1. Review Unit 4 (Chp 14): Chemical Kinetics (rates)
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Review Unit 4 (Chp 14): Chemical Kinetics (rates)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

2. Chemical Kinetics Chemical Kinetics is the study of:
The rate at which reactants are consumed or products are produced during a chemical rxn.
The factors that affect the rate of a reaction according to Collision Theory (temperature, concentration, surface area, & catalyst).
The mechanism, or sequence of steps, for how the reaction actually occurs.
The rate laws (equations) used to calculate rates, rate constants, concentrations, & time.
(M∙s–1) k (?∙s–1) [A] (mol∙L–1) t (s)

3. minimum E required to start reaction
The Collision Model
reactant bonds break, then product bonds form.
Reaction rates depend on collisions
between reactant particles by:
collision frequency
enough energy
proper orientation
activation energy:
minimum E required to start reaction
(Eact )
unsuccessful
successful
reactants
COLLISION
products

4. Potential Energy Diagram
demo
transition state
activated complex
…aka…
Energy
Profile
Eact
Potential Energy 
Demo: Transition State Ball Toss
∆Hrxn
Reaction progress 

5. 4 Factors That Affect Reaction Rates
Concentration
Temperature
Reactant Surface Area (particle size)
Catalyst
(↑ collisions)
(↑ collisions and ↑ energy)
(lowers Ea by changing mech.)
Boltzmann Distribution
more particles over the Eact
old foamy (elephant toothpaste)

6. Rate Ratios = ? aA + bB cC + dD Rate 2 HI(g)  H2(g) + I2(g)1 a
[A] t = b
[B] c
[C] d
[D] = − −
2 HI(g)  H2(g) + I2(g)
Initial rate of production of H2 is M∙s–1.
What is the rate of consumption of HI? = ? 2
mol HI
0.050 M∙s–1 H2 x =
0.10 M∙s–1 HI 1
mol H2
mol
L∙s
–0.10 M∙s–1 HI
mol ratio

7. Rate Laws rate = k[A]x[B]y[C]z recall…
Rate equations (or rate laws) have the form:
rate = k[A]x[B]y[C]z
rate constant
order with respect to reactants
A, B, & C
…or…
number of each particle involved
in collision that affects the rate
overall order of reaction = x + y +…
Example:
overall order
= ___ order
(4 particles in collision)
rate = k[BrO3–][Br–][H+]2
4th

8. Concentration and Rate Data
Initial Concentrations
Rate in M per unit time
Mixture
[BrO3–] / M
[Br–] / M
[H+] / M A
0.0050
0.25
0.30 10 B
0.010 20 C
0.50 40 D
0.60
160
Mathematically or conceptually show how to determine the orders and rate law.
Rate = k
[BrO3–]
[Br–]
[H+]2

9. Orders in Rate Laws …must be found experimentally (from data).
…do NOT come from the coefficients of reactants of an overall reaction.
…represent the number of reactant particles (coefficients) in the RDS of the mechanism.
…zero order reactants have no effect on rate b/c they do not appear in the RDS of the mechanism (coefficient of 0 in RDS).
…are typically 0, 1, 2, but be any # or fraction.

10. Integrated Rate Laws & Linear Graphs
zero-order
m = rate(M/s)
[A]
[A]0 = initial conc. at t = 0
[A]t = conc. at any time, t
1st or 2nd order t
first-order
second-order
m = k
m = –k 1
[A]
ln [A] t t
ln [A]t = –kt + ln [A]0 1
[A]t
= kt +
[A]0
on equation sheet (kinda)

11. Half-Life Half-life (t1/2): time at which half of
initial amount is reacted.
[A]t = 0.5 [A]0
Half-life (t1/2) is constant
for 1st order only.
= t1/2
0.693 k
given on exam 11

12. Mechanisms Step 1: NO + Br2 NOBr2 (fast)
Step 2: NOBr2 + NO  2 NOBr (slow)
RDS
From RDS Step 2: rate = k2 [NOBr2] [NO]
b/c step 1 is in equilibrium
From Step 1: rateforward = ratereverse
k1 [NO] [Br2] = k−1 [NOBr2]
solve for [NOBr2]
substitute for [NOBr2] k1
k−1
[NO] [Br2] = [NOBr2] 12

13. Mechanisms k2k1 k−1 rate = [NO] [Br2] [NO] rate = k [NO]2 [Br2]
Step 1: NO + Br2
NOBr (fast)
Step 2: NOBr2 + NO  2 NOBr (slow)
RDS
rate = k2 [NOBr2] [NO]
k2k1
k−1
rate =
[NO] [Br2] [NO]
rate = k [NO]2 [Br2]
substitute for [NOBr2] k1
k−1
[NO] [Br2] = [NOBr2] 13