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Chapter 16 — Chemical Equilibrium - 4
Nitrogen Dioxide Equilibrium
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Today’s Topics
• N2O4 ' NO2 equilibrium
N2O4(g) ' 2 NO2(g)
(colorless)
(brown)
P= (n/V).RT
• Le Chatelier’s Principle
© 2006 Brooks/Cole - Thomson
Writing and Manipulating K Expressions
Writing and Manipulating K Expressions
K using concentration and pressure units
Concentration Units
We have been writing K in terms of mol/L.
Kp = Kc (RT)∆n
These are designated by Kc
where ∆n is the change in the number of moles of gas during the reaction
But with gases, P = (n/V)•RT = [conc]•RT
For
P is proportional to concentration, so we can write K in terms of P.
These are designated by Kp.
For H2(g) + I2(g) ' 2 HI(g)
Kp =
S(s) + O2(g) ' SO2(g)
∆n = 0 and Kp = Kc
For SO2(g) + 1/2 O2(g) ' SO3(g)
P 2HI
PH 2 PI 2
∆n = –1/2 and Kp = Kc(RT)–1/2
Kc and Kp may or may not be the same.
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Nitrogen Dioxide Equilibrium
N2O4(g) ' 2 NO2(g)
aA + bB
cC + dD
Q
'
'
K
K
Q
K
Q
Kc =
[NO2 ]2
= 0.0059 at 298 K
[N2O4 ]
Reaction forms
products
Increase P in the system by reducing
the volume (at constant T).
© 2006 Brooks/Cole - Thomson
© 2006 Brooks/Cole - Thomson
Equilibrium
Reaction forms
reactants
2
Chapter 16 — Chemical Equilibrium - 4
7
Nitrogen Dioxide Equilibrium
N2O4(g) ' 2 NO2(g)
Kc =
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Nitrogen Dioxide Equilibrium
N2O4(g) ' 2 NO2(g)
[NO2 ]2
= 0.0059 at 298 K
[N2O4 ]
N2O4 + heat
'
(colorless)
Increase P in the system by reducing the volume.
In gaseous system the equilibrium will shift to the side
with fewer molecules (in order to reduce the P).
Kc =
Therefore, reaction shifts LEFT and P of NO2 decreases
and P of N2O4 increases.
2 NO2
∆Ho = + 57.2 kJ
(brown)
[NO2 ]2
[N2O 4 ]
Kc (273 K) = 0.00077
Kc (298 K) = 0.0059
© 2006 Brooks/Cole - Thomson
© 2006 Brooks/Cole - Thomson
Nitrogen Dioxide Equilibrium
N2O4(g) ' 2 NO2(g)
• ∆Ho = + 57.2 kJ (endothermic)
• Increase T. What happens to equilibrium
position and the value of K?
K increases as T goes up.
• Decrease T. Now what?
Equilibrium shifts left and K decreases.
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EQUILIBRIUM AND EXTERNAL EFFECTS
• Temperature and concentration
changes affect equilibria.
• The outcome is governed by
LE CHATELIER’S PRINCIPLE
• “...if a system at equilibrium is
disturbed, the system tends to shift its
equilibrium position to counter the
effect of the disturbance.”
© 2006 Brooks/Cole - Thomson
Le Chatelier’s Principle
• Change T
Le Chatelier’s Principle
“reactant”
“product”
Changes the value of K
This causes a change in equilibrium concentrations
• Add or take away reactant or product:
K does not change
Reaction adjusts to new equilibrium “position”
• Use a catalyst:
K does not change. A catalyst does not affect equilibrium.
Modify the kinetics of the reaction.
Adding a “reactant” to a chemical system.
reactants ' products
Blue line= initial state
Red line = new state
Water level= eq. state
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Chapter 16 — Chemical Equilibrium - 4
Le Chatelier’s Principle
“reactant”
Le Chatelier’s Principle
“product”
“reactant”
“product”
Removing a “reactant”
reactant” from a chemical system.
Adding a “product”
product” to a chemical system.
reactants ' products
reactants ' products
Blue line= initial state
Red line = new state
Water level= eq. state
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Le Chatelier’s Principle
“reactant”
Blue line= initial state
Red line = new state
Water level= eq. state
EQUILIBRIUM AND EXTERNAL EFFECTS
“product”
• Concentration changes:
–no change in K
–only the equilibrium
composition changes.
Removing a “product”
product” from a chemical system.
reactants ' products
Blue line= initial state
Red line = new state
Water level= eq. state
© 2006 Brooks/Cole - Thomson
CATALYSIS
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EQUILIBRIUM AND EXTERNAL EFFECTS
In auto exhaust systems — Pt, NiO
2 CO + O2 --->
---> 2 CO2
2 NO --->
---> N2 + O2
Catalytic exhaust system
• Add catalyst --->
---> no change in
• A catalyst only affects the RATE of
approach to equilibrium.
© 2006 Brooks/Cole - Thomson
© 2006 Brooks/Cole - Thomson
K