Equilibrium

1
Equilibrium
2
For equilibrium to occur

• System must be closed.
• Temperature must be constant.
• Reactions must be reversible (do not go to
  completion).
• H2(g) Cl2(g) 2HCl(g) energy
• No visible change A dynamic equilibrium
  exists.
• The rate of forward rx. the rate of the
  reverse rx.
• Homogeneous Equilibria all gaseous or aqueous
  phases.

3
Equilibrium in N2O4(g) q 2 NO2
(g)

• Initially, N2O4 large.

Rate f Rate r

• It decreases rapidly then more slowly.

• The NO2 starts at zero increases rapidly.

concentration
N2O4
NO2

• Both eventually, plateauno change

• Equilibrium reached.

Time
Teq
4
N2O4(g) q 2 NO2 (g)
What happened here?
concentration
Here?
Time
Teq
5
Predicting Changes
Le Chateliers Principle when a stress is
applied to a system at equilibrium. The reaction
will shift in a direction to minimize the
stress.
2H2(g) O2(g) 2H2O(g) energy
shift rxn speeds up in a particular
direction. E.g. shifts right. Forward rxn
speeds up. More products appear. We say the
reaction is favored to the right or the
position of the equilibrium shifts right.

• Types of stress change in concentrations
  change in volume in temperature

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2H2(g) O2(g) 2H2O(g) energy
H2
Make more
H2O
O2
Use it up!
H2O
H2
O2
Temp
Make more pressure
Volume
H2O
O2
H2
What would a catalyst do to the equilibrium
position?
Nothing! It would simply be reached sooner.
What would happen if we added He gas ?
Nothing! It does not affect the partial press of
other gases.
7
N2O4(g) q 2 NO2 (g)
colorless
orange
Describe what happens when your instructor
removes the tube from the freezer, containing the
system described above. Explain your
observation(s) using LeChateliers Principle, and
all of the appropriate terminology.
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Quantitative Aspect of Equilibrium

• Measurements in the N2O4 - NO2 system _at_ 100?C

If we divide Equilibrium product values by
reactant, each factor raised to the correct power
(explained in a moment). What do you notice????
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Quantitative Aspect of Equilibrium
N2O4(g) q 2 NO2(g)
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Mass Action Expression
Given the general equation aA(aq) bB(aq)
cC(aq) dD(aq) also if g gas
Cc Dd Aa
Bb
K eq
equilibrium constant expression
_at_ constant Temp. ( Keq can also be Kp , Kc ,
Ksp , Ka , Kb , etc.)
Product(s)
Right
K eq
0r
Reactant(s)
Left
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What does the K value tell us?
Prod.
large

• If K is gt 1

At Equilibrium there is more product than
reactant!
Reactant
small

• If K is lt 1

Prod.
At Equilibrium there is more reactant than
product!
small
Reactant
Large

• If K is 1 Products Reactants
• If K is very small..then practically no product
  
• is formed.

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Eg. Weak Acid HC2H3O2 ( H2O)
H (aq) C2H3O2- (aq)
H
C2H3O2-
1.8 x 10-5
Ka
HC2H3O2
What is there more ofreactant or product?
Reactant !
HC2H3O2 only 3 ionizes. There is almost no
product present. This is why it is a weak acid.
Strong acids 100 ionize.Ka is LARGE.
Insoluble salts PbI2 (s) ( H2O)
Pb2 (aq) 2 I- (aq)
Ksp
Pb2I-2 8.4 x 10-9 What does this
value tell us?
This solid is very insoluble mostly solid PbI2
present.
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Keq is used to calculate concentrations of
species at equilibrium. Given N2(g) O2(g)
2NO(g). At 25C the Kc 1.0 x 10-30. N2
0.040 O2 0.010. What is the concentration
of NO?
NO2
M.A.E. Kc
1.0 x 10-30
N2 O2
Small K value.very little product!
NO2 N2 O2 1.0 x 10-30
(0.040) (0.010) (1.0 x 10-30) 4.0 x
10-34 NO 2.0 x 10-17mol/L

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For the system CO2 H2 CO H2O
Kc 0.64 _at_ 900C. The initial concentrations
of reactants are both 0.100mol/L. When the system
reaches equilibrium what are the concentrations
of reactants and products?
Orig. Conc (mol/L) Change in Conc.
Equilib. Conc.
CO2 H2 CO H2O
0.100 0.100 0.000 0.000
-x
0.100-x
-x 0.100-x
x x
x x
CO H2O
x2
Kc 0.64
CO2 H2
(0.100-x)2
x 0.044 CO H2O 0.044mol/L
CO2 H2 0.100 - 0.044 0.056 mol/L
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Heterogeneous equilibrium

• Reactions in which one or more of the substances
  involved is a pure liquid or solid.

CO2(g) H2(g) CO(g)
H2O (l)

• Experimentally, we find, that the position of the
  equilibrium is independent of the amount of solid
  or liquid. Adding or removing a liquid or a
  solid has no effect on the equilibrium.

• We do not need to include terms for solids or
  liquids in the expression for Keq. They are
  totally ignored.

CO
Thus, Keq
CO2 H2
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Solubility K (read pgs.597-612)

• Calculating solubility from Ksp

PbSO4(s) Pb2(aq) SO4-2 (aq)
Given, PbSO4 , Ksp 1.6 x 10-8. Calculate its
solubility, Pb2 SO4-2. ,
Ksp Pb2 SO4-2 1.6 x 10-8
x2 1.6 x 10-8 (since x Pb2
SO4-2) x 1.3 x 10-4. Pb2 SO4-2
1.3 x 10-4M
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Try this problem Mg(OH)2(s) Mg2(aq)
2OH -1(aq) Ksp 1.5 x 10 -11 . Calculate the
Mg2, OH-1, and the Mg(OH)2 at
equilibrium.
If X Mg2
OH-1
2x
So how do we set up the mass action expression?
(x) (2x)2
1.5 x 10-11
x 1.6 x10 -4 Mg2 OH-1 3.2 x10 -4
Mg(OH)2 1.6 x10 -4
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More Problems
1. Bromine(I) chloride gas is formed in an
endothermic reaction. At 400C, after the
reaction reaches equilibrium, the mixture
contained 0.82M BrCl, 0.20M Br2 (g) and 0.48M
Cl2(g).
a. write the equation for this reaction
1 Cl2 (g) 1Br2 (g) q 2 BrCl (g)
b. Write the equilibrium Expression, the MAE
BrCl2
Keq
Cl2Br2
.822
7.0
c. Calculate the value for Keq
.20 .48
d. What direction is favored?
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2. Will a precipitate form when the following two
solutions are combined? (assume volumes are
additive.)
0.0025 M Pb(NO3)2 0.000036 M NaI
The Ksp for PbI2 8.4 x 10-9
Pb2 2.5 x 10-3 , I- 3.6 x
10-5 Reaction Quotient (2.5 x 10-3) (3.6 x
10-5)2 3.24 x 10-12
Since the rx. Quotient lt the Ksp , No ppt. Forms!