AP Chemistry Study Guide (Part 1 of 3)
Chapter 1 - Basics
Metric System
Prefixes
Conversions
Definitions
Methods of Separating Mixtures
Table of Contents
Chapter 2 - Nomenclature/History
Nomenclature
Binary
Greek (Molecular)
Roman
Latin
Acidic
Basic
Hydrates
Dead Guys
John Dalton
Guy-Lussac
Avagadro
J.J. Thompson
Rutherford
Robert Millikan
Origins of Radioactivity
Definitions
Chapter 3 - Stoichiometry
Stoichiometry
Empirical Formula
Molecular Formula
Gravimetric Analysis
Determining the Limiting Reactant
Percent Yield/Percent Error
Chapter 4 - Types of Reactions/Solutions
Parts of Solutions
Hydration
Three Types of Solutions
Acidic and Basic Solutions
Measuring Solutions
Making Solutions
Dilution
Types of Reactions
Titration
Redox Reactions
Definitions
Chapter 5 - Gas Laws
Table of Gas Laws
Boyles, Charles, Guy Lussac, and Avagadro's Law
Combined Gas Law
Ideal Gas Law
Dalton's Law
AP Chemistry Study Guide (Part 1 of 3)
Corrected Values - Van der Waal's Forces
Graham's Law of Effusion
Mole Fraction
Root Mean Square Velocity
Kinetic Molecular Theory
Stoichiometry of Gases
Chapter 6 - Thermochemistry
Basics of Thermodynamics
General Picture
Direction of Energy
Heat and Work (Basics)
The Laws of Thermodynamics
The Kinds of Energies
Entropy Changes
Spontaneity
Systems at Equilibrium
Putting =H>, =S>, and =G> Together
Heat and Work (Detailed)
Hess' Law
In-Depth Look at Hess' Law
Common Exothermic/Endothermic Reactions
Enthalpy of Hydration ΔHhyd
Definitions
Chapter 7 - Atomic Structure and Periodic Trends
Basics - Parts of a Wave
Equations/Constants
Electromagnetic Waves
Hydrogen Spectrum
The Bohr Model
Electron Configuration
Exceptions to Electron Configuration
Quantum Numbers
Polyelectronic Atoms and Effective Nuclear Charge
Heisenberg Uncertainty Principle
Quantum Energy
Einstein
Periodic Trends
Cations vs Anions
Chapter 8 - Bonding
Ions
Calculating Lattice Energy
Partial Ionic Character
Ionic Bonds
Energy of Ionic Attraction or Repulsion
Covalent Bonding
Dipole Moments
How It's Drawn
Covalent Bond Energies
Exceptions to Octet Rule
Resonance/Formal Charge
VSEPR
AP Chemistry Study Guide (Part 1 of 3)
Chapter 9 - Hybridization and Molecular Orbitals
Sigma and Pi Bonds
Bond Order
Pi Bonds
Hybridization
Magnetism
The Molecular Orbital Model
Homonuclear Diatomic Molecules
Let's Screw With Our Brains...
Chapter 10 - Liquids and Solids
The 2 Forces
Types of Intermolecular Forces
Dipole-Dipole Forces
Hydrogen Bonding Forces
London Dispersion Forces
Van der Waal's Forces
Liquids
Surface Tension
Beading
Capillary Action
Viscosity
Solids
Crystals and Types of Unit Cells
Special Atomic Solid (Carbon)
Molecular Solids
Molecules Without Dipole
Ionic Solids
Vapor Pressure
Condensation
Dynamic Equilibrium
What is This?
Wait...What?
What We Can Conclude
Phase Diagrams
Chapter 11 - Solutions and Solubility
Review
Energy of Making Solutions
Types of Solvent and Solutes
Structure and Solubility
Pressure Effects (Gases Dissolving in Liquids)
Everyday Example
Henry's Law
Temperature Effects
Vapor Pressure of Solutions
Raoult's Law
Deviation
Colligative Properties
Boiling Point Elevation/Freezing Point Depression
Electrolytes in Solution/Van Hoff Factor
Osmotic Pressure
Equations Involving the Van Hoff Factor
Other Definitions
AP Chemistry Study Guide (Part 1 of 3)
Chapter 12 - Chemical Kinetics
Factors That Affect the Rate of Chemical Reactions
Concentration
Temperature
Catalysts
Homogeneous Catalysts
Heterogeneous Catalysts
Reaction Mechanisms
Important Terms
Requirements
Integrated Rate Law
Rate Law Graphs
Arrhenius Equation/Derivation of Arrhenius Equation
Final Notes on Kinetics
Chapter 13 H Chemical Equilibrium
Beginning Notes
Special K Table
Relationship Between Keq and Kp
Reaction Quotient (Q)
Le Châtlier's Principle
Concentration
Pressure/Volume
Temperature
Catalysts
Reverse Reactions
What if it isn't at Equilibrium?!?!?!?
In-depth look at ICE Charts
Chapter 14/15 - Acids, Bases, Titration, and Equilibrium
Acids
Bases
Strong Acids
Strong Bases
Definitions
Molecularity of Acids/Bases
Electronegativity and Acids
Oxidation Numbers
Generalizations of Acids/Bases
Acid-Base Reactions and Equilibrium
Neutral Solutions
Acidic or Basic Favored
Common-Ion Effect
Acid-Base Reactions and Calculations
Major Species
Acid/Bases: Return of the Sig Figs
Titration and Equilibrium
Equivalence Point
In-depth look at Titration Reaction
3 Sets of Important Steps for Titration Reactions
Titration Curves
Buffer Solutions and the Hendersen-Hasselbach Equation
Example
Hendersen-Hasselbach Equation
Percent Ionization
AP Chemistry Study Guide (Part 1 of 3)
Solubility Product Constant
Q and Ksp
Relation to pH
Complex Ions (Ligands)
Complex Ion Nomenclature
Chapter 16 - Spontaneity, Entropy, and Free Energy
Equations (New and Old)
Relationship Between ΔG> and ΔG
Chapter 17 - Electrochemistry
Basics
Galvanic Cells (Cell Diagram)
Standard Reduction Potential (SRP) Chart
Calculating Voltage From a Picture
Spontaneity Table
Nernst Equation
Application
Batteries
Dry Cell or Leclanché Cell
Mercury Batteries
Lead Storage Batteries
Fuel Cell
Solid State Lithium
Electrolysis
Application
Chapter 18 - Nuclear Chemistry
Basics
Why Nuclei are Unstable
Nuclear Trends
Magic Numbers
Isotopes
Half-Life Calculations
Types of Radioactive Decay
Band of Stability
Synthesis of Nuclei
Binding Energy
Example
Chapter 1919-21 Aren't Important
Chapter 22 - Organic Chemistry
Basics
Nomenclature
Alkanes
Cycloalkanes
Alkanes Continued
Carbon Numbering
Oh Snap, 2 Numbers!
Relation Between Carbon Number and Substituents
But Wait! There's More! (Isomers)
Alkanes and Alkynes
Structural Formulas
Acetylene
AP Chemistry Study Guide (Part 1 of 3)
Additional Alkane Information
Addition Reactions of Alkenes
Markovnikov's Rule
Anti-Markovnikov's Rule
Hydro-boration Oxidation Reaction
Radical Addition Reaction
Exceptions to Radical Additions
Advanced Nomenclature
Cis- vs. TransExceptions to the Longest Chain Rule
Cycloalkenes
Special Alkene, Benzene
Benzene Nomenclature
Ortho-, Meta-, ParaReactions Involving Benzene
Basic Function Groups
Ester Nomenclature
Organic Chemistry Overview
Miscellaneous
AP Chemistry Study Guide (Part 1 of 3)
Metric System
Mass - kilograms (kg)
Length - centimeters (cm)
Time - seconds (s)
Volume (L)
Prefixes
Tera (T)
Giga (G)
Mega (M)
Kilo (k)
Deci (d)
Centi (c)
Milli (m)
Micro (μ)
Nano (n)
Conversions
Volume
1mL = 1cm3
1L = 1dm3
22.4L = 1 mole (at STP)
1L = 1.06qt
28.32L = 1ft3
Energy
1J = 1Nm (Newton-meter)
4.184 J= 1cal
4.184 kJ = 1Kcal
Chapter 1 - Basics
Temperature - Kelvin (K) or Celsius (>C)
Electric Current - ampere (amp, A)
Amount of a Substance - (mole)
1,000,000,000,000
1,000,000,000
1,000,000
1,000
.1
.01
.001
.000001
.000000001
1012
109
106
103
10-1
10-2
10-3
10-6
10-9
Mass
454g = 1lb
1kg = 2.205lbs
Other Important Conversions
1Latm = 101.3J
1J = kg P m2/s2
Pressure
1atm = 760 torr
1atm = 760mmHg
1atm = 101.325 kPa
1 atm = 14.7psi
Length (English-Metric)
1 in = 2.54cm
1m= 1.094yd
1m = 3.28ft
1.6km = 1mi
5280ft = 1mi
1 Angstrom (A0) = 1 P 10-10m
Temperature
K = >C + 273
> C = K - 273
R
>C = (>F T 32)
Speed
S
S
>F = >C + 32
Warp Factor 1.71= 15.00 x 108 m/s
R
STP: Standard Temperature and Pressure - 1.00atm
0°C
Room temperature: 25>C
Definitions
Precise - How repeatable something is
Accurate - How close/correct to the actual value
Mass - Resistance to change in motion
Weight - Force of gravity
Random Error - Results fluctuate each time
Systematic Error - Same thing wrong every time
Theory - Changeable. Explains how things behave the same in different systems
Law - Unchangeable. Summary of observations
Law of Conservation of Mass - The mass on the left is equal to the mass on the right
Homogeneous Mixture - A mixture that can be separated through chemical means only (filtration, etc.)
Heterogeneous Mixture - A mixture that can be separated by physical means
Substance - Something with constant composition (uncontaminated)
Physical Change - A change in the state of matter (solid - gas...Still contains same compounds)
Chemical Change - A change in the composition of the molecules (New compounds formed)
AP Chemistry Study Guide (Part 1 of 3)
Methods of Separating Mixtures
Distillation - Vaporizes a liquid (based on volatility*), condenses the most volatile vapors, that then condenses
goes through a condenser to change it back into a liquid.
*Volatility - How readily substances become gases
(The lower the temperature when the substance becomes gaseous, the more
volatile that substance is)
Filtration - Used to separate a mixture containing a solid and a liquid. Pour the mixture on a mesh (usually
filter paper) and the liquid will filter through, while the solid stays on the mesh.
Chromatography - Separates a system with 2 phases (mobile and stationary). Phases are based on the
affinities of the components (They move throughout the system at different rates).
- High affinity for mobile phase moves quickly through chromatography system, high affinity
for solid phase moves slowly though chromatography system.
Paper Chromatography - Uses a strip of porous paper (such as filter paper) for the stationary phase. Then a
drop of the mixture that needs to be separated is placed on the paper, and dipped in
liquid of the mobile phase which will travel up the paper.
- The components with the weakest attraction to the paper will travel up
faster than those that cling to the paper.
Nomenclature
Chapter 2 - Nomenclature/History
Binary
Greek (Molecular)
Roman
Latin
Acid
Base
Hydrates
IA, IIA, IIIA attached to VA, VIA, VIIA
IVA, VA, VIA, VIIA attached to IVA, VA, VIA, VIIA
Transition Metals
Transition Metals
Aqueous
Aqueous
Any compound with water attached
Binary
Metal name stay same and goes first
Nonmetal ending --> -ide
If Nonmetal is polyatomic ion, keep name the same
Greek (Molecular)
Prefixes: Mono, Di, Tri, Tetra, Penta, Hexa, Hepta, Octa, Nona, Deca (1-10)
Metal is placed first. If there is only 1 of the metal, prefix is not placed.
Nonmetal placed after, and always contains a prefix.
Roman
First find oxidation charge of the transition metal.
Name the metal
Place the charge (in Roman Numerals) in parenthesis after the metal
Place the nonmetal after, using binary rules.
Latin
Cuprous
Co+2
Cobaltous
Cu+1
+2
Cu
Cupric
Co+3
Cobaltic
Plumbous
Sn+2
Stannous
Pb+2
Plumbic
Sn+4
Stannic
Pb+4
Ferrous
Fe+3
Ferric
Fe+2
Metal is named first, followed by nonmetal using Binary rules
AP Chemistry Study Guide (Part 1 of 3)
Acidic
MUST be aqueous
Acidic if the H+1 ion is at the beginning of the compound
If the acid does not contain oxygen add the prefix hydroName the nonmetal by using Binary rules and changing -ide --> -ic
If the nonmetal is polyatomic and ends in -ate, change -ate --> -ic
If the nonmetal is polyatomic and ends in -ite, change -ite --> -ous
Basic
MUST be aqueous
Basic if an (OH)-1 ion is attached to the compound
Name the metal using Binary rules, or if transition metal Roman rules
Add hydroxide at the end
Hydrates
Used to account for water attached to some compounds
Name the compound and then select the Greek prefix that corresponds to the number of moles of water
attached and add -hydrate.
Dead Guys
John Dalton
Atomic theory stated:
All elements are made of atoms
Atoms of each element are identical
Chemical compounds are formed when atoms combine
Atoms are not changed in chemical reactions
Guy-Lussac
Under constant temperature and pressure, compounds react in whole number ratios by volume
Avagadro
Under constant temperature and pressure, volumes of a gas contain the same number of particles
J.J. Thomson
Cathode-Ray experiment, discovered electron
Rutherford
Gold foil experiment, discovered atom is mostly empty and contains a small, dense nucleus
Robert Millikan
Oil drop experiment, found exact mass of electron
Origins of Radioactivity
Discovered by Bequerel (accidently)
Alpha - Helium Nucleus, Beta - High-speed electron, Gamma - High energy light
Definitions
Law of Definite Proportions - Compounds have a constant composition by mass (Mass is proportional on both
sides)
Law of Multiple Proportions- 2 elements that form more than 1 compound can be reduced to a small whole
number ratio by combining the masses of the second element with 1 gram of the
first element.
-Water has 8g of oxygen per 1g of hydrogen
-Hydrogen peroxide has 16g of oxygen per 8g of hydrogen
-16:8----2:1
H2O: 1g H2 W
X YZ[\ ]^
_.`_a ]^
H2O2: 1g H2 W
bW
X YZ[\ ]^
_.`_a ]^
X YZ[\ c
X YZ[\ ]^
bW
bW
X YZ[\ c^
X YZ[\ ]^
Xd.``a c
X YZ[\ c
bW
b=8g O
e_.``a c
X YZ[\ c^
b= 16g O
AP Chemistry Study Guide (Part 1 of 3)
Chapter 3 - Stoichiometry
Stoichiometry
Mole - 6.022 x 1023 of anything
- Number of atoms in exactly 12g of C-12
Molar Mass - Add up mass of all elements on periodic table
Percent Composition - Percent of each element in a compound
fghh Zi j[\Y\kl
q 100
fghh Zi mZYnZokp
Empirical Formula
Based on mole ratios
Given: Percents of elements
Assume: 100%=100g
1.) Find the moles of each element
2.) Divide by lowest moles
- If given a .5 after dividing, multiply all moles by 2
- If given a .333 or .666 after dividing, multiply all moles by 3
- If given a .25 after dividing, multiply all moles by 4
Finds the lowest ratio
Molecular Formula
Finds the formula for actual molecule
Given: Molar mass of the molecule
1.) Find Empirical Formula
2.) Find molar mass of Empirical Formula
3.) Molar Mass/Empirical Formula molar mass=ratio
4.) Multiply all elements in Empirical Formula by ratio found
Gravimetric Analysis
Basically, just a fancy phrase for using
stoichiometry to find the mass of a ppt
Must write Net Ionic Equations to do these
problems
Example:
What mass of Fe(OH)3 would be produced by
reacting 75.0 mL of .105M Fe(NO3)3 with 125 mL
of .150M NaOH?
Fe(NO3)3(aq) + 3NaOH(aq) --> 3NaNO3(aq) +
Fe(OH)3(ppt)
NaOH is limiting reactant because it has higher
mole ratio
Find moles of NaOH by multiplying molarity and
volume (in liters)
Use mole to mole stoich to find out how many
moles of Fe(OH)2 there are
Use mole to mass stoich to find out how many
grams of Fe(OH)2 there are
Determining the Limiting Reactant of a Reaction
Quick way: Look at which reactant has higher mole ratio
- C3H8 + 5O2 --> 3CO2 + 4H2O [50.0g C3H8 75.0g O2 How many grams of CO2 ]
-O2 is the limiting reactant because O2 has 5 moles, while C3H8 has 1
Long way (safe): Do the stoichiometry for both reactants. Whichever gives the smallest number is the L.R.
X uvwx yz {|
e uvwx y
}}.`X~ y
b WX uvwx y {^ b WX uvwx y^ b = 150.g CO2
}}.XX~ yz {|
z |
^
50.0g C3H8 W
X uvwx ^
e uvwx y
}}.`X~ y
b W S uvwx ^ b WX uvwx y^ b = 61.9g O2
e_.``~ ^
^
^
75.0g O2 W
Percent Yield/Percent Error
Given: Theoretical Yield
log[ \[p
q 100
% Yield =
\Z\lg[ \[p
% Error = 100% - % Yield
Chapter 4 - Types of Reactions/Solutions
Parts of Solutions
Solution - homogenous mixture
Solute - What gets dissolved
Solvent - The stuff doing the dissolving
Soluble - Can be dissolved
Miscible - Liquids that dissolve each other
Hydration
The process of breaking ions of salts apart with water molecules
-Water is good solvent because it has polar molecules
AP Chemistry Study Guide (Part 1 of 3)
Three Types of Solutions
Strong electrolytes - Completely breaks apart into ions. Conducts electricity well.
Weak electrolytes - Partially breaks into ions. Conducts slight electricity.
Non-electrolytes - Does not fall apart. Does not conduct.
Acidic and Basic Solutions
Acids form H+ ions when they fall apart
Bases form (OH)- ions when they fall apart
Strong acids: H2SO4, HNO3, HCl, HBr, HI, HClO4
Strong bases: Group IA, IIA metal attached to an (OH)- ion
Measuring Solutions
Concentration - How much is dissolved
fZ[\h Zi hZ[ol\
Molarity - =
- =
l\h Zi hZ[olZk
fZ[\h Zi hZ[ol\
fZ[gl
- = q
Making Solutions
Describe how to make 250. mL of a 2.0M copper (II) sulfate dihydrate solution
2.0M =
uvwx
._S`
XRS.d}_~ y P _{^
b=
X uvwx y P_{^
.50mole CuSO4 P 2H2OW
98g CuSO4 P H2O
Weigh out 98g of CuSO4 dissolved in 250. mL of H2O to obtain a 2.0M CuSO4 solution
Dilution
The process of adding more solvent to a known solution
M1V1 = M2V2 (Moles are the same)
Stock solution is a solution of known concentration used to make more dilute solutions
18.5mL of 2.3M HCl is added to 250 mL of water. What is the concentration?
(2.3M)(.0185L) = (.250L)(X)
.04255 mole = .250X
M=.17M
Take 18.5 mL of 2.3M solution of HCl to 250. mL of H2O to obtain a 0.17M solution
Types of Reactions
Precipitation Reactions - 2 aqueous solutions of ionic compounds poured together to form a solid (ppt)
-Molecular: 3NaOH(aq) + FeCl3(aq) ---> 3NaCl(aq) + Fe(OH)3(s)
-Ionic: 3Na+1(aq) + 3(OH)-(aq) + Fe+3(aq) + 3Cl-1(aq) ---> 3Na+1(aq) + 3Cl-1(aq) + Fe(OH)3(s)
-Net Ionic: Fe+3(aq) + 3(OH)-(aq) ---> Fe(OH)3(s)
- All other ions are known as spectator ions
Acid-Base Reactions: Acid - proton donor; Base - proton acceptor (Transfer of H+ ions)
-Strong acid + Strong Base ----> Salt + water
Titration - A reaction where the indicator changes color = endpoint
# of H+ P MAVA = # of OH- P MBVB
Redox Reactions - The transfer of electrons
- Oxidation states
Element
0
VA, VIA, VIIA ion
-3, -2, -1 respectively
Oxygen
-2 (Peroxide -1)
Hydrogen
+1, -1 only when attached to alkali metal
VIIA
-1 up to +7 (Fluorine ALWAYS -1)
Total charge of compound
0
Total charge of ion
Charge of ion
AP Chemistry Study Guide (Part 1 of 3)
Titration
Volumetric analysis - Technique used for determining the amount of a certain substance through titration.
Titrant - The solution of a known concentration
Analyte - The solution containing the substance being analyzed
Equivalence point - The point where enough titrant has been added to react exactly with the analyte.
- Also known as the stoichiometric point
Indicator - A substance added before titration, which changes the color of the analyte at (or very near) the
equivalence point
Endpoint - The point where the indicator changes color
Equivalence point Endpoint
Redox Reactions
Acidic Medium: 1.) Write skeletal equation with oxidation numbers
2.) Write the Oxidation/Reduction half reaction
3.) Balance atoms that are NOT oxygen in each half reaction
4.) Balance the O2 atoms by adding H2O to the side that needs it
5.) Balance the Hydrogen's by adding H+ ions to the side that needs it
6.) Balance the charges by adding electrons
7.) Put the half reactions together and cancel
Basic Medium:
1.) Write skeletal equation with oxidation numbers
2.) Write the Oxidation/Reduction half reaction
3.) Balance atoms that are NOT oxygen in each half reaction
4.) Balance the O2 atoms by adding H2O to the side that needs it
5.) Balance the Hydrogen's by adding H+ ions to the side that needs it
6.) For every H+ ion added, add an OH- ion to BOTH sides*
7.) Balance the charges by adding electrons
8.) Put the half reactions together and cancel
Oxidation loses electrons/gains charge
Reduction gains electrons/loses charge
Oxidizing Agent - Electron acceptor
Reducing Agent - Electron donor
Oxidation 1/2 ALWAYS has electrons on products
Reduction 1/2 ALWAYS has electrons on reactants
*Wherever there is an H+ and OH- ion, it will condense into H O
2
Definitions
Aqueous solution - A solution where water is the solvent
Hydration - The process of using water to dissolve a salt into its ions
]^ c(¡)
NH4NO3(s) ¢££¤ NH4+(aq) + NO3-(aq)
Solubility - The capability of a substance to be dissolved
Electrical conductivity - A solutions ability to conduct an electric current
AP Chemistry Study Guide (Part 1 of 3)
Chapter 5 - Gas Laws
Remember that temperature is ALWAYS in Kelvin when dealing with gases
Boyle's Law:
Guy Lussac's Law:
Avogadro's Law:
Combined Gas Law:
§¨
¨
¥X ¦X = ¥_ ¦_
©¨
k¨
=
§¨ ©¨
©^
k^
¨
Dalton's Law:
Mole Fraction:
´X =
k¨
kµ¶·¸¡
=
§¨
§µ¶·¸¡
=
¬f­ agh _
¬f­ agh X
=
©^
^
Ideal Gas Law:
§^ ©^
PV = nRT
^
Corrected Pressure:
k _
V' = V - nb
Graham's Law of Effusion
«gl\ Zi \iiohZk Zi agh _
=
©¨
¨
Corrected Volume:
PTotal = P1 + P2... + Pn
«gl\ Zi \iiohZk Zi agh X
=
Charle's Law:
§^
^
P' = Pobs + aW b
©
"Real" Gas Law:
k _
®¥Z¯h + W b ° q (¦ T ±) = ²³
©
Root Means Square Velocity:
¹Yh = º
e«
f
Gases at STP:
Elements: H2, N2, O2, O3 (ozone), F2, Cl2, He, Ne, Ar, Kr, Xe, and Rn
Compounds: HF, HCl, HBr, HI, CO, CO2, NH3, NO, NO2, N2O, SO2, H2S, HCN
Note that the compounds are not in solution, so they are gases
Diffusion vs. Effusion
Diffusion - The process of gas molecules spreading throughout a room (cooking scents)
Effusion - The process of gas molecules escaping a narrow exit into a room (aerosol)
Boyles, Charles, Guy Lussac, and Avagadro's Law
P = Pressure (atm)
V = Volume (L)
T = Temperature (K)
n = Moles (mole)
When multiplied, there is an inverse relation
-Boyle's Law: Pressure increases, Volume decreases (vice versa)
When divided, there is a direct relation
-Guy Lussac: Pressure increase, Temperature increases (vice versa)
-Charles: Volume increase, Temperature increases (vice versa)
- Avagadros: Volume increase, moles increases (vice versa)
Ideal Gas Law
P: Pressure (atm)
V: Volume (L)
n: Moles (mole)
R: Universal Gas Law
ddConstant
.0821 Latm/Kmole
T: Temp. (K)
Corrected Values
Apostrophe is just
notation for corrected
value, not derivative
Values for a and b will be
given and are dependent
upon the gas
Pobs = Pobserved = Pressure
if it were ideal
Graham's Law
MW = Molecular Weight
Mole Fraction
Unitless
Root Mean Square
Velocity
R = 8.3145 J/Kmole
M = mass in kilograms
Units: J/kg
What's the Difference?
How can "R" be 2 different
values?
If we are working with
gases, we will use the
.0821 Latm/Kmole
If we are working with
Energy, we will use the
8.31 J/moleK
If you go back to Chapter
1, one of the important
conversions was that
1Latm = 101.3 J
So: .0821 x 101.3 = 8.31
AP Chemistry Study Guide (Part 1 of 3)
Combined Gas Law
Combination of Boyle's, Guy Lussac's, and Charle's Laws
Ideal Gas Law
In most cases, gases are not ideal and are real gases
Gases only approach ideal behavior at low pressure (< 1 atm) and high temperature
-Use law anyways, unless told otherwise
Dalton's Law
Either have to find PTotal when given the partial pressures, or find a partial pressure when given PTotal and
some of the partial pressures
PTotal can be found by altering ideal gas law equation:
( Zlg[ )²³
¥Zlg[ =
¦
Corrected Values - Van der Waals Forces
Real molecules interact with each other and take up space
The actual volume is less because of the particle sizes
More molecules will have a greater effect
Molecules will be attracted to each other which will make the observed pressure less than ideal
Depends on the number of molecules/liter
Graham's Law of Effusion
Gives us best estimate of the rate in which a gas will spread throughout a room
-As gases spread throughout a room, collisions with other molecules will slow down the rate
Mole Fraction
Gives us the ration of moles of the substance to the total number of moles
Units will cancel out
Root Mean Square Velocity
Gives us the average amount of kinetic energy exerted by the gas particles
Kinetic Molecular Theory
Assumptions (Only apply to ideal gases):
- Particles are so small, we can ignore volume
- Particles are in constant motion and their collisions cause pressure
- Particles do not affect each other, neither attracting or repelling
- Average kinetic energy is proportional to the Kelvin temperature
-Two gases at same temperature have same kinetic energy
Tells us the average kinetic energy:
3
¼½¾\ = ²³
2
R = Gas Law Constant 8.3145 J/Kmole
T= Temperature in Kelvin (K)
Stoichiometry of Gases
1 mole of any gas at STP = 22.4L
Just like regular stoichiometry, but may need to use gas laws in order to find a missing value
AP Chemistry Study Guide (Part 1 of 3)
Chapter 6 - Thermochemistry
Basics of Thermodynamics
Energy - The ability to do work
- Made of heat and work
Heat - Energy transferred between objects because of temperature differences
Work - A force acting over a distance
Exothermic Process - Releases heat/energy
Endothermic Process - Absorbs heat/energy
General Picture
The universe is divided into 2 halves: The system and the surroundings
System - Part of the world we want to study
Surroundings - Everything outside the system
Open system - Exchanges both matter and energy with surroundings
- Open flask - Chemicals (matter) capable of escaping, Temperature (heat - A kind of energy) of
reaction can affect be affected by stuff outside the flask
Closed system - Exchanges only energy with surroundings
- Sealed flask - Chemicals (matter) cannot escape, Temperature (heat - A kind of energy) of
reaction can affect or be affected by stuff outside flask
Isolated system - Exchanges neither matter or energy with surroundings
- Thermos flask - Chemicals (matter) cannot escape, Temperature (heat - A kind of energy) of
reaction cannot affect or be affected by stuff outside the thermos flask
Direction of Energy
Measured in Joules or calories
Number tells how many Joules or calories there are
Sign tell the direction
- Negative sign means the reaction is exothermic
- Positive sign mean the reaction is endothermic
Heat and Work (Basics)
Heat - Energy transferred from one object to another object (both at different at temperatures) through some
kind of contact
Work - Energy transferred that takes place when an object is moved against an opposing force
Thermal motion - Random molecular motion that is stimulated by heat energy
Uniform motion - Constant molecular motion that is stimulated by work energy
The Laws of Thermodynamics
First Law of Thermodynamics - Energy of the universe is constant (cannot be created or destroyed)
- Also known as the Law of Conservation of Energy
States that: =½ = ¿ + À
q is heat and w is work
Second Law of Thermodynamics - The disorder or entropy of a system tends to increase
- Low entropy = low disorder, high entropy = greater disorder
Third Law of Thermodynamics - Entropy of a pure crystalline solid at absolute zero is zero
The Kinds of Energies
Energy
Internal Energy (U/E)
Enthalpy (H)
Entropy (S)
Gibbs Free Energy
Bond Disassociation Energy*
Equations
=¹ = ¿ + À
=H>rxn = Á = H>Â (Products) T Á = H>Â (Reactants)
=S>rxn = Á S>§Zpolh T Á Ã>«\glgklh
=G>rxn = Á = G>Â (Products) T Á = G>Â (Reactants)
=H = Á ÄÅ ½Æ«\glgklh T Á ÄÅ ½Æ§Zpolh
*Remember that Bond Disassociation Energy is:
REACTANTS - PRODUCTS
AP Chemistry Study Guide (Part 1 of 3)
Internal Energy (U/E) - Energy that enters/leaves a system
- Some energy can return to surroundings as expansion work
- Ç =U < ¿
Enthalpy (H) - A state function: H = E + PV
- E = Energy, P = Pressure, V = Volume
- =H>i = Heat of formation
- =H = q
- Measured in kJ/mole
Entropy (S) - Measured in J/K mole
- If =S > 0 then disorder is increasing, becoming less organized, and more random
- Thermodynamically favorable
- If =S < 0 then disorder is decreasing, becoming more organized, and less random
=ÃhoZokpkah =
È=]
(Remember to convert to Joules)
-Depends directly on heat flow (=H) and inversely on temperature (T)
-If heat flows to the surroundings, Kinetic Energy is increased in the surroundings
and the entropy will increase. Entropy will decrease in the system.
- A constant heat flow at different temperatures will have a different impact
dependent upon a low or high temperature.
- High temperature will cause a lower entropy change on surroundings
-Low temperature will cause a higher entropy change on surroundings
Entropy Changes of Physical Changes and Chemical Reactions
S(S) < S(l) < S(g)
- Can quickly determine whether or not the entropy of a reaction is positive or negative by
looking at the phases
- If phases are the same, then look at what side has more moles
2O3(g) ---> 3O2(g)
More moles are produced so the entropy is increased (+)
-Aqueous to solids will have a negative entropy (become more ordered)
- Entropy will increase if: Volume increases Temperature increases, Kinetic Energy increases,
particles dissolve, or solid --->liquid--->gas
Gibbs Free Energy - Energy free to do useful work
=G = =H - T=S
- If =G> = negative number, reaction is spontaneous
- If =G> = positive number, reaction is non-spontaneous
- If =G> = 0, reaction is at equilibrium
Bond Disassociation Energy - The enthalpy change required to break a particular bond
- 2 useful things to remember: Draw the molecular structure for the compounds
REACTANTS - PRODUCTS
Spontaneity
A chemical reaction is spontaneous if there is an increase in the total entropy of the system and surroundings
Spontaneous change - A change that occurs without being driven by an external influence
Non-spontaneous change - A change that occurs only when driven by an external influence
Spontaneous exothermic reactions are common because they release heat, increasing entropy of
surroundings
Spontaneous endothermic reactions occur only when entropy of the system increases enough to overcome
the decrease in entropy of the surroundings
Systems at Equilibrium
A+BHC+D
No overall change in disorder
=S É 0 (No entropy change)
AP Chemistry Study Guide (Part 1 of 3)
Putting =Ê>, =Ë>, ÌÍÎ =Ï> ÐÑÒÓÔÕÓÖ
=×>
-
=Ã>
+
-
+
+
+
-
=Ø>
- at all temperatures
- at low temperatures
+ at high temperatures
+ at low temperatures
- at high temperatures
+ at all temperatures
Spontaneity
Spontaneous
Dependent upon *
=Ø>
Dependent upon **
=Ø>
Non-spontaneous
Using the equation =G = =H - T=S and the table we can assume:
=G = large negative - (T x small negative)
We know =H is a large negative because it is measured in kJ and needs to be converted to Joules. We
know =S is a small negative because it is already in Joules and no conversion is required.
So , if the temperature is low, meaning the temperature (in Kelvin) has a point in which when
multiplied by the small negative (=S) will result in a number that, when added to the large
negative (=H), the =G will be negative. If the temperature is above that point then we know that =G
will be positive.
-Hypothetically: =H = -1000J/mole and =S = -100J/Kmole T = 9K
=G = -1000J /mole - (9K x -100 J/Kmole)
=G = -1000J/mole - (-900J/mole)
=G = -1000J/mole + 900 J/mole = -100J/mole
-In this case we can see that a temperature above 9.99K will result in either
equilibrium or a positive =G
** Using the equation =G = =H - T=S and the table we can assume:
=G = large positive - (T x small positive)
We know =H is a large positive because it is measured in kJ and needs to be converted to Joules. We
know =S is a small positive because it is already in Joules and no conversion is required.
So , if the temperature is low, meaning the temperature (in Kelvin) has a point in which when
multiplied by the small positive (=S) will result in a number that, when subtracted to the large
positive (=H), the =G will be positive . If the temperature is above that point then we know that =G
will be negative.
-Hypothetically: =H = 1000J/mole and =S = 100J/Kmole T = 9K
=G = 1000J /mole - (9K x 100 J/Kmole)
=G = 1000J/mole - (900J/mole)
=G = 1000J/mole - 900 J/mole = 100J/mole
-In this case we can see that a temperature above 9.99K will result in either
equilibrium or a negative =G
*
Heat and Work (Detailed)
Work - A force acting over a distance
w = P=V
- If the volume increases (expands) then work is negative
- If the volume decreases (contracts) then work is positive
Heat - In a system that can't expand w = 0
=U = q (at constant volume)
w
q
+
+
+
+
+
-
=
=E
E
+
Depends on size of q and w
Depends on size of q and w
-
AP Chemistry Study Guide (Part 1 of 3)
Hess' Law
The process of using several reactions' =H> in order to find the =H> of one desired reaction
2 rules: If the reaction is reversed, flip the sign on that reactions =H>
If the reaction is multiplied or divided, so is the =H>
Hidden Rule: Don't be afraid of fractions
=H> = Standard Enthalpy (Enthalpy change for a reaction under standard conditions)
1.) Start by writing the "goal" equation
2.) Then write the other reaction and label (1, 2, 3, ...n)
3.) Look for unique compounds in the "goal" equation and find it in any of the reactions.
4.) Make those unique compounds match the "goal" equation by reversing the reaction or multiplying or
dividing.
5.) Cross out equations that have already been manipulated.
6.) Repeat steps 3-5 until all equations have been crossed out
7.) Rewrite the new equations and add up the =H> values
In-Depth Look at Example
Goal: C6H4(OH)2(aq) + H2O2(aq) ---> C6H4O2(aq) + 2H2O(l)
Eq 1.) C6H4(OH)2(aq) ---> C6H4O2(aq) + H2(g)
=H> = 177.4 kJ/mole
Eq 2.) H2(g) + O2(g) ---> H2O2(aq)
=H> = -191.2 kJ/mole
X
Steps 1 and 2
=H> = -241.8 kJ/mole
Eq 3.) H2(g) + O2(g) ---> H2O(g)
_
=H> = -43.8 kJ/mole
Eq 4.) H2O(g) ---> H2O(l)
Eq 1.) C6H4(OH)2(aq) ---> C6H4O2(aq) + H2(g)
=H> = 177.4 kJ/mole
Eq 2.) H2O2(aq) ---> H2(g) + O2(g)
=H> = 191.2 kJ/mole
Eq 4.) 2H2O(g) ---> 2H2O(l)
=H> = -87.6 kJ/mole
Unique compounds: C6H4(OH)2(aq) and C6H4O2(aq)
Both compounds are in Eq1 and nowhere else, making them unique compounds
Because they are already on the correct side, and have the correct number of
coefficients, nothing is required to manipulate equation 1
Unique compounds: H2O2(aq)
Located only in Eq2 which makes it a unique compound. However, looking at the
"goal" equation, we see that H2O2(aq) is on the reactants side, not the products, so
we need to reverse Eq2 AND flip the sign for the =H> of Eq2 (Rule 1)
Unique compounds: H2O(l)
Located only in Eq4 which makes it a unique compound. However, there are 2
moles of H2O(l) in the "goal" equation and only 1 mole in Eq4, which means we need
to multiply Eq4 by 2 AND multiply the =H> of Eq4 by 2 (Rule 2)
Eq 3.) 2H2(g) + O2(g) ---> 2H2O(g)
=H> = -483.6 kJ/mole
Unique compounds: none
When looking at this equation we see that there are no unique compounds (phases
matter). So, when there are no unique compound we must look at the other
equations to see what we will have to do to manipulate this equation. We can
approach this case 1 of 2 ways.
1.) We see that there is an excess O2(g) on equation 2 that is not in the
"goal" equation so we need to make it so the O2(g) will cancel, which means
multiply Eq3 by 2.
2.) We see that there is 2H2O(g) as a reactant in Eq4, but it is not located in
the "goal" equation. So we need to make it so the H2O(g) will cancel, which
means we must multiply Eq3 by 2.
Either way, we need to multiply Eq3 by 2 AND the =H> of Eq3 by 2 (Rule 2/Hidden)
Steps 3-6
AP Chemistry Study Guide (Part 1 of 3)
Eq 1.) C6H4(OH)2(aq) ---> C6H4O2(aq) + H2(g)
Eq 2.) H2O2(aq) ---> H2(g) + O2(g)
Eq 3.) 2H2(g) + O2(g) ---> 2H2O(g)
Eq 4.) 2H2O(g) ---> 2H2O(l)
=H> = 177.4 kJ/mole
=H> = 191.2 kJ/mole
=H> = -483.6 kJ/mole
=H> = -87.6 kJ/mole
Step 7
177.4kJ/mole + 191.2 kJ/mole - 483.6 kJ/mole - 87.6 kJ/mole = -202.6 kJ/mole
Common Exothermic/Endothermic Reactions
Vaporization - Endothermic
Melting - Endothermic
Freezing - Exothermic
Combustion - Exothermic
Photosynthesis - Endothermic
Breaking bonds - Endothermic
Forming bonds - Exothermic
Enthalpy of Hydration =Êéêë
The enthalpy change involving the hydration of gas-phase ions
Gas--->Aqueous
If ions dissolve and solution heats - Exothermic
If ions dissolve and solution cools - Endothermic
Definitions
Definitions
Potential Energy - Energy due to position and composition (water behind a dam)
Kinetic Energy - Energy due to motion of the object (dependent upon mass and velocity)
Pathway - The specific conditions in which energy is divided between work and heat
State function - Property of a system that depends only on the present state (Also known as state property)
Internal Energy - The sum of the Potential and Kinetic energies (represented as U or E)
Calorimetry - The science of measuring heat (Based on temperature change observations)
Calorimeter - Device used to measure the heat associated with a chemical reaction (Fancy word for 2
Styrofoam cups nestled within each other)
Heat capacity (C) - The energy required to raise a substance by 1ì (measured in J/ì)
Specific heat capacity (c, s) - The energy required to raise 1g of a substance by 1ì (measured in J/gì)
Molar heat capacity - The energy required to raise 1 mole of a substance by 1ì (measured in J/moleì)
Constant-pressure calorimetry - The process of measuring heat using a calorimeter under constant pressure
Constant-volume calorimetry - The process of measuring heat using a calorimeter under constant volume
Chapter 7 - Atomic Structure and Periodic Trends
Basics - Parts of a Wave
Quantity
Wavelength
Frequency
Symbol (NOT unit)
Unit of Measurement
Ú
1 cycle/sec (Hz, or 1/s)
Û
Meters (m)
Equations
ä = ÛÚ
Related. Plug 1st equation into 2nd
½ = Ùä/Û
equation to get the 3rd equation
½ = ÙÚ
Δ½ = ÙÚ
Ü = Ù/(Ûä)
Û = Ù/ÜÝ*
2.178 q 10ÈXá àâ _
å½ =
i _ T _
=æ P =(Üç) > Ù/4è
_
½ = T(6.02 Þ 10_e ßà/Ü)(2.18 Þ 10ÈXá à)(â ã_ )**
Definition
Distance between 2
crests of a wave
How many waves
produced in 1 second
Constants
c = 3.0 x 108 m/s
h = 6.626 x 10-34 J P s
Ý in this case is velocity, not
wavelength
**Energy required to remove 1
mole of electrons
*
AP Chemistry Study Guide (Part 1 of 3)
Electromagnetic Waves
Gamma raysàX raysàUV raysàvisibleàInfrared raysàmicrowavesàradio waves (Strongest - weakest)
-Visible Light Spectrum: Red
650-700nm
Orange
600-650nm
Yellow
580-600nm
Green
490-580nm
Blue
490-450nm
Indigo
450-410nm
Violet
410-400nm
Hydrogen Spectrum
Can only give off certain energies
Electrons attracted to the nucleus because of opposite charges, but didn't fall in because they were moving
Putting energy into the atom moved the electron away, when it returned it emitted light
The Bohr Model
n = Energy level
n=1 is called the ground state
Z = nuclear charge
2.178 q 10ÈXá àâ _
å½ =
i _ T _
å½ = ½ikg[ T ½klg[
Electron Configuration
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p….
s-2, p-6, d-10, f-14
Aufbäu Principle - As the protons are added by 1, the electrons fill up hydrogen-like orbitals in order of
energy level.
Hund's Rule - The lowest energy configuration for an atom is the one that has the max number of unpaired
electrons in the orbital.
Valence electrons - The outermost energy levels, not including the d orbital
Noble gases have filled energy levels
Left side of Periodic Table fills the s orbital
Right side of Periodic Table fills the p orbital
Transition metals fill the d orbital
Lanthanide or actinide series fills the f orbital
Elements in the same column have the same order shell electron configuration
Exceptions to Electron Configuration
Cr = [Ar] 4s1, 3d5
Cu = [Ar] 4s1, 3d10
La = [Xe] 6s2, 5d1
Ce = [Xe] 6s2, 4f1, 5d1
Pm = [Xe] 6s2, 4f3
Gd = [Xe] 6s2, 5d1 4f7
Lu = [Xe] 6s2, 5d1, 4f14
Quantum Numbers
Principal quantum number (n) - Size and energy of an orbital
- Integer values > 0
Angular momentum quantum number (ð) - Shape of orbital
- Integer values 0 to n-1
Magnetic quantum number (Üð ) - Tells direction in each shape
- Integer values -ð to ð
ñ Value
0
1
2
3
4
Orbital
s
p
d
f
g
AP Chemistry Study Guide (Part 1 of 3)
Electron spin quantum number (ms) - Tells direction electron is spinning
X
- Integer values ò
_
Pauli exclusion principle - No two electrons in a given atom can have the same set of quantum numbers
Only two electrons with opposite spins can occupy a given orbital
65.) Give the maximum number of electrons in an atom that can have these quantum numbers
a.) n = 4
b.) n = 5, mð = +1
X
c.) n = 5, ms = +
_
d.) n = 3, ð = 2
e.) n = 2, ð = 1
a.) n = 4. Because it doesn't give a specific ð value, we use the n-1 = ð to find that ð = 3. With this
we can see that there is a total of 16 orbitals when n = 4 (1 from 4s, 3 from 4p, 5 from 4d,
and 7 from 4f). And we also know through the Pauli exclusion principle that one orbital
can only carry a maximum of 2 electrons with opposite spins. So 16 x 2 = 32 electrons.
b.) n = 5, mð = +1. This problem goes into the imaginary orbital g. We also know that the mð value
ranges from + ð to - ð. And when n = 5, it will have all the previous ð and mð
values along with an additional row which will be:
ð
mð
n
Subshell
Orbitals in
Orbitals in
Subshell
Shell
5
0
4s
0
1
1
4p
1, 0, -1
3
2
4f
2, 1, 0 , -1, -2
5
3
4d
3, 2, 1, 0, -1, -2, -3
7
25
4
4g
4, 3, 2, 1, 0, -1, -2, -3, -4
9
We were also given that mð = +1. From this we can say that there are only 4
subshells capable of containing that specific mð value (4p, 4d, 4f, and 4g). 4 x 2= 8
electrons.
X
c.) n = 5, ms = + . Again, we go back to the imaginary g orbital. This time, we are not given any
_
specific ð value, so we again use n -1 = ð to find that ð = 4, With this we see that
there is a total of 25 orbitals when n = 5. So 25 x 2 = 50 electrons. However, it
X
told us ms = + which means to account for only the electrons spinning in a
_
positive direction. So we divide 50 by 2 to get a total of 25 electrons.
AP Chemistry Study Guide (Part 1 of 3)
d.) n = 3, ð = 2. Following the chart, we see that this takes us to the 3d row. And there is a total of 5 orbitals in
that subshell. 5 x 2 = 10 electrons.
e.) n = 2, ð = 1. Following the chart, we see that this takes us to the 2p row. And there is a total of 3 orbitals in
that subshell. 3 x 2 = 6 electrons.
Polyelectronic Atoms and Effective Nuclear Charge
Contains more than one electron
Has 2 energy contributions
-Kinetic Energy of moving electrons
-Potential Energy of attraction between nucleus and electrons
-Potential Energy from repulsion of electrons
Find energy by using the effective nuclear charge equation:
å½ =
_.Xóá qX`ô¨| õö ^
k÷ ^ È kø ^
Find energy required to remove 1 mole of electrons from a substance
_
½ = T(6.02 Þ 10_e ßà/Ü)(2.18 Þ 10ÈXá à)(â ã_ )
Heisenberg Uncertainty Principle
Basically, just accounts for what percent/how often the electrons will actually be in the position predicted.
Uses the equation:
=æ P =(Üç) > Ù/4è
Quantum Energy
Discovered by Max Planck who said ΔE came in chunks with a size hù
Use equation:
Δ½ = ÙÚ
-n is an integer
-h is Planck's Constant
Einstein
Stated Electromagnetic radiation is quantized in particles called photons
Equation to find mass of photon:
Ü = Ù/(Ûä)
Equation to find Energy of photon:
½ = Ùä/Û
½ = ÙÚ
Periodic Trends
Ionization Energy: Across a row - Increases; Down a column - Decreases
-Need to take into account the effective nuclear charge and when you are removing electrons form
half-filled or full orbitals (will be harder to remove)
Atomic Size: Across a row - Decreases; Down a group - Increases
-Measured in angstroms (1 x 10-10 meters). Influenced by shielding and charge on nucleus.
Electron Affinity: Across a row - More negative; Down a group - More positive
Ionic Size: Across a row - Decreases; Down a group - Increases
Electronegativity: Across a row - Increases; Down a group - Decreases
-The tendency for an atom to attract electrons to itself when chemically combined with another
element. Corresponds directly with electron affinity.
Metals: Low ionization energy, low effective nuclear charge, low electronegativity.
Nonmetals: More negative electron affinity, high electronegativity (except noble gases)
AP Chemistry Study Guide (Part 1 of 3)
Periodic Trends Continued
Metals lose electrons and have lowest ionization energy
Nonmetals gain electrons and have the most negative electron affinity
Alkali Metals: Decrease in ionization energy
Increase in atomic radius
Decrease in density
Decrease in melting point
Behave has reducing agents
React explosively with water (except Li)
Reducing Ability: The lower the ionization energy, the better the reducing agent
-Cs > Rb > K > Na > Li
-Only works for solids, not as solutions
-In solutions: Li > Na > K
- Reason is the water, when the solid metal is dissolved, there is an energy change known as
hydration energy
Oxidation Number Trends:
Group
IA
IIA
IIIB - IB
IIIA
IVA
VA
VIA
VIIA
VIIIA
Oxidation
Multiple
+2 or
Number
+1
+2
oxidation
+3
-3
-2
-1
0
+4
numbers
Cations Vs. Anions
Metals lost electrons, become smaller and are positively charged (Cation)
Nonmetals gain electrons, become large and are negatively charged (Anion)
