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Transcript of Chapter 15 Chemical Equilibrium. Copyright McGraw-Hill 2009 Double arrows () denote an equilibrium...
![Page 1: Chapter 15 Chemical Equilibrium. Copyright McGraw-Hill 2009 Double arrows () denote an equilibrium reaction. 15.1 The Concept of Equilibrium Most chemical.](https://reader031.fdocuments.us/reader031/viewer/2022012320/56649cff5503460f949cfe5a/html5/thumbnails/1.jpg)
Chapter 15
Chemical Equilibrium
![Page 2: Chapter 15 Chemical Equilibrium. Copyright McGraw-Hill 2009 Double arrows () denote an equilibrium reaction. 15.1 The Concept of Equilibrium Most chemical.](https://reader031.fdocuments.us/reader031/viewer/2022012320/56649cff5503460f949cfe5a/html5/thumbnails/2.jpg)
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Double arrows ( ) denote an equilibrium reaction.
15.1 The Concept of EquilibriumMost chemical reactions are reversible.
reversible reaction = a reaction that proceeds simultaneously in both directions
Examples:
)(NO 2 )(ON 242 gg
)(NH 2 )(H 3 )(N 322 ggg
)OH(CH )(H 2 )CO( 32 ggg
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Equilibrium)(NO 2 )(ON 242 gg 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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EquilibriumLet’s use 2 experiments to study the reaction
each starting with a different reactant(s).
)(NO 2 )(ON 242 gg
Exp #2
pure NO2
Exp #1
pure N2O4
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EquilibriumExperiment #1 )(NO 2 )(ON 242 gg
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EquilibriumExperiment #2 )(NO 2 )(ON 242 gg
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Equilibrium)(NO 2 )(ON 242 gg
Are the equilibrium pressures of NO2 and N2O4
related? Are they predictable?
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15.2 The Equilibrium Constant
rate forward ratereverse
At equilibrium,
2f 2 4 eq r 2 eq[N O ] [NO ]k k
)(NO 2 )(ON 242 gg
or2
2 eqfc
r 2 4 eq
[NO ]
[N O ]
kK
k
where Kc is the equilibrium constant
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The Equilibrium Constant
This constant value is termed the equilibrium constant, Kc, for this reaction at 25°C.
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The Equilibrium ConstantFor the NO2 / N2O4 system:
equilibrium constant expression
equilibrium constant
2
4
[NO ]
O ]K
2
2
0.143 at 25 C[N
)(NO 2 )(ON 242 gg
Note: at 100°C, K = 6.45
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The Equilibrium Constantreaction quotient = Qc = the value of the
“equilibrium constant expression” under any conditions.
)(COCl )(Cl )CO( 22 ggg For,
2 eqc
eq 2 eq
[COCl ]
[CO] [Cl ]K 2
c2
[COCl ]
[CO][Cl ]Q
Q > K reverse reaction favored
Q = K equilibrium present
Q < K forward reaction favored
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The Equilibrium Constant
The Law of Mass Action:Cato Maximilian Guldberg & Peter Waage, Forhandlinger: Videnskabs-Selskabet i Christiana 1864, 35.
K c d
c a b
[C] [D]
[A] [B]
For a reaction: D C B A dcba
C D
A BP
P PK
P P
c d
a b For gases:
For solutions: [ ] = mol/L
P in atm
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The Equilibrium ConstantNote:
• 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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15.3 Equilibrium Expressionshomogeneous equilibria = equilibria in which all
reactants and products are in the same phase.
• [CaO] and [CaCO3] are solids.
• Pure solids and liquids are omitted from equilibrium constant expressions.
)(CO )CaO( )(CaCO 23 gss Ex:
The equilibrium constant expression is,
K = [CO2]
heterogeneous equilibria = equilibria in which all reactants and products are not in the same phase.
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Exercise: Write the expressions for Kp for the
following reactions:)O(H 2 )O(N )(NONH (a) 2234 ggs
)(CuCl )(Cl )Cu( (b) 22 sgs
Solution:
2 2
2
N O H OPK P P (a)
2Cl
1PK P
(b)
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Equilibrium ExpressionsA. Reverse Equations
[1] )(NO 2 )(ON 242 gg For,
Conclusion:
C 1 6.99 at 25
0.143
C
PK
P 2
2 4
NO1
N O
2
0.143 at 25
[2] )(ON )(NO 2 422 gg For,
2 4
2
N O
NO
PK
P2 2
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Equilibrium ExpressionsB. Coefficient Changes
[1] )(NO 2 )(ON 242 gg For,
Conclusion:
C 0.143 0.378 at 25
For, [3] )(NO )(ON 21
242 gg
2
2 4
NO1/2
N O
PK
P3
C
PK
P 2
2 4
NO1
N O
2
0.143 at 25
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Equilibrium ExpressionsC. Reaction Sum (related to Hess’ Law)
[1] )(NO 2 )(ON 242 gg For,
For, [4] )(O )NO( 2 )(NO 2 22 ggg
2
2
NO O
NO
P PK
P
2
4 2
Add [1] + [4],
[5] )(O )NO( 2 )(ON 242 ggg
2
2 4
NO O
N O
P PK
P
2
5
42
2
ON
NO1 P
P K
2
K K 1 4
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Equilibrium Expressions
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Exercise: At 500ºC, KP = 2.5 1010 for,
)(SO 2 )(O )(SO 2 322 ggg
Compute KP for each of the following:
(a) At 500ºC, which is more stable, SO2 or SO3?
(g)O 21
(g)SO (g)SO (d) 223
(g)SO (g)O (g)SO (b) 322
21
(g)SO 3 (g)O (g)SO 3 (c) 322 23
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15.4 Using Equilibrium Expressions to Solve Problems
Q > K reverse reaction favored
Q = K equilibrium present
Q < K forward reaction favored
Predicting the direction of a reaction
Compare the computed value of Q to K
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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:
2 2
2HI
H I
P
QP P
0.47 )7.1()1.2(
)3.1(
2
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 = 2.59 102 for
At equilibrium, PH2S = 0.557 atm and PH2 = 0.173 atm,
calculate PS2 at 1130°C.
Solution:
2 2
2
2
H S 22
H S
2.59 10P P
KP
PS2 = 0.268 atm
2
2S 2
2
(0.173) 2.59 10
(0.557)
P
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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 2.00 atm 2.00 atm 0.00 atmChange x x +2xEquilibrium (2.00 – x) (2.00 – x) 2x
2 2
2ICl
I Cl
P
KP P
perfect square
2(2 ) 82.2
(2.00 )(2.00 )xx x
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(2 ) 9.066
(2.00 )xx
square root
2 x = 18.132 – 9.066 x
11.066 x = 18.132
x = 18.132 / 11.066 = 1.639
PI2 = PCl2
= 2.00 – x = 2.00 – 1.639 = 0.36 atm
PICl = 2x = (2)(1.639) = 3.28 atm
2(2 ) 82.2
(2.00 )(2.00 )xx x
Exercise #3: (cont.)
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Exercise #4: At 1280°C, Kc = 1.1 103 for
Initially, [Br2] = 6.3 102 M and [Br] = 1.2 102
M. What are the equilibrium concentrations of Br2
and Br at 1280°C?
Initial 6.3 102 M 1.2 102 MChange -x +2x
Equilibrium (6.3 102) - x (1.2 102) + 2x
Solution:
2 2 23
c 22
[Br] [(1.2 10 ) 2 ] 1.1 10
[Br ] (6.3 10 )
xK
x
4 x2 + 0.0491x + (7.47 105) = 0
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4 x2 + 0.0491x + (7.47 10-5) = 0
quadratic equation: a x2 + b x + c = 02 4
2
b b acx
a solution:
x = 1.779 103 and 1.050 102 Q: Two answers? Both negative? What’s happening?
Equilibrium Conc. x = 1.779 103 1.050
102
[Br2] = (6.3 102) – x = 0.0648 M 0.0735
M
[Br] = (1.2 102) + 2x = 0.00844 M 0.00900 M
impossible[Br2] = 6.5 102 M
[Br] = 8.4 103 M
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Exercise #5: A pure NO2 sample reacts at 1000 K,
KP is 158. If at 1000 K the equilibrium partial
pressure of O2 is 0.25 atm, what are the equilibrium
partial pressures of NO and NO2.
)(O )NO( 2 )(NO 2 22 ggg
Solution:
Initial ? 0 atm0 atm
Change Equilibrium
0.25 atm
+0.25+0.50+0.50 atm
0.50
2
2 2
22
NO O
2 2
NO NO
(0.50) (0.25) 158P
P PK
P P
rearrange and solve
PNO2
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2
2
NO
(0.50) (0.25) 158
P
2
2P
2
2
NO(0.50) (0.25)
158
P 2
4NO 3.956 10 0.01989
Exercise #5: (cont.)
= 3.956 104
PNO2= 0.020 atm
PNO = 0.50 atmsee 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 = 0.143 at 25°C. Calculate the equilibrium
partial pressures of N2O4 and NO2.
)(NO 2 )(ON 242 gg
2
2 4
2
NO
N O
0.143P
PK
P
PNO2 + PN2O4
= 1.30 atm
two equations and two unknowns – BINGO!
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PNO22 + 0.143 PNO2
0.1859 = 0
PN2O4 = 1.30 atm - PNO2
2
2
2
NO
NO
0.143(1.30 )
P
P
2
2 4
2
NO
N O
0.143P
PK
P PNO2
+ PN2O4 = 1.30 atm
Exercise #6: (cont.)
Use the quadratic formula,
PNO2 = +0.366 atm and 0.509 atm
PN2O4 = 1.30 atm - PNO2
= 1.30 0.366 = 0.934 atm
PN2O4 = 0.93 atm
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15.5 Factors That Affect Chemical Equilibrium
“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.
(continued on next 5 slides)
Le Châtelier’s Principle
kJ 92 H )(NH 2 )(H 3 )(N rxn322
ggg
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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]?
)(NH 2 )(H 3 )(N 322 ggg
N2 decreases
N2 increases
right
left
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Case II: Change: compress the system
Shift: ???to the right
Q: Why?
Ans: Total pressure has increased. Which direction will decrease the total pressure? Recall: P n
)(NH 2 )(H 3 )(N 322 ggg
N2 H2
NH3
(4 moles gas) (2 moles gas)
less gasless pressure
more gasmore 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.
Δ 92 kJH rxn
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
)(NH 2 )(H 3 )(N 322 ggg (4 moles gas) (2 moles gas)
less gasless pressure
more gasmore pressure
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Exercise: Hydrogen (used in ammonia production) is produced by the endothermic reaction,
)(H3 )(CO )(OH )(CH 224 gggg 750C
Ni
Assuming the reaction is initially at equilibrium, indicate the direction of the shift (L, R, none) if
(a) H2O(g) is removed.
(b) The temperature is increased.
(c) The quantity of Ni catalyst is increased.
(d) An inert gas (e.g., He) is added.
(e) H2(g) is removed.
(f) The volume of the container is tripled.
Left
Right
None
None
Right
Right