1. Chapter 9 Chemical Equilibrium
9.1
Rates of Reactions
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2. Collision Theory of Reactions
A chemical reaction occurs when
Collisions between molecules have sufficient energy to break the bonds in the reactants.
Molecules collide with the proper orientation.
Bonds between atoms of the reactants (N2 and O2) are broken, and new bonds (NO) form.
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3. Collision Theory of Reactions
A chemical reaction does not take place if the
Collisions between molecules do not have sufficient energy to break the bonds in the reactants.
Molecules are not properly aligned.
N O2
N O2
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4. Activation Energy The activation energy
Is the minimum energy needed for a reaction to take place upon proper collision of reactants.
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5. Reaction Rate and Temperature
Is the speed at which reactant is used up.
Is the speed at which product forms.
Increases when temperature rises because reacting molecules move faster, providing more colliding molecules with energy of activation.

6. Reaction Rate and Concentration
Increasing the
concentration of reactants
Increases the number of collisions.
Increases the reaction rate.
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7. Reaction Rate and Catalysts
A catalyst
Speeds up the rate of a reaction.
Lowers the energy of activation.
Is not used up during the reaction.
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8. Factors That Increase Reaction Rate
TABLE 9.1
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9. Chapter 9 Chemical Equilibrium
9.2
Chemical Equilibrium
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10. Reversible Reactions
In a reversible reaction, there is both a forward and a
reverse reaction.
Suppose SO2 and O2 are present initially. As they collide, the forward reaction begins.
2SO2(g) + O2(g) SO3 (g)
As SO3 molecules form, they also collide in the reverse reaction that forms reactants. The reversible reaction is written with a double arrow.
forward
2SO2(g) + O2 (g) SO3(g)
reverse

11. Chemical Equilibrium At equilibrium
The rate of the forward reaction becomes equal to the rate of the reverse reaction.
The forward and reverse reactions continue at equal rates in both directions.
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12. Chemical Equilibrium When equilibrium is reached
There is no further change in the amounts of reactant and product.
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13. Equilibrium At equilibrium The forward reaction of N2 and O2 forms NO.
The reverse reaction of 2NO forms N2 and O2.
The amounts of N2, O2, and NO remain constant.
N2(g) + O2(g) NO(g)
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14. Chapter 9 Chemical Equilibrium
9.3
Equilibrium Constants
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15. Equilibrium Constants
For the reaction
aA bB
The equilibrium constant expression, Kc, gives the concentrations of the reactants and products at equilibrium.
Kc = [B]b = [Products]
[A]a [Reactants]
The square brackets indicate the moles/liter of each substance.
The coefficients b and a are written as superscripts that raise the moles/liter to a specific power.

16. Writing a Kc Expression
In the Kc expression for the following reaction at equilibrium
2CO(g) + O2(g) CO2(g) STEP 1
The products are shown in the numerator and the reactants are shown in the denominator.
Kc = [CO2] [products] STEP 2
[CO] [O2] [reactants]
The coefficients are written as superscripts STEP 3
Kc = [CO2]2
[CO]2 [O2]

17. Guide to Writing a Kc Copyright © 2007 by Pearson Education, Inc.
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18. Heterogeneous Equilibrium
In heterogeneous equilibrium
Solid and/or liquid states may be part of a reaction.
2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g)
The concentration of solids and/or liquids is constant.
The Kc expression is written with only the gases.
Kc = [CO2][H2O]

19. Calculating Equilibrium Constants
What is the Kc for the following reaction?
H2(g) + I2(g) 2HI(g)
Equilibrium concentrations: H mole/L,
I2 1.2 mole/L, and HI 0.35 mole/L.
STEP 1 Write the Kc expression
Kc = [HI]2
[H2][I2]
STEP 2 Enter equilibrium concentrations.
Kc = (0.35) = x 10-2
(1.2)(1.2)

20. Chapter 9 Chemical Equilibrium
9.4
Using Equilibrium Constants
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21. Reaching Chemical Equilibrium
A container filled with SO2 and O2 or only SO3
Contains mostly SO3 and small amounts of O2 and SO3 at equilibrium
Reaches equilibrium in both situations.
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22. Equilibrium can Favor Product
If equilibrium is reached after most of the forward
reaction has occurred,
The system favors the product.
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23. Equilibrium with a Large Kc
At equilibrium,
A reaction with a large Kc produces a large amount of product. Very little of the reactants remain.
Kc = [NCl3]2 = 3.2 x 1011
[N2][Cl2]3
A large Kc favors the products.
N2(g) + 3Cl2(g) NCl3(g)
When this reaction reaches equilibrium, it will essentially consist of the product NCl3.

24. Equilibrium can Favor Reactant
If equilibrium is reached when very little of the
forward reaction has occurred,
The reaction favors the reactants.
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25. Equilibrium with a Small Kc
At equilibrium,
A reaction that produces only a small amount of product has a small Kc.
Kc = [NO] = 2.3 x 10-9
[N2] [O2]
A small Kc favors the reactants.
N2(g) + O2(g) NO(g)

26. Summary of Kc Values A reaction That favors products has a large Kc.
With about equal concentrations of products and reactants has a Kc close to 1.
That favors reactants has a small Kc.
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27. Large and Small Kc Values
TABLE 9.3
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28. Calculating Equilibrium Concentrations
When the Kc is known and the concentrations of reactants or products except for one are given
The concentration of the remaining substance can be calculated.
STEP 1 Write the Kc expression.
STEP 2 Solve the Kc expression for the unknown.
STEP 3 Substitute the known concentrations.
STEP 4 Check by using concentrations to solve for Kc.

29. Using Kc to Solve for Equilibrium Concentration
At equilibrium, the reaction
PCl5(g) PCl3(g) + Cl2(g)
has a Kc of 4.2 x 10-2 and contains [PCl3] = [Cl2] = 0.10 M.
What is the equilibrium concentration of PCl5?

30. Using Kc to Solve for Equilibrium Concentration
STEP 1 Write the Kc expression.
Kc = [PCl3][Cl2]
[PCl5]
STEP 2 Solve for the unknown.
[PCl5] = [PCl3][Cl2] Kc
STEP 3 Substitute known concentrations and solve.
[PCl5] = (0.10)(0.10) = M
4.2 x 10-2
STEP 4 Check by placing concentrations in Kc.
Kc = (0.10)(0.10) = x 10-2
(0.24)

31. Chapter 9 Chemical Equilibrium
9.5
Changing Equilibrium Conditions: LeChâtelier’s Principle
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32. LeChâtelier’s Principle
LeChâtelier’s principle states that
Any change in equilibrium conditions upsets the equilibrium of the system.
A system at equilibrium under stress will shift to relieve the stress.
There will be a change in the rate of the forward or reverse reaction to return the system to equilibrium.

33. Effect of Adding Reactant
Consider the following reaction at equilibrium
H2(g) F2(g) HF(g)
If more reactant (H2 or F2) is added, there is an increase in the number of collisions.
The rate of the forward reaction increases and forms more HF product until new equilibrium concentrations equal Kc again.
The effect of adding a reactant shifts the equilibrium towards the products.

34. Adding Reactant For the reaction A + B C at equilibrium
Adding more A upsets the equilibrium.
The rate of forward reaction increases to
re-establish Kc A + B C
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35. Effect of Adding Product
Consider the following reaction at equilibrium.
H2(g) + F2(g) HF(g)
When more HF is added, there is an increase in collisions of HF molecules.
The rate of the reverse reaction increases and forms more H2 and F2 reactants.
The effect of adding a product shifts the equilibrium towards the reactants.

36. Adding Reactant or Product
The equilibrium shifts towards
Products when H2 or F2 is added.
Reactants when HF is added.
H2(g) F2(g) HF(g)
Add H2 or F2
Add HF

37. Effect of Removing Reactant
Removing reactant, H2 or F2, from the following
reaction at equilibrium.
H2(g) + F2(g) HF(g)
Decreases the collisions between reactants.
Decreases the rate of the forward reaction.
Shifts the equilibrium towards the reactants.
H2(g) F2(g) HF(g)
Remove H2 or F2

38. Effect of Removing Product
When HF is removed from the following reaction at
equilibrium
H2(g) + F2(g) HF(g)
The rate of the reverse reaction decreases.
The rate of the forward reaction is greater.
Equilibrium shifts towards the products.
H2(g) F2(g) HF(g)
Remove HF

39. Concentration Changes and Equilibrium
TABLE 9.4
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40. Effect of Decreasing the Volume
When a reaction at equilibrium contains different numbers
of moles of reactants than products,
A decrease in volume
Increases the concentration (mole/L) upsetting the equilibrium.
Shifts the equilibrium towards the fewer number of moles.
N2(g) + 3H2(g) NH3(g)
Decrease volume
More moles
Fewer moles

41. Volume Decrease and Equilibrium
A volume decrease
Shifts the equilibrium towards the side with the smaller number of moles.
Shifts the equilibrium towards reactant A in this example.
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42. Effect of Increasing the Volume
When a reaction at equilibrium contains different
numbers of moles of reactants than products,
An increase in volume
Decreases the concentration (mol/L) upsetting the equilibrium.
Shifts the equilibrium towards the greater number of moles.
N2(g) + 3H2(g) NH3(g)
Increase volume
More moles
Fewer moles

43. Volume Increase and Equilibrium
A volume increase
Shifts the equilibrium towards the side with the greater number of moles.
Shifts towards reactants B and C in this example.
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44. Heat and Endothermic Reactions
For an endothermic reaction at equilibrium,
A decrease in temperature removes heat and the equilibrium shifts towards the reactants.
An increase in temperature adds heat and the equilibrium shifts towards the products.
CaCO3 (s) kcal CaO(s) + CO2(g)
Decrease T
Increase T

45. Temperature Change and Equilibrium
TABLE 9.5
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46. Heat and Exothermic Reactions
For an exothermic reaction at equilibrium,
A decrease in temperature removes heat and the equilibrium shifts towards the products.
An increase in temperature adds heat and the equilibrium shifts towards the reactants.
N2(g) + 3H2(g) NH3(g) kcal
Decrease T
Increase T

47. Temperature Change and Equilibrium
TABLE 9.6
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48. Changes and Equilibrium
TABLE 9.7
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