1. Chapter 13: Chemical Equilibrium
7/20/2018
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

2. Chapter 13: Chemical Equilibrium
7/20/2018
The Equilibrium State
Chemical Equilibrium: The state reached when the concentrations of reactants and products remain constant over time.
2NO2(g)
N2O4(g)
Colorless
Brown
Equilibrium has been mentioned a few times prior to this chapter. Forward and reverse rates for reaction mechanisms in Ch 12 (Chemical Kinetics), evaporation of liquids in Ch 10 (Liquids, Solids, and Phase Changes), and dissociation of weak acids in Ch 4 (Reactions in Aqueous Solution) to name three.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

3. Chapter 13: Chemical Equilibrium
7/20/2018
The Equilibrium State
2NO2(g)
N2O4(g)
At 25 °C.
Chemical Equilibrium is dynamic. It’s not that all chemical reactions stop but that the rates of change become constant.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

4. Chapter 13: Chemical Equilibrium
7/20/2018
This is a plot of how the rates of reactants and products change with time.
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5. The Equilibrium Constant Kc
Chapter 13: Chemical Equilibrium
7/20/2018
The Equilibrium Constant Kc
Why?
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

6. The Equilibrium Constant Kc
Chapter 13: Chemical Equilibrium
7/20/2018
The Equilibrium Constant Kc
For a general reversible reaction:
cC + dD
aA + bB
[A]a[B]b
[C]c[D]d
Products
Equilibrium equation:
Kc =
Reactants
Equilibrium constant
Equilibrium constant expression
“c” in Kc is for concentration.
[A] still means molarity of A.
No units for the equilibrium constant. Since the units would change depending on the coefficients of the reactants and products, they are left off.
For the following reaction:
2NO2(g)
N2O4(g)
[N2O4]
[NO2]2
Kc =
= 4.64 x 10-3 (at 25 °C)
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Copyright © 2008 Pearson Prentice Hall, Inc.

7. The Equilibrium Constant Kc
Chapter 13: Chemical Equilibrium
7/20/2018
The Equilibrium Constant Kc
Experiment 1
Experiment 5
[N2O4]
[NO2]2
0.0337
(0.0125)2
0.0429
(0.0141)2
Kc =
= 4.64 x 10-3
= 4.63 x 10-3
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8. The Equilibrium Constant Kc
Chapter 13: Chemical Equilibrium
7/20/2018
The Equilibrium Constant Kc
The equilibrium constant and the equilibrium constant expression are for the chemical equation as written.
[N2][H2]3
[NH3]2
2NH3(g)
N2(g) + 3H2(g)
Kc =
[NH3]2
[N2][H2]3 1 Kc
Kc =
N2(g) + 3H2(g)
2NH3(g) =
[N2]2[H2]6
[NH3]4
Kc = ´´
4NH3(g)
2N2(g) + 6H2(g)
= Kc 2
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Copyright © 2008 Pearson Prentice Hall, Inc.

9. The Equilibrium Constant Kp
Chapter 13: Chemical Equilibrium
7/20/2018
The Equilibrium Constant Kp
2NO2(g)
N2O4(g)
NO2 P 2
Kp =
P is the partial pressure of that component
N2O4 P
“p” in Kp means pressure.
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Copyright © 2008 Pearson Prentice Hall, Inc.

10. The Equilibrium Constant Kp
Chapter 13: Chemical Equilibrium
7/20/2018
The Equilibrium Constant Kp
K mol
L atm
Kp = Kc(RT)Dn
R is the gas constant,
T is the absolute temperature (Kelvin). n
is the number of moles of gaseous products minus the number of moles of gaseous reactants.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

11. Heterogeneous Equilibria
Chapter 13: Chemical Equilibrium
7/20/2018
Heterogeneous Equilibria
CaO(s) + CO2(g)
CaCO3(s)
Limestone
Lime
[CaCO3]
[CaO][CO2]
(1)
(1)[CO2]
Kc = =
= [CO2]
Pure solids and pure liquids are not included.
Heterogeneous refers to the reactants present in more than one phase.
This reaction is the thermal decomposition of solid calcium carbonate (manufacturing of cement).
Refer to section 13.4 for the mathematical reason for the simplification of the equilibrium expression.
Kc = [CO2]
Kp = P
CO2
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Copyright © 2008 Pearson Prentice Hall, Inc.

12. Heterogeneous Equilibria
Chapter 13: Chemical Equilibrium
7/20/2018
Heterogeneous Equilibria
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13. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
7/20/2018
Using the Equilibrium Constant
2H2(g) + O2(g)
2H2O(g)
(at 500 K)
Kc = 4.2 x 10-48
A very large value for K means that the numerator is much larger than the denominator.
A very small value for K means that the denominator is much larger than the numerator.
2HI(g)
H2(g) + I2(g)
2H2O(g)
2H2(g) + O2(g)
(at 500 K)
Kc = 57.0
(at 500 K)
Kc = 2.4 x 1047
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Copyright © 2008 Pearson Prentice Hall, Inc.

14. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
7/20/2018
Using the Equilibrium Constant
cC + dD
aA + bB
[A]ta[B]tb
[C]tc[D]td
Reaction quotient:
Qc =
The reaction quotient, Qc, is defined in the same way as the equilibrium constant, Kc, except that the concentrations in Qc are not necessarily equilibrium values.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

15. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
7/20/2018
Using the Equilibrium Constant
If Qc < Kc
net reaction goes from left to right (reactants to products).
If Qc > Kc
net reaction goes from right to left (products to reactants).
Emphasizing the relative magnitudes of the numerator and denominator may help students understand it.
If Qc = Kc
no net reaction occurs.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

16. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
7/20/2018
Using the Equilibrium Constant
At 700 K, mol of HI is added to a 2.00 L container and allowed to come to equilibrium. Calculate the equilibrium concentrations of H2, I2, and HI . Kc is 57.0 at 700 K.
2HI(g)
H2(g) + I2(g)
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

17. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
7/20/2018
Using the Equilibrium Constant
Copyright © 2008 Pearson Prentice Hall, Inc.

18. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
7/20/2018
Using the Equilibrium Constant
Set up a table:
2HI(g)
H2(g) + I2(g) I
0.250 C +x
-2x E x x
I = initial concentrations.
C = change in concentrations (use “x” for the change along with the stoichiometric coefficients).
E = equilibrium concentrations.
Substitute values into the equilibrium expression:
Kc =
[H2][I2]
[HI]2 x2
( x)2
57.0 =
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Copyright © 2008 Pearson Prentice Hall, Inc.

19. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
7/20/2018
Using the Equilibrium Constant
Solve for “x”: x2
( x)2
57.0 =
x =
Determine the equilibrium concentrations:
H2: M
There are a number of different problems that can be worked in addition to this one. Using the quadratic formula, successive approximations, etc.
A way to check the equilibrium concentrations is to plug them back into the equilibrium expression and see if you get the equilibrium constant (approximately).
I2: M
HI: (0.0262) = M
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Copyright © 2008 Pearson Prentice Hall, Inc.

20. Chapter 13: Chemical Equilibrium
7/20/2018
Copyright © 2008 Pearson Prentice Hall, Inc.

21. Le Châtelier’s Principle
Chapter 13: Chemical Equilibrium
7/20/2018
Le Châtelier’s Principle
Le Châtelier’s Principle: If a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves the stress.
The concentration of reactants or products can be changed.
The pressure and volume can be changed.
The temperature can be changed.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

22. Altering an Equilibrium Mixture: Concentration
Chapter 13: Chemical Equilibrium
7/20/2018
Altering an Equilibrium Mixture: Concentration
2NH3(g)
N2(g) + 3H2(g)
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23. Altering an Equilibrium Mixture: Concentration
Chapter 13: Chemical Equilibrium
7/20/2018
Altering an Equilibrium Mixture: Concentration
In general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Châtelier’s principle predicts that
the concentration stress of an added reactant or product is relieved by net reaction in the direction that consumes the added substance.
the concentration stress of a removed reactant or product is relieved by net reaction in the direction that replenishes the removed substance.
The iron(III)-thiocyanate complex equilibrium is a nice demonstration since it has nice color changes.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

24. Altering an Equilibrium Mixture: Concentration
Chapter 13: Chemical Equilibrium
7/20/2018
Altering an Equilibrium Mixture: Concentration
2NH3(g)
N2(g) + 3H2(g)
at 700 K, Kc = 0.291
An equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 is disturbed by increasing the N2 concentration to 1.50 M.
[N2][H2]3
[NH3]2
(1.50)(3.00)3
(1.98)2
Qc = =
= < Kc
Since Qc < Kc, more reactants will be consumed and the net reaction will be from left to right.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

25. Altering an Equilibrium Mixture: Pressure and Volume
Chapter 13: Chemical Equilibrium
7/20/2018
Altering an Equilibrium Mixture: Pressure and Volume
2NH3(g)
N2(g) + 3H2(g)
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

26. Altering an Equilibrium Mixture: Pressure and Volume
Chapter 13: Chemical Equilibrium
7/20/2018
Altering an Equilibrium Mixture: Pressure and Volume
In general, when an equilibrium is disturbed by a change in volume which results in a corresponding change in pressure, Le Châtelier’s principle predicts that
an increase in pressure by reducing the volume will bring about net reaction in the direction that decreases the number of moles of gas.
a decrease in pressure by enlarging the volume will bring about net reaction in the direction that increases the number of moles of gas.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

27. Altering an Equilibrium Mixture: Pressure and Volume
Chapter 13: Chemical Equilibrium
7/20/2018
Altering an Equilibrium Mixture: Pressure and Volume
2NH3(g)
N2(g) + 3H2(g)
at 700 K, Kc = 0.291
An equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 is disturbed by reducing the volume by a factor of 2.
[N2][H2]3
[NH3]2
(1.00)(6.00)3
(3.96)2
Qc = =
= < Kc
All we’re doing is decreasing the volume. Since the number of moles is not instantaneously changing, the concentration doubles and this is the stress.
Since Qc < Kc, more reactants will be consumed and the net reaction will be from left to right.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

28. Altering an Equilibrium Mixture: Temperature
Chapter 13: Chemical Equilibrium
7/20/2018
Altering an Equilibrium Mixture: Temperature
2NH3(g)
N2(g) + 3H2(g)
DH° = kJ
Note the size of Kc as the temperature increases. Mathematically, the magnitude of the denominator increases with respect to the numerator as the temperature increases.
As the temperature increases, the equilibrium shifts from products to reactants.
Copyright © 2008 Pearson Prentice Hall, Inc.

29. Altering an Equilibrium Mixture: Temperature
Chapter 13: Chemical Equilibrium
7/20/2018
Altering an Equilibrium Mixture: Temperature
In general, when an equilibrium is disturbed by a change in temperature, Le Châtelier’s principle predicts that
the equilibrium constant for an exothermic reaction (negative DH°) decreases as the temperature increases.
the equilibrium constant for an endothermic reaction (positive DH°) increases as the temperature increases.
A aqueous solution of cobalt(II) nitrate in a solution of hydrochloric acid will change colors from blue to pink as the temperature changes (it’s a reversible endothermic reaction) and makes for a nice demonstration.
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

30. The Effect of a Catalyst on Equilibrium
Chapter 13: Chemical Equilibrium
7/20/2018
The Effect of a Catalyst on Equilibrium
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