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
transition state
activated complex
…aka…
Energy
Profile
Eact
Potential Energy 
∆Hrxn
Reaction progress 

5. 4 Factors that Affect Rates
Concentration
Temperature
Exposed 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. Rate Law & Orders from Mechanisms
Step 1: NO + Br2
NOBr2
(slow)
(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] 8

9. Rate Law & Orders from Mechanisms
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] 9

10. Rate Law & Orders from Exp. 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
rate = k [BrO3–]x[Br–]y[H+]z
rate = k [H+]z
then… x = 1
if… 2 = [2]x
4 = [2]z
z = 2
Rate = k
[BrO3–]1
[Br–]1
[H+]2

11. Orders in Rate Laws …only 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).
…typically 0, 1, 2, but can be any # or fraction

12. Units of k (rate constant)
HW p. 619 #28
Units of k give info about order.
Rate is usually (M∙s–1) or (M∙min–1)
(next slide)
Order
Rate Law
k Units
0th
1st
2nd
3rd
rate =
k[A]0
rate =
k[A]
rate =
k[A]2
rate =
k[A]2[B]
M = ? s
M = ?∙M s
M = ?∙M2 s
M = ?∙M3 s M s 1 s 1
M∙s 1
M2∙s
M∙s–1
s–1
M–1∙s–1
M–2∙s–1

13. Model for Solving Orders
HW p. 619 #28
Model for Solving Orders
Determine the rate law for the reaction (from experimental data).
general rate law
rate = k [ClO2]x [OH–]y
(a)
rate1
rate2
(0.0248)
k (0.060)x(0.030)y = =
( )
k (0.020)x(0.030)y
(0.0248) x
0.020 y
0.030 =
( )
9 =
(3)x
(1)y
9 = (3)x
x = 2 (2nd order)

14. rate = k [ClO2]2 [OH–]y rate = k [ClO2]2 [OH–]1 rate3 rate2 (0.00828)
HW p. 619 #28 (cont.)
Determine the rate law for the reaction (from experimental data).
rate = k [ClO2]2 [OH–]y
(a)
rate3
rate2
( )
k (0.020)2(0.090)y = =
( )
k (0.020)2(0.030)y
3 = (3)y
y = 1 (1st order)
rate = k [ClO2]2 [OH–]1
OR rate = k [ClO2]2 [OH–]

15. rate = k [ClO2]2 [OH–] Exp 1: (0.0248) = k (0.060)2(0.030) (0.0248)
HW p. 619 #28 (cont.)
Calculate the rate constant (with units).
rate = k [ClO2]2 [OH–]
(b)
Exp 1:
(0.0248)
= k (0.060)2(0.030)
(0.0248)
k =
(0.060)2(0.030)
k = 230
M–2∙s–1
M = ?∙M2∙M s

16. rate = k [ClO2]2 [OH–] rate = (230)(0.010)2(0.025) rate = 0.00058
HW p. 619 #28 (cont.)
Calculate the rate when [ClO2] = M and [OH–] = M.
rate = k [ClO2]2 [OH–]
(c)
rate
= (230)(0.010)2(0.025)
rate =
M∙s–1

17. 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)

18. 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 18

19. M∙s–1 s–1 M–1∙s–1 Summary 0th Order 1st Order 2nd Order [A] ln[A] 1
Rate Law
Rate = k
Rate = k[A]
Rate = k[A]2
Integrated Rate Law
[A] = –kt + [A]0
ln[A] = –kt + ln[A]0
Linear plot
[A]
ln[A]
k & slope of line
Slope = rate
Slope = –k
Slope = k
M∙s–1
s–1
M–1∙s–1
Half-Life
depends on [A]0
y = mx + b
y = mx + b 1
[A] t t t
M = ? s
M = ?∙M s
M = ?∙M2 s
Units of k