# AP Chemistry ```Chemistry 9 Review
Reaction Rate

rate = [A]/t –example: 2 NO2(g)  2 NO(g) + O2(g)
if [NO2]/t = -10 M/s, then [NO]/t = + 10 M/s and
[O2]/t = +5 M/s

rate law: rate = k[A]n
o k = rate constant (depends on temperature)
o n = order of reaction (first, second, etc.)
o units for k are Mxt-1, where x = (1 – overall order)

summary chart
Order
0
1
2
k
k[A]
k[A]2
rate =
kt =
[A]o – [A]t
ln([A]o/[A]t)
1/[A]t – 1/[A]o
t&frac12; =
[A]o/2k
ln2/k
1/k[A]o
[A] vs. t
ln[A] vs. t
1/[A] vs. t
linear Plot
B. Collision Model

effective collision
o sufficient speed to weaken existing bonds
o correct orientation to allow new bonds to form

potential energy diagram
Name __________________________
Practice Problems
A.
1.
2.
3.
4.

reaction rate and temperature
o high temperature = faster rate (more reactants have
sufficient energy (Ea) to react
5.
The reaction, X + Y  Z, is diagramed below.
Which is...
I
II III
changed when a catalyst is added?
the activation energy for the forward reaction?
H for the forward reaction?
For the reaction whose rate law is Rate = k[A]2, a plot of what
values vs. time is a straight line?
[A]
ln[A]
1/[A]
A first-order reaction has a 10-minute half-life. How many
minutes are needed to reduce the concentration from 1 to &frac14;?
10 min
20 min
30 min
Consider the following reaction mechanism.
Ce4+(aq) + Mn2+(aq)  Ce3+(aq) + Mn3+(aq)
Ce4+(aq) + Mn3+(aq)  Ce3+(aq) + Mn4+(aq)
Mn4+(aq) + Tl+(aq)  Mn2+(aq) + Tl3+(aq)
a. What is the overall reaction?
b.
What are the intermediates?
c.
What is the catalysts?
d.
What is the rate law if the first step is the slow step?
I-(aq) + CIO-(aq)  IO-(aq) + Cl-(aq)
Three initial-rate experiments were conducted; the results are
shown in the following table.
Experiment
[ClO-]
[I-]
[IO-]/t
(M)
(M)
(M•s-1)
1
0.017
0.015
1.35 x 10-3
2
0.052
0.015
1.22 x 10-2
3
0.016
0.061
4.19 x 10-3
a. Determine the order of the reaction with respect to each
reactant listed below. Show your work.
I-
o
C.
Ae-Ea/RT
Arrhenius equation: k =

graph lnk vs. 1/T will generate a straight line
with slope = -Ea/R

2 data points: ln(k1/k2) = (Ea/R)(1/T2 – 1/T1)
o R = 8.31, Ea in J, and T in K
o k1/k2 = rate1/rate2
Reaction Mechanism (14.6 to 14.7)

complex reactions occur in steps (elementary reactions),
where each step has 1 or 2 reactants with low Ea
o intermediate forms in early step and consumed later
o catalyst consumed in slow step and produced later
step 1: 2 Br- + 2 H+ + H2O2(aq)  Br2(aq) + 2 H2O
step 2:
Br2(aq) + H2O2(aq)  2 Br- + 2 H+ + O2(g)
overall:
2 H2O2(aq)  2 H2O + O2(g)
o catalyst: H+ and Br-—included in rate law
o intermediate: Br2—not included in rate law
o rate = k[Br-]2[H+]2[H2O2]
ClOb.
For the reaction,
(1) write the rate law for the reaction.
(2) calculate rate constant, k, and specify units.
c.
What is the activation energy (Ea) for this reaction if the
rate doubles when the temperature increases from
20.0oC to 30.0oC? ln(k1/k2) = (Ea/R)[1/T2 – 1/T1]
```