1 / 12

120 likes | 194 Vues

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. . . .

Télécharger la présentation
## N 2 O 4(g) 2 NO 2(g)

**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

**Chemical Equilibrium**N2O4(g) 2 NO2(g) -x +2x**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**[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****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)**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**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**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**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**[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)**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**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)**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

More Related