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Ch.15 - Chemical Equilibrium
McMurry - Chemistry 8th Edition
McMurry8th EditionChemistryISBN: 9781292336145당신이 사용하는 게 아니라요?교과서 변경
모든 교과서McMurry 8th EditionCh.15 - Chemical Equilibrium문제 151b
15장, 문제 151b

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?

검증된 단계별 안내
1
Step 1: The given reaction is F2(g) ⇌ 2F(g). The equilibrium constant expression for this reaction is Kp = [F]^2 / [F2].
Step 2: Let's assume that 'x' is the fraction of F2 molecules that dissociate. Therefore, the initial concentration of F2 is (1-x) and the concentration of F at equilibrium is 2x.
Step 3: Substitute these values into the equilibrium constant expression. So, Kp = (2x)^2 / (1-x).
Step 4: You are given that Kp = 7.83 at 1500 K. Substitute this value into the equation from step 3.
Step 5: Solve the resulting equation for 'x'. This will give you the fraction of F2 molecules that dissociate at 1500 K.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
3m

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Bond Strength and Lone Pair Repulsion

The strength of a chemical bond is influenced by the presence of lone pairs of electrons. In the case of the F2 molecule, the lone pairs on each fluorine atom repel each other, leading to a weaker bond. This repulsion reduces the overall bond strength, making it easier for the F2 molecules to dissociate into individual fluorine atoms.
추천 영상:
가이드 코스
00:51
Electron Groups, Lone Pairs, and Bonding Groups Example

Equilibrium Constant (Kp)

The equilibrium constant (Kp) quantifies the ratio of the partial pressures of products to reactants at equilibrium for a gaseous reaction. In this case, Kp = 7.83 at 1500 K indicates that at this temperature, the products (dissociated F atoms) are favored over the reactants (F2 molecules), allowing us to calculate the extent of dissociation.
추천 영상:
가이드 코스
03:20
Equilibrium Constant Expressions

Dissociation Fraction

The dissociation fraction represents the proportion of a substance that has dissociated into its constituent parts at equilibrium. To find this fraction for the F2 molecules at 1500 K, one can use the equilibrium expression derived from Kp, which relates the concentrations of the dissociated species to the original concentration of F2, allowing for the calculation of how many molecules have broken apart.
추천 영상:
가이드 코스
00:36
Mole Fraction Formula
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
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.
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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?

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교과서 질문

The equilibrium constant Kc for the gas-phase thermal decomposition of cyclopropane to propene is 1.0 ⨉105 at 500 K:

(a) What is the value of Kp at 500 K?

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