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N 2 O 4(g) 2 NO 2(g)

Chemical Equilibrium. N 2 O 4(g) 2 NO 2(g). -x. +2x. Reaction being studied: CO (g) + 2 H 2(g) CH 3 OH (g). Experiment. [CO] mol/L. [H 2 ] mol/L. [CH 3 OH] mol/L. initial. . . . 1. change. change. change. . . .

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N 2 O 4(g) 2 NO 2(g)

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  1. Chemical Equilibrium N2O4(g) 2 NO2(g) -x +2x

  2. Reaction being studied: CO(g) + 2 H2(g) CH3OH(g) Experiment [CO] mol/L [H2] mol/L [CH3OH] mol/L initial    1 change change change          equilibrium    2 initial    equilibrium    3 initial    equilibrium    Chemical Equilibrium: Empirical Study

  3. [CH3OH ] [CH3OH ] [CO][H2]2 [CO][H2] [CH3OH ] (0.0247) = 1.09 (0.0753)2(0.151) (0.00892) = 0.596 (0.0911)2(0.0822) (0.0620) = 1.28 (0.138)2(0.176) (0.0247) = 2.17 (0.0753)(0.151) (0.0620) = 2.55 (0.138)(0.176) (0.00892) = 1.19 (0.0911)(0.0822) (0.0620) = 14.5 (0.138)(0.176)2 (0.00892) = 14.5 (0.0911)(0.0822)2 (0.0247) = 14.4 (0.0753)(0.151)2 [CO]2[H2] Trial relationships of the equilibrium data: Chemical Equilibrium: Empirical Study Experiment 1 2 3

  4.  2 H2O2(aq) 2 H2O + O2(g) Proposed mechanism: H2O2 + I– H2O + IO–slow H2O2 + IO– H2O + I–+ O2fast H2O + I–+ O2H2O2 + IO–fast Reverse reaction: 2 H2O + O2(g) 2 H2O2(aq Mechanism will be same except all steps reversed. At equilibrium: rateforward = ratereverse kforward [H2O2][I–] = kreverse [H2O]2[I–][O2]/[H2O2] H2O + IO–H2O2 + I–slow Rearranging: kforward [H2O]2[I–][O2] [H2O]2[O2] = = kreverse [H2O2]2 ][I–] [H2O2]2 Chemical reaction is at equilibrium when forward rate = reverse rate. Dynamic not static. Chemical Equilibrium: Kinetic Analysis Kinetic analysis of sample reaction: Iodide ion catalyzed decomposition of hydrogen peroxide. Rate law: rate forward = kforward [H2O2][I–] Rate law: ratereverse = kreverse [H2O]2[I–][O2]/[H2O2] = Kc(equilibrium constant)

  5. CO(g) + 2 H2(g) CH3OH(g) Replacing the pressures gives Kp= (PCH3 OH) (PCO )(PH2)2 Kc(RT)–2 = = Kp= (PCH3 OH) (PCO )(PH2)2 [CH3OH] (RT) [CO] (RT) [H2]2 (RT)2 Equilibrium Constants in Terms of Pressure, Kp Does Kp = Kc ? From the ideal gas law: PCO = nCORT/V = [CO] RT

  6. Change in direction or stoichiometry CO(g) + 2 H2(g) CH3OH(g) Reverse the reaction Kc = [CH3OH] = 14.5 [CO][H2]2 Kc = [CH3OH]2= (14.5)2 = 210 [CO]2[H2]4 Kc = [CO][H2]2 = 1/14.5= 0.069 [CH3OH] CO(g) + 2 H2(g) CH3OH(g) Double the reaction coefficients 2 CO(g) + 4 H2(g) 2CH3OH(g) Equilibrium Constant Relationships Significance of magnitude of K If K >> 1 Equilibrium lies to the right, products predominate If K << 1 Equilibrium lies to the left, reactants predominate

  7. Kc = [NO2] 2[Br2] = 0.014 [NOBr] 2 2 NOBr(g) 2 NO(g) + Br2(g) Find Kc for the sum of these two reactions: Kc = [BrCl]2= 7.2 [Br2][Cl2] Br2(g) + Cl2(g) 2 BrCl(g) (0.014)(7.2) = 0.10 = = X 2 NOBr(g) + Cl2(g) 2 NO(g) + 2 BrCl(g) Kc = [NO2] 2[BrCl]2 [NOBr]2[Cl2] [BrCl]2 [Br2][Cl2] [NO2] 2[Br2] [NOBr]2 Equilibrium Constant Relationships Adding equilibrium reactions: to find equilibrium constant of the net reaction

  8. From determined equilibrium concentrations: N2(g) + 3 H2(g) 2NH3(g) Equilibrium Concentrations at 500 K (0.439)2 = [N2]= 0.115 M [H2]= 0.105 M [NH3]= 0.439 M [NH3]2 (0.115)(0.105)3 Kc = [N2][H2]3 = 1450 Determination of K

  9. [SO2]mol/L [O2]mol/L [SO3]mol/L Initial: 1.000 1.000 0.000 Change: –0.926 –0.463 +0.926 2 SO2(g) + O2(g) 2 SO3(g) Equilibrium: 0.074 0.547 0.926 [SO3]2 Kc = [SO2]2[O2] Given all initial amounts and one equilibrium amount: Determination of K (0.926)2 = =  (0.074)(0.547)

  10. N2(g) + 3 H2(g) 2NH3(g) (PNH3)2 Kp = (PN2) (PH2)3 = Kp (PNH3)2 (PNH3)2 = (41)(0.10)(0.20)3 = 0.0328 (PN2) (PH2)3 (PNH3) = (0.0328)1/2 = 0.18 atm Finding Equilibrium Concentrations Kp = 41 at 400 K If PN2 = 0.10 atm and PH2 = 0.20 atm at equilibrium, find PNH3

  11. Br2(g) + Cl2(g) 2 BrCl(g) Kc = 7.0 at 373 K [Cl2] [BrCl] [Br2] Initial 0.20 0.20 0 Change Equilibrium 0.20–x 0.20–x 2x 4x2 [BrCl]2 [2x]2 = = Kc = 7.0 = (4x2)/ (0.040 – 0.40x + x2) [Br2][Cl2] 0.040 – 0.40x + x2 [0.20–x][0.20–x] Finding Equilibrium Concentrations If initial [Br2] = 0.20 M and [Cl2] = 0.20 M , find [BrCl] at equilibrium. Set up an equilibrium table: Let x = change in [Br2] –x +2x –x (7.0)(0.040 – 0.40x + x2) = 4x2 = 0.28 –2.8 x + 7x2 3x2– 2.8x + 0.28 = 0 (quadratic equation)

  12. Initial –x –x +2x Change Equilibrium Only the x = 0.12 will work, x = 0.82 will give negative concentration values. [Cl2] [Cl2] [BrCl] [BrCl] [Br2] [Br2] 0.20 0.20 0 x = –(–2.8)±[(–2.8)2–(4)(3)(0.28)]1/2 Initial 0.20 0.20 0 (2)(3) –0.12 Change –0.12 +0.24 0.20–x 0.20–x 2x x = 2.8 ± (4.48)1/2 = 2.8 ± 2.1 = 0.82 or 0.12 Equilibrium 0.08 0.08 0.24 6 6 Finding Equilibrium Concentrations [BrCl] = 0.24 M

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