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Ch.15 - Chemical Equilibrium
McMurry - Chemistry 8th Edition
McMurry8th EditionChemistryISBN: 9781292336145당신이 사용하는 게 아니라요?교과서 변경
15장, 문제 149

Halogen lamps are ordinary tungsten filament lamps in which the lamp bulb contains a small amount of a halogen (often bromine). At the high temperatures of the lamp, the halogens dissociate and exist as single atoms. (c) When the WBr41g2 diffuses back toward the filament, it decomposes, depositing tungsten back onto the fila- ment. Show quantitatively that the pressure of WBr4 from part (a) will cause the reaction in part (a) to go in reverse direction at 2800 K. [The pressure of Br1g2 is still 0.010 atm.] Thus, tungsten is continually recycled from the walls of the bulb back to the filament, allow-ing the bulb to last longer and burn brighter.

검증된 단계별 안내
1
Identify the reaction from part (a) and write the balanced chemical equation for the decomposition of \( \text{WBr}_4 \) into \( \text{W} \) and \( \text{Br}_2 \).
Use the given temperature (2800 K) and the pressure of \( \text{Br}_2 \) (0.010 atm) to calculate the equilibrium constant \( K_p \) for the reverse reaction using the equation \( K_p = \frac{P_{\text{products}}}{P_{\text{reactants}}} \).
Determine the partial pressure of \( \text{WBr}_4 \) from part (a) and use it to calculate the reaction quotient \( Q_p \) for the reverse reaction.
Compare \( Q_p \) with \( K_p \) to determine the direction of the reaction. If \( Q_p > K_p \), the reaction will proceed in the reverse direction.
Conclude that the pressure of \( \text{WBr}_4 \) is sufficient to drive the reaction in the reverse direction, allowing tungsten to be deposited back onto the filament.

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주요 개념

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Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. In the context of the halogen lamp, understanding how changes in pressure affect the equilibrium position is crucial. According to Le Chatelier's principle, increasing the pressure of a gaseous reactant will shift the equilibrium towards the side with fewer gas molecules, which is essential for analyzing the reaction involving WBr4.
추천 영상:
가이드 코스
04:21
Chemical Equilibrium Concepts

Gibbs Free Energy

Gibbs free energy (G) is a thermodynamic potential that helps predict the direction of chemical reactions. A negative change in Gibbs free energy (ΔG) indicates that a reaction is spontaneous. At high temperatures, the relationship between Gibbs free energy, enthalpy, and entropy becomes significant, as it can determine whether the decomposition of WBr4 is favorable under the given conditions, such as 2800 K.
추천 영상:
가이드 코스
01:51
Gibbs Free Energy of Reactions

Partial Pressure and Dalton's Law

Partial pressure refers to the pressure exerted by a single component of a gas mixture. Dalton's Law states that the total pressure of a gas mixture is the sum of the partial pressures of its individual components. In this scenario, understanding how the partial pressure of Br2 and WBr4 influences the overall reaction is vital for determining the conditions under which the tungsten recycling process occurs effectively.
추천 영상:
가이드 코스
00:27
Dalton's Law and Partial Pressure
관련 실천
교과서 질문

The F-F bond in F2 is relatively weak because the lone pairs of electrons on one F atom repel the lone pairs on the other F atom; Kp = 7.83 at 1500 K for the reaction F2(g) ⇌ 2 F(g). (a) If the equilibrium partial pressure of F2 molecules at 1500 K is 0.200 atm, what is the equilibrium partial pressure of F atoms in atm?

교과서 질문

Refining petroleum involves cracking large hydrocarbon molecules into smaller, more volatile pieces. A simple example of hydrocarbon cracking is the gas-phase thermal decomposition of butane to give ethane and ethylene: (c) A sample of butane having a pressure of 50 atm is heated at 500 °C in a closed container at constant volume. When equilibrium is reached, what percentage of the butane has been converted to ethane and ethylene? What is the total pressure at equilibrium?

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교과서 질문
At 1000 K, Kp = 2.1 * 106 and ΔH° = - 107.7 kJ for the reaction H21g2 + Br21g2 ∆ 2 HBr1g2. (b) For the equilibrium in part (a), each of the following changes will increase the equilibrium partial pressure of HBr. Choose the change that will cause the greatest increase in the pressure of HBr, and explain your choice. (ii) Adding 0.10 mol of Br2
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교과서 질문

The F-F bond in F2 is relatively weak because the lone pairs of electrons on one F atom repel the lone pairs on the other F atom; Kp = 7.83 at 1500 K for the reaction F2(g) ⇌ 2 F(g). (c) Why is the F-F bond in F2 weaker than the Cl-Cl bond in Cl2?

1354
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교과서 질문

The F-F bond in F2 is relatively weak because the lone pairs of electrons on one F atom repel the lone pairs on the other F atom; Kp = 7.83 at 1500 K for the reaction F2(g) ⇌ 2 F(g). (b) What fraction of the F2 molecules dissociate at 1500 K?

300
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