1. Chapter 15 Chemical Equilibrium

2. 15.1 The Concept of Equilibrium
Most chemical reactions are reversible.
reversible reaction = a reaction that proceeds simultaneously in both directions
Examples:
Double arrows ( ) denote an equilibrium reaction.
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Equilibrium
Consider the reaction
At equilibrium,
the forward reaction: N2O4(g)  2 NO2(g), and
the reverse reaction: 2 NO2(g)  N2O4(g)
proceed at equal rates.
Chemical equilibria are dynamic, not static – the reactions do not stop.
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Equilibrium
Let’s use 2 experiments to study the reaction
each starting with a different reactant(s).
Exp #1
pure N2O4
Exp #2
pure NO2
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Equilibrium
Experiment #1
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Equilibrium
Experiment #2
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Equilibrium
Are the equilibrium pressures of NO2 and N2O4 related? Are they predictable?
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8. 15.2 The Equilibrium Constant
At equilibrium, or
where Kc is the equilibrium constant
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9. The Equilibrium Constant
This constant value is termed the equilibrium constant, Kc, for this reaction at 25°C.
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10. The Equilibrium Constant
For the NO2 / N2O4 system:
equilibrium constant expression
equilibrium constant
Note: at 100°C, K = 6.45
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11. The Equilibrium Constant
reaction quotient = Qc = the value of the “equilibrium constant expression” under any conditions.
For,
Q > K  reverse reaction favored
Q = K  equilibrium present
Q < K  forward reaction favored
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12. The Equilibrium Constant
For a reaction:
For gases:
P in atm
For solutions:
[ ] = mol/L
The Law of Mass Action:
Cato Maximilian Guldberg & Peter Waage, Forhandlinger: Videnskabs-Selskabet i Christiana 1864, 35.
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13. The Equilibrium Constant
Note:
The equilibrium constant expression has products in the numerator, reactants in the denominator.
Reaction coefficients become exponents.
Equilibrium constants are temperature dependent.
Equilibrium constants do not have units. (pg. 622)
If K >>> 1, products favored (reaction goes nearly to completion).
If K <<< 1, reactants favored (reaction hardly proceeds).
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14. 15.3 Equilibrium Expressions
homogeneous equilibria = equilibria in which all reactants and products are in the same phase.
heterogeneous equilibria = equilibria in which all reactants and products are not in the same phase.
Ex:
The equilibrium constant expression is,
K = [CO2]
[CaO] and [CaCO3] are solids.
Pure solids and liquids are omitted from equilibrium constant expressions.
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Exercise: Write the expressions for Kp for the following reactions:
Solution:
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16. Equilibrium Expressions
A. Reverse Equations
For,
For,
Conclusion:
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17. Equilibrium Expressions
B. Coefficient Changes
For,
For,
Conclusion:
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18. Equilibrium Expressions
C. Reaction Sum (related to Hess’ Law)
For,
For,
Add [1] + [4],
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19. Equilibrium Expressions
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Exercise: At 500ºC, KP = 2.5  for,
(a) At 500ºC, which is more stable, SO2 or SO3?
Compute KP for each of the following: 1
(b) SO
(g) + O
(g) SO
(g) 2 2 3 2 3
(c) 3 SO
(g) + O
(g) 3 SO
(g) 2 2 3 2 1
(d) SO
(g) SO
(g) + O
(g) 3 2 2 2
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21. 15.4 Using Equilibrium Expressions to Solve Problems
Predicting the direction of a reaction
Compare the computed value of Q to K
Q > K  reverse reaction favored
Q = K  equilibrium present
Q < K  forward reaction favored
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Exercise #1: At 448°C, K = 51 for the reaction,
Predict the direction the reaction will proceed, if at 448°C the pressures of HI, H2, and I2 are 1.3, 2.1 and 1.7 atm, respectively.
Solution:
0.47 < 51  system not at equilibrium
Numerator must increase and denominator must decrease.
Consequently the reaction must shift to the right.
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Exercise #2: At 1130°C, K = 2.59  102 for
At equilibrium, PH2S = atm and PH2 = atm, calculate PS2 at 1130°C.
Solution:
PS2 = atm
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Exercise #3: K = 82.2 at 25°C for,
Initially, PI2 = PCl2 = 2.00 atm and PICl = 0.00 atm.
What are the equilibrium pressures of I2, Cl2, and ICl?
Solution:
Initial atm atm 0.00 atm
Change x  x +2x
Equilibrium (2.00 – x) (2.00 – x) 2x
perfect square
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Exercise #3: (cont.)
square root 
2 x = – x
x =
x = / =
PI2 = PCl2 = – x = – = atm
PICl = 2x = (2)(1.639) = atm
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Exercise #4: At 1280°C, Kc = 1.1  103 for
Initially, [Br2] = 6.3  102 M and [Br] = 1.2  102 M. What are the equilibrium concentrations of Br2 and Br at 1280°C?
Solution:
Initial 6.3  102 M 1.2  102 M
Change x x
Equilibrium (6.3  102) - x (1.2  102) + 2x
4 x x + (7.47  105) = 0
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4 x x + (7.47  10-5) = 0
quadratic equation: a x2 + b x + c = 0
solution:
x =   103 and   102
Q: Two answers? Both negative? What’s happening?
Equilibrium Conc. x =   103   102
[Br2] = (6.3  102) – x = M M
[Br] = (1.2  102) + 2x = M  M
[Br2] = 6.5  102 M
[Br] = 8.4  103 M
impossible
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Exercise #5: A pure NO2 sample reacts at 1000 K,
KP is If at 1000 K the equilibrium partial pressure of O2 is 0.25 atm, what are the equilibrium partial pressures of NO and NO2.
Solution:
Initial ? 0 atm 0 atm
Change
Equilibrium atm
0.50
+0.50
+0.25
PNO2
+0.50 atm
rearrange and solve
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Exercise #5: (cont.)
=  104
PNO2 = atm
PNO = atm
see ICE table
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Exercise #6: The total pressure of an equilibrium mixture of N2O4 and NO2 at 25°C is 1.30 atm. For the reaction:
KP = at 25°C. Calculate the equilibrium partial pressures of N2O4 and NO2.
two equations and two unknowns – BINGO!
PNO2 + PN2O4 = atm
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PNO2 + PN2O4 = atm
Exercise #6: (cont.)
PN2O4 = atm - PNO2
PNO PNO2  = 0
Use the quadratic formula,
PNO2 = atm and 0.509 atm
PN2O4 = atm - PNO2 =  = atm
PN2O4 = atm
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32. 15.5 Factors That Affect Chemical Equilibrium
Le Châtelier’s Principle
“If an equilibrium system variable is changed, the equilibrium will shift in the direction (right or left) that tends to reduce the change.”
Example: N2, H2, and NH3 are at equilibrium in a container at 500°C. kJ
92 H ) ( NH 2 3 N
rxn = D + o g
(continued on next 5 slides)
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Case I : Change: N2 is added
Shift: ???
to the right
Q: Why?
Ans: [N2] has increased. Which direction will decrease [N2]?
N2 decreases
N2 increases
right
left
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Case II: Change: compress the system
Shift: ???
to the right
N2 H2
NH3
Q: Why?
Ans: Total pressure has increased. Which direction will decrease the total pressure? Recall: P  n
(4 moles gas) (2 moles gas)
less gas
less pressure
more gas
more pressure
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Case III: Change: increase the temperature
Shift: ???
to the left
Q: Why?
Ans: Temperature has increased. Which direction decreases the temperature?
Recall, the reaction is exothermic.
endothermic
heat absorbed
right
left
exothermic
heat evolved
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Case IV: Change: add helium at constant volume
Shift: ???
none
Q: Why?
Ans: Helium is not a reactant or product. Adding helium (at constant V) does not change PN2, PH2 or PNH3. Hence the equilibrium will not shift.
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Case V: Change: add helium at constant total pressure
Shift: ???
to the left
Q: Why?
Ans: If the total pressure is constant, PN2 + PH2 + PNH3 must decrease. Which direction increases this sum?
Recall: P  n
(4 moles gas) (2 moles gas)
less gas
less pressure
more gas
more pressure
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Exercise: Hydrogen (used in ammonia production) is produced by the endothermic reaction, Ni
750C
Assuming the reaction is initially at equilibrium, indicate the direction of the shift (L, R, none) if
H2O(g) is removed.
The temperature is increased.
The quantity of Ni catalyst is increased.
An inert gas (e.g., He) is added.
H2(g) is removed.
The volume of the container is tripled.
Left
Right
None
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