1 / 20

Notes One Unit Eleven

This content explores the principles of dynamic equilibrium in chemical reactions, emphasizing the concepts of reversible reactions, mass action expressions, and calculating equilibrium constants (K). It discusses how changes in concentration and temperature affect reaction rates and equilibrium conditions. Through practical demonstrations and examples, including the use of a siphon and equilibrium calculations involving various reactions, this resource provides a comprehensive understanding of the behavior of chemical systems at equilibrium.

basia-cash
Télécharger la présentation

Notes One Unit Eleven

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related