# 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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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

Demo Dynamic Equilibrium

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

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

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

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 ] 1 2 3 4 5 0.0400 0.00000.03370.01254.64x10 -3 0.0000 0.08000.03370.01254.64x10 -3 0.0600 0.00000.05220.01564.66x10 -3 0.0000 0.06000.02460.01074.65x10 -3 0.0200 0.06000.04290.01414.63x10 -3 [0.0125] [0.0337] _______ 2 = [0.0156] [0.0522] _______ 2 =

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] 64 47 [CO 2 ] K eq = [Cl 2 ] K eq = [HF] [H2][H2][F2][F2] _______ K eq = 2 ____ 1

Calculating K 2) Mass Action Equation 1) Balanced Equation 3) Calculate 3) Calculate K Calculate the K eq at 74 0 C, if [CO] = 0.012 M, [Cl 2 ] = 0.054 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 =

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 = 0.400 atm and P NO = 0.270 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

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

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

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!

Writing Solubility Reactions Ag 3 PO 4 (s)  Ag +1 +PO 4 -3 3 ScF 3 (s)  Sc +3 + F -1 3 Sn 3 P 4 (s)  Sn +4 +P -3 34 Ag 3 PO 4 Dissolves: ScF 3 Dissolves: Sn 3 P 4 Dissolves: 1 cation1 ion 1 cation1 ion 1 cation1 ion

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!

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

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 ] 2 3 00 3X+2X+ 3X2X 2X [2X] 3X [3X] 23 108 X5X5X5X5 (2.2×10 -33 ) ________ 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 2.2×10 -33 = What is the concentration of the cation and anion for cadmium arsenate,Cd 3 (AsO 4 ) 2, if Ksp=2.2×10 -33 ? What is the concentration of the cation and anion for cadmium arsenate,Cd 3 (AsO 4 ) 2, if Ksp=2.2×10 -33 M 5 ?

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 10 -5 )27 4 8.20 x 10 -19 = M4M4 = K sp =

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 3 1.50 0 0.50 1.00 0.17 0.33 1.330.33 --+ [0.33] [1.00][1.33] ___________ K= 2 3 0.082M -2

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