 mole of \(\mathrm{B}(g)\) at 1.00 atm are mixed at \(25^{\circ} \mathrm{C} .\) c. Show by calculations that \(\Delta G=0\) at equilibrium.

In the text, the equation $$\Delta G=\Delta G^{\circ}+R T \ln (Q)$$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of mol/L for the quantities in \(Q\)specifically for aqueous reactions. With this in mind, consider the reaction $$\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)$$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) . Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C} .\) a. $[\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M}$ b. $[\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M$ c. $[\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} \mathrm{M}$ d. $[\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M$ e. $[\mathrm{HF}]=0.52 M,\left[\mathrm{F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} \mathrm{M}$ Based on the calculated DG values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?

The deciding factor on why HF is a weak acid and not a strong acid like the other hydrogen halides is entropy. What occurs when HF dissociates in water as compared to the other hydrogen halides?

Consider the reaction $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$$ a. Use \(\Delta G_{f}^{\circ}\) values in Appendix 4 to calculate $\Delta G^{\circ}$ for this reaction. b. Is this reaction spontaneous under standard conditions at 298 $\mathrm{K} ?$ c. The value of \(\Delta H^{\circ}\) for this reaction is 100 . kJ. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

True or false: High temperatures are favorable to a reaction both kinetically and thermodynamically. Explain.
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