Download 2. Covalent network

Chapter 1 – 3
1. Important people:
a. Rutherford-discovered the nucleus
b. Bohr- created the Bohr model and discovered how electron orbit
around the nucleus.
c. Heisenburg- asserted the uncertainty of principle of quantum
mechanics.
d. Pauli- created the Pauli Exclusion Priniciple (no two electrons in an
atom can have the same quantum number).
e. Levosier- discovered the role of oxygen in combustion, law of
conservation of mass and creation of the first textbook.
2. Stoichiometry:
a. Used to determine the moles of a molecule in a reaction when
given the moles of a different molecule.
i. Write a balanced equation.
ii. Set up a BCA table.
iii. Convert to moles.
iv. Fill in BCA table.
3. Percent Composition of Compounds:
a. The part, divided by the whole, multiplied by 100
4. Determining the empirical formula:
a. Determine the percentage of each element in your compound
b. Treat % as grams, and convert grams of each element to moles of
each element
c. Find the smallest whole number ratio of atoms
d. If the ratio is not all whole number, multiply each by an integer so
that all elements are in whole number ratio
5. Determining the molecular formula:
a. Find the empirical formula mass
b. Divide the known molecular mass by the empirical formula mass,
deriving a whole number, n
c. Multiply the empirical formula by n to derive the molecular formula
6. Calculating Percent Yield:
a. Actual yield - what you got by actually performing the reaction
b. Theoretical yield - what stoichiometric calculation says the reaction
SHOULD have produced
c. (Actual Yield/Theoretical Yield) x 100% = percent yield
7. Limiting Reactants:
a. Convert moles like normal.
b. Assume one reactant is the limiting reactant.
c. If there isn’t enough of the of the other reactant, then it isn’t the limiting
reactant.
Chapter 5

Boyle’s Law: PV=k
The product of pressure times volume is a constant— P is inversely related to k
Used to find new volume or pressure at a constant temp
P1V1=P2V2

Charles Law: V=bT
Temperature must be measured in Kelvin.
The volume of each gas is directly proportional to temperature.
V1/T1=V2/T2

Avogadro’s Law: V=an
Gas has to be at constant temperature and pressure. Volume is directly
proportional to number of moles.
V1n1/V2n2

Ideal Gas Law: PV=nRT
Constants from previous 3 laws (k,b,a) are combined to make a universal
constant R.
R=.0821 (L*atm)/(mol*k)
It can be used to solve for pressure, number of moles, volume, or temberature
when all other variables are held constant.
At STP (0C and 1atm), the molar volume of an ideal gas is 22.42 L. (Constant)
Dalton’s Law of Partial Pressures: Ptotal=ntotal(RT/V)
Used to find pressure of a gas that is in a mixture of gases in a container.
Only total number of moles matters.
Mole Fraction: X1=P1/Ptotal (Pressure is directly related to moles)


Kinetic Molecular Theory:
1. The particles are so small compared with the distances between them
that the volume of the individual particles can be assumed to be zero
2. The particles are in constant motion. Collisions of the particles with the
walls of the container cause pressure
3. Assume that the particles exert no forces on each other.
4. The average kinetic energy of a collection of gas particles is assumed to be
directly proportional to the Kelvin temperature of the gas.

Relationship between Temperature and Average Kinetic Energy:
(KE)avg=(3/2)RT

Root Mean Square Velocity: urms=(3RT/M)
M=mass of one mole of gas particles in kg.
R=8.345 j/k*mol
Equation gives you the average velocity of a gas particle.
Chapter 7 – 9 Part I
Relationships: As energy increases, frequency also increases, but wavelength
decreases.
Periodic Trends
-Ionization Energy: energy required to remove an electron from an atom. The further
the electron is from the nucleus, the easier it is to remove it.
*Increases across a period
*Decreases with increasing atomic # within a group
-First ionization energy:energy it takes to remove the first electron from an atom
*increases across the period
*decreases down a group
*there are some exceptions-p has a higher 1st IE then S, and copper is the other
exception
-Electron Affinity: energy associates with the addition of an electron
*Increases across a period
*Decreases down a group
-Atomic Radius: half of the distance between radius in a covalently bonded diatomic
molecule
*Decreases across a period
*Increases down a group
-Alkali Metals
Easily lose valence electrons
*react with halogens to form salts
*react violently with water
Large hydration energy
*positive ionic charge makes ions attractive to polar water molecules
WHY?:Because as you go across the period, one more proton is added which
creates a stronger attraction between the nucleus and the electrons. As you go down
a period, more electron orbitals are added, which weakens the attraction between
the electrons and the nucleus.
Ionic Bonds


An ion has a different size than its parent atom
o Anion is larger than its parent ion
o A cation is smaller than its parent atom
Lattice energy: the change in energy when ions are packed together to form
an ionic solid
o Lattice energy=k(Q1 Q2/r)
o K= constant
o Q1, Q2 = charges on the ions
o
r= distance between ion center
Covalent Bonds
A covalent bond is a form of chemical bonding that is characterized by the sharing of
pairs of electrons between atoms.
Electrons are shared equally
Atoms end up with fractional charges (positive side and negative side)
Can model covalent bonds using lewis dot structures.
Ex)
H, Li, Be, B forms stable molecules when they share two electrons
Elements Carbon and beyond stable when they are surrounded by eight molecules.
Second row elements (C, N, O, and F) should always obey the octet rule
B and Be often have fewer than 8 electrons around them - highly reactive molecules
Second row elements NEVER exceed octet rule, third row often satisfies or exceeds
octet rule
Multiple Bonds: single bond, double bond, triple bond (based on number of shared
electrons)
Bond energy: ΔH = bonds broken - bonds formed
Chapter 7-9 Part II
Shapes and Angles of molecules
Bonding
electron
pairs
Lone
pairs
Electron
domains
(Steric #)
2
0
2
linear
180°
CO2
3
0
3
trigonal planar
120°
BF3
2
1
3
bent
120° (119°)
SO2
4
0
4
tetrahedral
109.5°
CH4
3
1
4
trigonal
pyramidal
107.5°
NH3
2
2
4
angular
104.5°
H2 O
5
0
5
trigonal
bipyramidal
90°, 120°, 180°
PCl5
4
1
5
seesaw
180°, 120°, 90°
(173.1°, 101.6°)
SF4
3
2
5
T-shaped
90°, 180° (87.5°, <
180°)
ClF3
2
3
5
linear
180°
XeF2
6
0
6
octahedral
90°, 180°
SF6
5
1
6
square
pyramidal
90° (84.8°), 180°
BrF5
4
2
6
square planar
90° 180°
XeF4
7
0
7
pentagonal
bipyramidal
90°, 72°, 180°
IF7
Hybridization
Shape
Ideal bond angle
(example's bond Example Image
angle)
Steric Basic geometry
No.
0 lone pair
2
1 lone pair
2 lone pairs
3 lone pairs
Hybridization
SP
Linear
SP2
3
Trigonal planar
Bent
SP3
4
Tetrahedral
Trigonal pyramid
Bent
SP3d
5
Seesaw
Trigonal bipyramid
T-shaped
Linear
SP3d2
6
Octahedral
Square pyramid
Sigma bonds (bond)
Square planar
a. Bond in which the electron pair is shared in an area centered on a line
running between the atoms
b. Lobes of bonding orbital point toward each other
c. All bonds in methane are sigma bonds
Pi bonds (bonds)
a. Electron pair above and below the bond
b. Created by overlapping of nonhybridized 2p orbitals on each carbon
Double bonds
a. Double bonds always consist of one 
bond and one bond
Chapter 10 Part 1
1. Inter-molecular forces
a. Dipole - Dipole: (+) …... (-) interactions
b. Hydrogen Bonding: Hydrogen covalently bonded to N, O, F
c. London Dispersion Forces: random movement of electrons, exist between
all molecules
d. Polarizability increases with the number of electrons in a molecule
Ex: CCl4 experiences greater LDF than CH4
** Higher LDF = larger size
*** Higher Dipole - Dipole = smaller size
Greater IMFs: Higher BP , Higher FP, Lower VP
Weaker IMFs: Lower BP, Lower FP, Lower VP
Chapter 10 Part II
1. Metallic Bonding :
a. The chemical bond characteristic of metals, in which mobile
valence electrons are shared among atoms in a usually stable
crystalline structure
2. Covalent network:
a. Solid is a chemical compound in which the atoms are bonded by
covalent bonds in a continuous network. For example, diamond,
graphite, silica, ceramics and semiconductors.
3. Triple Point Graph
a. Triple Point
i.
Solid and liquid have identical vapor pressure
ii.
All three phases exist together in equilibrium
b. Critical temperature:
i.
The temperature above which the substance cannot exist
as a liquid, regardless of how great the pressure
c. Critical pressure:
i.
The pressure required to produce liquefaction at the critical
temperature
d. Critical point
i.
Point defined by the critical temperature and critical
pressure
Chapter 11 Part 1
Molality: Moles of solute
Kilograms of Solvent
Molarity: Moles of solute
Liters of Solution
Mole Fraction: Moles of Part A
Moles of EVERYTHING
Mass Percent: Mass of Solute
Mass of Solution
X 100!!!
Van’t Hoff Factor (i) : Moles of Particles in Solution
Moles of Solute Dissolved
For ionic compounds, the expected value of i is an integer greater than 1
a. NaCl, i = 2
b. BaCl2, i = 3
c. Al2(SO4)3, i = 5
Solution Conductivity
If the the solute is a molecular solid, there will be no metals so it will not
conduct.
Covalently bonded compounds are weak or nonelectrolytes which resist
conduction.
Example(C H O + H O...Nonconductive)
NO METAL, NO CONDUCT
If the Solute is a Ionic Compound, the lattice will break apart and individual
ions will retain their own properties and allow the solution to conduct.
Ionic Compounds are strong electrolytes which allow compounds to conduct.
Example(CaCO . + H O.....conductive)
METAL! Disassociate Conductive Properties! CONDUCT!!
Solute: is dissolved by a solvent (kool-aid powder)
Solvent: dissolves the solute (water)
Solution: solute + solvent (kool-aid)
6
12
6
3
2
2
Chapter 11 Part II
1. Vapor Pressure (http://www.unit5.org/chemistry/Vapor_2.jpg)
a. Pressure of a gas of a substance that is in equilibrium
b. Nonvolatile solutes lower vapor pressure
i.
Since they won’t evaporate, they block the surface and prevent
molecules of the solvent from escaping
c. Raoult’s Law
.
Psol = Xsolvent * Psolvent
1. Just Like PVNRT
d. Ionic Solid dissociate
.
Creates more particles that clog up surface.
2. Boiling Point
(http://chemwiki.ucdavis.edu/@api/deki/files/4082/=Entropy.JPG)
. Nonvolatile “get in the way” of things that are trying to evaporate.
0. Need more energy to get them to evaporate.
1. Therefore boiling point rises.
a. Change in Temp.=Kb*msolute
0. Delta T is the boiling point elevation
1. Kb is the molal boiling point elevation constant of the solvent
(given)
2. msolute is the molality of the solute in the solution
3. Freezing Point (http://chem.ps.uci.edu/~kcjanda/Group/clathrateweb/K12_files/water.jpg)
. Solutes get in the way of the freezing lattice
0. More energy must be taken out to “freeze” around the solutes
1. Freezing Point Decreases
a. Change in Temp.=Kf*msolute
0. Delta T is the boiling point elevation
1. Kf is the molal freezing point depression constant of the solvent
(given)
2. msolute is the molality of the solute in the solution
Chapter 12
Differential Rate Law



Represents how rate depends on concentration
Delta [A] / Delta t
Where [A] is concentration of reactant or product in a reaction
Reaction Mechanisms




Most chemical reactions occur by a series of steps, not one reaction
This series of steps is called the reaction mechanism
Two ways to determine reaction mechanism fits a reaction.
o Use rate determining step to see if the reaction occurs at the same
rate as the proposed mechanism
o Cancel intermediaries to see if reaction mechanism matches reaction
Example: 2NO + F → 2NO F
o Rate law = k[NO ] [F ]
o A suggested mechanism for this reaction is:
 NO +F → NO F+F SLOW
 F + NO → NO F
FAST
o Is this reaction mechanism acceptable?
o To determine, first cancel intermediaries
2
2
2
2
2
2
2
2
2
2
2

The equation ends up looking like
 2NO + F → 2NO F
Then, find if the slow step is equal to the rate law of the reaction
 The rate for the slow step is
 k[NO ][F ]
Since the rate law for the slow step doesn’t match the rate of the
actual reaction, the reaction mechanism that was proposed is not an
acceptable mechanism.
2
o
2
2
o
2
2
Collision Theory

Molecules must collide to react
o Collide enough to lead to activation energy to induce the reaction
Reaction profile



Exothermic reaction
o Products have less energy than reactants
Endothermic reaction
o Reactants have less energy than products
Change in energy from the initial reactants to the transition state is called the
activation energy
Chapter 13
1. When the rate of the forward reaction is equivalent to the rate of the reverse reaction
the system is at equilibrium.
2. jA+kB<-->lC+mD
a. j, j, l, and m are coefficient.
b.A and B are the concentrations of the reactants
c.
C and D are the concentrations of the products
k= [C]^l x [D]^m
[A]^j x [B]^k
3. K is the equilibrium constant and varies depending on temperature and the
coefficients of the balanced equation. Concentration does NOT effect k.
4. kp=kc(RT)^change in n
a. kc is the constant for concentration
b. kp is the constant for pressure
c. R=0.0821
d. T=temperature in Kelvin
e. change in n= difference in sums of coefficients for the products and the
reactants
5. The effect of change in pressure
a. When gaseous products or reactants are added pressure shifts away from
the added gas.
b. When gaseous products or reactants are removed pressure shifts towards
the removed gas.
c. Solids, liquids, and inert gases are not included in the equilibrium
equation.
d. When the volume of the container holding a gaseous system is reduced the
system responds by reducing its own volume.
e. When the container volume is increased the system will shift as to increase
its own volume.
6. Change in Temperature
a. When temperature increases the system will shift in the direction that
consumes the energy.
b. For an exothermic reaction energy is a product. The reaction will shift to
the left to use up the excess energy.
c. For an endothermic reaction energy is a reactant. The reaction will shift to
the right to use up the energy.
d. A decrease in temperature causes a system to shift in the direction that
replaces the lost energy.
7. Examples:
8. At a certain temperature, 4.0 mol NH3 is introduced onto a 2.0-L container, and
the NH3 partially dissociates by the reaction
a. 2NH3(g)<----->N2(g) + 3H2(g)
b. At equilibrium, 2.0 mol NH3 remains. What is the value of K for the
reaction?
c. (Answer: K=1.7)
9. At 25 C, K=0.090 for the reaction
a. H2O(g) + Cl2O(g)<------>2HOCl(g)
b. Calculate the concentrations of all species at equilibrium for each of the
following cases.
c. 1.0 mol pure HOCl is placed in a 1.0-L flask.
d. (Answer: 122 H2O+Cl2O; .06 HOCl)