1. Reaction Rates and Equilibrium
Chemical Kinetics
Reaction Rates and Equilibrium

2. Objectives Know and understand collision theory.
Be able to explain the role of an activated complex in a chemical reaction.
Be able to draw a potential energy diagram that includes activation energy.

3. Collision Theory
Not
enough KE
collision theory: particles must collide w/ enough KE to break bonds so that new bonds can form
many reactions require high T!
in car engine: N2 and O2 → NO2
Enough KE

4. Activated Complex activated complex: a high PE, short-lived molecule
H2 + Cl2
activated
complex
(H2Cl2)
2 HCl

5. Activation Energy
activation energy (Ea): the extra energy needed to produce an activated complex
example: a spark is needed to get fuel to burn

6. Activation Energy

7. Objectives Understand the concept of a rate.
Describe the factors that affect the rate of a chemical reaction.
Understand how catalysts and inhibitors affect reaction rates.
Explain the factors that affect explosions.

8. Reaction Rates
time ↓ rate ↑

9. Factors Affecting Reaction Rates
temperature (T↑ R↑)
concentration (C↑ R↑)
particle size (size↓ R↑)
catalyst: reduces Ea and increases rate (but NOT a reactant)
MnO2
2H2O2(l) → 2H2O(l) + O2(g)
inhibitor: increases Ea and decreases rate (example: preservatives)

10. The Catalytic Converter
A catalytic converter
reduces pollution.
IN: HC (unburned fuel),
CO, and NOx
OUT: CO2, H2O, and N2

11. Clock Reaction Step 1: IO3− + 3 HSO3− → I− + 3 HSO4−
Step 2: IO3− + 5 I− + 6 H+ → 3 I2 + 3 H2O
The iodine (I2) would cause the starch to turn purple, but the iodine in Step 2 is immediately consumed by a third reaction:
Step 3: I2 + HSO3− + H2O → 2I− + HSO4− + 2H+
The iodine cannot accumulate to stain the starch until all of the HSO3 − is consumed (in solution B) by this third step. Once all of HSO3− is gone, then the iodine accumulates quickly and stains the starch.

12. Objectives Understand the factors that make some reactions explosive.
Know the components of gunpowder.
Understand why high explosives always contain two particular elements.

13. Explosives Explosions are (1) VERY exothermic reactions that make
10 KNO3(s) + 3S(s) + 8C(s)
Explosions are
(1) VERY exothermic reactions that make
(2) LARGE AMOUNTS of gaseous products that occur at a
(3) VERY high rate.
black powder
-DH
2K2CO3(s) + 3K2SO4 (s) + 6CO2(g)+ 5N2(g)

14. High Explosives always contain O and N
decomposition reaction, forms H2O, CO2 and N2
high concentration of oxygen increases rate
N2 gas forms in very exothermic reactions
TNT
nitroglycerin

15. Zinc Pyrotechnics! “Negative-X”
ingredients: NH4NO3(s), Zn(s), NH4Cl(s), and H2O(l)
water added to the zinc produces heat to initiate
the reaction
water dissolves ammonium chloride (makes Cl-)…
NH4NO3(s) → N2O(g) + H2O(g) + 23 kJ
heat melts ammonium nitrate so 2nd reaction can occur…
Zn(s) + NH4NO3(l ) → N2(g) + ZnO(s) + 2H2O(g) kJ
Cl-

16. Reversible Reactions

17. Objectives Understand the concept of a reversible chemical reaction.
Understand the concept of chemical equilibrium and how systems respond to stresses to change the equilibrium position.
Be able to apply LéChâtelier’s principle to determine the effect that changes in concentration, temperature, or pressure have on the equilibrium position.

18. Reversible Reactions
many reactions can go “forward” or “reverse” 2NO2(g) ↔ N2O4(g)
red (warm) clear (cool)
reactants favored products favored
equilibrium position: tells which side is “favored”

19. Chemical Equilibrium chemical equilibrium forward rate = reverse rate
total amounts of reactant and product stay constant (but probably NOT equal)
2NO2(g) ↔ N2O4(g)

20. LéChâtelier’s Principle
LéChâtelier’s Principle: a system in chemical equilibrium will often respond to reduce an applied “stress”; a new equilibrium position will occur
Henry Louis LéChâtelier

21. LéChâtelier and Concentration
Add B B C A D A A
A + B → C + D, products favored C B B B A D B A
Remove A C
A + B ← C + D, reactants favored
A + B ↔ C + D
if a substance is added, the rxn is favored away from it
if a substance is removed, the rxn adjusts to replace it

22. LéChâtelier and Temperature
Just treat energy (# kJ) like a substance!
2NO2(g) ↔ N2O4(g) kJ
red clear
higher T (add energy) ← sky turns redder
lower T (remove energy) → sky appears clearer

23. LéChâtelier and Gases if V ↓ (P ↑), fewer molecules are favored
more V (less P)
less V (high P )
N2 (g) + 3H2(g) ← 2NH3(g)
N2 (g) + 3H2(g) → 2NH3(g)
reactant favored
product favored

24. Applying LéChâtelier N2(g) + 3H2(g) ↔ 2NH3(g) + energy
Add hydrogen gas?
products favored
Remove nitrogen?
reactants favored
Increase temperature?
Increase pressure?

25. Common Ion Effect
The addition of a “common ion” can affect an equilibrium…
PbCrO4(s) ↔ Pb2+ (aq) + CrO42-(aq)
adding PbCl2(aq), which contains Pb2+ , will favor the production of PbCr4(s)
adding K2CrO4 also produces more reactant

26. Objectives
Understand and apply the concept of an equilibrium constant (Keq).
Be able to calculate an equilibrium constant.

27. Equilibrium Constants (Keq)
“Equilibrium Position”…
If Keq > 1, products favored
If Keq < 1, reactants favored
[A] is concentration of A, etc.
aA + bB ↔ cC + dD
coefficient

28. Equilibrium Constants
At 500oC, the chemical equilibrium
N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)
has the following concentrations:
[N2] = M
[H2] = M
[NH3] = M
What is the equilibrium constant at 500oC?

29. Equilibrium Constants
The following chemical reaction has a
Keq = 5.63 x 1018 (at 25oC):
H2(g) + Br2(g) ↔ 2HBr(g).
If the concentration of each reactant is 2.11 x M
at 25oC, what is the concentration of HBr?

30. Solubility Constants: Extra-Credit
Insoluble compounds actually dissolve very slightly.
Ksp indicates the solubility of a saturated compound (usually at 25oC)
lower Ksp = lower solubility
Ca(OH)2 ↔ Ca2+(aq) + 2OH-(aq)
See page 562
Ca(OH)2  = 6.5 x 10-6  
ZnS  = 3.0 x 10-23
Ksp = [Ca2+][OH-]2 = 6.5 x 10-6
ZnS ↔ Zn2+(aq) + S2-(aq)
Ksp = [Zn2+][Cl2-] = 3.0 x 10-23

31. Objectives
Understand the concept of spontaneous and non-spontaneous processes.
Understand the concept of entropy and how it affects the spontaneity of a process.
Be able to predict the spontaneity of a process by considering the influence of enthalpy and entropy.

32. Spontaneous Processes
spontaneous: occurs naturally, favors products
C3H8 + 5O2 ↔ 3CO2 + 4H2O + energy
→ spontaneous
← non-spontaneous, why?
exothermic: tend to be spontaneous
endothermic: tend to be non-spontaneous
But endothermic processes do occur spontaneously:
H2O(s) + energy → H2O(l)
Why? Processes that increase ENTROPY tend to be spontaneous.

33. Entropy Does your room look like this? entropy: a measure of disorder
low entropy
high entropy
It takes a lot of effort and energy to reduce entropy!
Systems tend to become more disordered (increased entropy) due to probability.

34. solid ↔ liquid or (aq) ↔ gas
Entropy
tend to be
non-spontaneous
tend to be
spontaneous
LOWER
ENTROPY
HIGHER
ENTROPY
solid ↔ liquid or (aq) ↔ gas
fewer moles ↔ more moles

35. Predicting Reaction Outcomes
CH3CH2OH(l) + 2O2(g) → 2CO2(g) + 3H2O(g) + energy
exothermic = spontaneous
entropy? (liquid → gas, more moles) = spontaneous
spontaneous reaction!
2NaCl(s) + energy → 2Na(s) + Cl2(g)
endothermic = non-spontaneous
entropy? (solid → gas, more moles) = spontaneous
spontaneous or non-spontaneous?

36. Objectives Understand the concept of free energy.
Know the factors that affect the amount of Gibbs free energy that is either absorbed or released in a chemical reaction.
Be able to calculate the amount of Gibbs free energy involved in a chemical reaction (and determine the spontaneity of the reaction from this value).

37. Gibbs Free Energy
Gibbs free energy: energy that can do “work” (it can make things move)
Increasing entropy can do work—when a gas forms, it will do work as it expands
The heat released in an exothermic reaction can be used to do work as well.
Processes that release Gibbs free energy (G0) are spontaneous. Enthalpy, entropy, and temperature are all involved.
- DG0 = spontaneous reaction
+ DG0 = non-spontaneous
DG0 = DH0 – TDS0
enthalpy entropy
Gibbs free energy temperature

38. Gibbs Free Energy DG0 = DH0 – TDS0 2NaCl(s) + 800 kJ → 2Na(s) + Cl2(g)
If DH0 = +800 kJ and DS0 = +0.2 kJ/K, is the
reaction spontaneous at room temperature (300 K)?

39. Calculating Changes in Gibbs Free Energy
DG0 = S[DGf0 ·n] (products) - S[DGf0 ·n] (reactants)
Is the following reaction spontaneous at 25oC?
2NaCl(s) →2Na(s) + Cl2(g)
Is the following reaction spontaneous at 25oC?
2Ag+(aq) + 2Cl-(aq) → 2AgCl(s)