AP Chemistry Review
Oxidation-Reduction
• Oxidation means an increase in
oxidation state - lose electrons.
• Reduction means a decrease in
oxidation state - gain electrons.
• The substance that is oxidized
is called the reducing agent.
• The substance that is reduced
is called the oxidizing agent.
Half-Reactions
• All redox reactions can be thought
of as happening in two halves.
• One produces electrons Oxidation half.
• The other requires electrons Reduction half.
• Write the half reactions for the
following.
• Na + Cl2  Na+ + Cl• SO32- + H+ + MnO4- 
SO42- + H2O + Mn+2
Balancing Redox
Equations
• In aqueous solutions the key is
the number of electrons
produced must be the same as
those required.
• For reactions in acidic solution
an 8 step procedure.
 Write separate half reactions
 For each half reaction balance
all reactants except H and O
 Balance O using H2O
Acidic Solution
 Balance H using H+
 Balance charge using e Multiply equations to make
electrons equal
 Add equations and cancel
identical species
 Check that charges and
elements are balanced.
Practice
• The following reactions occur in
aqueous solution. Balance them
• MnO4 + Fe+2 Mn+2 + Fe+3
• Cu + NO3-  Cu+2 + NO(g)
• Pb + PbO2 + SO4-2  PbSO4
• Mn+2 + NaBiO3  Bi+3 + MnO4-
Basic Solution
• Do everything you would with
acid, but add one more step.
• Add enough OH- to both sides to
neutralize the H+
• Makes water
2• CrI3 + Cl2  CrO4 + IO4 + Cl
• Fe(OH) + H O  Fe(OH)
2
2 2
• Cr(OH)3 + OCl- + OH- 
CrO42- + Cl- + H2O
Thermochemistry
ENERGY AND WORK
(see definition of energy)
E = q(heat) + w(work)
Signs of q:
+q if heat absorbed
–q if heat released
w = -PV
NOTE: Energy is a state function. (Work and
heat are not.)
Signs of w (commonly related to work done by or to
gases)
+ w if work done on the system (i.e.,
compression)
-w if work done by the system (i.e.,
expansion)
When related to gases, work is a function of pressure
(pressure is force per unit of area) and volume
Exercise 1 Internal
Energy
Calculate ∆E for a system
undergoing an endothermic
process in which 15.6 kJ of
heat flows and where 1.4
kJ of work is done on the
system.
∆E = 17.0 kJ
ENTHALPY
Measure only the change in enthalpy, H
(the difference between the potential energies of the products
and the reactants)
H is a state function
H = q at constant pressure
(i.e. atmospheric pressure)
(true most of the time for us
and a very handy fact!)
Enthalpy can be calculated
from several sources including:
Stoichiometry
Calorimetry
From tables of standard
values
Hess’s Law
Bond energies
Calorimetry
The process of measuring heat
based on observing the
temperature change when a
body absorbs or discharges
energy as heat.
After all data is collected (mass
or volume; initial and final
temperatures) we can use the
specific heat formula to find the
energy released or absorbed.
We refer to this process as
constant pressure calorimetry.
** q = H @ these conditions.**
Specific heat capacity (Cp)
Same as specific heat but
specific to
1 gram of substance
Energy (q)
Released or gained -- q = mCpT
q = quantity of heat ( Joules or
calories)
m = mass in grams
ΔT = Tf - Ti
(final – initial)
Cp = specific heat capacity (
J/g°C)
Sample Problem B:
In a coffee cup calorimeter,
100.0 mL of 1.0 M NaOH
and 100.0 mL of 1.0 M HCl
are mixed. Both solutions
were originally at 24.6C.
After the reaction, the final
temperature is 31.3C.
Assuming that all solutions
have a density of 1.0 g/cm3
and a specific heat capacity of
4.184 J/gC, calculate the
enthalpy change for the
neutralization of HCl by NaOH.
Assume that no heat is lost to
the surroundings or the
calorimeter.
∆H = -5.6 kJ/mol
Now it is time to consider the
forces
that condense matter.
These can be due to ionic or
covalent bonding [intramolecular
forces — ionic stronger than
covalent] or much weaker attractive
forces we call intermolecular forces.
These are the forces between
(rather than within) molecules.
We briefly visited the IMF’s earlier
when discussing the nonideal behavior
of gases. These forces cause
changes of state by causing changes
among the molecules, NOT within
them.
Dipole-Dipole
Strongest IMF’s
Molecules with
dipoles orient
themselves so that
“+” and “-” ends of
the dipole are
close together.
Hydrogen Bonds
bonded H
H-“bond”
Dipole-dipole attraction in which
hydrogen on one molecule is attracted
to a highly electronegative atom on an
adjacent molecule. (F, O, N)
WHY is there such variation among
the covalent hydrides of groups IV
through VII?
Hydrogen bonding,
That’s why!
One would expect that BP would
increase with increasing molecular
mass [since the more electrons in a
molecule, the more polarizable the
cloud {more about that in the next
section}, the stronger the IMF’s, the
more E needed to overcome these
attractions and vaporize].
TWO Reasons
1. The lighter hydrides have the
highest En values which leads to
especially polar H-X bonds.
2. The small size of each dipole
allows for a closer approach of
the dipoles, further strengthening
the attractions.
London Dispersion
Forces
Weakest IMF’s
Relatively weak forces that
exist among noble gas atoms
and nonpolar molecules. (Ar,
C8H18)
Caused by instantaneous dipole
formation, in which electron
distribution becomes asymmetrical.
The newly formed dipoles now find
each other FAR more attractive
than before!
a.k.a.
Dipole-induced dipole if an ion or
polar molecule causes the distortion.
OR
Induced dipole-induced dipole if a
nonpolar moleule sets off the chain
reaction of induction like in iodine.
The ease with
which the electron
“cloud” of an
atom can be
distorted is called
polarizability.
You’ll want to
write about
polarizability when EXPLAINING
these concepts.
PHASE DIAGRAMS
Closed Systems
Represents phases as a function of
temperature and pressure.
Critical Temperature: Temperature
above which the vapor
cannot be liquefied.
Critical Pressure: Pressure
required to liquefy AT the
critical temperature.
Critical Point: Critical temperatue
and pressure (for water, Tc = 374° C
and 218 atm).
CHEMICAL
KINETICS:
THE RATES AND
MECHANISMS OF
CHEMICAL REACTIONS
THE COLLISION THEORY
OF REACTION RATES
•Particles must collide.
•Only two particles may collide
at one time.
•Proper orientation of colliding
molecules so that atoms can
come in contact with each
other to become products.
This new collision
product is at the
peak of the
activation energy
hump and is called
the activated
complex or the
transition state. At
this point, the
activated complex
can still either fall to
reactants or to
products.
Initial rxn rate =
k[A]m[B]n[C]p
k = rate constant
[A] = concentration of reactant A
[B] = concentration of reactant B
[C] = concentration of the catalyst—
won’t see
this too often in AP
m = order of reaction for reactant A
n = order of reaction for reactant B
p = order of reaction for the catalyst C
Adding the orders of each
reactant gives the overall
order of the reaction.
Experiment
Number
Initial Rate
mol/(Lhr)
Initial concentration
[A]o
Initial concentration
[B]o
1
0.50 x 10-2
0.50
0.20
2
0.50 x 10-2
0.75
0.20
3
0.50 x 10-2
1.00
0.20
4
1.00 x 10-2
0.50
0.40
5
1.50 x 10-2
0.50
0.60
The table below gives the results of
four experiments. Using these data,
determine the orders for all three
reactants, the overall reaction order,
and the value of the rate constant.
What is the value of k? What are the
units of k?
Experiment
Initial [BrO3-]
Initial [Br –]
Initial [H+]
1
2
3
4
0.10
0.20
0.20
0.10
0.10
0.10
0.20
0.10
0.10
0.10
0.10
0.20
Measured initial
rate (mol/Ls)
8.0 x 10-4
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
12.3 DETERMINING THE
FORM OF THE RATE LAW
--experimental convenience
Note the shape of this curve! It will save
you time in the future!
A SUMMARY
ELEMENTARY
STEP
MOLECULARITY
RATE EXPRESSION
A products
unimolecular
rate = k[A]
A + B  products
bimolecular
rate = k[A][B]
A + A  products
bimolecular
rate = k[A]2
2 A + B  products*
termolecular*
2
rate = k[A] [B]
• the slowest step is the rate
determining step
• reaction intermediate-produced in one step but
consumed in another.
• catalyst--goes in, comes out
unharmed and DOES NOT
show up in the final rxn.
Buffers
Determine the [H3O+] and
[C2H3O2-] in a solution that is
0.100 M
in both acetic acid and
hydrochloric acid.
Exercise 2:The pH of a Buffered
Solution
A buffered solution contains 0.50M
acetic acid (HC2H3O2, Ka = 1.8 X 105) and
0.50 M sodium acetate (NaC2H3O2).
Calculate the pH
A solution is 0.120 M in acetic acid
and 0.0900 M in sodium acetate.
Calculate the [H+] at equilibrium.
The Ka of acetic acid is 1.8 x 10-5.
[H+] = 2.4 x 10-5
Bonding
Bonding
Organic Chemistry
Alkanes
Organic Naming Prefixes
Alkyl Groups
Examples of Isomers
The formula C4H10 has two different
structures
CH3
CH3CH2CH2CH3
Butane
CH3CHCH3
2-methylpropane
When a CH3 is is used to form a branch,
it makes a new isomer of C4H10.
Alkenes
Alkynes
Carbon-carbon __________
bonds
Names end in -_________
HCCH
ethyne(acetylene)
HCC-CH3
propyne