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Notes One Unit Eleven

Notes One Unit Eleven. Dynamic Equilibrium Demo Siphon Demo Dynamic Equilibrium Rate of reaction Forward = Rate of Reaction Backward Chemical Concentration before and after equilibrium Mass Action Expression Calculating Equilibrium Constant K from Concentration. Demo Siphon.

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Notes One Unit Eleven

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  1. Notes One Unit Eleven • Dynamic Equilibrium • Demo Siphon • Demo Dynamic Equilibrium • Rate of reaction Forward = Rate of Reaction Backward • Chemical Concentration before and after equilibrium • Mass Action Expression • Calculating Equilibrium Constant K from Concentration.

  2. Demo Siphon

  3. Demo Dynamic Equilibrium

  4. Demo Dynamic Equilibrium

  5. Dynamic Equilibrium • Reversible reactions. • One reaction going forward. • One reaction going backward. • Temperature can change the Equilibrium

  6. Chemical Concentration before and after equilibrium • (a) Only 0.04 M N2O4 present initially • (b) Only 0.08 M NO2 present initially

  7. Mass Action Expression Mass Action Expressions d c [C] [D] Keq= b a [A] [B] Solids and liquids are left out

  8. Chemical Concentration before and after equilibrium [NO2] [0.0125] [0.0125] [0.0156] ______ 2 _______ 2 2 _______ _______ 2 Keq= = = = [N2O4] [0.0337] [0.0337] [0.0522] Experimental Data: 0.0000 0.0337 0.0125 4.64x10-3 0.0400 0.0800 0.0337 0.0125 4.64x10-3 0.0000 0.0000 0.0522 0.0156 4.66x10-3 0.0600 0.0600 0.0246 0.0107 4.65x10-3 0.0000 0.0600 0.0429 0.0141 4.63x10-3 0.0200 If the Temperature is the same, K will be the same.

  9. Mass Action Expression 4 6 [NO2] [H2O] __________ • 4NH3(g) + 7O2(g)4NO2(g) + 6H2O(g) • CaCO3(s)CaO(s) + CO2(g) • PCl3(l) + Cl2(g) PCl5 (s) • H2(g) + F2(g)  2HF(g) Keq= 4 7 [NH3] [O2] [CO2] Keq= 1 ____ Keq= [Cl2] 2 [HF] _______ Keq= [H2] [F2]

  10. Calculating K Calculate the Keq at 740C, if [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. 1) Balanced Equation CO(g) + Cl2(g)→ COCl2(g) 2) Mass Action Equation [COCl2] ________ Keq= [CO] [Cl2] 3) Calculate K [0.14 M1] _________________ Keq= 220M-1 Keq= [0.012M1] [0.054M1]

  11. Calculating Concentration From K The equilibrium constant K for the reaction is 158 atm at 1000K. What is the equilibrium pressure of O2, if the PNO2 = 0.400 atm and PNO = 0.270 atm? 1) Balanced Equation NO2(g) NO(g) + O2(g) 2 2 1 2) Mass Action Equation [NO] [O2] ________ 2 [NO] [O2] K= 2 2 [NO2] x K [NO2] 2 x[NO2] = 2 2 [NO2] K K 3) Solve for [O2]? [NO2] 2 _______ =[O2] K x [NO] 2 4) Calculate K (0.400atm) 2 _________ [O2 ]= (158atm) 347 atm =[O2 ] (0.270atm) 2

  12. Notes Two Unit Eleven Chapter Fourteen • Le Chatelier's Principle • Silver Chloride Demo • Calculating K from Initial Conditions • Calculating Concentration From Ksp

  13. Le Chatelier's Principle • If an external stress is applied, the system adjusts • An increased stress is reduced. • A decreased stress is increased.

  14. Le Chatelier's Lab In step 3, hydrochloric acid is used as a source of Cl-1 ions. Co(H2O)6+2(aq) + 4Cl-1(aq)CoCl4+2(aq) +6H2O(l) We see more blue! In step 5, why did adding H2O cause the change that it did? Co(H2O)6+2(aq) + 4Cl-1(aq)CoCl4+2(aq) +6H2O(l) We see more red! In step 6, silver ions from the AgNO3 react with Cl- ions to produce an insoluble precipitate. Co(H2O)6+2(aq) + 4Cl-1(aq)CoCl4+2(aq) +6H2O(l) We see more red! In step 7, acetone has an attraction for H2O. Co(H2O)6+2(aq) + 4Cl-1(aq)CoCl4+2(aq) +6H2O(l) We see more blue!

  15. Writing Solubility Reactions Ag3PO4 Dissolves: 1 cation 1 ion Ag3PO4 (s) 3 Ag+1 + PO4-3 ScF3 Dissolves: 1 cation 1 ion 3 F-1 ScF3 (s)  Sc+3 + Sn3P4 Dissolves: 1 cation 1 ion Sn3P4 (s)  3 Sn+4 + 4 P-3

  16. Silver Chloride Demo AgCl(s) Ag+1(aq) + Cl-1(aq) NaCl is added We see a cloudy solid:AgCl(s)! Le Chatelier's Principle!

  17. Writing Solubility Reactions NaCl Dissolves: NaCl(s) Na+1 + Cl-1 CaF2 Dissolves: CaF2(s) Ca+2+ 2F-1

  18. Calculating Concentration From Ksp What is the concentration of the cation and anion for cadmium arsenate,Cd3(AsO4)2, if Ksp=2.2×10-33 M5? 1) Balanced Equation AsO4-3(aq) Cd3(AsO4)2(s) 3 Cd+2(aq)+ 2 2) Mass Action Equation [Cd+2] [AsO4-3] Ksp = 2 3 3) What do we know? 0 0 Before Eq + 3X + 2X Change 3X 2X At Eq 4) Calculate X [3X] [2X] 3 2 2.2×10-33= 108 X5 2.2×10-33= X=1.2x10-7M1 ^(1/5) (2.2×10-33) [Cd+2] =3(1.2x10-7M ) ________ X= [AsO4-3] =2(1.2x10-7M ) 108

  19. Calculating Ksp from Concentration If the molar solubility of BiI3 is 1.32 x 10-5, find its Ksp. 1) Balanced Equation BiI3(s) Bi+3 + 3 I-1 2) Mass Action Equation [Bi+3] [I-1] Ksp = 3 3) What do we know? 0 0 Before Eq + X + 3X Change X 3X At Eq 4) Calculate Ksp [Bi+3] [I-1] 3 Ksp= [X] [3X] 4 X 27 = 3 Ksp= 27 (1.32 x 10-5) = 4 8.20 x 10-19 M4 Ksp=

  20. Calculating K from Initial Conditions In a flask 1.50M H2 and 1.50M N2 is allowed to reach equilibrium. At equilibrium [NH3] =0.33M. Calculate K. 1) Balanced Equation 3 H2(g) + N2 (g)→ NH3(g) 1 2 2) Mass Action Equation [NH3] ________ 2 K= [N2] [H2] 3 3) What do we know? H2 N2 NH3 1.50 0 1.50 Before Eq - - + 0.33 0.50 0.17 Change 1.00 1.33 0.33 At Eq 4) Calculate K [0.33] ___________ 2 K= 0.082 M-2 K= [1.00] [1.33] 3

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