1. Chapter 6
Chemical Equilibrium

2. Chemical Equilibrium 1. Reaction Rates 2. The Equilibrium Condition
3. The Equilibrium Constant
4. Heterogeneous Equilibria
5. Equilibrium Expressions Involving Pressures
6. Applications of the Equilibrium Constant
7. Solving Equilibrium Problems
8. Le Chatelier’s Principle

3. Reaction Rates Some chemical reactions proceed rapidly.
The precipitation reactions , where the products form practically the instant the two solutions are mixed.
Other reactions proceed slowly.
Like the decomposition of dye molecules of a sofa placed in front of a window.
The rate of a reaction is measured in the amount of reactant that changes into product in a given period of time.
Chemists study ways of controlling reaction rates.

4. 2 N2O5 (g) → 4 NO2(g) + O2(g) Over time, the concentrations
of reactants decrease
as products increase.

5. 2 N2O5 (g) → 4 NO2(g) + O2(g): Rate vs. Time
Because reactant concentrations
decrease, the rates of reactions
slow down over time.

6. Factors Effecting Reaction Rate: Reactant Concentration
The higher the concentration of reactant molecules, the faster the reaction will generally go.
Increases the frequency of reactant molecule collisions.
Since reactants are consumed as the reaction proceeds, the speed of a reaction generally slows over time.

7. Effect of Concentration on Rate
Low concentrations of reactant molecules leads to fewer effective collisions, therefore a slower reaction rate.
High concentrations of reactant molecules lead to more effective collisions, therefore a faster reaction rate.

8. Factors Effecting Reaction Rate: Temperature
Increasing the temperature increases the number of molecules in the sample with enough energy so that their collisions can overcome the activation energy.
Increasing the temperature also increases the frequency of collisions.
So the rate increases because the frequency of effective collisions increases.
Both these mean that increasing temperature increases the reaction rate.

9. Reaction Dynamics, Continued
The forward reaction slows down as the amounts of reactants decreases.
At the same time, the reverse reaction speeds up as the concentration of the products increases.
Eventually, the forward reaction is using reactants and making products as fast as the reverse reaction is using products and making reactants. This is called chemical equilibrium.
Dynamic equilibrium is reached when the rates of two opposite processes are the same.

10. Effect of Temperature on Rate
Low temperatures lead to fewer molecules with enough energy to overcome the activation energy, and less frequent reactant collisions, therefore a slower reaction rate
High temperatures lead to more molecules with enough energy to overcome the activation energy, and more frequent reactant collisions, therefore, a faster reaction rate.

11. Chapter 13: Chemical Equilibrium
4/14/2017
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. 7 (Reactions in Aqueous Solution), to name three.
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12. Chemical Equilibrium
When a reaction reaches equilibrium, the amounts of reactants and products in the system stay constant.
The forward and reverse reactions still continue.
Because they go at the same rate, the amounts of materials do not change.

13. Equilibrium
As the reaction proceeds, the forward reaction slows down as the reactants get used up. At the same time, the reverse reaction speeds up as product concentration increases.
Initially, we only have reactant molecules in the mixture. The reaction can only proceed in the forward direction, making products.
Once equilibrium is established, the concentrations of the reactants and products in the final mixture do not change, (unless conditions are changed).
Eventually, the forward and reverse rates are equal. At this time equilibrium is established.

14. Equilibrium, Continued
Because the reactant concentration
decreases, the forward reaction slows.
As the products accumulate, the
reverse reaction speeds up.
As the forward reaction proceeds
it makes products and uses reactants.
Once equilibrium is established,
the forward and reverse reactions
proceed at the same rate, so the
concentrations of all materials
stay constant.
Initially, only the forward
reaction takes place.
Eventually, the reaction proceeds
in the reverse direction as fast as
it proceeds in the forward direction.
At this time equilibrium is established.
Rate
Rate forward
Rate reverse
Time

15. Hypothetical Reaction 2 Red  Blue
Time
[Red]
[Blue]
0.400
0.000 10
0.208
0.096 20
0.190
0.105 30
0.180
0.110 40
0.174
0.113 50
0.170
0.115 60
0.168
0.116 70
0.167
0.117 80
0.166 90
0.165
0.118
100
110
0.164
120
130
140
150
The reaction slows over time,
but the red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the red and
the blue molecules no longer change—
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of red molecules turning into blue is the
same as the rate of blue molecules turning
into red.

16. Hypothetical Reaction 2 Red  Blue, Continued

17. Equilibrium  Equal
The rates of the forward and reverse reactions are equal at equilibrium.
But that does not mean the concentrations of reactants and products are equal.
Some reactions reach equilibrium only after almost all the reactant molecules are consumed—we say the position of equilibrium favors the products.
Other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed—we say the position of equilibrium favors the reactants.

18. Equilibrium Constant
Even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them.
For the reaction H2(g) + I2(g)  2HI(g) at equilibrium, the ratio of the concentrations raised to the power of their coefficients is constant.

19. The Equilibrium Constant Keq
Chapter 13: Chemical Equilibrium
4/14/2017
The Equilibrium Constant Keq
For a general reversible reaction:
cC + dD
aA + bB
[A]a[B]b
[C]c[D]d
Products
Equilibrium equation:
Keq =
Reactants
Equilibrium constant
Equilibrium constant expression
“c” in Kc is for concentration.
No units for the equilibrium constant. See the explanation in section 13.4 for the reason.
For the following reaction:
2NO2(g)
N2O4(g)
[N2O4]
[NO2]2
Keq =
= 4.64 x 10-3 (at 25 °C)
Copyright © 2010 Pearson Prentice Hall, Inc.

20. Equilibrium Constant, Continued
For the general equation aA + bB  cC + dD, the relationship is given below:
The lowercase letters represent the coefficients of the balanced chemical equation.
Always products over reactants.
The constant is called the equilibrium constant, Keq.

21. Writing Equilibrium Constant Expressions
For aA + bB  cC + dD, the equilibrium constant expression is:
So for the reaction
2 N2O5  4 NO2 + O2, the
equilibrium constant
expression is:

22. Equilibrium Constants for Heterogeneous Equilibria
Pure substances in the solid and liquid state have constant concentrations.
Adding or removing some does not change the concentration because they do not expand to fill the container or spread throughout a solution.
Therefore, these substances are not included in the equilibrium constant expression.
For the reaction CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l):

23. Heterogeneous Equilibria
Chapter 13: Chemical Equilibrium
4/14/2017
Heterogeneous Equilibria
CaO(s) + CO2(g)
CaCO3(s)
Limestone
Lime
[CaCO3]
[CaO][CO2]
(1)
(1)[CO2]
Keq = =
= [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.
Keq = [CO2]
Kp = P
CO2
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24. Write the Equilibrium Constant Expressions, Keq, for Each of the Following:
2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba+2(aq) + SO4-2(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l)

25. Write the Equilibrium Constant Expressions, Keq, for Each of the Following, Continued:
2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba+2(aq) + SO4-2(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(g)
Keq =
[CO]2•[O2]
[CO2]2
Keq =
[Ba+2]•[SO4-2]
Keq =
[CO2] •[H2O]2
[CH4]•[O2]2

26. What Does the Value of Keq Imply?
When the value of Keq > > 1, we know that when the reaction reaches equilibrium, there will be many more product molecules present than reactant molecules.
The position of equilibrium favors products.
When the value of Keq < < 1, we know that when the reaction reaches equilibrium, there will be many more reactant molecules present than product molecules.
The position of equilibrium favors reactants.

27. A Large Equilibrium Constant

28. A Small Equilibrium Constant

29. Write the Equilibrium Constant Expressions, Keq, and Predict the Position of Equilibrium for Each of the Following:
2 HF(g) Û H2(g) + F2(g) Keq = 1 x 10-95
2 SO2(g) + O2(g) Û 2 SO3(g) Keq = 8 x 1025
N2(g) + 2 O2(g) Û 2 NO2(g) Keq = 3 x 10-17

30. Write the Equilibrium Constant Expressions, Keq, and Predict the Position of Equilibrium for Each of the Following, Continued:
2 HF(g) Û H2(g) + F2(g) Keq = 1 x 10-95
2 SO2(g) + O2(g) Û 2 SO3(g) Keq = 8 x 1025
N2(g) + 2 O2(g) Û 2 NO2(g) Keq = 3 x 10-17
Favors reactants.
Favors products.
Favors reactants.

31. Calculating Keq
The value of the equilibrium constant may be determined by measuring the concentrations of all the reactants and products in the mixture after the reaction reaches equilibrium, then substituting in the expression for Keq.
Although you may have different amounts of reactants and products in the equilibrium mixture, the value of Keq will always be the same.
The value of Keq depends only on the temperature.
The value of Keq does not depend on the amounts of reactants or products with which you start.

32. Initial and Equilibrium Concentrations for H2(g) + I2(g)  2HI(g)
Constant
[H2]
[I2]
[HI]
0.50
0.0
0.11
0.78
0.055
0.39
0.165
1.17
1.0
0.5
0.53
0.033
0.934

33. Example 1—Find the Value of Keq for the Reaction from the Given Concentrations: 2 CH4(g)  C2H2(g) + 3 H2(g).
Given:
Find:
[CH4] = M, [C2H2] = M, [H2] = M
Keq
Solution Map:
Relationships:
[CH4], [C2H2], [H2]
Keq
Solve:
Check:
Keq is unitless.

34. Practice—Calculate Keq for the Reaction 2 NO2(g)  N2O4(g) at 100 C if the Equilibrium Concentrations Are [NO2] = M and [N2O4] = M.

35. Practice—Find the Value of Keq for the Reaction from the Given Concentrations: 2 NO2(g)  N2O4(g).
[NO2] = M, [N2O4] = M
Keq
Solution Map:
Relationships:
[NO2], [N2O4]
Keq
Solve:
Check:
Keq is unitless.

36. Example 3 —Find the Value of [HI] for the Reaction at Equilibrium from the Given Concentrations and Keq: H2(g) + I2(g)  2HI(g)
Given:
Find:
[I2] = M, [H2] = M, Keq = 69
[HI]
Solution Map:
Relationships:
[I2], [H2], Keq
[HI]
Solve:

37. A Sample of PCl5(g) Is Placed in a 0
A Sample of PCl5(g) Is Placed in a L Container and Heated to 160 °C. The PCl5 Is Decomposed into PCl3(g) and Cl2(g). At Equilibrium, Moles of PCl3 and Cl2 Are Formed. Determine the Equilibrium Concentration of PCl5 if Keq =

38. A Sample of PCl5(g) Is Placed in a 0
A Sample of PCl5(g) Is Placed in a L Container and Heated to 160 °C. The PCl5 Is Decomposed into PCl3(g) and Cl2(g). At Equilibrium, Moles of PCl3 and Cl2 Are Formed. Determine the Equilibrium Concentration of PCl5 if Keq = , Continued.
PCl5 Û PCl Cl2
Equilibrium
Concentration, M ?
if in addition you calculate Kp from Kc you find that it is 2.26, slightly greater than 1 – showing that the equilibrium constant may be unreliable for predicting the position of equilibrium when it is close to 1

39. The Equilibrium Constant Keq
Chapter 13: Chemical Equilibrium
The Equilibrium Constant Keq
4/14/2017
Even though 4 of the 5 trials have different equilibrium concentrations, why is there some apparently constant value assigned for each trial?
Why?
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40. The Equilibrium Constant Keq
Chapter 13: Chemical Equilibrium
4/14/2017
The Equilibrium Constant Keq
For a general reversible reaction:
cC + dD
aA + bB
[A]a[B]b
[C]c[D]d
Products
Equilibrium equation:
Keq =
Reactants
Equilibrium constant
Equilibrium constant expression
“c” in Kc is for concentration.
No units for the equilibrium constant. See the explanation in section 13.4 for the reason.
For the following reaction:
2NO2(g)
N2O4(g)
[N2O4]
[NO2]2
Keq =
= 4.64 x 10-3 (at 25 °C)
Copyright © 2010 Pearson Prentice Hall, Inc.

41. The Equilibrium Constant Keq
Chapter 13: Chemical Equilibrium
4/14/2017
The Equilibrium Constant Keq
Experiment 1
Experiment 5
[N2O4]
[NO2]2
0.0337
(0.0125)2
0.0429
(0.0141)2
Keq =
= 4.64 x 10-3
= 4.63 x 10-3
Copyright © 2010 Pearson Prentice Hall, Inc.

42. The Equilibrium Constant Keq
Chapter 13: Chemical Equilibrium
4/14/2017
The Equilibrium Constant Keq
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)
Keq =
[NH3]2
[N2][H2]3 1
Keq
Keq =
N2(g) + 3H2(g)
2NH3(g) =
[N2]2[H2]6
[NH3]4
Keq = ´´
4NH3(g)
2N2(g) + 6H2(g)
= Keq 2
Copyright © 2010 Pearson Prentice Hall, Inc.

43. The Equilibrium Constant Kp
Chapter 13: Chemical Equilibrium
4/14/2017
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.
Copyright © 2010 Pearson Prentice Hall, Inc.

44. The Equilibrium Constant Kp
Chapter 13: Chemical Equilibrium
4/14/2017
The Equilibrium Constant Kp
K mol
L atm
Kp = Keq(RT)∆n
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 © 2010 Pearson Prentice Hall, Inc.

45. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
4/14/2017
Using the Equilibrium Constant
Keq > 103
Products predominate over reactants.
Keq < 10-3
Reactants predominate over products.
10-3 < Keq < 103
Appreciable concentrations of both reactants and products are present.
Copyright © 2010 Pearson Prentice Hall, Inc.

46. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
4/14/2017
Using the Equilibrium Constant
cC + dD
aA + bB
[A]ta[B]tb
[C]tc[D]td
Reaction quotient:
Q =
The reaction quotient, Q, is defined in the same way as the equilibrium constant, Keq, except that the concentrations in Q are not necessarily equilibrium values.
Copyright © 2010 Pearson Prentice Hall, Inc.

47. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
4/14/2017
Using the Equilibrium Constant
Students have difficulties with the reaction quotient. Both the graphic, Figure 13.5 p504, on this slide and a text-based description in the next slide have been included.
Copyright © 2010 Pearson Prentice Hall, Inc.

48. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
4/14/2017
Using the Equilibrium Constant
If Q < Keq
net reaction goes from left to right (reactants to products).
If Q > Keq
net reaction goes from right to left (products to reactants).
Students have difficulties with the reaction quotient. Both the graphic, Figure 13.5 p504, on the previous slide and a text-based description in this slide have been included.
Emphasizing the relative magnitudes of the numerator and denominator may help students understand it.
If Q = Keq
no net reaction occurs.
Copyright © 2010 Pearson Prentice Hall, Inc.

49. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
4/14/2017
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 . Keq is 57.0 at 700 K.
2HI(g)
H2(g) + I2(g)
Copyright © 2010 Pearson Prentice Hall, Inc.

50. Chapter 13: Chemical Equilibrium
4/14/2017
Copyright © 2010 Pearson Prentice Hall, Inc.

51. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
4/14/2017
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 (volume is 2.00 L).
C = change in concentrations (use “x” for the change along with the stoichiometric coefficients).
E = equilibrium concentrations.
Substitute values into the equilibrium expression:
Keq =
[H2][I2]
[HI]2 x2
( x)2
57.0 =
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

52. Using the Equilibrium Constant
Chapter 13: Chemical Equilibrium
Using the Equilibrium Constant
4/14/2017
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
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

53. Disturbing and Re-Establishing Equilibrium
Once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same.
However, if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re-established.
The new concentrations will be different, but the equilibrium constant will be the same.
Unless you change the temperature.

54. Le Châtelier’s Principle
Chapter 13: Chemical Equilibrium
4/14/2017
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.
Stress means a change in concentration, pressure, volume, or temperature that disturbs the original equilibrium.
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

55. Le Châtelier’s Principle
Le Châtelier’s principle guides us in predicting the effect on the position of equilibrium when conditions change.
“When a chemical system at equilibrium is disturbed, the system shifts in a direction that will minimize the disturbance.”

56. The Effect of Concentration Changes on Equilibrium
Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found.
That has the same Keq.
Removing a product will increase the amounts of the other products and decrease the amounts of the reactants.
You can use to this to drive a reaction to completion!
Remember: Adding more of a solid or liquid does not change its concentration and, therefore, has no effect on the equilibrium.

57. Altering an Equilibrium Mixture: Concentration
Chapter 13: Chemical Equilibrium
4/14/2017
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 © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

58. The Effect of Concentration Changes on Equilibrium, Continued
When NO2
is added,
some of it
combines
to make
more N2O4.

59. The Effect of Concentration Changes on Equilibrium, Continued
When N2O4
is added,
some of it
decomposes
to make
more NO2.

60. Practice—Predict the Effect on the Equilibrium When the Underlined Substance Is Added to the Following Systems:
2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba2+(aq) + SO42-(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l)

61. Practice—Predict the Effect on the Equilibrium When the Underlined Substance Is Added to the Following Systems, Continued:
2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba2+(aq) + SO42-(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l)
Shift right, removing some of the added CO2 and
increasing the concentrations of CO and O2.
Shift left, removing some of the added Ba2+ and
reducing the concentration of SO42-.
Shift right, removing some of the added CO2 and decreasing the O2,
while increasing the concentration of CO2.

62. Effect of Volume Change on Equilibrium
For solids, liquids, or solutions, changing the size of the container has no effect on the concentration.
Changing the volume of a container changes the concentration of a gas.
Same number of moles, but different number of liters, resulting in a different molarity.

63. Effect of Volume Change on Equilibrium, Continued
Decreasing the size of the container increases the concentration of all the gases in the container.
This increases their partial pressures.
If their partial pressures increase, then the total pressure in the container will increase.
According to Le Châtelier’s principle, the equilibrium should shift to remove that pressure.
The way to reduce the pressure is to reduce the number of molecules in the container.
When the volume decreases, the equilibrium shifts to the side with fewer molecules.

64. The Effect of Volume Change on Equilibrium, Continued
Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products.
When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules—the reactant side.

65. Practice—Predict the Effect on the Equilibrium When the Volume Is Reduced.
2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba2+(aq) + SO42-(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l)

66. Practice—Predict the Effect on the Equilibrium When the Volume Is Reduced, Continued.
2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba2+(aq) + SO42-(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l)
Shift left because there are fewer gas molecules
on the reactant side than on the product side.
No effect because none of the substances are gases.
Shift right because there are fewer gas molecules
on the product side than on the reactant side.

67. The Effect of Temperature Changes on Equilibrium
Exothermic reactions release energy and endothermic reactions absorb energy.
If we write heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s principle to predict the effect of temperature changes.
However, heat is not matter and not written in a proper equation.

68. The Effect of Temperature Changes on Equilibrium for Exothermic Reactions
For an exothermic reaction, heat is a product.
Increasing the temperature is like adding heat.
According to Le Châtelier’s principle, the equilibrium will shift away from the added heat.
The concentrations of C and D will decrease and the concentrations of A and B will increase.
The value of Keq will decrease.
How will decreasing the temperature effect the system?
aA + bB  cC + dD + heat

69. The Effect of Temperature Changes on Equilibrium for Endothermic Reactions
For an endothermic reaction, heat is a reactant.
Increasing the temperature is like adding heat.
According to Le Châtelier’s principle, the equilibrium will shift away from the added heat.
The concentrations of C and D will increase and the concentrations of A and B will decrease.
The value of Keq will increase.
How will decreasing the temperature effect the system?
Heat + aA + bB  cC + dD

70. The Effect of Temperature Changes on Equilibrium

71. Practice—Predict the Effect on the Equilibrium When the Temperature Is Reduced.
Heat + 2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba2+(aq) + SO42-(aq) (endothermic)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l) (exothermic)

72. Practice—Predict the Effect on the Equilibrium When the Temperature Is Reduced, Continued.
Heat + 2 CO2(g) Û 2 CO(g) + O2(g)
Heat + BaSO4(s) Û Ba2+(aq) + SO42-(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l) + Heat
Shift left, reducing the value of Keq.
Shift left, reducing the value of Keq.
Shift right, increasing the value of Keq.