1. Chapter 13
Chemical Equilibrium

2. 13.1 Describing Chemical Equilibrium
Reactants  Product Reactants  Products When substances react, they eventually form a mixture of reactants and products in dynamic equilibrium
forward
reverse
For a general reversible reaction:
c C + d D
a A + b B
Double arrows show reversible reaction

3. Chapter 13: Chemical Equilibrium
The Equilibrium State
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Many reactions do not go to completion instead of reaching
Chemical Equilibrium:
The state reached when the concentrations of reactants and products remain constant over time.
The rate forward and reverse have become equal
2NO2(g)
N2O4(g)
Brown
Colorless
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.
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4. Chapter 13: Chemical Equilibrium
The Equilibrium State
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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.
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5. Chapter 13: Chemical Equilibrium
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This is a plot of how the rates of reactants and products change with time.
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6. 13.2 The Equilibrium Constant Kc
Chapter 13: Chemical Equilibrium
13.2 The Equilibrium Constant Kc
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For a homogeneous reaction
cC + dD
aA + bB
Coefficient
[A]a[B]b
[C]c[D]d
Products
Equilibrium equation:
Kc = =
Reactants
Equilibrium constant
Equilibrium constant expression
when concentrations are used
Kc is temperature dependent
“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.
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7. 13.3The Equilibrium Constant Kp
Chapter 13: Chemical Equilibrium
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13.3The Equilibrium Constant Kp
When writing an equilibrium expression for a gaseous reaction in terms of partially pressure, we call it equilibrium constant, Kp
2NO2(g)
N2O4(g)
NO2 P 2
Kp =
N2O4 P
P is the partial pressure of that component
“p” in Kp means pressure.
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8. Examples
Write the equilibrium constant, Kc and K’c for the following reactions:
a. CH4(g) + H2O(g) CO(g) H2(g)
b. 2 SO2(g) + O2(g) SO3(g)

9. Examplea Write Kp and K’p for the following reactions:
CO(g)+ 2H2(g)   CH3OH(g)

10. The Equilibrium Constant Kc
Chapter 13: Chemical Equilibrium
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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) = ´´
4NH3(g)
2N2(g) + 6H2(g)
K c = ??????
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11. Relating the Equilibrium Constant Kp and Kc
Chapter 13: Chemical Equilibrium
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Relating the Equilibrium Constant Kp and Kc
Kp = Kc(RT)Dn
K mol
L atm
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.
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12. Example Phosphorous pentachloride dissociate on heating
PCl5(g) PCl3(g) + Cl2(g)
If Kc equals 3.26 x 10-2 at 191oC, what is Kp at this temperature?

13. Example Consider the following reaction 2NO (g) + O2 (g) 2NO2 (g)
If Kp = 1.48 x 104 at 184 C, what is Kc?

14. 13.4Heterogeneous Equilibria
Chapter 13: Chemical Equilibrium
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13.4Heterogeneous 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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15. Heterogeneous Equilibria
Chapter 13: Chemical Equilibrium
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Heterogeneous Equilibria
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16. Example
In the industrial synthesis of hydrogen, mixtures of CO and H2 are enriched in H2 by allowing the CO to react with steam. The chemical equation for this so-called water-gas shift reaction is
CO(g) + H2O(g) CO2(g) + H2(g)
What is the value of Kp at 700K if the partial pressures in an equilibrium mixture at 700K are 1.31 atm of CO, 10.0 atm of H2O, 6.12 atm of CO2, and atm of H2?

17. Examples Consider the following unbalanced reaction
(NH4)2S(s) NH3(g) + H2S(g)
An equilibrium mixture of this mixture at a certain temperature was found to have [NH3] = M and [H2S] = M. What is the value of the equilibrium constant (Kc) at this temperature?

18. 13.5 Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
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13.5 Using the Equilibrium Constant
When we know the numerical value of the equilibrium constant, we can make certain judgments about the extent of the chemical reaction
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.
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19. Predicting the Direction of Reaction
Chapter 13: Chemical Equilibrium
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Predicting the Direction of Reaction
For a chemical system whether in equilibrium or not, we can calculate the value given by the law of mass action
We called it reaction quotient, Qc
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.
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20. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
Using the Equilibrium Constant
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The equilibrium between reactant A and product B:       A  B
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).
If Qc = Kc
Emphasizing the relative magnitudes of the numerator and denominator may help students understand it.
no net reaction occurs.
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21. Finding Equilibrium concentrations from Initial Concentrations
Chapter 13: Chemical Equilibrium
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Finding Equilibrium concentrations from Initial Concentrations
Steps to follow in calculating equilibrium concentrations from initial concentration
Write a balance equation for the reaction
Make an ICE (Initial, Change, Equilibrium) table, involves
The initial concentrations
The change in concentration on going to equilibrium, defined as x
The equilibrium concentration
Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x
Calculate the equilibrium concentrations form the calculated value of x
Check your answers
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22. Calculating Equilibrium Concentrations
Sometimes you must use quadratic equation to solve for x, choose the mathematical solution that makes chemical sense
Quadratic equation
ax bx + c = 0

23. Finding Equilibrium concentrations from Initial Concentrations
[products]
For Homogeneous Mixture
Kc =
[reactants]
aA(g) bB(g) cC(g)
Initial concentration (M) [A]initial [B]initial [C]initial
Change (M) ax bx cx
Equilibrium (M) [A]initial – ax [B]initial + bx [C]initial + cx

24. Finding Equilibrium concentrations from Initial Concentrations
[products]
For Heterogenous Mixture
Kc =
[reactants]
aA(g) bB(g) cC(s)
Initial concentration (M) [A]initial [B]initial ……..
Change (M) ax bx ……..
Equilibrium (M) [A]initial – ax [B]initial + bx ……….

25. Examples
The value of Kc for the reaction is 3.0 x Determine the equilibrium concentration if the initial concentration of water is M
C(s) + H2O(g) CO(g) + H2(g)

26. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
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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)
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27. Examples Consider the following reaction
I2(g) + Cl2(g) ICl(g) Kp = 81.9 (at 25oC)
A reaction mixture at 25oC initially contains PI2 = atm, PCl2 = atm, and PICl = atm. Find the equilibrium pressure of I2, Cl2 and ICl at this temperature.
In which direction does the reaction favored?

28. Example
A flask initially contained hydrogen sulfide at a pressure of 5.00 atm at 313 K. When the reaction reached equilibrium, the partial pressure of sulfur vapor was found to be 0.15 atm. What is the  Kp for the reaction?
2H2S(g) H2(g) + S2(g)

29. 13.6 Le Châtelier’s Principle
Chapter 13: Chemical Equilibrium
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13.6 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.
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30. 13.7Altering an Equilibrium Mixture: Concentration
Chapter 13: Chemical Equilibrium
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13.7Altering an Equilibrium Mixture: Concentration
2NH3(g)
N2(g) + 3H2(g)
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31. Altering an Equilibrium Mixture: Concentration
Add reactant – denominator in Qc expression becomes larger
Qc < Kc
To return to equilibrium, Qc must be increases
More product must be made => reaction shifts to the right
Remove reactant – denominator in Qc expression becomes smaller
Qc > Kc
To return to equilibrium, Qc must be decreases
Less product must be made => reaction shifts to the left

32. Altering an Equilibrium Mixture: Concentration
Chapter 13: Chemical Equilibrium
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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.
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33. Altering an Equilibrium Mixture: Concentration

34. Altering an Equilibrium Mixture: Concentration
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.
Which direction will the net reaction shift to re-establish the equilibrium?
2NH3(g)
N2(g) + 3H2(g)
at 700 K, Kc = 0.291

35. Example
The reaction of iron (III) oxide with carbon monoxide occurs in a blast furnace when iron ore is reduced to iron metal:
Fe2O3(s) CO(g) Fe(l) + 3CO2(g)
Use Le Chatellier’s principle to predict the direction of net reaction when an equilibrium mixture is disturbed by:
a. adding Fe2O3
b. Removing CO2
c. Removing CO; also account for the change using the reaction quotient Qc

36. Example Consider the following reaction at equilibrium
CO(g) Cl2(g) COCl2(g)
Predict whether the reaction will shift left, shift right, or remain unchanged upon each of the following reaction mixture
a. COCl2 is added to the reaction mixture
b. Cl2 is added to the reaction mixture
c. COCl2 is removed from the reaction mixture: also account for the change using the reaction quotient Qc

37. 13.8 Altering an Equilibrium Mixture: Changes in Pressure and Volume
Chapter 13: Chemical Equilibrium
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13.8 Altering an Equilibrium Mixture: Changes in 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. Which direction will the net reaction shift?
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38. Altering an Equilibrium Mixture: Pressure and Volume
Chapter 13: Chemical Equilibrium
4/21/2017
Altering an Equilibrium Mixture: Pressure and Volume
2NH3(g)
N2(g) + 3H2(g)
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39. Altering an Equilibrium Mixture: Pressure and Volume
Chapter 13: Chemical Equilibrium
4/21/2017
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.
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40. Altering an Equilibrium Mixture: Pressure and Volume
If reactant side has more moles of gas
Denominator will be larger
Qc < Kc
To return to equilibrium, Qc must be increased
Reaction shifts toward fewer moles of gas (to the product)
If product side has more moles of gas
Numerator will be larger
Qc > Kc
To return to equilibrium, Qc must be decreases
Reaction shifts toward fewer moles of gas (to the reactant)

41. Altering an Equilibrium Mixture: Pressure and Volume
Reaction involves no change in the number moles of gas
No effect on composition of equilibrium mixture
For heterogenous equilibrium mixture
Effect of pressure changes on solids and liquids can be ignored
Volume is nearly independent of pressure
Change in pressure due to addition of inert gas
No change in the molar concentration of reactants or products
No effect on composition

42. Example Consider the following reaction at chemical equilibrium
2 KClO3(s) KCl(s) O2(g)
a. What is the effect of decreasing the volume of the reaction mixture?
b. Increasing the volume of the reaction mixture?
c. Adding inert gas at constant volume?

43. Examples
Does the number moles of products increases, decreases or remain the same when each of the following equilibria is subjected to a increase in pressure by decreasing the volume?
PCl5(g) PCl3(g) Cl2(g)
CaO(s) CO2(g) CaCO3(s)
3 Fe(s) H2O(g) Fe3O4(s) H2(g)

44. Altering an Equilibrium Mixture: Temperature
Chapter 13: Chemical Equilibrium
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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.
Contains more reactant than product
Kc decreases with increasing temperature
the equilibrium constant for an endothermic reaction (positive DH°) increases as the temperature increases.
Contains more product than reactant
Kc increases with increasing temperature
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.
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45. 13.9 Altering an Equilibrium Mixture: Temperature
Chapter 13: Chemical Equilibrium
4/21/2017
13.9 Altering an Equilibrium Mixture: Temperature
2NH3(g)
N2(g) + 3H2(g)
DH° = kJ (exothermic )
As the temperature increases, the equilibrium shifts from products to reactants.
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.
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46. Example The following reaction is endothermic CaCO3(s) CaO(s) + CO2(g)
a. What is the effect of increasing the temperature of the reaction mixture?
b. Decreasing the temperature?

47. Examples
In the first step of Ostwald process for the synthesis of nitric acid, ammonia is oxidized to nitric oxide by the reaction:
2 NH3(g) O2(g) NO(g) H2O(l)
ΔHo = -905 kJ
How does the temperature amount of NO vary with an increases in temperature?

48. The Effect of a Catalyst on Equilibrium
Catalyst increases the rate of a chemical reaction
Provide a new, lower energy pathway
Forward and reverse reactions pass through the same transition state
Rate for forward and reverse reactions increase by the same factor
Does not affect the composition of the equilibrium mixture
Does not appear in the balance chemical equation
Can influence choice of optimum condition for a reaction

49. The Effect of a Catalyst on Equilibrium
Chapter 13: Chemical Equilibrium
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The Effect of a Catalyst on Equilibrium
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