the basics
Download
Skip this Video
Download Presentation
The Basics

Loading in 2 Seconds...

play fullscreen
1 / 159

The Basics - PowerPoint PPT Presentation


  • 51 Views
  • Uploaded on

The Basics. Elements, Molecules, Compounds, Ions Parts of the Periodic Table How to Name. Classification of Elements. Metalloids (or semi-metals) – along the stair-step line Properties are intermediate between metals and nonmetals.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' The Basics' - leala


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
the basics
The Basics
  • Elements, Molecules, Compounds, Ions
  • Parts of the Periodic Table
  • How to Name
classification of elements
Classification of Elements
  • Metalloids (or semi-metals) – along the stair-step line
    • Properties are intermediate between metals and nonmetals
  • Nonmetals – found on the right-hand side of the Periodic Table
  • Metals – found on the left-side of the Periodic Table
elements molecules compounds and ions
Elements , Molecules, Compounds, and Ions
  • Element – one single type of atoms
    • Al Cu He
    • Naturally occurring elements that are Diatomic are still elements
    • N2 O2 F2 Cl2 Br2 I2 H2
    • How many elements are in Mn(SO4)2 ? 3
    • How many Atoms? 9
  • Molecule - smallest electrically neutral unit of a substance that still has the properties of that substance, 2 or more different elements
slide4

Elements , Molecules, Compounds, and Ions

  • Compounds – groups of atoms
    • Ionic and Molecular
  • Molecular Compounds – share electrons typically 2 or more non-metals (hydrocarbons)
  • Example H2S CO2 C5H10
  • Ionic Compound (salts) – transfer electrons typically metal and non metal (watch for poly atomic ions)
  • FeS Mg(OH)2 (NH4)3PO4
slide5
Ions
  • Ions have either lost or gained electrons
  • Typically Metals lose electrons to become positive
  • Example cations
  • Mono-valent Mg -> Mg2+ + 2e-
    • Group 1A = Metal1+, 2A =Metal2+ and 3A = metal3+
    • Multi-valent Fe --> Fe2+ + 2e- and Fe --> Fe3+ + 3e-
  • Non-metals gain electrons - anion
  • S + 2e- -> S2-
poly atomic ions
Poly Atomic Ions
  • A molecular compound with a charge
  • NH4+
  • CO32-
  • SO42-
  • NO3-
  • OH-
  • H3O+
  • Ammonium
  • Carbonate
  • Sulfate
  • Nitrate
  • Hydroxide
  • Hydronium
acids and bases from ions h or oh
Acids and Bases From Ions H+ or OH-
  • Acids look for hydrogen up front (HA) or as COOH
  • Example HF H3PO4 C4H6COOH
  • Strong Acids
  • HCl, H2SO4, HBr, HI, HClO3,HNO3
  • Base look for hydroxide or NHgroup
  • Example KOH C4H6NH2
naming compounds
Naming Compounds

1. Ionic or Covalent

2. Ionic – two ions or

Poly atomic ions

Covalent

2 non-metals

Or a hydrocarbon

Type of Metal

Mono-valent metals groups

1A 2A 3A

Name the Metal

Name the Nonmetal + ide

(if PAI use its name)

CaF2 calcium fluoride

RbNO3 rubidium nitrate

Al(OH)3 aluminum hydroxide

Multi-valent Metal

Transitions metals and under the stairs

Find the Charge on the Metal

To make the compound neutral

Write the Charge with roman numeral

Name nonmetal + ide (if PAI use its name)

Ni+Cl- nickel(I) chloride

Pb2+SO42- lead(II) sulfate

Pb4+(SO4)22- lead(IV) sulfate

ionic naming
Ionic naming
  • Name to Formula– Final compound must be neutral based on subscripts and charges
  • Magnesium Fluoride Mg2+ + F- MgF2
  • Ammonium Sulfide NH4+ + S2- (NH4)2S
  • Tin(II) Carbonate Sn2+ + CO32- SnCO3
  • Iron(III) Oxide Fe3+ + O2- Fe2O3
  • Iron(II) Oxide Fe2++ O2- FeO
slide10

Ionic or Covalent

Ionic – two ions or

Poly atomic ions

Covalent

2 non-metals

Or a hydrocarbon

Hydrocarbons

Look for how many carbons

One – methane CH4

Two – ethane C2H6

Three – propane C3H8

Four – butane C4H10

Use the prefix to tell how many of each atom there are

Mono is never used with the first element

Example

PBr3 Phosphorous tribromide

CCL4 Carbon tetrachloride

P2O5 diphosphorouspentoxide

CO carbon monoxide

Hydrogen up front

Most likely an Acid you should have memorized

HCl - Hydrochloric acid

HI - Hydroiodic acid

HBr - Hydrobromic acid

HNO3 Nitric Acid

H2SO4 - Sulrufic acid

HClO- Hypochlorous acid

molecular compounds
Molecular Compounds
  • Name to formula – charge does not matter this time, just use the prefixes or memorize
  • Tetraarsenichexoxide As4O6
  • Sulfur hexafluroide SF6
  • Butane C4H10
  • Nitric acid HNO3
lewis dot structures
Lewis Dot Structures
  • Rules
  • Fewest number of atoms goes in the middle or C if present
  • Connect remaining elements with single bonds
  • Make sure all elements have 8 electrons (H only 2)
  • Count the number of electrons in structure
  • Add up valence electrons from PT
    • Too many e- in structure: remove 2 adjacent pairs fill in with one bond
    • Too few e- in structure: add to central atom
examples
EXAMPLES:
  • CH4
  • 1. (1) C + (4) H

(1)(4e-) + (4)(1e-) = 8e-

  • 2. Spatial order
  • 3. Draw bonds
  • 4. Octet rule satisfied?
  • 5. # of e- match?

H

C

H

H

H

examples1
EXAMPLES:
  • CO2
  • 1. (1) C + (2) O

(1)(4e-) + (2)(6e-) = 16e-

  • 2. Spatial order
  • 3. Draw Bonds
  • 4. Octet rule satisfied?
  • 5. # of e- match?

O

C

O

examples2
EXAMPLES:
  • NH3
  • 1. (1) N + (3)H

(1)(5e-) + (3)(1e-) = 8e-

  • 2. Spatial order
  • 3. Draw bonds
  • 4. Octet rule satisfied?
  • 5. # of e- match?

N

H

H

H

examples3
EXAMPLES:
  • CCl4
  • 1. (1) C + (4) Cl

(1)(4e-) + (4)(7e-) = 32e-

  • 2. Spatial Order
  • 3. Draw bonds
  • 4. Octet rule satisfied?
  • 5. # of e- match?

Cl

Cl

Cl

C

Cl

examples4
EXAMPLES:
  • NH4+
  • 1. (1) N + (4) H - (1)(+)

(1)(5e-)+ (4)(1e-) - (1)(1e-) = 8e-

  • 2. Spatial order
  • 3. Draw bonds
  • 4. Octet rule satisfied?
  • 5. # of e- match?

H

[ ]+

N

H

H

H

examples5
EXAMPLES:
  • SO42-
  • 1. (1) S + (4) O + (2)(-)

(1)(6e-)+ (4)(6e-) + (2)(1e-) = 32e-

  • 2. Spatial Order
  • 3. Draw bonds
  • 4. Octet rule satisfied?
  • 5. # of e- match?

O

[ ]

2-

O

S

O

O

examples6
EXAMPLES:
  • CN-
  • 1. (1) C + (1) N + (1)(-)

(1)(4e-) + (1)(5e-)+ (1)(1e-) = 10e-

  • 2. Spatial order
  • 3. Draw Bonds
  • 4. Octet rule satisfied?
  • 5. # of e- match?

[ ]-

C

N

examples7
EXAMPLES:
  • CO32-
  • 1. (1) C + (3) O + (2)(-)

(1)(4e-)+ (3)(6e-) + (2)(1e-) = 24e-

  • 2. Spatial Order
  • 3. Draw bonds
  • 4. Octet rule satisfied?
  • 5. # of e- match?

[ ]

2-

O

C

O

O

vsepr
VSEPR:
  • Regions of electron density (where pairs of electrons are found) can be used to determine the shape of the molecule.
  • CO2
  • Here there are two regions of electron density.
  • The regions want to be as far apart as possible, so it is linear.

O

C

O

examples8
EXAMPLES:

1

4

H

  • CH4
  • There are four electron pairs.
  • You would expect that the bond angles would be 90° but…
  • Because the molecule is three-dimensional, the angles are 109.5°.
  • The molecule is of tetrahedralarrangement.

C

H

H

2

H

3

examples9
EXAMPLES:

1

4

  • NH3
  • Four regions of electron density
  • But one of the electron pairs is a lone pair
  • The shape is called trigonal pyramidal

N

H

H

2

H

3

examples10
EXAMPLES:

1

4

  • H2O
  • Four regions of electron density
  • But two are lone pairs
  • This structure is referred to as bent

O

H

H

2

3

examples11
EXAMPLES:

1

3

[ ]

2-

  • CO32-
  • Three regions of electron density
  • This structure is referred to as trigonal planar

C

O

O

2

O

practice determining molecular shape
Practice determining molecular shape:
  • H2S
    • 4 regions of e- density
    • 2 lone pairs
    • bent

S

H

S

H

H

H

practice determining molecular shape1
Practice determining molecular shape:
  • SO2
    • 3 areas of e- density
    • 1 lone pair
    • bent

S

O

O

S

O

O

practice determining molecular shape2
Practice determining molecular shape:
  • CCl4
    • 4 areas of e- density
    • tetrahedral

Cl

3d

Cl

Cl

C

Cl

practice determining molecular shape3
Practice determining molecular shape:
  • BF3
    • 3 areas of e- density
    • trigonal planar

F

B

F

F

B

F

F

F

practice determining molecular shape4
Practice determining molecular shape:
  • NF3
    • 4 areas of e- density
    • 1 lone pair
    • pyramidal

3d

N

F

F

F

16 3 polar bonds and molecules
16.3Polar Bonds and Molecules
  • In covalent bonds, the sharing of electrons can be equal
  • or it can be unequal.
nonpolar covalent bonds
Nonpolar Covalent Bonds
  • Nonpolar covalent bond - This is a covalent bond in which the electrons are shared equally.
  • EXAMPLES:
  • H2
  • Br2
  • O2
  • N2
  • Cl2
  • I2
  • F2

Br

Cl

H

N

F

I

O

O

Br

Cl

H

N

F

I

polar bonds and molecules
Polar Bonds and Molecules
  • If the sharing is unequal, the bond is referred to as a dipole.
  • A dipole has 2 separated, equal but opposite charges.
  • “∂” means partial

_

+

polar bonds and molecules1
Polar Bonds and Molecules
  • Polar covalent bond - a covalent bond that has a dipole
  • It usually occurs when 2 different elements form a covalent bond.
  • EXAMPLE:
  • H + Cl  H Cl
polar bonds and molecules2
Polar Bonds and Molecules
  • Electronegativity - This is the measure of the attraction an atom has for a shared pair of electrons in a bond.
  • Electronegativity values increase across a period and up a group.
examples12
Examples:
  • Identify the type of bond for each of the following compounds:
  • HBr

Br = 2.8

H = 2.1

0.7

.1 < < 1.9

Polar Covalent

H

Br

examples13
Examples:
  • NaF

F = 4.0

Na = 0.9

3.1

  • N2

N = 3.0

N = 3.0

0.0

Na

F

> 2

Ionic

Non-Polar

Covalent

N

N

molecular polarity
Molecular Polarity
  • If there is only one bond in the molecule, the bond type and polarity will be the same.
  • If the molecule consists of more than 2 atoms, you must consider the shape. To determine its polarity, consider the following:
    • Lone pairs on central atom
      • If so… it is polar
    • Spatial arrangement of atoms
      • Do bonds cancel each other out (symmetrical)?
        • If so… nonpolar
      • Do all bonds around the central element have the same difference of electronegativity?
        • If so… nonpolar
polar molecules
Polar Molecules
  • If the molecule is symmetrical it will be nonpolar.
    • Exception hydrocarbons are nonpolar
  • If the molecule is not symmetrical it

will be polar, with a different atom or with lone pair(s)

Br

Cl

Cl

C

Cl

attractions between molecules
Attractions Between Molecules
  • Van der Waals forces – the weakest of the intermolecular forces. These include London dispersion and dipole-dipole forces.
    • London dispersion forces – between nonpolar molecules and is caused by movement of electrons
attractions between molecules1
Attractions Between Molecules
  • van der Waals forces(cont.)-
    • Dipole interactions – between polar molecules and is caused by a difference in electronegativity.
attractions between molecules2
Attractions Between Molecules
  • Hydrogen bonds – attractive forces in which hydrogen, covalently bonded to a very electronegative atom (N, O, or F) is also weakly bonded to an unshared (lone) pair of electrons on another electronegative atom.

O

O

H

H

H

H

O

H

H

attractions between molecules3
Attractions Between Molecules
  • Ionic Bonding-occurs between metals and nonmetals when electron are transferred from one atom to another.
  • These bonds are very strong.
summary of the strengths of attractive forces
Summary of the Strengths of Attractive Forces

Ionic bonds

hydrogen bonds

dipole-dipole attractions

LDF

writing and balancing chemical equations
Writing and Balancing Chemical Equations

Example:

Write the equation for the formation of sodium hydroxide and hydrogen, from the reaction of sodium with water.

slide46
Write the equation for the formation of sodium hydroxide and hydrogen, from the reaction of sodium with water.
  • Write the formulas of all reactants to the left of the arrow and all products to the right of the arrow.

Sodium + water

sodium hydroxide + hydrogen

Translate the equation and be sure the formulas are correct.

Na + H2O NaOH + H2

slide47
Write the equation for the formation of sodium hydroxide and hydrogen, from the reaction of sodium with water.
  • Once the formulas are correctly written, DO NOTchange them. Use coefficients (numbers in front of the formulas), to balance the equation.DO NOT CHANGE THE SUBSCRIPTS!

_____Na + _____H2O ____NaOH + _____H2

slide48

Begin balancing with an element that occurs only once on each side of the arrow.

  • Ex: Na

2

2

2

_____Na + _____H2O ____NaOH + _____H2

When you are finished, you should have equal numbers of each element on either side of the equation

Na

H

O

2

Na

H

O

2

4

4

2

2

slide49

To determine the number of atoms of a given element in one term of the equation, multiply the coefficient by the subscript of the element.

Ex: In the previous equation (below), how many hydrogen atoms are there?

4

2

2

2

____Na + _____H2O ____NaOH + _____H2

slide50

Helpful Hints:

  • Balance elements one at a time.
  • Balance polyatomic ions that appear on both sides of the equation as single units. (Ex: Count sulfate ions, not sulfur and oxygen separately)
  • Balance H and O last. Save the one that is in the most places for last…
  • Use Pencil!

(NH4)2SO4 (aq) + BaCl2 (aq)  BaSO4 (s) + 2NH4Cl (aq)

practice
Practice:
  • Balance the equation for the formation of magnesium nitride from its elements.

Mg2+

N3-

Mg3N2

____Mg + ____N2

3

____Mg3N2

ex nh 3 o 2 no 2 h 2 o
Ex: NH3 + O2 NO2 + H2O
  • H can be balanced by placing a 2 in front of NH3 and a 3 in front of H2O. Then put a 2 in front of NO2 for nitrogen to balance.

3

2

_____NH3 + _____O2____NO2 + ____H2O

2

slide53

_____NH3 + _____O2____NO2 + ____H2O

2

7/2

2

3

  • Now all that is left to balance is the oxygen. There are 2 O on the reactant side and 7 on the product side. Our only source of oxygen is the O2. Any whole number we place in front of the O2 will result in an even number of atoms. The only way to balance the equation is to use a coefficient of 7/2.
slide54

2Mg + O2 2MgO

Stoichiometry – study of calculations of quantities in chemical reactions using balanced chemical equations.

2 moles Mg + 1mole O2 2 moles MgO

slide55

2Mg + O2 2MgO

  • The mole ratios can be obtained from the coefficients in the balanced chemical equation.
  • What are the mole ratios in this problem?
  • Mole ratios can be used as conversion factors to predict the amount of any reactant or product involved in a reaction if the amount of another reactant and/or product is known.
slide56

What’s that mean?

Well, a stoichiometry problem gives you an amount of one chemical and asks you to solve for a different chemical.

To get from one type of chemical to another, a MOLE RATIO must be found between the two chemicals. You get the MOLE RATIO from the BALANCED CHEMICAL EQUATION!

slide57

A balanced chemical equation tells the quantity of reactants and products as well as what they are.

2 mol 1 mol 2 mol

*the coefficients are*

Mole Ratios

slide58

The MOLE RATIO is your mechanism of transition between the chemical that is your starting given and the chemical you are solving for.

The MOLE RATIO is the bridge between the two different chemicals

moles

moles

(given)

(want)

Mole

Ratio

? want

given

example
EXAMPLE
  • How many grams of KClO3 must decompose to produce KCl and 1.45 moles O2?

2KClO3→ 2KCl + 3O2

2 mol KClO3

122.6 g KClO3

1.45 moles O2

=

3 mol O2

1 mol KClO3

119 g KClO3

mol-mol ratio

GFM

slide60
CaCO3, limestone, is heated to produce calcium oxide, CaO, and CO2. What mass of limestone is required to produce 156.0 g of CaO?

CaCO3 (s)  CaO (s) + CO2 (g)

156.0 g CaO

1 mol CaO

1 mol CaCO3

100.1 g CaCO3

56.1 g CaO

1 mol CaO

1 mol CaCO3

GFM

mol-mol ratio

GFM

278.4 g CaCo3

=

example1
EXAMPLE
  • Calculate the number of joules released by the oxidation of 5.00 moles of Na completely react with oxygen gas. ΔH = -416 kJ/mol

4Na + O2 2Na2O

1000 J

5.00 mol Na

2 mol Na2O

-416 kJ

1mol Na2O

1 kJ

4 mol Na

mol-mol ratio

enthalpy

1.04x106 J

atomic orbital the region in space where the electron is likely to be found
Atomic orbital – the region in space where the electron is likely to be found

A quantum mechanical model of a hydrogen atom, which has one electron, in its state of lowest energy. The varying density of the spots indicates the relative likelihood of finding the electron in any particular region.

electrons can be described by a series of 4 quantum numbers you must be familiar with all of these
Electrons can be described by a series of 4 quantum numbers.You must be familiar with all of these!
slide64
1. Principle quantum number (n)-describes the principal energy level an electron occupies-values of 1,2,3,4,etc

1

2

3

4

5

6

7

slide65

2. Azimuthal quantum number (l)-describes the shape of atomic orbitals-s orbitals are spherical, p orbitals are peanut shaped, d orbitals are daisies and f orbitals are fancy-designates a sublevel-values of 0 up to and including n-1 0=s, 1=p, 2=d, 3=f

slide66

s orbital

Spheres

slide67

p orbitals

Peanuts

slide68

d orbitals

Daisies or Doughnuts

slide69

Fancy

f orbitals

slide71

3. magnetic quantum number (ml)-Designates the spatial orientation of an atomic orbital in space-values of -l to +l so s has 1 orbital p has 3 orbitals (x, y, and z) d had 5 orbitals (xy, xz, yz, x2-y2 and z2) f has 7 orbitals

slide72

4. spin quantum number -describes the orientation of the individual electrons; values of +1/2 and -1/2-each orbital can hold 2 electrons with opposite spins

electron spins

quick and easy electron configuration

Now, lets try to do an electron configuration for carbon.

Begin with the first quantum number and use the periodic table to write the configuration.

Now, try the three in your notes.

He 1s2

Na 1s2 2s2 2p6 3s1

Ti 1s2 2s2 2p6 3s2 3p6 4s2 3d2

Quick and Easy Electron Configuration

1

2

3

4

5

6

7

S

3

4

5

6

P

D

45

F

slide75

For a shorter way to write electron configuration, write the nearest noble gas and then continue. AKA “Shorthand Notation”.

Ex: Ti can be written as

OR

1s2 2s2 2p6 3s2 3p6 4s2 3d2

[Ar] 4s2 3d2

electron configuration of ions
Electron Configuration of Ions
  • For the loss of an electron remove electrons from the last orbital

Ge 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2

Ge2+1s2 2s2 2p6 3s2 3p6 4s23d10

For the gain of electrons add them in the last orbital filled

At [Xe] 6s2 4f14 5d10 6p5

At-[Xe] 6s2 4f14 5d106p6

slide77

Atomic Radius

SMALLEST

Increasing Atomic Radius

BIGGEST

Increasing Atomic Radius

slide78
Ion Size
  • Anions are larger than the atoms from which they were formed.
  • The negative charge means more electrons are present causing the size of the ion to be larger.
  • Cations are smaller than the atoms from which they were formed.
  • The positive charge means fewer electrons are surrounding the nucleus, thus pulling the existing electrons closer and causing the ion to be smaller.
  • Trend: ionic radius increases from right to left and top to bottom
slide79

Ionic Radius

Increasing Ionic Radius

Increasing Ionic Radius

slide80
Ionization energy - energy required to overcome the attraction of the nuclear charge and remove an electron from a gaseous atom
  • 1st ionization energy: the energy required to remove the first electron
  • 2nd ionization energy: the energy required to remove the second electron
  • 3rd ionization energy: the energy required removing the third electron
  • Trend: ionization energy increases from left to right and bottom to top
slide81

Increasing Ionization Energy

Increasing Ionization Energy

Ionization Energy

slide82
Electronegativity – the tendency for the atoms of the element to attract electrons when they are chemically combined with atoms of another element
  • Note: Noble gases don’t have values for electronegativity because their outer orbitals are full and they do not need to gain or lose electrons to be stable.
  • Trend: electronegativity increases from left to right and bottom to top
slide83

Increasing Electronegativity

HIGHEST

Increasing Electronegativity

Electronegativity

slide85

We can measure not only the length of each wave of light, but also its frequency of occurrence.

slide87

amplitude- height of the wave from the origin to the crestwavelength -  - distance between the crestsfrequency -  - the number of wave cycles to pass a given point per unit of time.The units of frequency are 1/s, s-1, or Hertz (Hz)

slide89
Ex. A certain wavelength of yellow light has a frequency of 2.73 x 1016s-1. Calculate its wavelength.

 = c/

= c/

= 3.00 x 108m/s

2.73 x 1016s-1

= 1.10 x 10-8 m

slide90

Spectrum- series of colors produced when sunlight is separated by a diffraction gradient.ROYG. BIVRed: has the longest wavelength, lowest frequency  Lowest energyViolet: has the shortest wavelength, highest frequency  highest energy

slide92

Kinetic theory- The tiny particles in all forms of matter are in constant motion.

  • A gas is composed of particles, usually molecules or atoms. We treat them as, Hard spheres, Insignificant volume, and Far from each other
  • The particles in a gas move rapidly in constant random motion.
  • All collisions are perfectly elastic.
gas laws
Gas Laws
  • One single set of conditions
  • PV = nRT
  • Two Sets of conditions
  • P1 V1 = P2 V2

T1 T2

slide94
Ex. If a helium-filled balloon has a volume of 3.40 L at 25.0oC and 120.0 kPa, what is its volume at STP = 0°C, 1 ATM?

V1 = 3.40L V2 = ?

P1 = 120.0 kPa P2 = 1 atm = 101.3 kPa

T1 = 25.0oC + 273 = 298K

T2 = 0oC + 273 = 273K

P1V1 = P2V2 (120.0 kPa)(3.40L) = (101.3 kPa)V2

T1 T2 298K 273K

V2 = 3.69L

slide95
Ex. A 5.0 L flask contains 0.60 g O2 at a temperature of 22oC. What is the pressure (in atm) inside the flask?

P=? V= 5.0L R = 0.0821 (L.atm/mol.K)

T = 22°C +273= 295K

n = 0.60g O2 1 mol O2 = 0.01875 mol

32.0g O2

PV = nRT

P(5.0L) = (0.01875mol)(0.0821Latm/molK)(295K)

P = 0.091 atm

factors affecting solubility
Factors affecting solubility:
  • The nature of the solute and the solvent
    • “like dissolves like”
slide97
Miscible - liquids that are soluble in each other
    • Ex. ethanol and water
  • Immiscible- liquids that are not soluble in each other
    • Ex. oil and water
slide98
Molarity(M) = moles of solute

liters of solution

  • A 1 M solution contains 1 mole of solute per 1 L of solution. A 0.5 M NaCl solution has 0.5 mol NaCl in 1 L total solution.
examples14
EXAMPLES:
  • What is the concentration in molarity of a solution made with 1.25 mol NaOH in 4.0 L of solution?

# mol

# L

M =

M = ?

#mol = 1.25 mol

V = 4.0 L

M = 1.25 mol

4.0 L

= 0.3125 M

= 0.31 M

enthalpy h the amount of heat in a system at a given temperature enthalpy change h q m c t
Enthalpy (H)- the amount of heat in a system at a given temperatureEnthalpy change: H = q = m C  T
slide102

Ex. A 25 g sample of a metal at 75.0oC is placed in a calorimeter containing 25 g of H2O at 20.0oC. The temperature stopped changing at 29.4oC. What is the specific heat of the metal?

standard heat of formation of a compound h f o
Standard Heat of Formation of a compound( Hfo)

*Hfo of a free element in its standard state is zero.H = Hfo products - Hforeactants

ex 1 calculate h for the following reaction endo or exo thermic caco 3 s cao s co 2 g
Ex. 1. Calculate  H for the following reaction:(endo or exo thermic?)CaCO3(s) CaO(s) + CO2(g)

∆Hfo values:CaCO3 = -1207.0 kJ/molCaO = -635.1 kJ/molCO2 = -393.5 kJ/mol

∆H = [-635.1 + (-393.5)] – [-1207.0]

∆H = 178.4 kJ

endothermic, heat going in

calculate the heat of reaction for the following reaction endo or exo thermic 2h 2 g o 2 g 2h 2 o g
Calculate the heat of reaction for the following reaction: (endo or exothermic?) 2H2(g) + O2(g)  2H2O(g)

∆Hfo values:H2O(g) = -241.8 kJ/mol

∆H = [2(-241.8)] – [2(0) + 0]

∆H = -483.6 kJ

Exothermic

Heat leaving

Remember to multiply heat values by coefficients!!!

entropy calculation example mn s 2o 2 g mno 4 s s j mol 1 k 1 mn s 32 8 o 2 205 0 mno 4 120 5
Entropy Calculation Example: Mn(s) + 2O2(g)  MnO4(s)S° J mol-1 K-1Mn (s) = 32.8 O2 = 205.0 MnO4 = 120.5

∆S= [S° Products] – [S° Reactants]

∆S = [120.5] – [2(205.0) + 32.8]

∆S = -322.3 J/mol K

entropy s a measure of randomness or disorder
Entropy, S - a measure of randomness or disorder
  • associated with probability (There are more ways for something to be disorganized than organized.)
  • Entropy increases going from a solid to a liquid to a gas.
  • Entropy increases when solutions are formed.
  • Entropy increases in a reaction when more atoms or molecules are formed.
  • The entropy of a substance increases with temperature.
gibbs free energy g
Gibbs free energy, G
  • Energy available to do work
  • Go = standard free energy change
    • change in free energy that occurs if the reactants in their standard states are converted to products in their standard states

Go =Ho -T So

When Gois negative the reaction is spontaneous is the forward direction

When Gois positive the reaction is nonspontaneous is the forward direction

slide109

What temperature would a reaction be spontaneous if ΔH = 9500 J/mol and ΔS = 6.5J/mol K?

Go =Ho -T So

9500 = T(6.5)

T = 1461 K

Above 1461 K this reaction will be spontaneous and below it will not

slide110

Will this reaction be spontaneous at 100°C?

ΔH = -18 KJ/moland ΔS = 94.3 J/mol K

Go =Ho -T So

Yes at all temperatures

Go=(Ho) -T (So)

Go=(-) - (+)(+)

Go= can only be negative

Don’t believe me try putting in the values and craze high/low number for temperature

g o h o t s o
Go =Ho -T So
  • A spontaneous reaction has a negative G. For example, when ice melts H is positive (endothermic), S is positive and G = 0 at 0oC.
  • If...
collision theory
Collision Theory:
  • In order to react, two or more particles must collide with sufficient energy (called the activation energy) and with the proper molecular orientation. If the colliding particles do not have either of these two prerequisites, no product is formed.
factors that affect reaction rate
Factors that affect reaction rate:
  • Temperature- Reactions go faster at higher temperatures. Particles have more kinetic energy. More colliding particles have enough energy to overcome the activation energy barrier.
factors that affect reaction rate1
Factors that affect reaction rate:
  • Concentration- Increasing the concentration of reactants usually increases the reaction rate. If there are more particles to collide, there should be a greater number of collisions that produce products.
factors that affect reaction rate2
Factors that affect reaction rate:
  • Catalysts- A catalyst is a substance that speeds up a reaction by lowering the activation energy barrier. It is not a product or reactant and it is not used up or changed itself.
rate law
Rate Law
  • equation that is written that expresses how the reaction rate of a particular reaction is dependent upon the concentrations of its reactants.
  • For the reaction aA + bB  cC + dD, the general form of the rate law would be:

Rate = k [A]a[B]b

slide117

Rate = k [A]x[B]y

  • Rate is usually expressed as mol/L∙time.
  • k is the specific rate constant. It is constant for a given reaction at a given temperature. The faster a reaction, the larger the k value.
  • [A] and [B] represent the concentrations of reactants A and B in moles per liter (M).
  • x and y are the order of the reactant. They can only be determined by analyzing experimental data. These exponents are usually positive integers.
examples15
EXAMPLES

2A + B 2C

Ex.[A][B]Rate

1 0.1 0.2 0.10

2 0.1 0.4 0.20

3 0.2 0.4 0.80

Determine the rate law:

[A]

x

0.1

0.2

0.1

0.8

( )

=

( ) 23 =8

[B]

x

0.2

0.4

0.1

0.2

=

( )

3

1

rate = k

[A]

[B]

half life
Half-life

Based on how much time does it take ½ of the substance to change into products

x = time / half-life (number of half-lives)

mf = (mi)(1/2)x

slide120

If you start with 2.00 g of nitrogen-13 how many grams will remain after 4 half lives?

mf = 2.00 (1/2)4

mf=0.125 g

Phosphorous-32 has a half-life of 14.3 yr. How many grams remain after 57.2 yr from a 4.0 g sample

mf = 4.00 (1/2)(57.2/14.3)

mf = 0.25 g

slide121

BrØnsted-Lowry

  • Acid: hydrogen-ion donor (protondonor)
  • Base: hydrogen-ion acceptor (protonacceptor)

NH3 + H2O  NH4+ + OH-

base acid

slide122

BrØnsted-Lowry

  • Acid: hydrogen-ion donor (protondonor)
  • Base: hydrogen-ion acceptor (protonacceptor)

NH3 + H2O  NH4+ + OH-

base acid

working ka and kb problems
Working Ka and Kb problems

R = reaction

I = initial

C = change

E = equilibrium

  • 1st: Write the equation
  • 2nd: Set up a reaction diagram (RICE diagram)
  • 3rd: Set up Ka or Kb expression
  • 4th: Substitute values into Ka expression
  • 5th: Solve Ka expression for X.
  • 6th: Calculate pH from H+ or OH- concentration.
example2
Example:

Calculate the pH of a 0.10 M solution of acetic acid. The Ka for acetic acid is 1.8 x 10-5.

R HC2H3O2 H+ + C2H3O2-

I

C

E

.1

0

0

-x

+x

+x

.1- x

x

x

[H+]

[C2H3O2-]

[x]

[x]

Ka =

1.8 x 10-5

=

=

[HC2H3O2]

[0.1-x]

slide125

[x]

[x]

1.8 x 10-5

=

pH = -log [H+]

[.1-x]

pH = -log(1.3 x 10-3)

x2

1.8 x 10-5

=

.1-x

pH= 2.87

x2

1.8 x 10-5

=

.1

x2

1.8 x 10-6

=

= [H+]

x

=

1.3 x 10-3

slide126

Example:

Calculate the pH of a 0.25 M solution of HCN. The Ka for HCN is 6.2 x 10-10.

R HCN  H+ + CN-

I

C

E

.25

0

0

-x

+x

+x

.25 - x

x

x

[H+][CN-]

[HCN]

[x][x]

.25 – x

=

Ka = 6.2 x 10-10 =

slide127
Ka = 6.2 x 10-10 = [H+][CN-] = [x][x]

[HCN] .25 – x

X2

.25

= 6.2 x 10-10

X2 =

1.55 x 10-10

x = 1.24 x 10-5

= [H+]

pH = -log [H+]

pH = -log(1.24 x 10-5)

pH= 4.9

slide128
Calculate the pH of a 0.15 M solution of ammonia. The Ka of ammonia is 1.8 x 10-9.

R NH3 + H2O (l)  NH4+ + OH-

I

C

E

.15

0

0

-x

+x

+x

.15 –x

x

x

Ka= [NH4+] [OH-]

[NH3]

[x][x]

.15 – x

=

slide129
Ka = [NH4+] [OH-] = [x][x]

[NH3] .15 – x

X2 = 1.8 x 10-9

.15

X2 = 2.7 x 10-10

X = 0.000016

pH = -log (H+)

pH = -log (0.000016)

pH = 4.78

= [H+]

lechatelier s principle
LeChatelier\'s Principle
  • When a stress is applied to a system, the equilibrium will shift in the direction that will relieve the stress.

Henry Louis Le Chatelier

changes in concentration
Changes in concentration
  • An increase in concentration of:
    • a reactant will cause equilibrium to shift to the right to form more products.
    • a product will cause equilibrium to shift to the left to form more reactants.
  • A decrease in concentration of:
    • a product will cause equilibrium to shift to the right to form more products.
    • a reactant will cause equilibrium to shift to the left to make more reactants.
a b c d
A + B  C + D
  • Remove A or B ←
  • Add C or D ←
  • Remove C or D →
  • Add A or B →
changes in temperature
Changes in temperature
  • Treat energy as a product or reactant and temperature changes work just like changes in concentration!
  • An increase in temperature of an exothermic reaction (H is negative) will cause equilibrium to shift to the left. A decrease in temperature of an exothermic reaction will cause equilibrium to shift to the right.
  • An increase in temperature of an endothermic reaction (H is positive) will cause equilibrium to shift to the right. A decrease in temperature of an endothermic reaction will cause equilibrium to shift to the left.
example3
Example:

N2 + O2 2NO

H = 181 kJ (endothermic)

  • addition of heat
  • lower temperature

181 kJ +

example4
Example:

2SO2 + O2 2SO3

H= -198 kJ (exothermic)

  • increase temperature
  • remove heat

+ 181 kJ

example5
Example:

CaCO3 + 556 kJ  CaO + CO2

  • decrease temperature
changes in pressure
Changes in pressure
  • Changes in pressure only affect equilibrium systems having gaseous products and/or reactants.
  • Increasing the pressure of a gaseous system will cause equilibrium to shift to the side with fewer gas particles.
  • Decreasing the pressure of a gaseous system will cause equilibrium to shift to the side with more gas particles.
addition of a catalyst
Addition of a catalyst
  • Adding a catalyst does not affect equilibrium. Catalysts speed up the forward and reverse reactions equally.
example6
Example:

P4(s) + 6Cl2(g)  4PCl3(l)

  • increase container volume
    • Shifts to side with more gas
  • decrease container volume
    • Shifts to side with less gass
  • add a catalyst
    • Inert gases have no effect on equilibrium

N/C

example7
Example:

Consider the reaction:

2NO2(g)  N2(g) + 2O2(g)

which is exothermic

  • NO2 is added
  • N2 is removed
  • The volume is halved
  • He (g) is added
  • The temperature is increased
  • A catalyst is added

+ HEAT

N/C

N/C

mechanisms
Mechanisms
  • Uni molecular

A → B

AB → A + B

A2→ 2A

  • Bi molecular

A + B → AB

AB + C → AC + B

Only one thing changing at a time

overall reactions
Overall reactions

NO2 + F2 NO2F + F (slow step)

NO2 + F  NO2F (fast step)

Simplify F on both sides

2NO2 + F2  NO2F

Intermediate F

Rate = k[NO2]1[F2]1

Rate depends on both NO2 and F2

overall reactions1
Overall reactions

PO2 + Cl PClO2 (fast step)

PClO2 + PO2 P2O4 + Cl (slow step)

Simplify Cland PClO2 on both sides

2PO2 P2O4

Intermediate PClO2

Catalyst Cl

Rate = k[PO2]2

Rate depends on only PO2

oxidation reduction reactions
Oxidation Reduction Reactions
  • Oxidation-reduction reactions- chemical changes that occur when electrons are transferred between reactants.
  • Also called REDOX reactions
slide145

Oxidation Reduction Reactions

  • Oxidation
  • Modern definition - loss of electrons
  •  Examples
  • 4Fe + 3O2 2Fe2O3 (rusting of iron)
  • C + O2 CO2 (burning of carbon)
  • C2H5OH + 3O2 2CO2 + 3H2O (burning of ethanol)
slide146

To help remember these definitions, use one of these mnemonic devices:

Oxidation

Is

Loss

Reduction

Is

Gain

LEO (Lose Electrons-Oxidation)

the lion goes

GER (Gain Electrons-Reduction)

formation of ions
Formation of Ions
  • Ex. 2Na + S  Na2S
  • Sodium goes from the neutral atom to the 1+ ion. Therefore, it has lost an electron (It was oxidized). Sulfur goes from the neutral atom to the 2- ion. Therefore, it has gained two electrons. (It was reduced)
question
Question
  • Lead loses four electrons. It take on a charge of ___. Does this mean that it is oxidized or reduced?
redox reaction examples
REDOX Reaction Examples
  • Identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent for each of the following:

MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O

+4

-2

+1

-1

+2

-1

0

+1

-2

Mn

+4

+2

Gained 2 e-

Reduced

If it was Reduced, then the reactant that contains Mn acts as the “Oxidizing Agent”. (MnO2)

Cl

-1

0

Lost 1 e-

Oxidized

If it was Oxidized, then the reactant that contains Cl acts as the “Reducing Agent”. (HCl)

batteries
Batteries
  • Electrochemical cells that convert chemical energy into electrical energy are called voltaic cells.
  • The energy is produced by spontaneous redox reactions.
  • Voltaic cells can be separated into two half cells.
  • A half cell consists of a metal rod or strip immersed in a solution of its ions.
batteries1
Batteries
  • We write half reactions to show what happens in each part of the cell.
    • Example Write the half reactions that occur in the Fe2+/Ni2+ cell.
    • Oxidation Fe  Fe2+ + 2e-
    • Reduction Ni2+ + 2e- Ni
diagram of voltaic cell for the reaction of zinc and copper
Diagram of voltaic cell for the reaction of zinc and copper.
  • Diagram of voltaic cell for the reaction of zinc and copper.
  • Oxidized: Zn  Zn2+ + 2e-
  • Reduced: Cu2+ + 2e- Cu

Direction of electron flow

From anode to cathode

anode

cathode

Solution of

Solution of

Zinc ions

Copper ions

half cells
Half-Cells
  • The half cells are connected by a salt bridge. A salt bridge is a tube containing a solution of ions.
  • Ions pass through the salt bridge to keep the charges balanced.
  • Electrons pass through an external wire.
  • The metal rods in voltaic cells are called electrodes.
  • Oxidation occurs at the anode and reduction occurs at the cathode. (An Ox and Red Cat)
  • The direction of electron flow is from the anode to the cathode. (FAT CAT )
calculating the charge of a battery
Calculating the Charge of a Battery
  • The potential charge of a battery can be calculated with a set of values from a table of reduction potentials.
    • To do this, write the oxidation and reduction half reactions.
    • Look up the cell potentials from the data table.
    • Flip the sign of the cell potential for oxidation.
    • Add the potentials together.
slide155

Example A common battery is made with nickel and cadmium. What is the cell potential of this battery? (E0Cd = -0.40V, E0Ni = -0.25V)

Oxidation Cd Cd2+ + 2e- E0 = 0.40V

Reduction Ni2+ + 2e- Ni + E0 = -0.25V

Total = E0 = 0.15V

spontaneous electrochemical reactions
Spontaneous Electrochemical Reactions

ΔG = -nFE

Recall the NiCd battery

Total = E0 = 0.15V

Positive Voltage, -ΔG, spontaneous reaction

Negative Voltage, +ΔG, nonspontaneous reaction

slide157

monomers

Polymer

The word, polymer, implies that polymers are constructed from pieces (monomers) that can be easily connected into long chains (polymer). When you look at the above shapes, your mind should see that they could easily fit together.

slide158

There are two types of polyethylene polymers (plastics). One is when the polyethylene exists as long straight chains. The picture here shows the chains of one carbon with two hydrogen atoms repeating. The chain can be as long as 20,000 carbons to 35,000 carbons. This is called high density polyethylene (HDPE).

slide159

Low density polyethylene (LDPE) is made by causing the long chains of ethylene to branch. That way they cannot lie next each other, which reduces the density and strength of the polyethylene. This makes the plastic lighter and more flexible.

ad