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Chap15-part3 - EXAM3:April15thFriday2011(notApril11th...

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EXAM‐3: April 15 th Friday 2011 (not April 11 th ) Syllabus: Chapters 15, 16 and 17 Mastering Chemistry Available for Ch‐15
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Homogeneous equilibria involve reactants and products in the same phase. Examples: N 2 O 4 (g) ↔ 2 NO 2 (g) CO(g) + Cl 2 (g) ↔ COCl 2 (g) CO(g) + 2 H 2 (g) ↔ CH 3 OH(g) Heterogeneous equilibria involve at least one reactant or product in a different phase. Examples: PbCl 2 (s) ↔ Pb 2+ (aq) + 2 Cl (aq) H 2 O(l) + CO 3 2‐ (aq) ↔ OH (aq) + HCO 3 (aq) CO 2 (g) + H 2 (g) ↔ CO(g) + H 2 O(l) SnO 2 (s) + 2 CO(g) ↔ Sn(s) + 2 CO 2 (g) 2
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When a pure solid or a pure liquid is involved in a heterogeneous equilibrium, its concentraJon is not included in the equilibrium‐ constant expression. The concentraJon of a pure liquid or pure solid remains constant during the reacJon. Doubling the amount of a pure liquid or solid doubles the volume but does not change the concentraJon. Equilibrium constants are concerned with changes in concentraJons. Also, there is the noJon of acJviJes and reference values. Heterogeneous Equilibrium
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DecomposiJon of calcium carbonate: CaCO 3 (s) ↔ CaO(s) + CO 2 (g) The equilibrium pressure of CO 2 (g) is the same even though the amounts of CaCO 3 (s) and CaO(s) are much different.
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Examples: Homogeneous Equilibria
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Examples: Heterogeneous Equilibria
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The reacJon has an equilibrium constant K c = 7.1 x 10 ‐4 . What are the equilibrium concentraJons of Ca 2+ and CrO 4 2‐ in a saturated soluJon of CaCrO 4 ?
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The reacJon has an equilibrium constant K c = 7.1 x 10 ‐4 . What are the equilibrium concentraJons of Ca 2+ and CrO 4 2‐ in a saturated soluJon of CaCrO 4 ? 1) Write K c = [ Ca 2 + ][ CrO 4 2 ] = 7.1 x 10 4
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The reacJon has an equilibrium constant K c = 7.1 x 10 ‐4 . What are the equilibrium concentraJons of Ca 2+ and CrO 4 2‐ in a saturated soluJon of CaCrO 4 ?
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