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Notes One Unit Eleven Dynamic Equilibrium Demo Siphon Demo Dynamic Equilibrium Rate of reaction Forward = Rate of Reaction Backward Chemical Concentration.

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Presentation on theme: "Notes One Unit Eleven Dynamic Equilibrium Demo Siphon Demo Dynamic Equilibrium Rate of reaction Forward = Rate of Reaction Backward Chemical Concentration."— 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

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 N 2 O 4 present initially(b) Only 0.08 M NO 2 present initially

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

8 Chemical Concentration before and after equilibrium Experimental Data: If the Temperature is the same, K will be the same. [NO 2 ] [N 2 O 4 ] ______ K eq = 2 [0.0125] [0.0337] _______ 2 = Trial Number Initial Concentration Equilibrium Constant [N 2 O 4 ][NO 2 ][N 2 O 4 ][NO 2 ][NO 2 ] 2 /[N 2 O 4 ] x x x x x10 -3 [0.0125] [0.0337] _______ 2 = [0.0156] [0.0522] _______ 2 =

9 Mass Action Expression 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O(g) CaCO 3 (s)  CaO(s) + CO 2 (g) PCl 3 (l) + Cl 2 (g)  PCl 5 (s) H 2 (g) + F 2 (g)  2HF(g) [NO 2 ] [NH 3 ][O2][O2] __________ K eq = [H2O][H2O] [CO 2 ] K eq = [Cl 2 ] K eq = [HF] [H2][H2][F2][F2] _______ K eq = 2 ____ 1

10 Calculating K 2) Mass Action Equation 1) Balanced Equation 3) Calculate 3) Calculate K Calculate the K eq at 74 0 C, if [CO] = M, [Cl 2 ] = M, and [COCl 2 ] = 0.14 M. CO(g) + Cl 2 (g)→ COCl 2 (g) [COCl 2 ] [CO][Cl 2 ] ________ K eq = [0.14 M 1 ] [0.012M 1 ][0.054M 1 ] _________________ K eq = 220M -1 K eq =

11 Calculating Concentration From K 2) Mass Action Equation 3) Solve for [O 2 ]? 1) Balanced Equation 4) Calculate 4) Calculate K The equilibrium constant K for the reaction is 158 atm at 1000K. What is the equilibrium pressure of O 2, if the P NO2 = atm and P NO = atm? 212 NO 2 (g)  NO(g) + O 2 (g) [NO] [NO 2 ] [O 2 ] ________ K= 2 2 [NO] [NO 2 ] =[O 2 ] _______ K x 2 2 (0.270atm) (0.400atm) =[O 2 ] _________ (158atm) 2 2 [O 2 ]= 347atm K [NO][O 2 ] [NO 2 ] 2 K = K x[NO 2 ] 2 [NO 2 ] 2 x 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 Co(H 2 O) 6 +2 (aq) + 4Cl -1 (aq)  CoCl 4 +2 (aq) +6H 2 O(l) In step 3, hydrochloric acid is used as a source of Cl -1 ions. We see more blue! Co(H 2 O) 6 +2 (aq) + 4Cl -1 (aq)  CoCl 4 +2 (aq) +6H 2 O(l) In step 5, why did adding H 2 O cause the change that it did? We see more red! Co(H 2 O) 6 +2 (aq) + 4Cl -1 (aq)  CoCl 4 +2 (aq) +6H 2 O(l) In step 6, silver ions from the AgNO 3 react with Cl - ions to produce an insoluble precipitate. We see more red! Co(H 2 O) 6 +2 (aq) + 4Cl -1 (aq)  CoCl 4 +2 (aq) +6H 2 O(l) In step 7, acetone has an attraction for H 2 O. We see more blue!

15 Writing Solubility Reactions Ag 3 PO 4 (s)  Ag +1 +PO ScF 3 (s)  Sc +3 + F -1 3 Sn 3 P 4 (s)  Sn +4 +P Ag 3 PO 4 Dissolves: ScF 3 Dissolves: Sn 3 P 4 Dissolves: 1 cation1 ion 1 cation1 ion 1 cation1 ion

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 CaF 2 Dissolves: CaF 2 (s)  Ca F -1

18 Calculating Concentration From Ksp Cd 3 (AsO 4 ) 2 Cd 3 (AsO 4 ) 2 (s)  Cd +2 (aq) Cd +2 (aq) + AsO 4 -3 (aq) 23 AsO 4 -3 [AsO 4 -3 ] Cd +2 [Cd +2 ] X+2X+ 3X2X 2X [2X] 3X [3X] X5X5X5X5 (2.2× ) ________ 108 ^(1/5) X=1.2x10 -7 M 1 [Cd +2 ] =3(1.2x10 -7 M ) [AsO 4 -3 ] =2(1.2x10 -7 M ) X= Ksp = 2) Mass Action Equation 3) What do we know? Before Eq Change At Eq 1) Balanced Equation 4) Calculate X 2.2× = What is the concentration of the cation and anion for cadmium arsenate,Cd 3 (AsO 4 ) 2, if Ksp=2.2× ? What is the concentration of the cation and anion for cadmium arsenate,Cd 3 (AsO 4 ) 2, if Ksp=2.2× M 5 ?

19 Calculating Ksp from Concentration 00 X+3X+ X3X Ksp = 2) Mass Action Equation 3) What do we know? Before Eq Change At Eq 1) Balanced Equation 4) Calculate 4) Calculate K sp = K sp = If the molar solubility of BiI 3 is 1.32 x 10 -5, find its K sp. BiI 3 (s)  Bi +3 +I -1 3 [I -1 ][Bi +3 ] 3 [I -1 ][Bi +3 ] 3 [3X][X] 3 27 X 4 = (1.32 x ) x = M4M4 = K sp =

20 Calculating K from Initial Conditions 2) Mass Action Equation 3) What do we know? Before Eq Change At Eq 1) Balanced Equation 4) Calculate 4) Calculate K In a flask 1.50M H 2 and 1.50M N 2 is allowed to reach equilibrium. At equilibrium [NH 3 ] =0.33M. Calculate K. 321H 2 (g) + N 2 (g)→ NH 3 (g) [NH 3 ] [H 2 ] [N 2 ] ________ K= 2 3 H2H2 N2N2 NH [0.33] [1.00][1.33] ___________ K= M -2


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