N 2 O 4(g) 2 NO 2(g)
description
Transcript of N 2 O 4(g) 2 NO 2(g)
N2O4(g) 2 NO2(g)
Chemical Equilibrium
-x
+2x
Chemical Equilibrium: Empirical Study
Reaction being studied: CO(g) + 2 H2(g) CH3OH(g)
Experiment [CO] mol/L [CH3OH] mol/L[H2] mol/L
equilibrium
1 initial
equilibrium
2 initial
equilibrium
3 initial
change
change
change
Chemical Equilibrium: Empirical StudyTrial relationships of the equilibrium data:
[CO][H2]
[CH3OH ]
(0.00892) = 1.19(0.0911)(0.0822)
(0.0247) = 2.17(0.0753)(0.151)
(0.0620) = 2.55(0.138)(0.176)
[CO]2[H2]
[CH3OH ]
(0.00892) = 0.596(0.0911)2(0.0822)
(0.0247) = 1.09(0.0753)2(0.151)
(0.0620) = 1.28(0.138)2(0.176)
[CO][H2]2
[CH3OH ]
(0.00892) = 14.5(0.0911)(0.0822)2
(0.0247) = 14.4(0.0753)(0.151)2
(0.0620) = 14.5(0.138)(0.176)2
Experiment
1
2
3
Chemical Equilibrium: Kinetic AnalysisChemical reaction is at equilibrium when forward rate = reverse rate.Dynamic not static.Kinetic analysis of sample reaction: Iodide ion catalyzed decomposition of hydrogen peroxide.
2 H2O2(aq) 2 H2O + O2(g)
Proposed mechanism: H2O2 + I– H2O + IO– slow
H2O2 + IO– H2O + I– + O2 fast
Rate law: rate forward = kforward [H2O2][I–]
Reverse reaction: 2 H2O + O2(g) 2 H2O2(aq
Mechanism will be same except all steps reversed.
H2O + I– + O2 H2O2 + IO– fast H2O + IO– H2O2 + I– slow
Rate law: ratereverse = kreverse [H2O]2[I–][O2]/[H2O2]
At equilibrium: rateforward = ratereverse kforward [H2O2][I–] = kreverse [H2O]2[I–][O2]/[H2O2]
= Kc (equilibrium constant)[H2O2]2 ][I–]
Rearranging: kforward [H2O]2[I–][O2] [H2O]2[O2] [H2O2]2kreverse
==
Equilibrium Constants in Terms of Pressure, Kp
CO(g) + 2 H2(g) CH3OH(g)
Does Kp = Kc ?
From the ideal gas law: PCO = nCORT/V = [CO] RT
Kp = (P CH3 OH)
(PCO )(PH2)2
Replacing the pressures gives
Kp = (P CH3 OH)
(PCO )(PH2)2
= [CH3OH] (RT)
[CO] (RT) [H2]2 (RT)2
= Kc (RT)–2
Equilibrium Constant RelationshipsSignificance of magnitude of K If K >> 1 Equilibrium lies to the right, products predominate If K << 1 Equilibrium lies to the left, reactants predominate
Change in direction or stoichiometry
CO(g) + 2 H2(g) CH3OH(g) Kc = [CH3OH] = 14.5 [CO][H2]2
Reverse the reaction
CO(g) + 2 H2(g) CH3OH(g) Kc = [CO][H2]2 = 1/14.5 = 0.069
[CH3OH]
2 CO(g) + 4 H2(g) 2CH3OH(g) Kc = [CH3OH]2 = (14.5)2 = 210 [CO]2[H2]4
Double the reaction coefficients
Equilibrium Constant Relationships
Adding equilibrium reactions: to find equilibrium constant of the net reaction
2 NOBr(g) 2 NO(g) + Br2(g) Kc = [NO2] 2[Br2] = 0.014
[NOBr] 2
Br2 (g) + Cl2(g) 2 BrCl(g) Kc = [BrCl]2 = 7.2 [Br2][Cl2]
Find Kc for the sum of these two reactions:
2 NOBr(g) + Cl2(g) 2 NO(g) + 2 BrCl(g)
Kc = [NO2] 2[BrCl]2
[NOBr]2[Cl2]=
[NO2] 2[Br2] [NOBr]2
[BrCl]2 [Br2][Cl2] X = 0.014)(7.2) = 0.10
Determination of K
[N2] = 0.115 M [NH3] = 0.439 M[H2] = 0.105 M
Equilibrium Concentrations at 500 K
[NH3]2
[N2][H2]3
c =
N2(g) + 3 H2(g) 2NH3(g)
From determined equilibrium concentrations:
Determination of K
2 SO2(g) + O2(g) 2 SO3(g)
Given all initial amounts and one equilibrium amount:
[SO2]mol/L [O2]mol/L [SO3]mol/L
Equilibrium:
Initial:
Change:
[SO3]2
c =
[SO2]2[O2]
1.000 1.000 0.000
0.5470.074 0.926
+0.926–0.926 –0.463
Finding Equilibrium Concentrations
N2(g) + 3 H2(g) 2NH3(g)
If PN2 = 0.10 atm and PH2
= 0.20 atm at equilibrium, find PNH3
Kp = 41 at 400 K
Kp =
(PNH3)2
(PH2)3(PN2
)
(PNH3)2 K
p (PH2)3(PN2
)
(PNH3)2 0.20)3 = 0.0328
(PNH3) 0.0328)1/2 = 0.18 atm
Finding Equilibrium Concentrations
If initial [Br2] = 0.20 M and [Cl2] = 0.20 M , find
[BrCl] at equilibrium.
Br2(g) + Cl2(g) 2 BrCl(g)Kc = 7.0 at 373 K
Set up an equilibrium table: Let x = change in [Br2]
[Cl2] [BrCl][Br2]Initial
EquilibriumChange
0.20 0.20 0–x–x +2x
0.20–x 0.20–x 2x
Kc = [Br2][Cl2]
[BrCl]2
=[0.20–x][0.20–x]
[2x]2
=0.040 – 0.40x + x2
4x2
7.0 = (0.040 – 0.40x + x2) (4x2)/
3x2 – 2.8x + 0.28 = 0 (quadratic equation)
(7.0)(0.040 – 0.40x + x2) = 4x2 = 0.28 –2.8 x + 7x2
Finding Equilibrium Concentrations
x = –(–2.8)±[(–2.8)2 –(4)(3)(0.28)]1/2
(2)(3)
x = 2.8 ± (4.48)1/2 = 2.8 ± 2.1 = 0.82 or 0.12 6 6
Only the x = 0.12 will work, x = 0.82 will give negative concentration values.
[Cl2] [BrCl][Br2]Initial 0.20 0.20 0Change –x–x +2xEquilibrium 0.20–x 0.20–x 2x
[Cl2] [BrCl][Br2]Initial 0.20 0.20 0
Equilibrium 0.08 0.08 0.24Change –0.12 –0.12 +0.24
[BrCl] = 0.24 M