1. 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

2. Collision Theory of Reactions (continued)
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

3. Activation Energy The activation energy
is the minimum energy needed for a reaction to take place upon proper collision of reactants

4. 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, thereby providing more colliding molecules with energy of activation

5. Reaction Rate and Concentration
Increasing the
concentration of reactants
increases the number of collisions
increases the reaction rate

6. 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

7. Reversible Reactions
In a reversible reaction, there are both forward and
reverse reactions.
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. This reversible reaction is written with a double arrow.
forward
reverse

8. 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

9. Chemical Equilibrium (continued)
When equilibrium is reached,
there is no further change in the amounts of reactant and product

10. 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)

11. 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.

12. Guide to Writing the Kc Expression

13. Writing a Kc Expression
Write the Kc expression for the following:
STEP 1 Write the balanced equilibrium equation:
2CO(g) + O2(g) CO2(g)
STEP 2 Write the product concentrations in the numerator and the reactant concentration in the denominator:
Kc = [CO2] [products]
[CO][O2] [reactants]
STEP 3 Write the coefficients as superscripts:
Kc = [CO2]2
[CO]2 [O2]

14. Heterogeneous Equilibrium
In heterogeneous equilibrium,
Gases and solid and/or liquid states are part of the reaction.
2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g)
The concentration of solids and liquids is constant.
The Kc expression is written with only the compounds that are gases.
Kc = [CO2][H2O]

15. Guide to Calculating the Kc Value

16. Example of Calculating Equilibrium Constants
What is the Kc for the following reaction?
H2(g) + I2(g) 2HI(g)
Equilibrium concentrations: [H2] = 1.2 moles/L
[I2] = 1.2 moles/L [HI] = 0.35 mole/L
STEP 1 Write the Kc expression:
Kc = [HI]2
[H2][I2]
STEP 2 Substitute the equilibrium concentrations in Kc: Kc = [0.35] = x 10-2
[1.2][1.2]

17. Reaching Chemical Equilibrium
A container initially filled with SO2(g) and O2(g) or only
SO3(g)
contains mostly SO2(g) and small amounts of O2(g) and SO3(g) at equilibrium
reaches equilibrium in both situations

18. Equilibrium Can Favor Product
If equilibrium is reached after most of the forward
reaction has occurred,
the system favors the products

19. 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 = 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.

20. Equilibrium Can Favor Reactant
If equilibrium is reached when very little of the
forward reaction has occurred,
the reaction favors the reactants

21. 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)
When this reaction reaches equilibrium, it will essentially consist of the reactants N2 and O2.

22. 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

23. Guide to Using the Kc Value

24. Using Kc to Solve for an 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?

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

26. Le Châtelier’s Principle
Le Châtelier’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

27. What happens when a system is at equilibrium and you upset the balance?

28. Le Chatelier’s Principle
If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress.

29. Le Chatelier’s Principle
Used to predict how a equilibrium system will react to changes in concentration, pressure (volume) and temperature.

30. Effects of Changes on Equilibrium

31. Heat and Endothermic Reactions
For an endothermic reaction at equilibrium,
a decrease in temperature (T) removes heat, and the equilibrium shifts toward the reactants
an increase in temperature adds heat, and the equilibrium shifts toward the products.
CaCO3(s) kcal CaO(s) + CO2(g)
Decrease T
Increase T

32. Heat and Exothermic Reactions
For an exothermic reaction at equilibrium,
a decrease in temperature removes heat, and the equilibrium shifts toward the products
an increase in temperature adds heat, and the equilibrium shifts toward the reactants.
N2(g) + 3H2(g) NH3(g) kcal
Decrease T
Increase T

33. Saturated Solution A saturated solution
contains the maximum amount of dissolved solute
contains solid solute
is an equilibrium system:
solid ions in solution

34. Solubility Product Constant
The solubility product constant for a saturated solution
gives the ion concentrations at constant temperature
is expressed as Ksp
does not include the solid, which is constant
Fe(OH)2(s) Fe2+(aq) + 2OH−(aq)
Ksp = [Fe2+] [OH−]2

35. Guide to Calculating Ksp

36. Example of Calculating Solubility Product Constant
Calculate the Ksp of PbSO4 (solubility 1.4 x 10–4 M).
STEP 1 Write the equilibrium equation for dissociation:
PbSO4(s) Pb2+(aq) + SO42−(aq)
STEP 2 Write the Ksp expression:
Ksp = [Pb2+][SO42−]
STEP 3 Substitue molarity values and calculate:
Ksp = (1.4 x 10–4) x (1.4 x 10–4)
= 2.0 x 10–8

37. Molar Solubility (S) The molar solubility (S) is
the number of moles of solute that dissolve in 1 L of solution
determined from the formula of the salt
calculated from the Ksp

38. Calculating Molar Solubility (S)

39. Solubility Calculation
Determine the solubility (S)2 of SrCO3 (Ksp = 5.4 x 10–10).
STEP 1 Write the equilibrium equation for dissociation:
SrCO3(s) Sr2+(aq) + CO32−(aq)
STEP 2 Write the Ksp expression:
Ksp = [Sr2+][CO32−]
STEP 3 Substitute S for the molarity of each ion into Ksp:
Ksp = [Sr2+][CO32−] = [S][S]
= S2 = 5.4 x 10−10
STEP 4 Calculate the solubility, S:
S = = x 10−5 M