# Mole-Volume Hypothesis: Avagadro hypothesized

advertisement ```Mole-Volume Hypothesis: Avagadro hypothesized that equal volume of gases of
the same temperature and pressure contain equal number of particles.
At ST
P, 1 mol (6.02*1023 particles) of any gas occupies a volume of 22.4 L
(Molar Volume)
At STP, Volume of gas = (#mol) * (22.4 L/mol)
At STP, molar mass = (density) * (molar volume)
Mixture of Gases
Air- 78% nitrogen, 21% oxygen, 0.9% argon, 0.03% CO2 (% by volume)
Partial Pressure- pressure due to any individual component in a gas mixture
Total Pressure- sum of partial pressure
Ptotal = Pa + Pb + Pc …
Pa = Xa Ptotal
Xa = (moles of a) / (total # moles of gas) “mole fraction of a gas”
Diffusion / Effusion
Diffusion- tendency of molecules to move toward areas of low concentration until
the concentration is uniform.
Effusion- gas escapes through a tiny hole in a container (lower molar mass effuses
faster)
Graham’s Law of Effusion- the rate of effusion is inversely proportional to the
square root of a gas’s molar mass.
Ex) (RateA / RateB) = Square root of (molar mass B / molar mass A)
Real Gases
At high pressures or low temperatures, gases don’t behave ideally, so we use the
Vander waals Equattion: (P + a(n/V)2)
( V-nb ) = nRT
Isaac Newton- particles
Waves
Amplitude- The maximum distance from equilibrium
Wavelength (“lambda” )- distance from point on wave to identical point on
next wave
Frequency (“nu” )- number of waves to pass a point in one second (unit:
Hertz, Hz)
1 Hz = (1/sec) = S-1
C = 
C= the speed of light = (frequency) X (wavelength) = 3* 108 m/s
Dual Nature of LightActs as waves + particles (photons)
Atomic Spectra:
Pass electric current through a gas in a tube
This energizes electros of atoms,
The electrons absorb energy and move to higher energy levels,
Then the electrons lose energy and come back, and come back to original position,
and emit light
Ground State- lowest possible energy state of e- in atom
E = quantum of energy
E = h
h = Planck’s constant = 6.63 * 10-34 J*S
E = (hC) / 
Bohr Model (1913)
Neils Bohr
E- only found in orbits around nucleus each orbit has a fixed amount of energy
celled “energy level”
Like a ladder:
Lowest rung  energy level closest to nucleus (lowest energy)
Person climbs from rung to rung  e- travel from energy level to another
To move from energy level to enegy level,
e- must gain/lose certain amount of energy
Quantum of energy- amount of energy required to move an e- from one energy
level to another
 Higher energy levels are closer together
Schrodinger (1887- 1961):
Atom: probability of finding an electron within a certain region of space
surrounding the nucleus is represented (in his model, the Quantum Mechanical
Model,) as a fuzzy cloud.
Cloud is more dense where there is greater probability of finding eAtomic Orbitals: regions of space with highest probability of finding e- in an
energy level
Energy levels represented by principle quantum numbers: n = 1, 2, 3, 4, …
Energy sublevels: correspond to orbitals of different shape describing where e- are
likely to be found
Sublevels:
Represented by letters ( S, P, D, F)
Labeled by angular momentum numbers (L).
L
Sublevel
0
S (spherical)
1
p (dumbbell shaped)
2
d
3
f
Principal energy
# of sublevels
Type of sublevels # of e-
N=1
1
1s
2
N=2
2
2s, 2p
8
N=3
3
3s, 3p, 3d
18
N=4
4
4s, 4p, 4d, 4f
32
level
2n2 = maximum number of e- an energy level can hold
Magnetic Quantum Number (Ml)- specifies orientation of orbital
Heisenberg Uncertainty Principle- impossible to know both the velocity and
position of e- at the same time (may know one, but never both).
Electron Configuration- way in which electrons are arranged in orbitals around
the nucleus
Rules for writing e- configurations:
Aufbau Principle- e- occupy orbitals of lowest energy level
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 5d, 4f, 6p
Puali Exclusion Principle- to occupy the same orbital, two electrons must have
opposite spins.
Hund’s Rule- electrons occupy orbitals of the same energy in a way that makes
the number of electrons with the same spin direction as large as possible.
*Exceptions to the Aufbau Principle*
Cr (column)- 1s2 … 4s1 3d5
Cu (column)- 1s2 … 4s1 3d10
Atomic Trends
Atomic Radius
One half of he distance between the nuclei of 2 atoms of the same element when
atoms are joined
Measured in picometers ( (pm) * 10 -12)
Ionic Size
Size of cation < size of atom
Size of anion > size of atom
Ionic Size decreases from left to right, but get larger with first anion, then continue
to decrease.
As you go down the periodic table, and add energy levels, the ionic size gets
bigger.
Ionization Energy- energy required to remove an e- from atom.
Decreases as you go down the periodic table.
Increases as you go across.
Isoelectronic = same # of e-
Ionic Compound:
Crystalline solids at room temperature
Conduct electric current when melted or dissolved in water
Metallic Bonding:
Metals: made of closely packed cations
Valence electrons of metal atoms can be modeled as a sea of ee- are mobile
Alloys: mixture of two or more elements (brass- copper + zinc)
Covalent Bonding:
Lower melting and boiling points than ionic bonded compounds
Weaker than ionic bonds
Bond two non-metals
Molecular
Ionic
Collection of molecules
Array of cations and anions
Rules for Drawing Lewis Structures:
Count number of valence electrons in compound
Use dashes between central atom and surrounding atoms
Draw lone pairs on surrounding atoms, then on central
Check that every atom has eight and that used correct number of electrons
Resonance- actual bonding is a hybrid, or mixture, of the extremes represented by
the resonance structures.
Exceptions to Octet Rule (for central atom only!):
Odd number of valence electrons:
NO2 (odd number of electrons)
SO3Expanded Ocktet:
PCl5 (more than four bonds each)
Se6
ICl5
***********End Of Material For Test**********************
VSEPR Theory- Valence Shell Electron Pair Repulsion
Explains 3D shape of molecules
Repulsion between electron pairs causes molecular shapes to adjust so that valence
electron pairs stay as far apart as possible.
Bent Shape (because of the two lone pairs)  H2O
Linear  CO2 (O=C=O)
Polar Bonds + Molecules
Non-polar covalent bond:
Diatomic elements
Polar Covalent Bond:
Between atoms where electrons are shared unequally.
The more EN an atom has, the more an atom attracts electrons more strongly and
gains a slightly negative charge
Less EN an atom gains slightly positive charge
Delta EN
Type of Bond
EX
0 - 0.4
Non-polar
H2
0.4 - 1.0
Moderately polar
HCl
1.0 -2.0
Very polar
HF
2.0 – and higher
ionic
NaCl
Polar Molecule- one end is slightly positive, and one end is slightly negative (EX
H2O). (Dipole).
Non-Polar Molecule- bond polarities cancel each other out (symmetrical
structure).
Classifying Reactions:
Combination Reaction- when 2 or more substances react to produce 1 new
substance.
Decomposition Reaction- 1  2 substances
Single Replacement- one element replaces a second element in a compound
Double Replacement- exchange of cations between two compounds
Combustion- element/compound reacts with O2, usually products are CO2 + H2O
Density
Solid
Liquid
Gas
highest
medium
lowest
Not shape Not
Takes shape, not
Takes shape +
volume
volume
volume
Attractions Between Molecules
Inter molecular forces- weaker than covalent or ionic bonds
(in order weakest  strongest)
Van der waals (London Dispersion Forces)
Weakest
Caused by motion of electrons
Occur between non-polar and polar molecules  occur in ALL compounds
Dipole-Dipole forces
Between two polar molecules
Slightly negative region of one polar molecule is attracted to the slightly positive
region of another polar molecule
Hydrogen Bonds
Attractive forces in which a hydrogen atom covalently bonded to a highly electro
negative atom (N, O, F)
Properties of Water
Surface Tension- resistance of a liquid to an increase in its surface area
 energy required to overcome the intermolecular forces in liquid.
Viscosity- resistance to flow
Capillary Reaction- spontaneous rising of a liquid in narrow tube, two things
responsible for this:
CohesiveAdhesive-
∆Hfus
- molar heat of fusion
Heat absorbed by 1 mol of a solid as it melts to liquid at melting point
Water: 6.01 KJ/mol
∆Hsolid = - ∆Hfus
How many grams of ide at 00 C will melt if 2.25 KJ of heat is added?
[2.25 KJ] / [6.01 (KJ/ mol)] = (.37 mol) * (18) = 6.7g ice
∆Hvap
-molar heat of vaporization = amount of heat to vaporize 1 mol of
a liquid to vapor at boiling point
∆Hvap = -∆Hcond
Water: 40.7 KJ/mol
Heat: q
Measured in joules or calories
4.184 J = 1 cal
Heat Capacity- amount of heat required to increase temperature of a substance by
10 C
Specific Heat (c)- amount of heat required to increase temperature of 1 g of a
substance by 10 C
J / q 0C
Cal / g0 C
Water
4.184
1
Ice
2.1
.5
Iron
.46
.11
q = mc∆T
q: heat absorbed/ released
c: specific heat
∆T: chance in temp (TF – Ti)
Heat required to raise 27 g H2O from 10 to 90
q= (27g) * (4.184 J
Vaporization- liquid  vapor
Evaporation- when vaporization occurs at surface of liquid that is not boiling
Solubility= maximum quantity of a substance that will dissolve in a certain
quantity of water at a specified temperature.
Unsaturated Solution- contains less dissolved solute than the amount that the
solvent can normally hold at that temperature.
Supersaturated Solution- solution that contains more solute than could be
dissolved at that temp
Molality (m)- (mol Solute) / (Kg Solvent)
Mole Fraction- (mol Solute) / (total mol solution)
Mass percent- [(mass Solute) / (mass Solution)] * 100
Colligative Properties:
Properties that depend only on amount of solute, and NOT on their identity
The three properties (we’ll discuss) are…
Vapor pressure lowering- solution that has lower VP than its pure solvent
Boiling point elevation- difference in temperature between boiling point of
solution and boiling point of pure solvent.  ∆Tb = Kb * m Kb: molal boiling
point elevation constant (water: .512 oC / m)
m: molality
Freezing point depression-  ∆TF = KF * m
KF: molal freezing pt depression constant (water: 1.86 oC/m)
Suspension- mixture that particles begin to settle out upon standing
Greater than 1000 nm diameter
Colloid- particles 1nm – 1000 nm
Acid
Base
Sour taste Dissolve metal Litmus
Bitter Feel slippery Blue NaOH,
turns red HCl, H2SO4, HNO3
NH3
Acid
Arrhenius
Base
Produce H+ in solutions Produces OH- in
solutions
Bronsted Lowry
H+ donor
H+ acceptor *
Lewis
Accept e- pair
Donates e- pair
NH4+
+
OH-
 NH3
(acid
(base
(conjugate
H+ donor)
H+ acceptor)
base)
+
H2O
(conjugate acid)
Pure water
[H+] = 1 * 10-7 molar
[OH-] = 1 * 10-7
[H+] * [OH-] = 1 * 10-14 = KW (Ion Constant Product for H2O
[H+] > [OH-]  Acidic
Ex) [H+] = 1 * 10-5
[OH-]= ??  1 * 10-9
 Acidic
pH scale:
pH = -Log [H+]
(log 10 =1, log 100 = 2  Log X = Y, 10Y = X)
Ranges from 1 (most acidic) to 14 (most basic)
pOH = -Log [OH-]
pH + pOH = 14
Log Scale:
pH = -Log [H+]
Given pH, find H+
Method 1
Method 2
pH = -LogX -pH = LogX Log-1(-
pH = -LogX -pH = LogX 10-pH = X
pH) = X
Acid base indicator
Its acid form and base form have different
color in solution Strength of acid + bases ·
Strong acids –
completely ionize in solution o Ex. HCl, H2SO4, HNO3 § HCl
+ H2O  H3O+ + Cl- - 100 % ionized ·
Weak acid – partially
ionized in solution o Ex. Acetic acid § CH3COOH + H2O 
H3O+ + CH3COO- (partial ionization – can tell by double
arrow) ·
Strong Base o Ex. Mg(OH)2 , Ca(OH)2 ·
Weak
base o Ex. NH3 ·
Concentration / strength o
Concentration § How much acid is dissolved in solution §
Concentrated/ dilute o Strength § Extend of ionization ú
Strong weak Neutralization reaction ·
Acid +Base à Salt +
Water o Mix solution of strong acid and strong base and we get
a neutral solution results o Salt = compound consisting of anion
from aid and cation from base ·
Ex. HCl + NaOH à H2O +
NaCl ·
Ex. H2SO4 + 2KOH à 2H2O + K2SO4 ·
Q: How
many moles of potassium hydroxide are needed to completely
neutralize 1.56 mol of phosphoric acid? o H3PO4 + 3KOH à
3H2O + K3PO4 § 1.56 mol x (3 KOH / 1 H3PO4) Titration ·
The process of adding a known amount of solution of known
concentration to determine the concentration of another
solution · Standard solution (know concentration of) we kept
titrating until neutralization occurred, which is called the end
point ·
Ex. a 25 ml solution of H2SO4 is neutralized by adding
18 ml of 1 M NaOH what is concentration H2SO4? o H2SO4 +
2NaOH à 2H2O + Na2SO4 § 1M = x mol / .018 L ú .018 mol
NaOH ú .009 mol H2SO4 ú .009 mol / .025 L = .36 M
Buffer:
A solution in which pH remains relatively the same.
Made of weak acid and its salt OR a weak base and its salt
Ex) CH3OOH (acetic acid) + NaCH3OO (salt) = acid buffer
Thermochemistry:
Potential energy- energy that is stored in chemical bonds
Kinetic Energy- energy of motion
Heat transfer or work
Heat
Reaction absorbs heat  endothermic
Reaction releases heat  exothermic
Work = -P∆V
At a constant P…
If…
∆V > 0
W<0
 Work is being done by system
And if…
∆V < 0
W>0
 Work is being done on the system
∆E = q + W
+
-
q
Endothermic
Exothermic
W
Done on system
By system
E
Endogenic
Exergenic
Ex) q= +214 KJ
W= +110 KJ
∆E = 214 +110 = 324 KJ
2NaHCO3  Na2CO3 + H2O
∆H = 129 KJ
Calculate heat required to decompose 2.24 mol NaHCO3
* (129 KJ / 2 mol) = 144.5 KJ
Heat of Combustionheat of reaction for complete burning of one mole of substance
Ex) CH4 + 2O2  CO2 + H2O (∆H = 890 KJ)
Hess’s Law:
2P + 5Cl2  2PCl5
∆H = ?
Given:
PCl3 + Cl2  PCl5 ∆H = -88kJ
2PCl3  2P + 3Cl2 ∆H = 574 kJ
 flip the second reaction and get 2P + 3Cl2  2PCl3 ∆H = -574
 multiply first reaction by 2 and get 2PCl3 + 2Cl2  2PCl5 ∆H = -176
 combine them and get 2P + 5Cl2  2PCl5
∆H = -750
Standard Heats of formation:
 change in enthalpy that accompanies the formation of one mole of a compound
from its elements
∆HFo = 0 for free elements and diatomics.
∆Hº (heat of reaction) = ∑ (np∆Hfº, products) - ∑(nr∆Hfº, reactants)
Gibbs Free Energy- energy available to do work
Spontaneous Reaction
Occurs naturally, releases free energy
If ∆G is negative, the reaction is spontaneous
Non-spontaneous Reaction
Doesn’t occur naturally
Requires energy input
If ∆G is positive, the reaction is Non-spontaneous
Entropy (∆S)- measure of disorder (randomness) of a system
Law of Disorder- natural tendency of system is in direction of more disorder.
Increase in S  favors spontaneous reaction
For a give substance, the S solid < S liquid < S gas
S increases when divide substance
In a reaction: when # mol product > # mol reactant, S increased
When Temperature increases, S increases
Gibbs Helneholtz Equation:
∆G = ∆H – T∆S
*T- in Kelvin
If ∆G is negative, the reaction is spontaneous
If ∆G is positive, the reaction is Non-spontaneous
Rate:
Measure of the speed of any change that occurs in any interval of time
Usually (∆ [ ]) / Time
Activation Energy:
Minimum energy that colliding particles must obtain in order to react
Factors that affect Rate:
Temperature- If temperature increases, rate increases
Particle Size- If you have smaller particle, rate increases
Concentration- If concentration increases, rate increases
Catalyst
Speeds up reaction without being used
Lowers activation energy
Rate Law- expression for rate of reaction in terms of concentration of reactants
Rate Constant (K)-
Stresses: *** solids cannot change concentration!!!
Temp
Pressure, volume
Concentration
Concentration- Changing the concentration of any reactant or product disturbs the
equilibrium and the system adjusts to minimize the effect of change.
H2CO3  CO2 + H2O
At equilibrium, I add H2CO3, Forward reaction favored, Shifts to the right
At equilibrium, remove H2O, shift to the right
Changing the cncetration of any reactant or product disturbs the equilibrium and
the system adjusts to minimize the effect of change.
Temperature- increasing the temperature causes equilibrium to shift in the
direction that absorbs heat.
2SO2 + O2  2SO3 + Heat, ∆H = -X (exothermic)
If Temperature goes up, reaction shifts to the left
Pressure- If we increase pressure, system wants to decrease it, so it shifts in
direction of fewer moles of gas.
N2 (g) + 3H2 (g)  2NH3 (g)
If pressure goes up, reaction goes to the right
If pressure goes down, reaction goes to the left
VolumeIf volume goes up, pressure goes down
If Volume goes down, pressure goes up
Equilibrium Constant (Keq)
Ratio of product concentrations to reactant concentration, with each concentration
raised to a power equal to the number of moles of substance in balanced equation.
Keq > 1,  products favored over reactants
Keq < 1,  reactants are favored
Redox:
Oxidation- loss of eReduction- gain of e- or reduction of charge
Ex) Mg + S  Mg2+ + S2Mg  Mg2+ + 2e- (oxidation)
S + 2e-  S2- (reduction)
Mg: reducing agent
S: oxidizing agent
Silver nitrate reacts with copper to form copper (II) nitrate and silver.
What is oxidized?
What is reduced
Identify oxidizing + reducing agents
2AgNO3 + Cu  CuNO3 + 2Ag
Cu is oxidized
Ag is reduced
Oxidizing agent = Ag
Reducing agent = Cu
Oxidation #:
Positive or negative number assigned to an element to indicate its degree of
oxidation or reduction
Ex) Charge: S2- Oxidation #: -2
Rules for assigning Oxidation #s:
Oxidation number of monatomic ion is equal to its charge
Ex) Br- : -1, Fe3+ : +3
Oxidation number of hydrogen in a compound is +1, except when in metal
hydrides, such as NaH, where it’s -1
Oxidation number of oxygen in a compound is -2, except in peroxides such as
H2O2, where it is -1
Oxidation number of an uncombined element is zero
Ex) Cu, N2
For a neutral compound, the sum of the oxidation numbers is zero
For polyatomic ions, the sum of the oxidation numbers is the charge of the ion
When calculating oxidation numbers in molecules with three or more elements,
find the oxidations numbers of the elements from the outside in.
2AgNO3 + Cu 
(Ox agent)
Cu(NO3)2
+
2Ag
(Red agent)
Balancing Redox Reactions in Acidic Solution:
Use oxidation numbers to write out half reactions
Ex) MnO4- + Fe2+  Fe3+ + Mn2+
Mn = +7, O = -2, Fe (#1) = 2+, Fe (#2) = 3+, Mn (#2) = 2+
Half Reactions: MnO4-  Mn2+
Fe2+  Fe3+
Balance each half reaction:
Balance all elements except H and O
Balance O by adding H2O
MnO4-  Mn2+ + 4H2O
Balance H by adding H+
8H+ + MnO4-  Mn2+ + 4H2O
Balance each reaction’s charge by adding e5e- + 8H+ + MnO4-  Mn2+ + 4H2O
Fe2+  Fe3+ + 1eMultiply Half reactions so that they have equal number of eFe2+  5Fe3+ + 5eAdd half reaction and cancel/simplify
Check that charges and elements are balanced
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