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Consider these electron transitions in a hydrogen atom: I. n = 2 n = 1 II. n = 3 n = 1 III. n= 1n = 4 Which of the electron transitions would release the most energy? A) I B) I| C) II D) I, II, and III release the same amount of energy

Consider these electron transitions in a hydrogen atom: I. n = 2 n = 1 II. n = 3 n = 1 III. n= 1n = 4 Which of the electron transitions would release the most energy? A) I B) I| C) II D) I, II, and III release the same amount of energy

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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### Electron Transitions in a Hydrogen Atom **Consider these electron transitions in a hydrogen atom:** 1. **n = 2 → n = 1** 2. **n = 3 → n = 1** 3. **n = 1 → n = 4** **Question:** Which of the electron transitions would release the most energy? **Options:** - **A) I** - **B) II** - **C) III** - **D) I, II, and III release the same amount of energy** #### Explanation: In the context of electron transitions in a hydrogen atom, energy is released when an electron transitions from a higher energy level (higher n value) to a lower energy level (lower n value). The amount of energy released is proportional to the difference in the energy levels. - **Option I (n = 2 → n = 1):** This transition releases energy because the electron moves from a higher energy level (n = 2) to a lower energy level (n = 1). - **Option II (n = 3 → n = 1):** This transition also releases energy, and the energy released will be more than in Option I since the difference between the energy levels (n = 3 and n = 1) is greater. - **Option III (n = 1 → n = 4):** This transition would require the absorption of energy since the electron is moving from a lower energy level (n = 1) to a higher energy level (n = 4). Therefore, **Option II (n = 3 → n = 1)** would release the most energy.
Transcribed Image Text:### Electron Transitions in a Hydrogen Atom **Consider these electron transitions in a hydrogen atom:** 1. **n = 2 → n = 1** 2. **n = 3 → n = 1** 3. **n = 1 → n = 4** **Question:** Which of the electron transitions would release the most energy? **Options:** - **A) I** - **B) II** - **C) III** - **D) I, II, and III release the same amount of energy** #### Explanation: In the context of electron transitions in a hydrogen atom, energy is released when an electron transitions from a higher energy level (higher n value) to a lower energy level (lower n value). The amount of energy released is proportional to the difference in the energy levels. - **Option I (n = 2 → n = 1):** This transition releases energy because the electron moves from a higher energy level (n = 2) to a lower energy level (n = 1). - **Option II (n = 3 → n = 1):** This transition also releases energy, and the energy released will be more than in Option I since the difference between the energy levels (n = 3 and n = 1) is greater. - **Option III (n = 1 → n = 4):** This transition would require the absorption of energy since the electron is moving from a lower energy level (n = 1) to a higher energy level (n = 4). Therefore, **Option II (n = 3 → n = 1)** would release the most energy.
### Determining the Heat of Formation for PbS Using the provided table and the chemical equation below, we aim to determine the heat of formation (ΔH°f) of PbS in kJ/mol. #### Chemical Reaction \[ 2 \text{PbS (s)} + 3 \text{O}_2 \text{(g)} \rightarrow 2 \text{SO}_2 \text{(g)} + 2 \text{PbO (s)} \quad \Delta H^\circ = -828.4 \text{ kJ/mol} \] #### Given Data | Substance | \( \Delta H_f^\circ \) (kJ/mol) | |-----------|-------------------------------| | \( \text{O}_2 \text{(g)} \) | 0 | | \( \text{SO}_2 \text{(g)} \) | -296.9 | | \( \text{PbO (s)} \) | -217.3 | ### Explanation This equation corresponds to the enthalpy change for the reaction given. To find the standard enthalpy change of formation (ΔH°f) for PbS, you can use Hess's Law, which involves the sum of the enthalpies of the products minus the reactants. #### Steps to Determine \( \Delta H_f^\circ \) for PbS: 1. **Total Enthalpy Change for Products:** \[ \Delta H_{\text{products}} = 2 \times \Delta H_f^\circ (\text{SO}_2) + 2 \times \Delta H_f^\circ (\text{PbO}) \] Plug in the given values: \[ \Delta H_{\text{products}} = 2 \times (-296.9 \text{ kJ/mol}) + 2 \times (-217.3 \text{ kJ/mol}) \] \[ \Delta H_{\text{products}} = -593.8 \text{ kJ/mol} + (-434.6 \text{ kJ/mol}) = -1028.4 \text{ kJ/mol} \] 2. **Using the Total Enthalpy Change for the Reaction:** \[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants
Transcribed Image Text:### Determining the Heat of Formation for PbS Using the provided table and the chemical equation below, we aim to determine the heat of formation (ΔH°f) of PbS in kJ/mol. #### Chemical Reaction \[ 2 \text{PbS (s)} + 3 \text{O}_2 \text{(g)} \rightarrow 2 \text{SO}_2 \text{(g)} + 2 \text{PbO (s)} \quad \Delta H^\circ = -828.4 \text{ kJ/mol} \] #### Given Data | Substance | \( \Delta H_f^\circ \) (kJ/mol) | |-----------|-------------------------------| | \( \text{O}_2 \text{(g)} \) | 0 | | \( \text{SO}_2 \text{(g)} \) | -296.9 | | \( \text{PbO (s)} \) | -217.3 | ### Explanation This equation corresponds to the enthalpy change for the reaction given. To find the standard enthalpy change of formation (ΔH°f) for PbS, you can use Hess's Law, which involves the sum of the enthalpies of the products minus the reactants. #### Steps to Determine \( \Delta H_f^\circ \) for PbS: 1. **Total Enthalpy Change for Products:** \[ \Delta H_{\text{products}} = 2 \times \Delta H_f^\circ (\text{SO}_2) + 2 \times \Delta H_f^\circ (\text{PbO}) \] Plug in the given values: \[ \Delta H_{\text{products}} = 2 \times (-296.9 \text{ kJ/mol}) + 2 \times (-217.3 \text{ kJ/mol}) \] \[ \Delta H_{\text{products}} = -593.8 \text{ kJ/mol} + (-434.6 \text{ kJ/mol}) = -1028.4 \text{ kJ/mol} \] 2. **Using the Total Enthalpy Change for the Reaction:** \[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants
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