The Basics Elements, Molecules, Compounds, Ions Parts of the Periodic Table How to Name 1.

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The Basics

• Elements, Molecules, Compounds, Ions• Parts of the Periodic Table• How to Name

Classification of Elements• Metals – found on the left-side of the

Periodic Table

•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

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

• 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

Elements , Molecules, Compounds, and Ions

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

• 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 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 NH group

• Example KOH C4H6NH2

Naming Compounds1. 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 MetalName the Nonmetal + ide(if PAI use its name)

CaF2 calcium fluorideRbNO3 rubidium nitrateAl(OH)3 aluminum hydroxide

Multi-valent MetalTransitions metals and under the stairs

Find the Charge on the MetalTo make the compound neutralWrite the Charge with roman numeralName nonmetal + ide (if PAI use its name)

Ni+Cl- nickel(I) chloridePb2+SO4

2- lead(II) sulfatePb4+(SO4)2

2- lead(IV) sulfate

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

Covalent 2 non-metals

Or a hydrocarbon

Ionic or Covalent

Ionic – two ions or Poly atomic ions

Use the prefix to tell how many of each atom there areMono is never used with the first element

ExamplePBr3 Phosphorous tribromideCCL4 Carbon tetrachlorideP2O5 diphosphorous pentoxideCO carbon monoxide

Hydrogen up frontMost likely an Acid you should have memorized

HCl - Hydrochloric acidHI - Hydroiodic acidHBr - Hydrobromic acidHNO3 Nitric AcidH2SO4 - Sulrufic acidHClO - Hypochlorous acid

HydrocarbonsLook for how many carbons

One – methane CH4

Two – ethane C2H6

Three – propane C3H8

Four – butane C4H10

Molecular Compounds

• Name to formula – charge does not matter this time, just use the prefixes or memorize

• Tetraarsenic hexoxide As4O6

• Sulfur hexafluroide SF6

• Butane C4H10

• Nitric acid HNO3

Lewis Dot Structures• Rules1. Fewest number of atoms goes in the middle or

C if present2. Connect remaining elements with single bonds3. Make sure all elements have 8 electrons (H only 2)

4. Count the number of electrons in structure5. 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:

• 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?

EXAMPLES:

• CO2

• 1. (1) C + (2) O (1)(4e-) + (2)(6e-) = 16e-

• 3. Draw Bonds

• 4. Octet rule satisfied? • 5. # of e- match?

EXAMPLES:

• NH3

• 1. (1) N + (3)H(1)(5e-) + (3)(1e-) = 8e-

• 3. Draw bonds

EXAMPLES:• CCl4

• 1. (1) C + (4) Cl(1)(4e-) + (4)(7e-) = 32e-

• 2. Spatial Order

• 3. Draw bonds

ClClCl

EXAMPLES:

• NH4+

• 1. (1) N + (4) H - (1)(+) (1)(5e-)+(4)(1e-) - (1)(1e-) = 8e-

• 3. Draw bonds

EXAMPLES:• SO4

• 1. (1) S + (4) O + (2)(-) (1)(6e-)+ (4)(6e-) + (2)(1e-) = 32e-

• 3. Draw bonds

EXAMPLES:

• CN-

• 1. (1) C +(1) N + (1)(-)

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

• 3. Draw Bonds

• 4. Octet rule satisfied? • 5. # of e- match?

C N[ ]-

EXAMPLES:• CO3

• 1. (1) C + (3) O + (2)(-) (1)(4e-)+ (3)(6e-) + (2)(1e-) = 24e-

• 3. Draw bonds

OO[ ]2-

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.

EXAMPLES:

• 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 tetrahedral arrangement.

EXAMPLES:

• NH3

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

EXAMPLES:

• H2O• Four regions of electron density• But two are lone pairs

• This structure is referred to as bent

EXAMPLES:

• CO32-

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

Practice determining molecular shape:

• H2S– 4 regions of e- density– 2 lone pairs– bent

SHHS HH

• SO2

– 3 areas of e- density– 1 lone pair– bent

• CCl4

– 4 areas of e- density– tetrahedral

ClClCl

• BF3

– 3 areas of e- density – trigonal planar

• NF3

– 4 areas of e- density – 1 lone pair– pyramidal

16.3 Polar Bonds and Molecules

• In covalent bonds, the sharing of electrons can be equal

• or it can be unequal.

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

H HBr BrO ON NCl ClI IF F

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 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

• 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.

Examples:

• Identify the type of bond for each of the following compounds:

• HBrBr = 2.8H = 2.1

0.7 Polar Covalent.1 < < 1.9

Examples:

• NaFF = 4.0Na = 0.9

3.1• N2

N = 3.0N = 3.0

Non-PolarCovalent

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

• If the molecule is symmetrical it will be nonpolar.– Exception hydrocarbons are nonpolar

• If the molecule is not symmetrical itwill be polar, with a different atom or with lone pair(s)

ClClBr

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

• van der Waals forces(cont.)-– Dipole interactions – between polar molecules

and is caused by a difference in electronegativity.

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.

• 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

Ionic bonds hydrogen bonds dipole-dipole attractions LDF

Writing and Balancing Chemical Equations

Example:

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

1. Write the formulas of all reactants to the left of the arrow and all products to the right of the arrow.

Sodium + water

Translate the equation and be sure the formulas are correct.

Na + H2O NaOH + H2

sodium hydroxide + hydrogen

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

2. Once the formulas are correctly written, DO NOT change them. Use coefficients (numbers in front of the formulas), to balance the equation. DO NOT CHANGE THE SUBSCRIPTS!

_____Na + _____H2O ____NaOH + _____H2

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

Ex: Na

_____Na + _____H2O ____NaOH + _____H2

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

4. 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?

____Na + _____H2O ____NaOH + _____H22 2 2

• 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)

Helpful Hints:

Practice:• Balance the equation for the formation of

magnesium nitride from its elements.

____Mg + ____N2

Mg2+ N3-

3 ____Mg3N2

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.

_____NH3 + _____O2 ____NO2 + ____H2O2 32

• 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.

_____NH3 + _____O2 ____NO2 + ____H2O2 327/2

Stoichiometry – study of calculations of quantities in

chemical reactions using balanced chemical equations.

2Mg + O2 2MgO

2 moles Mg + 1mole O2 2 moles MgO

• 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.

2Mg + O2 2MgO

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!

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*

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

? want(given) (want)

MoleRatio

EXAMPLE• How many grams of KClO3 must decompose to

produce KCl and 1.45 moles O2?

2KClO3 → 2KCl + 3O2

1.45 moles O2

3 mol O2

2 mol KClO3

1 mol KClO3

122.6 g KClO3 =

119 g KClO3

mol-mol ratio

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

156.0 g CaO56.1 g CaO

1 mol CaO1 mol CaO

CaCO3 (s) CaO (s) + CO2 (g)

mol-mol ratio

1 mol CaCO3

100.1 g CaCO3

278.4 g CaCo3=

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

5.00 mol Na

4 mol Na

2 mol Na2O

mol-mol ratio

1 mol Na2O-416 kJ

1000 J

enthalpy 1.04x106 J

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

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!

1. Principle quantum number (n)

-describes the principal energy level an electron occupies

-values of 1,2,3,4,etc

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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

Spheres

67Peanuts

Daisies or Doughnuts

Where do I find the orbital shapes?

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

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

Symbol Shape Orbitals Electrons

s sphere 1 2

p peanut 3 6

d Daisy/ donut 5 10

f fancy 7 14

Quick and Easy Electron Configuration

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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

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

1s2 2s2 2p6 3s2 3p6 4s2 3d2 [Ar] 4s2 3d2

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 4s2 3d10

For the gain of electrons add them in the last orbital filledAt [Xe] 6s2 4f14 5d10 6p5

At- [Xe] 6s2 4f14 5d10 6p6

Atomic Radius

Increasing Atomic Radius

BIGGEST

SMALLEST

• 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

Ionic Radius

Increasing Ionic Radius

• 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

81Ionization Energy

Increasing Ionization Energy

• 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

83Electronegativity

Increasing Electronegativity

vityHIGHEST

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

Parts of a wave:

Amplitude

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)

= c/ where c = speed of light

c= 3.00 x 108 m/s

As increases, decreases.

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

Spectrum- series of colors produced when sunlight is separated by a diffraction gradient.

ROY G. BIVRed: has the longest wavelength, lowest frequency Lowest energyViolet: has the shortest wavelength, highest frequency highest energy

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

1. A gas is composed of particles, usually molecules or atoms. We treat them as, Hard spheres, Insignificant volume, and Far from each other

2. The particles in a gas move rapidly in constant random motion.

3. All collisions are perfectly elastic.

Gas Laws

• One single set of conditions• PV = nRT• Two Sets of conditions• P1 V1 = P2 V2 T1 T2

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

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?

P=? V= 5.0L R = 0.0821 (L.atm/mol.K)T = 22°C +273= 295Kn = 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

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?

Factors affecting solubility:

• The nature of the solute and the solvent– “like dissolves like”

• 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

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.

EXAMPLES:• What is the concentration in molarity of a

solution made with 1.25 mol NaOH in 4.0 L of solution? # mol

#mol = 1.25 molV = 4.0 L

M = 1.25 mol 4.0 L

= 0.3125 M = 0.31 M

Exothermic reactions have - H

Endothermic reactions have + H

Enthalpy (H)- the amount of heat in a system at a given temperature

Enthalpy change:

H = q = m C T

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( Hf

* Hfo of a free element in its standard state is

zero. H = Hf

o products - Hf

oreactants

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

2. Calculate the heat of reaction for the following reaction: (endo or exo thermic?) 2H2(g) + O2(g) 2H2O(g)

∆Hfo values:

H2O(g) = -241.8 kJ/mol

∆H = [2(-241.8)] – [2(0) + 0] ∆H = -483.6 kJExothermicHeat leaving

Remember to multiply heat values by coefficients!!!

Entropy Calculation Example: Mn(s) + 2O2(g) MnO4(s)

S° J mol-1 K-1 Mn (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

• 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• 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 Go is negative the reaction is spontaneous is the forward directionWhen Go is positive the reaction is nonspontaneous is the forward direction

• 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 KAbove 1461 K this reaction will be spontaneous and below it will not

• Will this reaction be spontaneous at 100°C?ΔH = -18 KJ/mol and ΔS = 94.3 J/mol K

Go =Ho -T So Yes at all temperatures Go =(Ho) -T (So) Go =(-) - (+)(+)Go = can only be negativeDon’t believe me try putting in the values and craze high/low number for temperature

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...

Entropy, ΔS Enthalpy, ΔH Spontaneity

Positive Positive Yes at high temp

Negative Positive Never spontaneous

Positive Negative Always spontaneous

Negative Negative Yes at low temp

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:

• Temperature- Reactions go faster at higher temperatures. Particles have more kinetic energy. More colliding particles have enough energy to overcome the activation energy barrier.

• 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.

• 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

• 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

• 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.

Rate = k [A]x[B]y

( )0.20.4

0.10.8( ) 23 =8

0.10.2

EXAMPLES

2A + B 2CEx. [A] [B] Rate1 0.1 0.2 0.102 0.1 0.4 0.203 0.2 0.4 0.80Determine the rate law:

( ) =x

rate = k

[A] [B]3 1

0.10.2

Half-life

Based on how much time does it take ½ of the substance to change into productsx = time / half-life (number of half-lives)mf = (mi)(1/2)x

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

mf = 0.25 g

• Acid: hydrogen-ion donor (proton donor)• Base: hydrogen-ion acceptor (proton acceptor) NH3 + H2O NH4

+ + OH-

base acid

BrØnsted-Lowry

• Acid: hydrogen-ion donor (proton donor)• Base: hydrogen-ion acceptor (proton acceptor) NH3 + H2O NH4

+ + OH-

base acid

BrØnsted-Lowry

Working Ka and Kb problems

• 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.

R = reaction

I = initial

C = change

E = equilibrium

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-

Ka =[H+] [C2H3O2

-] [HC2H3O2]

=[x] [x]

[0.1-x]= 1.8 x 10-5

[x] [x][.1-x] = 1.8 x 10-5

.1-x = 1.8 x 10-5

.1 = 1.8 x 10-5

x2 = 1.8 x 10-6

x = 1.3 x 10-3 = [H+]

pH = -log(1.3 x 10-3)

pH = -log [H+]

pH= 2.87

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-

.25 - x

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

[x][x].25 – x

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

[HCN] .25 – x X2

.25= 6.2 x 10-10

= [H+]1.55 x 10-10X2 =

x = 1.24 x 10-5

pH = -log(1.24 x 10-5)pH= 4.9

pH = -log [H+]

• 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-

.15 –x

Ka = [NH4+] [OH-]

[NH3] [x][x].15 – x=

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

[NH3] .15 – x

X2 = 1.8 x 10-9

X2 = 2.7 x 10-10

X = 0.000016

pH = -log (H+)

pH = 4.78

= [H+]

pH = -log (0.000016)

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

• 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• Remove A or B ← • Add C or D ←• Remove C or D → • Add A or B →

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.

Example:

N2 + O2 2NO

H = 181 kJ (endothermic)

• addition of heat• lower temperature

181 kJ +

Example:

2SO2 + O2 2SO3

H= -198 kJ (exothermic)

• increase temperature• remove heat

+ 181 kJ

Example:

CaCO3 + 556 kJ CaO + CO2

• decrease temperature

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

• Adding a catalyst does not affect equilibrium. Catalysts speed up the forward and reverse reactions equally.

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

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

Mechanisms

• Uni molecularA → BAB → A + BA2 → 2A

• Bi molecular A + B → AB AB + C → AC + B Only one thing changing at a time

Overall reactionsNO2 + F2 NO2F + F (slow step)

NO2 + F NO2F (fast step)

Simplify F on both sides2NO2 + F2 NO2F

Intermediate FRate = k[NO2]1[F2]1

Rate depends on both NO2 and F2

Overall reactionsPO2 + Cl PClO2 (fast step)

PClO2 + PO2 P2O4 + Cl (slow step)

Simplify Cl and PClO2 on both sides

2PO2 P2O4

Intermediate PClO2

Catalyst ClRate = k[PO2]2

Rate depends on only PO2

Oxidation Reduction Reactions

• Oxidation-reduction reactions- chemical changes that occur when electrons are transferred between reactants.

• Also called REDOX 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)

Oxidation Reduction Reactions

OxidationIsLoss

ReductionIsGain

LEO (Lose Electrons-Oxidation)

the lion goes

GER (Gain Electrons-Reduction)

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

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

• Lead loses four electrons. It take on a charge of ___. Does this mean that it is oxidized or reduced?

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 0-1 +1 -2

+4 +2 Gained 2 e-

Oxidized

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

-1 0 Lost 1 e-

Reduced

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

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.

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.• Oxidized: Zn Zn2+ + 2e-

• Reduced: Cu2+ + 2e- CuDirection of electron flow

Solution of Solution of

From anode to cathode

anode cathode

Zinc ions Copper ions

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

• 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.

Example A common battery is made with nickel and cadmium. What is the cell

potential of this battery? (E0Cd = -0.40V, E0

Ni = -0.25V)

Oxidation Cd Cd2+ + 2e- E0 = 0.40VReduction Ni2+ + 2e- Ni + E0 = -0.25V

Total = E0 = 0.15V

Spontaneous Electrochemical Reactions

ΔG = -nFERecall the NiCd batteryTotal = E0 = 0.15V

Positive Voltage, -ΔG, spontaneous reaction Negative Voltage, +ΔG, nonspontaneous reaction

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.

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).

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.

The Basics Elements, Molecules, Compounds, Ions Parts of the Periodic Table How to Name 1.

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