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