Given the two reactions
PbCl2(s)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.81×10−10, and
AgCl(s)⇌Ag+(aq)+Cl−(aq), K4 = 1.22×10−4,
what is the equilibrium constant (K final) for the following reaction?
PbCl2(s)+2Ag+(aq)⇌2AgCl(s)+Pb2+(aq)
Answers I have tried that are incorrect are 3.72⋅1015, 2.69⋅10−18, 8.17⋅105
Chemistry Equilibrium?
2016-10-23 6:22 am
回答 (1)
2016-10-23 2:21 pm
PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq) …. K₃ = [Pb²⁺][Cl⁻]² = 1.81 × 10⁻¹⁰
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) …. K₄ = [Ag⁺][Cl⁻] = 1.22 × 10⁻⁴
PbCl₂(s) + 2Ag⁺(aq) ⇌ 2AgCl(s) + Pb²⁺(aq) …. K
K = [Pb²⁺] / [Ag⁺]²
= [Pb²⁺][Cl⁻]² / [Ag⁺]²[Cl⁻]²
= [Pb²⁺][Cl⁻]² / ([Ag⁺][Cl⁻])²
= K₃ / K₄²
= (1.81 × 10⁻¹⁰) / (1.22 × 10⁻⁴)²
= 1.22 × 10⁻²
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) …. K₄ = [Ag⁺][Cl⁻] = 1.22 × 10⁻⁴
PbCl₂(s) + 2Ag⁺(aq) ⇌ 2AgCl(s) + Pb²⁺(aq) …. K
K = [Pb²⁺] / [Ag⁺]²
= [Pb²⁺][Cl⁻]² / [Ag⁺]²[Cl⁻]²
= [Pb²⁺][Cl⁻]² / ([Ag⁺][Cl⁻])²
= K₃ / K₄²
= (1.81 × 10⁻¹⁰) / (1.22 × 10⁻⁴)²
= 1.22 × 10⁻²
收錄日期: 2021-04-18 15:41:40
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