Structure and Properties Early Atomic Theories Scientist Model Atomic Theory Flaws Dalton (1805)

Little hard indestructible balls

Postulates Statements reflecting experimental evidence of the time.

Thomson (1897)

Raisin Bun

Negative particles (electrons) embedded in positive material. Based on studies with cathode ray tubes.

Rutherford (1911)

Nuclear atom Nucleus containing protons with electrons traveling around the nucleus.

Bohr (1913) Solar System (Planetary)

Nucleus containing protons electrons orbit nucleus in shells representing different energy levels.

Sub-atomic Particles

Name Symbol Discoverer Mass Charge LocationProton

(p) p1

1 Goldstein 1.00727252 u 1.6726430x10-24g

+1 In nucleus

Electron (e)

e01− Thomson 0.0005485712u 9.1093897x10-28g

-1 1.60217733x10-19C

Orbiting nucleus

Neutron (n)

n10 Chadwick

(1932) 1.008665u

1.674954 x10-24g 0 In

nucleus Isotopes: atoms of the same element with the same number of

protons (atomic number, Z) but with a different number of neutrons (mass number, A)

Ex. Cl Cl 3717

3517

Modern Atomic Theory (based on quantum mechanics) Quantum Number

Symbol Meaning

Principal

n n=1,2,3,.

Energy level of the electron.

Secondary l l=0!n-1

Shape of the orbital inside which the electron is traveling. l=0 l=1 l=2 l=3

Magnetic ml -l ! +l

Orientation of the orbital in 3-D space. l=0 , ml =0 l=1 , ml = -1, 0, +1 l=2 , l=3,

Spin ms

+ ½ ,- ½

Spin of the electron (opposite directions)

Filling Orbitals Aufbau principle: Hund�s Rule: Pauli Exclusion Principle: Order for Filling orbitals 1s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f

Element Configuration Na 1s2 2s2 2p6 3s1 P 1s2 2s2 2p6 3s2 3p3 (3px

1 3py1 3pz

1 ) Ca 1s2 2s2 2p6 3s2 3p6 4s2

Br 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

Bi 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p3

Mn 1s2 2s2 2p6 3s2 3p6 4s2 3d5 Abbreviated Electron Configurations (Noble Gas Method)

Element Configuration Ca [Ar] 4s2 Bi [Xe] 6s2 4f14 5d10 6p3 Ra [Rn] 7s2

Exceptions Cu and Cr and others

Lewis Dot Diagrams Showing the valence shell electrons for representative elements.

I A

II A III A IV A V A VI A VII A VIII A

Method For Drawing Lewis Structures (Covalently bonded molecules)

1) Decide which atoms are bonded. 2) Count all valence electrons. (include charges) 3) Place two electrons in each bond. 4) Complete the octets of atoms attached to the central atom, by adding e- in pairs. 5) Place the remaining electrons on the central atom.

If the central atom does not have an octet, form double bonds. If necessary, form triple bonds.

H2O PO4

3- NO31- CO2

Lewis Dot Diagrams: Molecules Containing Only Single Covalent Bonds

1) Decide which atoms are bonded. 2) Count all valence electrons. (include charges) 3) Place two electrons in each bond with the central atom. Complete the octets of atoms attached to the central atom, by adding e- in pairs. 4) Add the remaining electrons to the central atom in pairs.

Note: the central atom may have more or less than 8 electrons

H

NCB Be Li

Si

F O

Al Mg Na

Ne

He

Br

I

KrAs

Sb

Ca K

ArCl SP

Bi

Trends in the Periodic Table Property Description Trend Pattern Force of attraction for the outer electron

Net effect of the protons in the nucleus attracting the outermost electron.

Atomic radius

Distance from the nucleus to the outermost part of the electron cloud.

Ionization energy

Quantity of energy required to remove an electron from an atom or ion. First, second and third ionization energies.

Electron affinity

Quantity of energy given off when an atom gains an electron

Electronegativity

Net tendency of an atom to gain an electron.

Electron Configuration and the Periodic Table 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn As Br Kr Rb Sr Y Ag Cd Sn Sb I Xe Cs Ba Au Hg Pb Bi Rn Fr Ra La Ce Yb Lu Ac Th No Lr Terminology: alkali metals, alkaline earth metals, halogens, transition elements, representative elements, period, group 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1s1 He 2s1 2s2 2p1 2p6 3s1 3s2 3p1 3p6 4s1 4s2 3d1 3d10 4p1 4p6 5s1 5s2 4d1 4d10 5p1 5p6 6s1 6s2 6p1 6p6 7s1 7s2 La

5d1 4f1 4f14

5d1 5f1 5f14

Structure of the Atom Review Scientist Contribution Democritus proposed the existence of the atom (atomos-

indivisible) Alchemy pattern of observation and experimentation Roger Bacon developed conclusions through experimentation William Gilbert studied electrostatics Ben Franklin found lightning to be electricity Coulomb Coulomb's Law Antoine Lavosier Law of Conservation of Mass Joseph Proust Law of Definite Proportions John Dalton First Atomic Theory Fraunhofer (1814) Solar spectrum Bunsen and Kirchoff spectroscopy(1859) Maxwell (1860�s) Electromagnetic spectrum Mendeleev (1872) Periodic table Crookes (1872) Cathode rays Balmer (1875) H-spectral lines Hertz (1887) Photoelectric effect Faraday Determined that binding force was electrical in

nature ( Expt. electrolysis of water ) Worked with discharge tubes to determine cathode ray properties

Thomson Used cathode ray tube to determine the charge to mass ratio of the electron ( e/m )

Millikan Oil drop experiment- to determine the minimum charge of the electron

Planck (1900) quantum J.J. Thomson (1904) Raisin Bun Model of Atom H. Becqueral Discovered radiation Einstein (1905) photon Rutherford Discovered radioactive particles Rutherford (1911) Nuclear Atom Model Neils Bohr (1913) Solar System model Wave Mechanical Model

De Broglie -wave characteristics Schrodinger -wave equations Quantum Mechanics ( n,l,m,s )

Bonding A) Ionic - involves a metal donating electrons to a non-metal thereby

creating an electrostatic FOA between the anion and the cation

Formula and name

Lewis Dot Bonding Electron configuration bonding

Properties

NaCl

MP= 801oC

BP=1413oC

D= 2.165g/cm3 state- solid

MgF2

MP= 1266oC

BP= 2239oC state- solid

Li2O

MP= >1700oC

BP= 1200oC at 600 mmHg

D=2.013g/cm3

B) Covalent Bond-attraction due to the mutual sharing of a pair of electrons between two non-metal atoms

Formula and name

Lewis dot bonding Electron configuration bonding Properties

H2 MP= -259oC

BP= -252.5oC D= 0.0899 g/L state - gas

HF MP= -83.1oC

BP= 19.54oC D= 0.991 g/L State - gas

H2O MP= 0oC

BP= 100oC

D= 1 g/cm3

State - liquid

Empirical Rules for Polar and Non-polar Molecules Type Description of Molecule Examples Polar AB Diatomic with different

atoms (compound) Shape: linear

HCl(g) CO(g)

NxAy Containing nitrogen and other atoms. Shape: triangular pyramidal

NH3(g) NF3(g)

OxAy Containing oxygen and other atoms. Shape:bent or V

H2O(l) OCl2(g)

CxAyBz Containing carbon and two other kinds of atoms.

CHCl3(l) C2H5OH(l)

Non-Polar Ax All elements Cl2(g)

N2(g)

CxAy Containing carbon and only one other kind of atom. Polar bonds that cancel due to symmetry.

CO2(g) CH4(g)

Strength of the polarity can be observed by the degree to which a stream of the liquid is attracted to charged rods (electrostatic)

VSEPR: Valence Shell Electron Pair Repulsion Theory Valence shell electron pair, being negatively charged, stay as far from each other as possible so that the repulsions between them are minimized. ( repulsion comes from bonded pairs and non-bonded pairs)

Hybridization and Electron Promotion A) Electron promotion: an electron in a filled orbital gains energy and jumps to a slightly

higher (energy) orbital, thereby resulting half filled orbitals which take on new shapes and through repulsion orientate themselves to new positions in space ( at maximum distances apart ).

B) Hybrid Electron Configurations

Element Group Electron configuration

Hybrid electron configuration

Hybrid Orbitals

Bond angle

4Be II A (2) 1s2 2s2 1s2 2s1 2p1 sp 180o

5B III A (13)

1s2 2s2 2p1 1s2 2s1 2p1 2p1 sp2 120o

6C IV A (14)

1s2 2s2 1s2 2s1 2p1 2p1 2p1 sp3 109.5o

15P V A (15)

1s2 2s2 s2 p3 3 at 120o

2 at 90o

16S VI A (16)

1s2 2s2 s2 p3 d 90o

C) Shapes and Polarities of Molecules (single covalent bonds) Shape/ name General

formula Molecular example

Hybrid orbitals

Bond angles Polar or non-polar

linear AX HF, H2 either linear* AX2 BeH2 sp 1800 Non-polar

Bent, V AX2 H2O , OCl2 >90o Polar

planar triangular

AX3 BI3 sp2 120o Non-polar

triangular pyramidal

AX3 NH3 , PCl3 >90o Polar

tetrahedral AX4 CH4 sp3 109.5o Non-polar

trigonal bipyramidal

AX5 PCl5 s2 p3 120o 90o

Non-polar

octahedral AX6 SF6 s2 p3 d Non-polar

Shapes and Polarities of Molecules Draw the Lewis dot bonding for each of the following. Use the number of electron pairs and VESPR rules to determine the molecular shape. Use the electronegativities to determine the polarity of each bond in the molecule. Use the shape of the molecule to determine the polarity of the molecule.

CCl4

NI3 H2S SF6

BeI2

PF5 BCl3 Br2

2) Indicate whether the following molecules are polar or non-polar

H-C-C-H

H H

H H H-C-C-C-H

F H H

H H H

Cl-C-C-C-C-C-C-Cl

H H H H H H

H H H H H H

O=C=O

H-C-O-H

H

H

C

Cl

HH

H

Aggregates ( solids ) (and what bonds them together) Property Polar

molecular solids

Non-polar molecular solids

Covalent Network solids

Metallic solids

Ionic Solids

Units that occupy the lattice points of the crystal

Binding force within the crystal

Solid model

Appearance

Brittleness

Malleability

Relative melting point

Solubility in water

Electrical conductivity solid liquid solution

Examples of elements with this type of bonding

Examples of compounds with this type of bonding

Matching Properties to the type of bonding. Match the following list of substances to their properties and

name the type of bonding holding the aggregate together: HBr , N2 , Fe , KCl , Si Property/ name

MP (oC) 776 -209.86 1410 1535 -11

BP (oC) subl 1500 -195.8 2355 3000 126

Density 1.984 g/cm3

1.25 g/L 2.32 g/cm3

7.86 g/cm3

Electrical conductivity

liquid or solution

does not conduct

conducts in all states

does not conduct

State white solid

colourless gas

steel gray solid

silver coloured solid

pale yellow gas

Brittle or malleable

B B B M B

Bonding -explain how you identified the type of bonding

Aggregate Bonding: macro structures (crystals) 1. Ionic Crystals ( solids ) Ionic solids are held together by the electrostatic force of attraction between anions and cations . Since each anion is attracted (bonded) by many cations ( and vice versa ) there is often a very large total force of attraction.

Model Resulting Properties

2 Metallic Solids: Metallic bonding The outer electrons of a metal atom are loosely attracted to its nucleus and each outer electron is attracted by all other nuclei in the general area. Since each outer electron is attracted by many nuclei, this results in a large total force of attraction. �Nuclei attract a sea of electrons�

Model Resulting Properties

Cl-

Na+

Na+

Na+ Na+

Na+

Na+

+ + +

+ + +

e e

3 Covalent Network Solids All the atoms in the solid are linked together by an array of covalent bonds. Since each covalent bond requires large quantities of energy to break these crystals are very difficult to break apart.

Model Resulting Properties

examples: diamond and graphite ( allotropes of C) , sand or quartz SiO2 , red phosphorus P , granite , asbestos 4. Polar Molecular Solids (Covalent) Polar molecules use their partial charges ( dipoles ) to attract other polar molecules. This Intermolecular force of attraction is called dipole-dipole attraction. Since it is based on partial charges it is a weak force of attraction.

Model Resulting Properties

5. Non-polar Molecular Solids (Covalent) Non-polar molecules induce temporary dipoles that are called London Dispersion Forces. This is the weakest type of force of attraction.

Model Resulting Properties

δ-Cl---Hδ+

δ+H---Clδ- δ+H---Clδ-

δ-Cl---Hδ+

F---F F---F

F---F F---F δ-

δ+

Shapes and Polarities of Molecules 1. For each of the following molecules/molecular ions: draw the Lewis dot diagram to show the

covalent bonding , name the shape of the molecule and tell whether the molecule is polar or non-polar.

All of these molecules are to be drawn with only single covalent bonds. The central atom does not have to have 8 electrons ( an octet ) OCl2

BrF AsF5 SbCl61- BrF41-

XeF31+

SBr6 FCl21+ ICl41- POCl3

XeOF4

PI3 TeF4 IO41- HBr

IF5

BrCl3 BeI2 BBr3 H2S

Answers: Shapes and Polarities of Molecules OCl2 Bent -polar

BrF linear -polar

AsF5 Triangular bipyramidal -non-polar

SbCl61- Octahedron -non-polar

BrF41- Square planar -non-polar

XeF31+

T-shape -polar

SBr6 Octahedral -non-polar

FCl21+ Bent -polar

ICl41- Square planar -nonpolar

POCl3 Tetrahedron -polar

XeOF4 Square pyramid -polar

PI3 Triangular pyramidal -polar

TeF4 Irregular tetrahedron -polar

IO41- Tetrahedron -non-polar

HBr Linear -polar

IF5 Square pyramid -polar

BrCl3 T-shape -polar

BeI2 Linear -non-polar

BBr3 Planar triangular -non-polar

H2S Bent -polar

Multiple Choice Questions: Atomic Structure and Molecular Architecture

1. How many sigma ( σ ) bonds and pi ( π ) bonds are there in the following molecule

C C C

? sigma ___ pi ___ A) 8 ,1 B) 8, 7 C) 1, 1 D) 3, 0 2. The strength of a dipole in a polar covalent bond depends on the:. A) number of neutrons C) charge of the ions B) covalent mass D) difference in electronegativity between the atoms 3. Which of the following elements has the highest FOA for its outer electron? A) Ne B) Al C) Na D) Li 4. The type of aggregate with the set of properties: very low melting point, brittle, does not conduct electricity in any state. A) metal B) ionic C) polar molecular D) non-polar molecular 5. The solid with dipole-dipole attraction holding the lattice units together. A) I2 B) NaF C) HI D) Au 6. The overlap orbital bond diagram for a "p"and a "p"orbital is:

. .

A). .

B). .

C). .

D)

7. Which of the following compounds has the highest degree of ionic bonding? A) CsCl B) CO2 C) CCl4 D) H2O 8. Which compound exhibits the greatest covalent character? A) NO B) MgO C) NaCl D) Rb2S 9. The strength of intermolecular bonds due to London dispersion forces , increases in a given family (i.e. halogens ) as the A) number of electrons in the atom increases C) temperature increases B) molecular size decreases D) atomic mass decreases 10. Given that element X will form diatomic molecules with each of the elements P, Q, S, T and that the ionization energies of

the elements are; X = 588 kJ/mol , P = 1008 kJ/mol , Q = 840 kJ/mol , S = 1764 kJ/mol, T = 1344 kJ/mol Which of the following, lists the compounds in increasing ionic character? A) XP, XQ, XS, XT B) XQ, XP, XT, XS C) XQ, XT, XP, XS D) XS, XT, XP, XQ 11. Highest electronegativity is a characteristic of: A) metals B) non-metals C) metalloids D) noble gases 12. The molecular shape that results between an oxygen atom and fluorine atoms is: A) Linear B) Tetrahedral C) Bent D) Trigonal pyramidal 13. A nitrogen atom produces at ____ hybrid that bonds to form a _____ shape:. A) sp3 , tetrahedral B) s2p3 , trigonal bipyramidal C) sp2 , planar triangular D) s2 , linear 14. The quantum number that indicates the orbital cloud shape in three dimensional space is the ____ quantum number. A) Principle B) Spin C) Secondary D) Magnetic 15. Which of the following pairs of elements is most likely to form an ionic bond? A) N + S B) Cl + C C) Br + Li D) Mg + Cu 16. Which of the following is the electron configuration for an alkali metal? A) 1s22s22p5 B) 1s2 2s2 2p6 3s1 C) 1s2 2s2 2p6 D) 1s2 2s2 17. The amount of energy required when an atom loses an electron is a definition for: A) Electron affinity B) Atomic radius C) Ionization potential D) Electronegativity 18. The molecule exhibiting the greatest amount of hydrogen bonding is: A) HF B) NaCl C) CH4 D) H2 19. The formula BCl3, has a ____ shape and it would be _______. A) planar triangular, polar B) planar triangular, non-polar C) tetrahedral, polar D) tetrahedral, non-polar 20. Which of the following structural formulae represents a polar molecule?

O=C=OA)

H-C-C-H

H H

H H

B) Cl

CH H

H

C) O=OD)

21. A molecule with an octahedral shape would have bond angles of: A) 104.5o B) 90o C) 109.5o D) 120o 22. The atom that could form a "sp2" hybrid when it bonds is: A) Be B) B C) C D) S 23. The elements A and B form an ionic bond. What coordination number would result in a higher melting point? A) 7 B) 5 C) 6 D) 4 24. Which set of formulae contains, an ionic compound, a polar molecule and a network solid? A) LiOH, CaCl2, HBr B) MgO, HI, SiO2 C) H2O, CO2, SiO2 D) Mg, KCl, S8 25. The binding force ( intermolecular ) within solid oxygen , O2(s), is A) electrostatic FOA B) dipole-dipole attraction C) hydrogen bonding D) London forces 1-A 6-A 11-D 16-B 21-B 2-D 7-A 12-C 17-C 22-B 3-A 8-A 13-B 18-A 23-A 4-D 9-A 14-C 19-B 24-B 5-C 10-B 15-C 20-C 25-D

Chemical Change:

Thermochemical Equation

A chemical reaction including the enthalpy.

A) Exothermic reactants ! products + heat

Examples: 2Fe(s) +3CO2(g) ! Fe2O3(s) + 3CO(g) ∆H = -26.38 kJ

C3H8(g) + 5 O2(g) ! 3CO2(g) + 4H2O(l) +1411.1 kJ

B) Endothermic reactants + heat ! products

Examples: 2HgO(s) + 181.9 kJ ! 2 Hg(l) + O2(g)

C(s) + H2O(g) ! CO(g) + H2(g) ∆H = +1313 kJ

Net Ionic Equation Review 3Na2CO3(aq) + 2H3PO4(aq) ! 2Na3PO4(aq) + 3H2O(l) + 3CO2(g) Mg(s) + 2HCl(aq) ! MgCl2(aq) + H2(g) CuSO4(aq) + H2S(aq) ! CuS(s) + H2SO4(aq) Ba(OH)2(aq) + HNO3(aq) ! Ba(NO3)2(aq) + H2O(l)

Experiment: Heat of Combustion ∆Hcombustion

Purpose: to determine the heat of combustion for paraffin Apparatus:

glass rod

inner can

water

outer can

candle

Method:

1. Determine the initial mass of the candle. 5. Light the candle and burn it for 10 minutes. 2. Set up the apparatus as in the diagram. 6. Measure the final temperature of the water. 3. Place 100 mL of cold water in the can. 7. Measure the final mass of the candle. 4. Take the initial temperature of the water. 8. Calculate the heat of combustion.

Observations: 1. Table of measurements

measurement description measurement uncertainty %uncertainty initial candle mass final candle mass mass wax consumed initial water temperature final water temperature temperature change mass of water specific heat capacity water 4.2 J/(goC)

Molar mass paraffin 250 g/mol

2. Calculate the heat of combustion (∆H) for the paraffin. ( include uncertainty ) Conclusions: 1. State the heat of combustion for paraffin in kJ/mol Include uncertainty. 2. Describe 6 possible errors that may have affected your heat of combustion answer.

Error Effect on heat of combustion 3. How would you know if your experimental results were valid? 4. Solve the following problems given ∆H

combustion = 5.25x106 J/mol as the heat of combustion

for paraffin.

A. What mass of candle wax would be needed to be burned to warm 800g of water from 10.oC to 30.oC ?

cwater = 4.2 J/(goC) ( 3.2 g ± 0.16g)

B). In a student experiment the value 5068350J /mol with a 5% uncertainty was found. Express the answer and the uncertainty in standard form. ( 5.06835 x 106 J/mol ± 2.534175 x 105 J/mol )

C). A sample of the same wax is burned in an atmosphere enriched with oxygen to give complete

combustion. If 1.4 g ± 0.05 g of wax warms 300 g ± 1 g of water from 278 K ± 0.5 K to 308 K ± 0.5 K

on burning, calculate the number of kJ/mol of wax burned with uncertainty. (6.8 x 103 kJ/mol ± 4.9 x

102 kJ/mol)

D).The water in question #C was held in a metal container with a mass of 100g. The specific heat of the

metal is 0.84 J/(goC). Calculate the effect on the answer to 4(a) if the heat absorbed by the

container was included. ( 2.9 x 104 J/g ) E).The heat produced by complete combustion of a mole of ethane ( C2H6 ) is 1432.3kJ.

a) Write the equation showing the heat term in the equation. b) Calculate the mass of ethane which must be burned to; i) release 20 kJ of heat

(0. 42 g )

ii) warm 500 g of water from 0.0oC to 50.0oC. ( 2.2 g )

Experiment: Heat of Reaction Purpose: to determine the relationship between heat and reaction path. Apparatus: thermometer, styrofoam cup Method: Part I 1. Put 100 mL of cool tap water in a styrofoam cup. Stir carefully with the

thermometer and read the temperature. ( include uncertainty ) 2. Measure approximately 2 g of solid NaOH pellets. (record actual mass) 3. Put the solid NaOH into the water and stir with the thermometer. Read and

record the temperature. Part II 1. Rinse a styrofoam cup with some 0.50 M HCl solution. Then put 100 mL of

0.50 M HCl into the cup. Measure the temperature. 2. Repeat steps 2 and 3 as in part I Part III 1. Measure 50 mL of 1.0 M HCl into a styrofoam cup and measure the

temperature. 2. Measure 50 mL of 1.0 M NaOH into a 250 mL beaker and measure the

temperature. 3. Add the NaOH solution to the HCl solution, mix and read the highest

temperature. Observations: Part I Part II Part III initial temperature final temperature temperature change ∆T

# moles NaOH

m= MM= n = m/MM

m = MM = n = m/MM

M = V = n = M V

Heat in Joules Q = m∆tc Heat of reaction ∆H = Q/n NaOH

Conclusions: 1. Write the Net Ionic Equation for each of the three reactions. Part I , II, III 2. Compare reaction I + reaction III with reaction II. Do they add up? 3. Compare ∆HI + ∆HIII with ∆HII.

Lab. Test: Heat of Solution Name_______________________________ Purpose: To determine the quantity of heat absorbed from the surroundings ( in

kJ/mol ) when a solute dissolves in water. To determine the validity of the experimental results Apparatus: Method: Observations:

measurement value uncertainty %uncertainty specific heat capacity H2O 4.2 J/(goC)

molar heat of solution Calculations: ∆H soln = % Error = Conclusions: Data: Molar Heat of Solution ( ∆Hsoln ) for some salts NH4Cl 14.82 kJ/mol KNO3 34.93 kJ/mol

KCl 17.23 kJ/mol NaCH3COO. H2O 19.73 kJ/mol

Sample Hess's Law Problems 1) Calculate the heat of reaction for the reaction; 2NaHCO3(s) ---> Na2CO3(s) + H2O(g) + CO2(g) ∆Ho = [n∆Ho

f Na2CO3 + n∆Hof H2O + n∆Ho

f CO2 ] - [ n∆Hof NaHCO3 ]

2) Calculate the heat of reaction for the reaction; C2H5OH (l) + 3 O2(g) ---> 2 CO2(g) + 3 H2O (g) (-1234.77 kJ ) 3) Calculate the heat of reaction for the reaction; 2 NO (g) + O2 (g) ---> 2 NO2 (g) (-113.14 kJ) 4) Calculate the heat of reaction for the reaction; NaOH (s) + HCl (g) --->NaCl (s) + H2O (g) (-133.7kJ) 5) Use the ∆Hocombustion to calculate ∆Ho

f for glycine

4 C2H5NO2 (s) + 9 O2 (g) ---> 8 CO2 (g) + 10 H2O (g) + 2 N2 (g) (-418.01 kJ/mol) ∆Hocombustion = -973.49 kJ/mol of glycine (amino acid) n∆Hocombustion =[ n∆Ho

f CO2 + n∆Hof H2O + n∆Ho

f N2 ] - [ n∆Hof C2H5NO2 + n∆Ho

f O2 ]

6) Calculate the heat of formation for sucrose: C12H22O11 (s) + 12 O2 (g) ---> 12 CO2 (g) + 11 H2O (l) (-1144.7 kJ/mol) ∆Hocombustion = -5640.9 kJ/mol

Enthalpy Changes: An Application of the Law of Additivity of Reaction Heats For many chemical reactions it is difficult to measure directly the change in enthalpy, ∆H .

However , the value for ∆H for a given reaction is independent of the series of steps involved in getting from reactants to products. Furthermore, whenever a reaction can be expressed as the algebraic sum of the series of 2 or more simpler reactions, the heat of reaction is the algebraic sum of the heats of these simpler reactions. Thus ∆H for a given reaction is often determined by breaking the overall reaction down into a series of steps and determining the ∆H value for each step either experimentally or from standard tables of enthalpy changes.

Magnesium metal combines very rapidly with oxygen gas. The equation for the reaction is: Mg (s) + 1/2 O2 (g) ! MgO (s) #4

Although much energy is evolved in the form of heat and light it is difficult to measure this energy directly. But this same equation is obtained when the following three equations are combined algebraically.

MgO (s) + 2HCl (aq) ! MgCl2 (aq) + H2O (l) #1

Mg (s) + 2HCl (aq) ! MgCl2 (aq) + H2 (g) #2

H2 (g) + 1/2 O2 (g) ! H2O (l) #3

In this experiment the values of ∆H for equations #1 and #2 are determined experimentally. The value for equation #3 is obtained from the table of enthalpies of formation in your text book. By combining these three equations appropriately, the enthalpy change for the overall reaction ( #4 ) can be calculated. A styrofoam cup is used as the calorimeter.

Method: Reaction #1 1. Add 100 mL of 1.00 M HCl solution to a styrofoam cup and measure the

temperature of the solution. 2. Measure out accurately about 1.00 g of magnesium oxide solid, MgO, and add

this to the HCl solution. Stir the contents of the cup during this reaction and measure the highest temperature that the solution reaches.

Reaction #2 1. Repeat procedures 1 and 2 but add 0.50 g of magnesium metal, Mg to the acid

instead of magnesium oxide. Observations: Calculations: 1. Calculate the heat of each reaction in kiloJoules/mol and find the heat of reaction

for equation #4 with its uncertainty. 2. Calculate the percent error if the ∆H for equation #4 is -601.7 kJ/mol. Conclusions: 1. Combine equations #1, #2, and #3 in such a way that their algebraic sum gives the

overall equation #4. 2. Discuss the sources of error and compare the uncertainty with the %error. Are your results valid? Explain.

Bond Energy: is the quantity of energy 1) released when a bond forms , or 2) is absorbed when a bond is broken

Bond Type Bond Energy Bond Type Bond Energy C-H 413 kJ/mol C-N 292 kJ/mol C-O 351 C=N 615 C=O 715 C N 891 C-C 348 C-F 439 C=C 615 C-Cl 328 C?C 812 H-O 464 O=O 494 H-N 391

Sample Calculation: The energy released from respiration of glucose D-glucose + oxygen ! carbon dioxide + water + energy C6H12O6 + 6 O2 ! 6 CO2 + 6 H2O + energy

C

H-C-O-H

H-C-O-H

H-C-O-H

H-O-C-H

H-O-C-H

O H

H

+

O=O

O=O

O=O

O=O

O=O

O=O

O=C=O

O=C=O

O=C=O

O=C=O

O=C=O

O=C=O

+

H-O-H

H-O-H

H-O-H

H-O-H

H-O-H

H-O-H

+ energy

Energy required for the breaking of bonds Energy released when new bonds form 5 C-C = 5(348) = 1740 kJ/mol 12 C=O = 12(715) = 8580 kJ/mol 1 C=O = 715 12 H-O = 12(464) = 5568 7 C-H = 7(413) = 2891 5 C-O = 5(351) = 1755 5 H-O = 5(464) = 2320 6 O=O = 6(494) = 2470 Absorbed energy 11891 kJ/mol Released Energy 14148 kJ/mol

Net energy change = 14128 kJ - 11891 kJ = 2237 kJ released ( EXOTHERMIC )

Problems: 1. Calculate the quantity of heat that would be released in formation of each of the

following from atoms. C3H8 (4000 kJ) CH3COOH (3117kJ) 2. Calculate the quantity of heat absorbed in breaking apart the following molecules

into atoms. CH3CH2NH2 (3487kJ) benzene C6H6 (5367kJ) 3. Calculate the enthalpy change for the complete combustion of CH4. 4. Using their bond energies , explain why Cl in CFC's reacts to destroy the ozone and

not the F.

Enthalpy of Formation of Ammonium Chloride In this experiment you will use Hess� Law to determine the heat of a reaction that is difficult to determine directly:

½ N2(g) + 2H2(g) + ½ Cl2(g) ! NH4Cl(s) Eq. A

This equation can be obtained by combining the following six equations: NH3(aq) + H+(aq) ! NH4+(aq) Eq I NH4Cl(s) ! NH4+(aq) + Cl-(aq) Eq II ½ N2(g) + 3/2 H2(g) ! NH3(g) Eq III NH3(g) ! NH3(aq) Eq IV ½ H2(g) + ½ Cl2(g) ! HCl(g) Eq V HCl(g) ! H+(aq) + Cl-(aq) Eq VI

Thus by combining the heats of reaction of these six reactions the heat of formation of ammonium chloride may be calculated. The heats of reactions for reactions I and II will be determined experimentally while the values for reactions III to VI will be obtained from heat of formation tables and heat of solution tables. Method: Reaction I Measure the initial temperatures of 100 mL 1.0 M HCl(aq) and 100 mL of 1.0 M NH3(aq) , mix them in a double styrofoam cup and record the highest temperature reached. Reaction II Place 200 mL of tap water in a double styrofoam cup and add about 8.0 g of solid NH4Cl. (record the actual mass). Mix and record the lowest temperature reached. Observations: Prepare a suitable table of measurements for the experiments.

Calculations: A) Calculate the amount of heat ( Q ) produced or consumed for the two reactions. Assume that

the solutions are basically water with a density of 1.00 g/mL and a specific heat capacity of 4.18 J/goC

B) Calculate the moles of ammonia and the moles of ammonium chloride. C) Calculate the heat of reaction ( ∆H , kJ/mol ) of ammonia in reaction I and for reaction II the

heat of reaction of ammonium chloride. Conclusions and Questions 1. Write equations I - VI as thermochemical equations. Obtain the heat of reaction for each

equation from tables. Combine equations I - VI to obtain equation A and therefore the heat of formation for ammonium chloride.

2. Obtain the accepted value for ammonium chloride from the tables and calculate the percent error.

3. Briefly discuss some experimental sources of error . 4. Describe in what ways your coffee cup calorimeter is different from a �Bomb Calorimeter�.

Heats of Formation Substance ∆Ho

f kJ/mol Substance ∆Hof kJ/mol

Ag(s) 0 H2O(l) -286

Ag2SO4(aq) -698 H2O(g) -242

Al(s) 0 H2O2(l) -186

Al2O3(s) -1670 H3PO4(aq) -1280

Br2(g) 30.9 H2SO4(l) -814

Br2(l) 0 H2SO4(aq) -908

CH4(g) -74.8 KBr(s) -392

C2H4(g) 52.3 K2SO4(aq) -1409

C2H6(g) -84.7 Mg(s) 0

C2H5OH(l) -278.0 Mg3N2(s) -461

C4H10(g) -125 Mg(NO3)2(aq) -875

CO2(g) -393.5 Mg(OH)2(s) -925

Ca(s) 0 N2(g) 0

CaCl2(aq) -878 NH3(g) -46.2

CaO(s) -635 NH4Cl(aq) -300

Ca(OH)2(s) -987 NH4NO3(s) -366

Cl2(g) 0 NO(g) 90.4

Cu(s) 0 NO2(g) 33.8

Cu(NO3)2(aq) -350 N2O(g) 82.1

CuSO4(aq) -679 NaCl(s) -411

Fe(s) 0 NaCl(aq) -466

Fe2O3(s) -822 NaNO3(aq) -477

H(g) 218 NaOH(aq) -427

H2(g) 0 Na2SO4(aq) -1380

HBr(g) -36.2 O2(g) 0

HCl(g) -92.3 P4O10(s) -2980

HCl(aq) -167 Zn(s) 0

HNO2(aq) -119 Zn(NO3)2(aq) -569

HNO3(aq) -207

Bond Energies Bond Type Bond Energy Bond Type Bond Energy

C-H 413 kJ/mol C-N 292 kJ/mol

C-O 351 C=N 615

C=O 715 C≡N 891

C-C 348 C-F 439

C=C 615 C-Cl 328

C≡C 812 H-O 464

O=O 494 H-N 391

Uncertainty Rules and calculations Most measurements have a certain degree of accuracy that is related to the smallest unit that the instrument measures. The �UNCERTAINTY� is generally taken as half the smallest measurable unit. Examples

Measurement Reading Smallest measured unit Uncertainty 14g measured on a balance 1 g +/- 0. 5 g 34.8 cm with metre stick 0.1 cm +/- 0.05 cm

22 ½ oC ½ oC ¼ oC 12.065 g 17.2 cm

50. g 25 oC

1.3567 km 0.23 kg

Uncertainty calculations

1. % Uncertainty

0.5%10.0

1000.05 y uncertaint %

tmeasuremen g 10.0 ex.urementactualmeas

100yuncertaintty%uncertain

=×=

×=

2. Converting % Uncertainty back to the original unit

g 0.25-/ g 50.0 isanswer

0.25g100

0.550.0 y Uncertaint

0.5% /ex.50.0g100

yuncertaint % answer t measuremenyUncertaint

+∴

=×=

−+

×=

Uncertainty rules for Calculations

1. Addition or Subtraction: the individual uncertainties are added was ∆t (t1=67 oC +/- ¼ oC, t2=64 oC +/- ¼ oC) ∆t= ∆m (m1=2.4 g +/- 0.05 g, m2=2.2 g +/- 0.05 g ∆m=

2. Multiplying or Dividing: add the percent uncertainties

Ex. Calculate density if 40.o g occupies 20.0 mL

0.0075g/mL -g/mL 2.00 0.375% 2.00g/mL0.25% -mL 20.0

0.125% - 40.0g

0.05mL -20.0mL

0.050g- 40.0g

Vm

D +=−+=

+

+=

+

+==

Problems

1. Determine the temperature change if water initially at 25.1oC was heated to 28.6 oC. Include uncertainty. 2. A clean empty test tube had a mass of 15.26 g. Salt was added to the test tube until the mass of the test tube plus the

salt was 17.36 g. Calculate the mass of the salt with its uncertainty. 3. Find the final volume of a gas from the following set of data.

V1=500 mL +/- 5 mL P1=750 mmHg +/- 10 mmHg T1= 300K +/- 1 K V2= P2=900 mmHg +/- 10 mmHg T2=250K +/- 1 K

4. Find the uncertainty in kJ/mol of NaOH when 4.00g +/- 0.01 g of NaOH is dissolved in 100. g of water causing a temperature change of 8.5 oC +/- 0.5 oC. Q=m∆tc, ∆H=Q/n and c=4.2 J/goC

EXPERIMENT: Factors Affecting Rate of Reaction Reaction Equation: Replacement

Zn(s) + 2 HCl (aq) ! ZnCl2(aq) + H2 (g)

Mg(s) + 2 HCl (aq) ! MgCl2(aq) + H2 (g)

Net Ionic Equation: Problems: To properly design and perform experiments that demonstrate

the factors that affect the rate of this reaction. (stoichiometry calculations using 0.01 mol metal with 2 times the needed acid to provide excess))

To fully explain as part of your conclusions how varying these

factors affected the reaction rate ( on the particle level) To pay attention to the controls and the variable in these

experiments To estimate the rate of increase in the reaction rate due to

each factor (variable). i.e. 2x 2 times the rate Record the times Materials Available: Mg and Zn in various sizes HCl solutions of various concentrations glassware and thermometers

Rates of Chemical Reactions Not all matter reacts at the same rate ex. Newspaper yellows ( slow oxidation ) over a number of years.

Newspaper burns ( rapid oxidation ) in a few seconds

Conditions Necessary for a Chemical Reaction to occur: 1.

2.

3.

Explanation of the Factors Affecting Rate of Reaction 1) Chemical Nature of the Reactants

Therefore the reaction rate will depend on:

The Forces of Attraction

Repulsive Forces

Therefore the rate of reaction decreases as the force of attraction within the reactants increases because :

2) Ability of the Reactants to Come Into Contact with Each Other

Collision Theory Reactant molecules translating toward each other will gradually slow down as the repulsive forces increase ( p-p or e-e ). The molecule then collides with the other reacting molecule. The molecules stop for an instant and then fly apart. If they react, the particles will break their existing bonds and chemically bond in a different pattern. Within the reacting medium ( liquid or gas ) there will be an enormous number of collisions, but very few actual reactions. ( there are estimated to be 10

36 collisions per second in a room )

There are two basic reasons for the lack of reactions:

A. The orientation of the molecules at collision ( p755)

B. Activation Energy ( Ea ) Is the minimum EK

3) Concentration of the reactants

As the concentration [mol/L] of the reactants decreases

As the slope of the reactant concentration curve decreases,

4) Effect of Temperature The rate of Reaction increase as the temperature increases. The reaction rate generally

Temperature ( K) is the measure of the

There is a certain amount of energy necessary to cause the breaking of bonds in a collision. This energy is called the ACTIVATION ENERGY Ea

product

[conc] mol/L

Time !

reactant

T T

T T>

Number of particle

EK !

T T

T T>

Number of

ti l

Ea

1. 2. Conservation of Energy All energy put into a reaction must be accounted for. Therefore; ET = EP + EK

As the particles approach collision Draw the EP vs. reaction path for reversible reactions

Ereactant

products

∆H

EP

Rxn !

Exothermic reaction ∆H = -ve

EP

Rxn!

EP

Rxn!

ReactionNo i

5) Effect of Catalysts of Rate of Reaction Reaction 2H O 2H O O2 2 (aq) 2 (l) 2 (g) → +

+ Catalyst 2H O 2H O O2 2 (aq) 2 (l) 2 (g)MnO2 → +

Catalyst: 2. Reaction Mechanism with catalyst Slow reaction A-A + C ! A2C

Fast reaction A2C + B-D ! A2B + D + C

RXN!

A-A

A-A + B-EP

A2B + D

Activated complex or transition state

Reaction A-A + B-D ! A2B + D

A-A + C EP A2C

RXN!

A2C + B-D EP A2B + D + C

RXN!

Reactions and Energy

(kJ)

0

10

20

30

40

50

60

70

80

reaction co-ordinates

A + B

C + D

reactants

products

Ea

A*B activated complex ( transition st

Ep

1. Complete the chart below from the data on the above graph

Equation Type of Reaction ∆Hrxn

Ea

A+ B ! C + D

C+ D ! A + B

2. Does the temperature of a system affect the Activation Energy? Explain

your answer. 3. Describe significance of an activated complex. 4. Complete the chart below

Reaction A Reaction B Reaction C

P.E. P.E. P.E.

reaction --> reaction --> reaction --> Most likely to React Explain Least Likely to React. Explain Produces most energy? Consumes most energy Give an example of each type of reaction

Experiment: A Study of Reaction Rates ( Iodine Clock ) Purpose: to determine the extent that concentration ( Part I ) and temperature control

( Part II ) the rate of a chemical reaction ( Iodine Clock ) Background Information on this Reaction The clock reaction is performed by mixing the two solutions described below. Solution A : is a dilute solution of potassium iodate, KIO3 , which is the source of the

reacting species, the iodate ion , IO3-(aq) .

Solution B: contains some starch and the other reacting species, the hydrogen sulphite

ion , HSO32-(aq)

The initial step in the reaction is represented by the equation: IO HSO I 3SO 3H3 (aq)

-3 (aq)2-

(aq)++ → + +− −3 3

2( ) ( )aq aq

When the hydrogen sulphite ions (HSO32-(aq)) are consumed , the iodide ions (I-(aq)),

react with the remaining iodate ions (IO3-(aq)) to produce iodine, I2(s)

5I 6H IO 3I 3H O(aq) (aq) 3 (aq) 2(s) 2 (l)− + −+ + → +

The molecular iodine reacts with the starch to produce a blue substance, which indicates the completion of the reaction. Apparatus: test tubes , stopwatch, burette , and beaker Part I : Effect of Concentration Changes on Reactivity Method:

Dilute solutions of solution A will be prepared to vary the [IO3-] while keeping the

[HSO32-] constant. The temperature of all the solutions should be kept at room

temperature. 1. Use a burette to measure 10.0 mL of solution A and pour it into a clean test tube.

Use a volumetric pipet to place 10.0 mL of solution B into a second clean test tube. 2.Pour solution A into the test tube containing solution B, then pour the mixture back

and forth 3 times to obtain uniform mixing. Time the reaction from the instant that the two solutions first make contact. Record.

3.Prepare different concentrations of the solution A (KIO3) . Do the dilutions as directed by your teacher.

Solution A volume (mL)

9.0 8.0 7.0 6.0 5.0

Plus Distilled water (mL)

1.0 2.0 3.0 4.0 5.0

Note: the total volume is always 10.0 mL. Mix the solutions well. 4.Prepare 5 test tubes each containing 10.0 mL solution B. Repeat step #2 for each

solution in #3. Record the times of reaction.

Observations: two groups results Soln A! 10.0 mL 9.0 mL 8.0 mL 7.0 mL 6.0 mL 5.0 mL Time of reaction #1

Time of reaction # 2

Calculations: 1.The initial concentration of KIO3 (A) is 0.020 mol/L. Calculate the number of moles

per millilitre of KIO3 in the initial solution A. 2. Calculate the initial concentration of [ KIO3 ] in each of the 6 solutions. 3. Pot a graph of concentration [KIO3 ] Vs time of reaction. Conclusions: 1. What generalizations can you make concerning the affect of varying the

concentration on the time of reaction. 2. How is the TIME of the reaction related to the RATE of the reaction? 3. Why is it important to keep the total volume of 10.0 mL during the dilutions of

solution A? Part II: Affect of Temperature Method: 1.Place 10.0 mL of solution A into a clean test tube and place 10.0 mL of solution B

into a second clean test tube. Place the two solutions in the water bath and allow the contents to reach the temperature of the water bath. The water bath must be at the temperature indicated by you teacher for your group.

2. Mix the two solutions (3 times) and record the time for the reaction. The test tube should be put back into the water bath after mixing.

3. Obtain the other temperature results from the remainder of the class. Observations:

Temperature 5oC 10oC 15oC room oC 30oC 35oC 40oC

Time #1

Time #2 Plot the graph of time Vs temperature for this reaction

. Allow for extrapolation to 0 oC and 50 oC. Conclusions: 1. What generalizations can you derive from the graph of time Vs temperature? 2. What are the extrapolated time values for 0 oC and 50 oC?

Rate of Chemical Reactions

Rate of Reaction = change in concentration

change in time

r [A]t

= ∆∆

ex. C H 5O 3CO 4H Orate of disappearance rate of appearance(- ) + ( ) 0.90 ) + ( )

3 8(g) 2(g) 2(g) 2 (g)

molL s

+ → +

→

→ + ⋅(

ex. 4 2 2 53 2 2 2KNO K O N Os s g g( ) ( ) ( ) ( )

( )

→ + +

→ ⋅

) ( ) + ( ) + (0.30 molL s

Graphing Concentration Vs Time to Determine Reaction Rate Rate of reaction can be determined by taking the slope of the curve Average Rate (slope of a secant) and Instantaneous Rate ( slope of tangent )

[conc]

Time

[conc]

Time

secant

tangent

Heat and Rates of Reaction Multiple Choice 1. Which statement about a homogeneous catalyst is false? A) It is found in the same phase as the reactants and products. B) It causes a reaction to go faster by making the reaction occur by a one-step mechanism. C) It combines with the reactants in one step of the mechanism and is produced again in a later step. D) It lowers the activation energy of both the forward and reverse reactions. E) It speeds up the reaction without affecting the enthalpy change for the overall reaction. 2. According to collision theory, which of the following factors does not influence the rate of reaction? A) collision energy C) collision rebound direction E) collision frequency B) collision orientation D) molecular speeds 3. Complete combustion of 14 g of propane caused 100 g of water to increase in temperature from

20oC to 40oC. The heat of combustion for propane in J/g was ________ J/g A) 600 B) 26400 C) 8400 D) 143 E) 0.0600 Use the following enthalpy diagram to answer questions 4 & 5

20

60

70

∆ H

rxn --->

NH Cl (s)

NH Cl (aq)

4

4

4. The activation energy for the endothermic reaction is: ____ kJ/mol A) 40 B) 50 C) -40 D) 10 E) -10 5. The correct thermochemical equation for this dissolving is: A) NH4Cl (s) → NH4Cl (aq) ∆H = - 40 kJ/mol B) NH4Cl (s) → NH4Cl (aq) + 40 kJ/mol C) NH4Cl (s) +40 kJ/mol → NH4Cl (aq) D) NH4Cl (s) - 40 kJ/mol → NH4Cl (aq) E) NH4Cl (aq) +40 kJ/mol → NH4Cl (s) 6. What is the heat of reaction for the equation: 1/2 N2 + O2 → NO2 Given: N2 + O2 → 2NO ∆H = 180 kJ NO2 → NO + 1/2 O2 ∆H = 57 kJ A) 123 kJ B) 237 kJ C) 147 kJ D) 33 kJ E) 15.4 kJ 7. The rate order for the following rate law is: r=k[X]3[Y]2 A) 6 B) 2 C) 4 D) 3 E) 5 8. For the equation: C5H12 + 8O2 ! 5CO2 + 6H2O

The rate of disappearance of O2 is 4.8 x 10-4 mol/L s, therefore the rate of appearance of CO2 is: ____ mol/L s

A) 5 B) 4.8 x 10-4 C) 3 x 10-4 D) 2.4671 E) 1.2 x 10-5

Use the following graph to answer questions 9 & 10

[A] vs time

time (sec)

[A] mol/L

0

10

20

30

40

50

60

0 1 2 3 4 5 6 7

0

1

23 4 5 6 7

9. The average rate of consumption of [A] for 0-7 sec was: _____ mol/L s A) -7.7 B) 2.4 C) 44 D) 308 E) -6.3 10. The instantaneous rate of consumption at 3 sec is: ____ mol/L s A) 7.67 B) -5.12 C) -21.7 D) 14.6 E) 1.56 11. The following data were collected for the reaction : 2 M + 2 N ---> 2 P + Q

initial [M] mol/L initial [N] mol/L Initial Rate (mol/L s) 0.10 0.10 2.3 x 10-6 0.10 0.20 9.2 x 10-6 0.20 0.20 9.2 x 10-6

What is the Rate Law for this reaction? r = k __________ A) [M][N]2 B) [M]2[N]2 C) [M]2 D) [N]2 E) [M][N] 12. The rate law constant for the reaction in question #11 is: A) 2.3 x 10-4 B) 2.3 x 10-6 C) 2.3 x 10-8 D) 9.2 E) 4.2 x 10-5 13. The rate order for the reaction in question #11 is: A) 2 B) 3 C) 4 D) 5 E) 6 14. Which of the following will not increase the rate of reaction for: HCl(aq) + Mg(s) → H2(g) + MgCl2(aq) A) heating the HCl to a higher temperature D) agitating the reaction mixture B) increasing the concentration of the HCl E) replacing Mg filings with Mg ribbon C) adding a catalyst 15. C2H6 collided with O2 , but they did not react. The lack of reaction could have been caused by: A) the O2 having more than the necessary Ea B) these molecules will never react with each other under any conditions C) the molecules did not collide at the proper angle to cause reaction D) the catalyst inhibited the reaction E) the Ea for the reaction was very small 16. Given the following data about bond energies

C-H 413 kJ/mol C=O 494 kJ/mol C-C 348 kJ/mol C-O 351 kJ/mol H-O 464 kJ/mol

How much heat would be released in the breaking apart of carbon dioxide?

A) 988 kJ B) 478 kJ C) -988 kJ D) 4750 kJ E) 123 kj 17. The rate order for the following rate law is: r=k[A]2[B] A) 1 B) 2 C) 4 D) 0 E) 3 18. For the equation: 2C3H8 + 9O2 → 6CO2 + 6H2O The rate of appearance of CO2 is 0.40 mol/L s, therefore the rate of disappearance of O2 is: ____

mol/L s A) -0.10 B) -0.90 C) -0.60 D) -0.30 E) -1.2 19. For the equation: A + 2B → 3C The rate of appearance of C is 9.0 x 10-7 mol/L s, therefore the rate of disappearance of B is: ____

mol/L s A) -3.0x10-7 B) 1.2 x 10-6 C) 6.0 x 10-7 D) 9.0 x10-7 E) 6.0 x 10-7 20. Which of the following does not influence the speed of a chemical reaction? A) concentration of reactants C) nature of reactants E) formula mass of reactants B) temperature D) presence of catalyst 21. A first order reaction starts with a concentration of reactant equal to 0.040 M. After two hours, the

molarity is 0.020 M. What is the molarity after three hours? A) 0.012 M B) 0.010 M C) 0.017 M D) 0.014 M E) 0.0050 M 22. The acid catalyzed decomposition of hydrogen peroxide is a first order reaction with the first order

rate constant given below. For an experiment in which the starting concentration of hydrogen peroxide is 0.0805 M, what will the concentration be 295 minutes after the reaction begins?

2H2O2 → 2H2O + O2 k = 1.33 x 10-4 min-1

A) 0.0802 M B) 0.00129 M C) 0.0676 M D) 0.0156 M E) 0.0774 M 23. The rate of reaction increases with increasing temperature primarily because: A) an endothermic reaction is "helped along" by the greater amount of heat in the system at the higher

temperature. B) at the higher temperature a greater fraction of molecules possesses the activation energy. C) the activation energy becomes smaller as the temperature is increased. D) changing the temperature usually alters the mechanism. E) the number of collisions per second increases with increasing temperature. 24. A first order reaction, M → N , has t½ = 30.0 s. If the initial concentration of M is 0.200 mol/L, what

will be the concentration of M after 2.00 minutes? A) 0.0125 M B) 0.0500 M C) 0.0250 M D) 0.00625 M E) 0.100 M 25. A first order reaction, X → B , has k = 3.00 x 10-3 s-1. If the initial concentration of X is 0.400 M, how

many seconds will it take for the concentration to drop to 0.150 M? A) 264 s B) 157 s C) 327 s D) 97 s E) 558 s 26. Which of the following is false? Increasing the temperature of a reaction mixture... A) increases the average kinetic energy of both the reactant and product molecules. B) increases the number of effective collisions per second for the reactant molecules. C) usually increases the rate of the reaction. D) decreases the fraction of reactant molecules that have relatively low kinetic energies. E) usually decreases the activation energy for the reaction 27.. The following data were collected for the reaction : A + 2B → C

initial [A] mol/L initial [B] mol/L Initial Rate (mol/L s) 0.10 0.10 3.6 x 10-4 0.10 0.20 7.2 x 10-4

0.20 0.10 7.2 x 10-4 What is the Rate Law for this reaction? r = k __________ A) [A][B]2 B) [A] C) [B] D) [A]2[B] E) [A][B]

28. Below are some data for the reaction: 2NO(g) +2H2(g) → N2(g) +2H2O(g). What is the rate law for the reaction?

Experiment [NO]o mol/L [H2]o mol/L Rate (mol/L s) 1 0.02 0.02 8.0 2 0.04 0.02 32.0 3 0.02 0.04 16.0

A) r= k[NO] [H2]2 B) r= k[NO]2 [H2] C) r= k[NO] [H2] D) r= k[NO]/ [H2] E) r= k[NO]2[H2]2

29. The initial concentration of a reactant in a first order reaction is 0.620 M. What will its concentration

be after 3 half lives? A) 0.0865 B) 0.207 M C) 0.310 M D) 0.103 M E) 0.0775 M 30. The following data apply to the reaction between A and B at a constant temperature:

initial [A] mol/L initial [B] mol/L Initial Rate (mol/L s) 0.020 0.030 0.0396 0.060 0.030 0.0396 0.060 0.060 0.158

The value of the rate constant in the expression r = k [A]m[B]n is: A) 44 B) 215 C) 1100 D) 733 E) 653 31. The rate law for a reaction is: r = k[A]2 . Give the units for k if rate is expressed in mol l-1 s-1. A) s-1 B) mol2L-2s-1 C) L mol-1 s-1 D) L2 mol-1 s-l E) none of the other four answers are correct 32. A first order reaction has t1/2 = 4.0 x 10 3 s-1. The value of the rate constant is: ____ s-1 A) 3.5x10-4 B) 8.7x10-5 C) 1.2x10-4 D) 4.8x10-4 E) 1.7x10-4 33. Why is a reaction mechanism that involves a step such as : A + B + C → products , not a good

choice if alternatives exist ? A) Three reactants require too much activation energy. B) The rate law is too complicated. C) the reactants would tend to be consumed faster than a process involving two-body collisions. D) None of the other choices is correct. E) Three-body collisions are infrequent. 34. The rate constants , at two different temperatures, for the reaction: CH3I + Br- → CH3Br + I-

are given below: t=30oC k= 1.38 x 10-4 L/mol s t=40oC k= 1.21 x 10-3 L/mol s

What is the activation energy for this reaction? R =8.314 J/mol K A) 200 kJ/mol B) 40.3 kJ/mol C) None of the other choices is correct D) 92.7 kJ/mol E) 343 kJ/mol 35. The reaction A + 2B → 4C is proceeding with the rate of disappearance of A equal to

0.10 mol/L s. What is the rate of disappearance of B ? A) 0.40 mol/L s B) 0.10 mol/L s C) 0.20mol/L s D) 0.01mol/L s E) 0.05mol/L s 36. A reaction has the rate law: r = k[A]2[B]2 . The overall order of the reaction is: A) 4 B) 3 C) 5 D) 2 E) 1 37. Below is some data for the hypothetical reaction, 2A + B → C. What is the rate expression for this

reaction? initial [A] mol/L initial [B] mol/L Initial Rate (mol/L s)

0.10 0.10 0.20

0.10 0.20 0.80 0.20 0.10 0.20

A) r=k[A]2[B]2 B) r=k[A]2[B] C) r=k[A][B]2 D) r=k[A][B] E) r=k[B]2 38. The following data were collected for the reaction : 2 M + 2 N ---> 2 P + Q

initial [N] mol/L initial [M] mol/L Initial Rate (mol/L s) 0.10 0.10 1.2 x 10-2 0.10 0.20 1.2 x 10-2 0.30 0.20 1.08 x 10-1

What is the Rate Law for this reaction? r = k __________ A) [M][N]2 B) [N]2 C) [M]2 D) [N] E) [M][N] 39. Which of the following is not true about a catalyst? A) it lowers the activation energy of the reverse reaction. B) A homogeneous catalyst is consumed in one step in the reaction mechanism. C) It decreases ∆H for the reaction. D) It changes the nature of the activation complex in the rate determining step of the reaction. E) It increases the number of effective collisions. 40. The rate law for a certain reaction is r = k[A]2. If the concentration of A doubled, the rate of the

reaction will increase by a factor of: A) 1 B) 8 C) 2 D) 4 E) 16 41. A reaction and its rate expression are given below. When [C4H6] = 2.0 mol/L, the rate is

0.106 mol/L s. What is the rate when [C4H6]=4.0 mol/L?

2C4H6(g) → C8H12(g) r=k[C4H6]2 A) 0.053 mol/L s C) 0.022 mol/L s E)0.212 mol/L s B) 0.106 mol/L s D) 0.424 mol/L s 42. A reaction has the rate law: r = k[M]2[N] . The overall order of the reaction is: A) 2 C) 1 E) unable to be determined from information available B) 3 D) 4 43. If the reaction 2NO + 2H2 → N2 + 2H2O occurred in just a single step, what would the overall order

of the reaction be? A) 4 C) 1 E) impossible to be determined from information available B) 2 D) 3 44. A certain substance decomposes by a first order process. The half-life for the reaction at 25oC is

20.0 minutes. If 0.020 mol of this substance is present 1.00 hour after the start of the reaction, how many moles were present initially?

A) insufficient data are available to answer the question B) 0.08 C) 0.32 D) 0.06 E) 0.16 45. The first order reaction P → Q has k = 3.00x10-3 s-1. If the initial concentration of P is 0.400 M ,

what will its concentration be after 500 seconds? A) 0.136 M B) 0.0893 M C) 0.00828 M D) 0.249 M E) 0.0763 M 46. Which statement does not apply to a heterogeneous catalyst? A) It is found in a different phase than the reactants B) It lowers the activation energy of the reaction. C) It changes the mechanism of the reaction by providing an easier path to the products. D) It is poisoned when its surface becomes covered by reactants. E) Reactants are adsorbed on its surface where they react with each other.

47. What is the half-life of a first order reaction for which k= 2.05x10-2 s-1? A) 48.7 s B) 33.8 s C) 0 0.00205 s D) 0.00142 s E) 0.225 s

48. Which of the following is not true about a catalyst ? A) it may be homogeneous D) it speeds up the forward reaction B) it speeds up the reverse reaction E) it may be heterogeneous C) it acts as an inhibitor 49. Given the following potential energy diagram for the reaction: A + B → C + D

P.E.

A + B

C + D

reaction coord ---> Which of the following is the best interpretation of the graph? The graph is; A) exothermic and may be spontaneous C) endothermic and may be spontaneous B) exothermic and is not spontaneous D) endothermic and is not spontaneous 50. Which type of radiation is potentially the most dangerous ? A) comic B) alpha C) beta D) gamma 51. Given the following data about bond energies:

C-H 413 kJ/mol C=O 494 kJ/mol C-C 348 kJ/mol C-O 351 kJ/mol H-O 464 kJ/mol

How much heat would be released in the formation of butanoic acid?

C C C C O H

H H H

HHH

H

O

A) 5244 kJ B) 4780 kJ C) 3849 kJ D) 4750 kJ 52. What does the following diagram show you about reaction: X + Y →Z

reaction coord -->

X + Y

Z

EP

A) it is rapid C) it is exothermic B) it is endothermic D) it is spontaneous 53. Which of the following partial reactions is an example of Nuclear Fusion ?

A) 88

22624Ra He→ + C) 6

1410C e→ +−

B) 24

24He He+ → D) 92

23501

56141

01U n Ba 3 n+ → + +

54. Which partial reaction equation in question #53 would complete as a nuclear fission reaction? 55. 5.61g of potassium hydroxide was dissolved in 200 mL of water ( c=4.2 J/(goC). The temperature of

the water rose by 7.0oC, which means that the molar heat of solution ( J/mol ) was A) 104.8 B) 5800 C) 1048 D) 58800

56. When NH4NO3(s) dissolves in water the resulting solution feels cold. Which of the following

statements best explains this observation? A) NH4NO3(s) ! NH4NO3(aq) + 28.56 kJ B) NH4NO3(s) ! NH4NO3(aq) ∆H = +28.56 kJ C) The reaction is exothermic. D) Heat is released by the dissolving, causing the temperature of the temperature to drop. Heat of Vaporization is the amount of energy required to change one mole of a substance

from a liquid to a gas at its boiling point. Use this information to answer questions 57 and 58 57. Use the following data to determine the∆H of vaporization for SiCl4 Si(g) + 2Cl2(g) ! SiCl4(g) ∆H = -611.94 kJ Si(g) + 2Cl2(g) ! SiCl4(l) ∆H = -642.60 kJ A) -30.66kJ B) +30.66kJ C) +57.60kJ D) -170.1kJ 58. How many kilojoules of heat are required to change 85.0 g of SiCl4 from a liquid to a gas at its

boiling point? A) 15.33 kJ B) 61.32 kJ C) 365 kJ D) 30.66 kJ A small 2.0 g candle is consumed in heating 100 g of water from 25.0oC to 75.0oC. Use this

information to answer questions 59 to 60. 59. If 200 g of water are heated by a 2.0 g candle, the temperature change from 25.0oC would be

about: A) 25oC B) 75oC C) 150oC D) 50oC 60. If two of the 2.0 g candles are consumed in the heating of 500 g of water whose initial temperature is

25oC, the temperature of the water after heating would be: A) 55oC B) 30oC C) 45oC D) 20oC 61. Ignoring the heat losses to the air, the amount of heat per gram of candle would be ____ J/g. A) 420 B) 21000 C) 2010 D) 10500 62. Acetylene (C2H2) burns in the presence of the air to form carbon dioxide and water according to the

first of the following equations. Use the other three equations to determine the ∆H for the combustion in kJ/mol.

C2H2(g) + 5/2 O2(g) → 2 CO2(g) + H2O(g) 2C(s) + H2(g) → C2H2(g) ∆H = -227.64 kJ C(s) + O2(g) → CO2(g) ∆H = - 395.22 kJ H2(g) + 1/2 O2(g) → H2O(g) ∆H = - 242.76 kJ A) -410.34 kJ B) -865.62 kJ C) -805.56 kJ D) -1260.84 kJ 63. Ethanol ( C2H5OH ) both burns and evaporates. Use this information to identify which of the

following statements is false. A) The burning of ethanol is an exothermic process B) The bonds between adjacent molecules of alcohol are weaker than those between the atoms

within the molecule C) More heat is involved in burning than in evaporating a mole of ethanol D) The evaporation of ethanol is an exothermic process 64. The molecular enthalpy of a substance is comprised of the following combination of energy forms. 1. chemical bonding energy 2. nuclear energy

3. potential energy 4. vibrational kinetic energy 5. translational and rotational kinetic energy A) 1 and 3 B) 3 and 5 C) 1,2,4 and 5 D) 1,3,4 and 5

65. Which of the following reactions represents Beta decay?

A) H H He n

B) U n Sr Xe 3 n

C) Sr Y e

D) H H e

12

13

24

01

92235

01

3890

54143

01

3890

3990

10

+

+ → +

+ → + +

→ +

→ +

−

−

1-B 14-E 27-E 40-D 53-B 2-C 15-C 28-B 41-D 54-D 3-A 16-A 29-E 42-B 55-D 4-B 17-E 30-A 43-A 56-B 5-C 18-C 31-C 44-E 57-D 6-D 19-E 32-E 45-C 58-A 7-E 20-E 33-E 46-D 59-A 8-C 21-D 34-IGNORE 47-B 60-C 9-E 22-E 35-C 48-C 61-D 10-B 23-B 36-A 49-A 62-C 11-D 24-A 37-E 40-D 63-D 12-A 25-C 38-B 51-A 64-D 13-A 26-E 39-C 52-B 65-C

Acid Base Equilibria Definitions for Acids and Bases

1. Operational (Working) Definitions

A) Acids (Corrosive): HCl , H2SO4 , HNO3 , HC2H3O2 , H2CO3 , H3PO4 Taste: sour Reaction with indicators

litmus : red stays red , blue turns red bromthymol: yellow phenolphthalein : colourless

Chemical Reactivity

1 Acids react with active metals to produce hydrogen and an ionic compound Zn(s) + 2HCl(aq) ! ZnCl2(aq) + H2(g) NIE Zn(s) + 2H+

(aq) ! Zn2+ (aq) + H2(g)

2 Acids react with carbonates/bicarbonates to produce CO2 , H2O , ionic compound CaCO3(s) + 2HNO3(aq) ! Ca(NO3)2(aq) + CO2(g) +H2O(l) NIE NaHCO3(aq) + HCl(aq) ! NaCl(aq) + CO2(g) +H2O(l) NIE

3 Acids Neutralize bases to make an ionic salt and water H2SO4(aq) + KOH(aq) ! H2O(l) + K2SO4(aq) NIE HC2H3O2(aq) + Mg(OH)2(s) ! NIE

B) Bases (Caustic ): NaOH , KOH , Ca(OH)2 , Mg(OH)2 , Al(OH)3

Taste : bitter Feel : oily Reaction with indicators

litmus : blue stays blue , red turns blue bromthymol: blue phenolphthalein : pink

Chemical reactivity Neutralizes acids 2. Arrhenius Definitions (based on these substances as electrolytes)

Acid Is a substance that produces hydrogen ions ( hydronium ions) in a water solution ( aqueous ) HCl(aq) ! H+

(aq) + Cl-(aq) HCl(aq) + H2O(l) ! H3O+

(aq) + Cl-(aq) Base Is a substance that produces hydroxyl ions ( hydroxide ions ) in a water

solution NaOH(aq) ! Na+

(aq) + OH-(aq)

Review Electrolyte Is a substance that will conduct an electric current when dissolved in

water. (produces free ions ). Therefore solute must be either a soluble ionic or soluble polar covalent compound.

Non-electrolyte Is a substance that when dissolved in water does not conduct an electric current. (does not produce free ions , non-polar covalent)

Dissociation Is the process by which the ions in an ionic compound are pulled free of the crystal lattice by the attraction of the dipoles in polar water molecules. The ions are then hydrated (solvated) NaCl(aq) ! Na+

(aq) + Cl-(aq) Ionization Is the process by which polar water molecules attract the dipoles in a

polar molecule and break the polar bond to produce free ions. The ions are then hydrated . HF(aq) ! H+

(aq) + F-(aq)

Types of Electrolytes Type Description Examples Strong In the solution the solute is highly dissociated/ionized to form

large concentrations of ions. The large concentration of ions means that the solution can conduct large electric currents. Solutes have high ksp , ka

H2SO4 , HCl , NaOH , HNO3 , NaCl , KNO3

Weak Solutions with low [ions] and therefore conduct only small amounts of electric current. Low ksp , ka

Mg(OH)2 , CH3COOH , citric acid

3. Bronsted-Lowry Acid Is a molecule or ion that can give up a hydrogen ion (proton donor )

HCl(g) + H2O(l) ⇔ H3O+(aq) + Cl-(aq)

Base Is a molecule or ion that can react with a hydrogen ion ( a proton acceptor) NH3(g) + H2O(l) ⇔ NH4

+(aq) + OH-

(aq)

Acid-Base Equilibrium 2HCl(aq) + Ca(OH)2(aq) ⇔ CaCl2(aq) + 2H2O(l) Bronsted acid Bronsted base Conjugate Base Conjugate acid

A strong acid produces a weak conjugate base. (equilibrium shifts right ) A weak acid produces a strong conjugate base. equilibrium shifts left )

Amphiprotic: substances that can act either as Bronsted acids or as bases Ex: NH3 + H+ ! NH4

+ ( base) NH3 ! H+ + NH2

- ( acid )

Identify the acid , base, conjugate base , conjugate acid in: HNO3(aq) + NaOH(aq) ⇔ H2O(l) + NaNO3(aq)

NaHCO3(aq) + HCl(aq) ⇔ NaCl(aq) + H2CO3(aq)

What is the conjugate base for each of the following? a. H2SO4 b. H2CO3 c. NH3 d. HF

4. Lewis Acid Is an electron pair acceptor in the formation of a coordinate covalent bond Base Is an electron pair donor in the formation of a coordinate covalent bond

H3N: + BF3 ! H3N:BF3 Lewis base Lewis acid Neutralization: is the formation of a coordinate covalent bond between the donor (base) and the acceptor (acid) H+ + OH- ! H-OH

Ka and pH Problems 1.Possible Calculations

HA H A

k[H ] [A ]

[ HA ]

pH log[H ] pOH = -log[OH ]pH pOH 14[H ] 10 [OH ] 10

(aq) (aq) (aq)

a(aq) (aq)

(aq)

-

pH - pOH

↔ +

=

= −+ =

= =

+ −

+ −

+

+ − −

2.Sample Questions i). Write the ka expression for each of the following:

HBr , HNO2 (nitrous acid) , HPO42- (monohydrogen phosphate) ii) Write the ka expressions for the following polyprotic acids H2SO4 , H3PO4 iii) A) Formic acid (HCHO2), is a monoprotic acid. In a 0.100 M solution of formic acid,

the pH is 2.38 at 25oC. Calculate the ka for this temperature. B) When butter turns rancid, its foul odour is mostly that of butyric acid, a weak acid.

A 0.0100 M solution of butyric acid has a pH of 3.40 at 20oC. Calculate the ka for this temperature.

C) At 60oC, the pH of 0.0100 M butyric acid is 2.98 . Calculate the ka at this temperature.

iv) Simplifications for weak acids

[HA] [HA] [HA]if weak then : [HA] [HA]

eq int ionided

eq int

= −=

A) The concentration of a sample of vinegar was found to be 0.75 M acetic acid. Calculate the values of [H+] and pH of this sample. ka=1.8x10-5

B) Nicotinic acid, HC2H4NO2, is a B vitamin. it is also a weak acid with ka=1.4x10-5.

What is the [H+] and the pH of a 0.010 M solution? v) Percent Ionization: %ionization [HA] x100

[HA]ionized

initial=

A) Calculate the % ionization in 0.10 M acetic acid. B) Calculate the % ionization of 0.010 M and 0 0010 M acetic acids.

ka and pH Problems

1. 6.1 g of benzoic acid, C6H5COOH, is dissolved in 5.0L of distilled water. The pH of the

solution is 5.3. Find the ka. (2.5x10-9)

2. An acid , HA has a ka=4.6x10-2. If a solution of HA has a pH of 3.3, find the concentration of the HA solution. A) Using the assumption that very little HA ionizes. (5.4x10-6) B) Since ka is actually fairly large, this assumption is not valid. Calculate the [HA] without

the simplification. (5.0x10-4) 3. What is the pH of a 0.010 M NH3 solution.( NH NH OH )3 (aq) 4 (aq)

+(aq)↔ + −

Assume only 4.0% ionizes. (pH=10.6) 4. What is the pH of a 0.0010 M solution of carbonic acid? (diprotic)

Note: H CO H HCO k 3.5x10HCO H CO k 4.4x10

2 3 3 a7

3 32

a11

↔ + =↔ + =

+ − −

− + − −

Hint: One of these reactions may be ignored. (pH=4.7) 5. What is the pH of a 0.0010 M solution of phosphoric acid ? ka values are 1.1x10-2 , 7.4x10-8 , 4.8x10-13 (pH=3) 6. Acid HA has a molar mass of 120 g/mol and ka=1.0x10-4.

Acid HB has a molar mass of 190 g/mol and ka= 1.0x10-6.

A 1.00 g sample is made up to 50.0 mL of solution. Find the pH of this solution given that it takes 16.7 mL of 0.50 M NaOH to neutralize the 50.0 mL of the acid. (pH=2.4)

Concentration Problems

1. A) Write the balanced equation for the neutralization of sodium hydroxide with sulphuric acid.

B) How many moles of NaOH is required to neutralize 0.40 moles of H2SO4 (0.80) C) What volume of 0.50 M NaOH will neutralize 30 mL of 0.15 M H2SO4? (56.3g)

2. A) Calculate the molarity of a solution containing 100 g of hydrogen sulphate (H2SO4 ) in 250 mL of solution. (4.08M)

B) Calculate the molarity of a solution containing 2.8 g of potassium hydroxide in 250 mL of solution. (0.20M)

C) What volume of the solution referred to in "A" will be required to neutralize 45 mL of the solution referred to in "B" ? (1.1mL)

3. 20.0 mL of 3.0 M NaOH solution neutralized 300 mL of phosphoric acid solution. A) Calculate the molarity of the H3PO4 solution. (0.067M) B) The [H3PO4 ] in g/L . (6.57g/L)

4. What volume of 0.10 M NaOH solution would be required to neutralize 0.196 g of hydrogen sulphate ? (0.040L)

5. What is the concentration of acid obtained by diluting 20 mL of 3.5 M acid to 250 mL ? (0.28M)

6. A solution of NaOH is 20% by mass. The density of the solution is 1.2 g/cm3. Calculate the molarity. (6.0M)

7. A bottle of concentrated nitric acid is labelled: 70% by mass , specific gravity =1.42. A) What is the [HNO3] ? (15.8M) B) What volume of this concentrated acid is required to make , 5.0 L of 1.5 M solution ? (0.47L)

8. 20.0 mL of stock HCl ( S.G.=1.18 , 35% by mass ) is made up to 0.500 L by adding water. What volume of this acid will be required to neutralize 60.0 mL of 0.50 M KOH?(66.7mL)

9. What volume of 0.50 M NaOH is needed to neutralize a 1.20 g sample of a monoprotic acid with a molar mass of 240 g/mol ? (0.010L)

10. A sample of monoprotic solid acid with a mass of 0.99 g is neutralized by 14.6mL of 0.88 M NaOH solution. Calculate the molar mass of the solid acid. (77g/mol)

11. A 0.450 g sample of impure CaCO3 was reacted with 50 mL of 0.0985 M HCl. The excess acid was neutralized with 6.00 mL of 0.105 M NaOH. Calculate the % CaCO3 in the sample.

Brønsted-Lowrey Acid Base Theory An acid is a proton donor A base is a proton acceptor

Write the formulas of the conjugate acid for each of the following Brønsted bases

Base HO21- SO42- CO32- CN1- NH21- NH3 H2PO41- HPO42-

Acid Write the formulas of the conjugate base for each of the

following Brønsted acids Acid H2O HI HNO2 HPO42- H2 NH41+ H3PO4 H2PO41-

Base Identify the conjugate acid-base pairs in a Brønsted Acid-Base reaction

HSO41-(aq) + PO43-(aq) ! SO42-(aq) + HPO42-(aq)

CN1-(aq) + H3O+(aq) ↔ HCN(aq) + H2O

Amphiprotic ( Amphiphoteric ) 1) H2O + HCl(aq) ! H3O+(aq) + Cl1-(aq)

H2O + NH3(aq) ↔ NH41+(aq) + OH1-(aq) 2) HCO31-(aq) + OH1-(aq) ! CO32-(aq) + H2O

HCO31-(aq) + H3O1+(aq) ! H2CO3(aq) + H2O

HC2H3O2+ H2O↔H3O1+ + C2H3O21-

Acid

AcidBase

Base

conjugate pair

conjugate pair

Buffers: are solutes that protect against large changes in pH Buffered Solution: is the name given to the solution that contains the solutes ( Buffer) Buffered solutions are composed of two solutes 1) A weak Bronsted acid ( HA ) (very little ionization )

2) A conjugate base ( proton acceptor ) ( high degree of dissociation )

Acid Equilibrium HA(aq) ↔ H+(aq) + A-

(aq) Equilibrium Shifts ( Le Chatlier ) 1) Addition of an Acid ( hydrogen ions ) pH drop H+(aq) + A-(aq) ! HA(aq) 2) Addition of a Base ( hydroxyl ions ) pH increase OH-(aq) + HA(aq) ! A-(aq) + H2O(l) Buffering in Human Blood ( H2CO3 and HCO3-) 1) If the pH is too high H2CO3(aq) + OH-(aq) ! HCO3(aq)- + H2O(l) 2. If the pH is too low HCO3(aq)- H+(aq) ! H2CO3(aq)

Buffered solutions are used to standardize a pH meter

Types of Buffered Solutions pH < 7 1) Weak acid ( HA ) plus an ionic salt ( NaA ) with high ksp

HA + OH- ! H2O + A- weak Bronsted acid consumes OH- A- + H+ ! HA + H2O conjugate base consumes H+

pH>7 2. Weak base ( MOH and its salt ( high ksp ) MOH + H+ !. M+ + H2O M+ + OH- ! MOH Examples Phosphate buffer (inside body cells) NaH2PO4 , Na2HPO4

HPO42- + H1+ ! H2PO41- + OH1- ! + H2O

Ammonia/ammonium ion buffer

NH41+ + OH1- ! NH3 + H1+ !

Acetic acid/acetate ion

HC2H3O2 + OH1- ! C2H3O21- + H1+ !

Hydrolysis : This means a reaction with water Hydrolysis of ionic salts will result in an equilibrium shift in the water ionization 1) Anions ( -ve ) can hydrolyze water as follows A-(aq) + H2O(l) ↔ 2) Cations ( +ve ) can hydrolyze water as follows B+(aq) + H2O(l) ↔ Bronsted-Lowry Acids and Bases Rule#1 Anions Anions that form weak conjugate bases hydrolyze to make ____________________ Equilibrium shifts to the production of acid molecules. CO32-(aq) + H2O(l) !

Anions of strong acids ( 100% ionization ) do not hydrolyze. HCl , HI , HNO3 , HBr , HClO4 are strong monoprotic acids ( large ka )

Rule #2 Cations

Metal ions ( cations ) from group IA ( alkali ) or IIA ( alkaline earth except Be ) do not hydrolyze. Other metal ions may hydrolyze to generate H+ ions.

Cu2+(aq) + H2O(l) !

HCN(aq) + H2O(l) ↔ H3O+(aq) + CN-(aq)

Conjugate Pair

Conjugate Pair

Rule #3 Cations

The ammonium ion of an ammonium salt hydrolyzes and tends to make an aqueous solution that is slightly acidic.

NH4+(aq) + H2O(l) ↔ Hydrolysis and pH of Ionic Salt Solutions Hydrolysis Summary

Cation from Anion from Resulting solution pH

Examples

Strong base NaOH KOH

Strong acid HCl

HNO3

Neutral Neither ion hydrolyzes

NaCl KNO3

Strong base NaOH

Weak acid HC2H3O2

Basic

Weak base Fe(OH)2

Strong acid HCl

Acidic

Weak base

Weak acid H2CO3

Varies depending of strength

Chemical systems and Equilibrium Equilibrium

Systems:

Dynamic Equilibrium

Concentration

Types of Equilibrium Systems 1) Physical 2) Chemical 1) Physical Systems ( state changes or dissolving ) ex. water and water vapour

in a bottle ex. mothball and air in a box ex. solute precipitate in water

A) Water in equilibrium with water vapour in a sealed jar. H2O(l) + heat H2O(g) initially: rate of evaporation > rate of condensation eventually rate of evaporation = rate of condensation

EQUILIBRIUM H2O(l)

rate of evaporation

rate of condensation H2O(g)

Factors that cause a shift in this equilibrium Factor Direction of equilibrium shift Temperature Volume

2) Chemical system ( forward and reverse reactions) 2 22 2 2H O H Ol( ) ↔ +

liquid

vapour

liquid vapour solid vapour

solid

vapour

solid

liquid

liquid solid

Equilibrium in Chemical Reactions Consider the following thermochemical equation

H2(g) + I2(g) ! 2HI(g) + Heat

H2(g) colour I2(g)colour HI is colourless

2. EP Vs reaction coordinate diagram for this reaction 3. Explain the expected observations for this experiment if initially only

reactants were present: I) explanation of each observation

[H2]

[I2]

[HI]

4. Explain what happens to the following reaction rates over time. A) FORWARD reaction: production of HI B) REVERSE reaction: consumption of H2 and I2 5. Write the Equilibrium Chemical Equation for this reaction Reactions of Nitrogen Oxides

NO2 red-brown gas N2O4 colourless gas 1. Demonstration: Effect of Temperature on tube containing gaseous mixture

H2 +I2

2HI

∆H

Ea forward

Ea reverse

Ep

Reaction coordinates

hot water cold water

II) Write an equation for the hot and cold stress showing the shift in the equilibrium hot: N2O4(g) + heat 2NO2(g) cold: N2O4(g) + heat 2NO2(g) III) Draw a labelled Ep diagram Vs reaction coordinates for this reaction. 2) Experiment: Pressure as a stress on the equilibrium syringe filled with nitrogen oxides

Observation:

original increasing pressure decreasing pressure

Conclusions: Explain the effect of using pressure as a stress on this equilibrium. Write the

equations to show the shift in the equilibrium Describe the possible stresses that can affect an equilibrium.

Le Chatelier's Principle ( 1850-1936 )

Sources of Stress

1. Addition or removal of either a reactant or a product ( [conc] changes )

ex. N2O4(g) + 59.0 kJ 2NO2(g) 2. Changing the volume of gaseous products ( Pressure changes) 3H2(g) + N2(g) 2NH3(g) 3 vol 1 vol 2 vol

Reducing the volume of the system causes the equilibrium to shift toward the lower volume of gases. _______________________.

Increasing the pressure always drives the reaction in the direction of the fewest number of gaseous molecules. ____________________

[ ] mol/L

time

N2O4

NO2

addition removal

[H2]

[N2]

[NH3]

[conc]

time

3. Changes in temperature 3H

2(g) + N2(g) 2NH

3(g) + heat exothermic ∆Hf=-46.19 kJ/mol

4. Effect of a Catalyst

5. Addition of an inert gas at constant volume

There is a pressure increase within the container but it does not affect the reactants because it is inert.

time

[H2]

[N2]

[NH3]

[conc]

reactant Ep

RXN!

Without catalyst

product

Le Chatelier Questions 1. For the system: heat + COCl2 (g) CO (g) + Cl2 (g) Explain the effect on the equilibrium concentration of chlorine when: A) Adding heat to the system. B) Decreasing the pressure of the system. C) Removing CO from the system. D) Adding a catalyst to the system. E) Graph the concentration changes if [CoCl2] started at 10 mol/L and went

through changes described in parts A-D. 2. Explain the effect on the equilibrium [ SO2 ] of:

2SO2 (g) + O2 (g) 2SO3 (g) + heat A) Removing heat from the system B) Removing SO2 from the system C) Decreasing the pressure on the system D) Increasing the pressure on the system E) Adding O2 to the system 3. For the system: 3A (g) + 3B (g) 4C (g) + D (g) + heat Explain the effect of the [ D ] of: A) Increasing the concentration of A B) Adding more B C) Increasing the pressure D) Removing heat from the system E) Removing C from the system 4. Explain the effect on the equilibrium of applying the named stress to each of

the following: A) Water expands when freezing: H2O (s) + pressure H2O (l) Stress: increase the pressure on the ice B) Equilibrium: 2H2 (g) + 2NO (g) N2 (g) + 2H2O (g) Stress: increase system pressure C) Equilibrium: SO2 (g) + 1/2 O2 (g) SO3 (g) + 65 kJ Stress: increasing temperature D) Equilibrium: P4 (g) + 6H2 (g) 4PH3 (g) Stress: adding hydrogen gas E) Equilibrium: FeO (s) + CO (g) Fe (s) + CO2 (g) Stress: removal of some Fe (s) F) Equilibrium: H2O (l) + 40 kJ H2O (g) Stress: increased pressure on the system

Solving Kc Problems 1.At 900oC, H2(g) + CO2(g) H2O(g) + CO(g)

The equilibrium concentrations of reactants and products are: [H2]=0.24 mol L-1, [CO2]= 1.80 mol L-1, [H2O]=[CO]=0.88 mol.L-1 Calculate the Kc for this reaction. H2(g) + CO2(g) H2O(g) + CO(g)

[initial] [change] [equilibrium] 0.24 mol.L-1 1.80 mol.L-1 0.88 mol.L-1 0.88 mol.L-1

page 311 #15 15. At equilibrium for the following reaction: PCl5(g) PCl3(g)+Cl2(g) the following concentrations are obtained. [PCl5]=0.010 mol.L-1 , [PCl3]=0.15 mol.L-1 ,[Cl2]= 0.37 mol.L-1

PCl5(g) PCl3(g) +Cl2(g)

[initial] [change] [equilibrium] 0.010 mol.L-1 0.15 mol.L-1 0.378 mol.L-1

2.For the reaction: NH4Cl(s) NH3(g) + HCl(g)

At 400K the equilibrium constant, Kc, for this reaction is 6.0x10-9. Calculate the equilibrium concentration of each gas at this temperature.

Kc=6.0x10-9 NH4Cl(s) NH3(g) + HCl(g)

[initial] [change] [equilibrium] X mol.L-1 X mol.L-1

Perfect Squares Simplification 3. At 430oC , Kc=1.84x10-2 for : 2HI(g) H2(g) + I2(g) If 0.100 mol of HI is placed in a 1.00L vessel what are the concentrations at equilibrium.

Kc=1.84x10-2 2HI(g) H2(g) + I2(g)

[initial] 0.100 mol.L-1 0 0

[change] -2X X X [equilibrium] 0.100-2X X mol.L-1 X mol.L-1

Page 316 #17 Kc=0.212

N2O4(g) 2NO2(g)

[E] 0.155mol.L-1 2X mol.L-1 #18

SO2(g) + NO2(g) NO(g) + SO3(g)

[I] 0.100mol.L-1 0.100mol.L-1 0 0

[C] -x -x +x +x [E] 0.100-x 0.100-x x mol.L-1 x mol.L-1

Assumption when Kc is very small

An approximation can be made if the concentration from which x is subtracted, or to which x is added, is at least 1000 times the value of the equilibrium constant.

3. At 727oC the equilibrium constant, Kc, for the dissociation of molecular iodine to iodine

atoms is 3.80x10-5. I2(g) 2I(g)

The original concentration of molecular iodine is 0.200 mol.L-1 , calculate the concentration of atomic iodine at equilibrium.

.Kc=3.80x10-5 I2(g) 2I(g)

[I] 0.200mol.L-1 0

[C] -x +2x [E] 0.200-x 2x mol.L-1

#19. Kc=2.2x10-10 COCl2(g) CO(g) + Cl2(g)

[initial] 0.100mol.L-1 0 0

[change] -X X X [equilibrium] 0.100-X X mol.L-1 X mol.L-1

Chemical Equilibrium Law Experiment Equation: Fe3+(aq) + SCN1- (aq) --> FeSCN2+ (aq) red Prelab Calculations 1. Colour Spectroscopy to determine [FeSCN2+] I) [A] is 0.6 M II) solution B colour intensity matches A III) depth A = 3 cm , depth B = 6 cm IV) [B]

[A]depth Adepth B

or [B] = [A] depth Adepth B

= •

[B] = 0.6M x 3cm/6cm =0.3 M 2. Dilution calculations A) [KSCN] : 5 mL KSCN diluted to 10 mL ( 5mL KSCN + 5mL Fe(NO3)3 )

M M VV

x M mLmL

x M21

2

331 2 0 10 5

1010 10= = • =

−−. .

B) [Fe(NO3)3 ] Vial # Dilution to 25 mL Dilution to 10 mL 1

M MVV

M mLmL

M21 1

2

0 200 510

0100= = • =. .

2 M MV

VM mLmL

M21 1

2

0 200 1025

0 080= = • =. . M MVV

M mLmL

M21 1

2

0 080 510

0 040= = • =. .

3 M M mL

mLM2

0 080 1025

0 032= • =. . M M mLmL

M20 032 5

100 016= • =. .

4 5 3. Completion of Concentration chart initial [Fe3+ ]

initial [SCN1]

Depth Ratio standard/vial#

equilibrium [FeSCN2+ ]

equilibrium [Fe3+ ]

equilibrium [SCN1-]

0.100 M 0.0010 M 0.0010 M 0.099 M 0 M 0.040 M 0.0010 M 0.016 M

A B

Chemical Equilibrium Experiment : Quantitative Background: Fe3+(aq) + SCN-(aq) ---> FeSCN2+(aq) colourless colourless red colour Assumption:

The intensity of the colour in the solution at equilibrium indicates the relative concentration of the FeSCN2+ ion. By matching the colour intensity of a vial of known concentration with the colour intensity of a vial of unknown concentration, the unknown concentration can be determined by depth comparison. depth solution known = [ unknown ] depth solution unknown [known ]

Purpose: to determine a quantitative relationship between [reactants] and [products] in a chemical

equilibrium. ( to find the Equilibrium Law relationship ) Apparatus: vials, graduated cylinders, beakers, light source Method: 1. Line up 5 dry flat-bottomed vials numbered 1 to 5. To each vial add 5.00 mL of 0.00200M KSCN 2. To vial #1 add 5.00 mL of 0.200 M Fe(NO3)3 . This vial is your standard that will be used for

comparison. 3. Measure out 10.0 mL of 0.200 M Fe(NO3)3 in a 25.0 mL graduated cylinder and then fill the cylinder

with distilled water to the 25.0 mL mark. Pour the diluted solution into a clean dry beaker to mix the solution. Measure out 5.00 mL of this solution and add it to vial # 2. Save the remaining solution for further dilutions. Next step.

4. Pour 10.0 mL of the solution from step #3 into the 25 mL graduated cylinder and discard the remainder. Fill the graduated cylinder to the 25.0 mL mark with distilled water. Pour this solution into a clean dry beaker. Take 5.00 mL of this solution and place it in vial #3 and save the rest for future dilutions.

5. Continue the dilution pattern as in step #4 until the remaining vials have 5.00 mL of a successively more dilute ferric nitrate solution added to them.

6. Wrap each vial with white paper leaving the top and bottom open. 7. Place vial #2 and vial #1 ( standard) over the light source so that you can look down from above.

Remove ( and save ) the drops from the standard until the colour intensity matches in the two vials. Record the depth of both vials when the colour intensity matches. Repeat this procedure with vial #3 to #5 until the colour intensity matches the standard.

Observations: Pre-lab calculations 1. Calculate the concentration of SCN- in each vial ( 5 mL diluted to 10 mL when solutions are mixed )

Record the [SCN-] M1V1 = M2V2

2. Calculate the concentration of Fe3+ in each of the vials. Vial# Dilution to 25 mL Dilution to 10 mL 1 M 2 = M V1 1

V2 2 M 2 = M V1 1

V2

3

4 5

3. Complete the following Chart

initial [Fe3+]

initial [SCN-]

Height Ratio standard/vial#

Equil. [FeSCN2+]

Equil. [Fe3+]

Equil. [SCN-]

0.100 M 0.00100M 0.00100M

4. Calculate the value of each of the following by using the equilibrium concentrations from the above

chart. Equilibrium Constant Expression Vial #

[ Fe ] [ FeSCN ]+3 +2

[ SCN ]-

[ FeSCN ]

[ Fe ] [ SCN ]

2+

3+ -

[ FeSCN ]2+

3+[ Fe ] + [ SCN ]

-

2 3 4 5 Conclusions; 1. Which of the calculations give the most constant numerical value for the Equilibrium Constant

Expression? Explain your answer is the most constant. 2. Re-write the equilibrium Constant Expression using the names ( not formulae ) of the ions. 3. List and explain any errors in your experiment.

Dissolving : Energy Considerations 1. Energy must be added to overcome the attractive forces holding the ions

together in the crystal. ( solute) The magnitude of the lattice energy is reflected in melting points, boiling points, and heat of formation. The attractions depend upon the sizes of the ion charges and the ionic diameters.

2. Energy needs to be added to separate the particles of the liquid from one another. ( solvent ) ( polar . non-polar )

3. Energy is released as the ions and solvent molecules interact. This "SOLVATION ENERGY� is greater for smaller more highly charged ions and more polar solvents.

Consider The Following Data Substance M.P. oC B.P. oC ∆H

Formation

Heat of Solution

Solubility (g/100g H2O

NaF 980 1700 -136.3 -0.48 (12oC) 4.22 (18oC) NaCl 801 1413 -98.4 -1.28 (18oC) 35.7 (0oC) NaBr 755 1390 -86.3 -0.19 (18oC) 79.5 NaI 651 1300 -69.5 +1.41 (18oC)

1. From considerations of ionic radii and Coulomb's Law which would you expect to have the

greater lattice energy, NaF or NaCl? Is your answer supported by the values for their melting points?

2. Given the melting point of CaO is 2580oC, arrange the compounds CaO, SrO, MgO, and BaO in the expected order of decreasing M.P.

3. Would you expect NaCl or NaF to have the larger Solvation Energy? Why? 4.

Heat of solution = Lattice energy (energy put in to pull ions apart)

+ Energy put in to pull solvent molecules apart

+ Solvation energy (energy released when ions and solvent interact)

Consider your answers to questions 1 and 3 for NaCl and NaF. Which factor appears to predominate in explaining their relative heats of solution?

5. Considering the tendencies to minimum energy and maximum randomness, explain the fact NaCl is more soluble than NaF.

6. What would be the effect on the solubility of NaCl if CCl4 were used as a solvent rather than water ? Explain your answer.

7. Predict a value for the solubility of NaI in water.

Equilibrium Problems 1. A mixture of H2 and I2 is allowed to react at 448oC. When equilibrium is established, the concentrations of the

participants are found to be [H2] = 0.46 mol/L , [I2] = 0.39 mol/L, [HI] = 3.0 mol/L. Calculate the Kc value at

448oC from this data. answer: 50

2. Assume that in the analysis of another equilibrium mixture at 448oC, the equilibrium [ ] of H2 and I2 are both 0.50 M. What is the equilibrium concentration of HI ? : 3.5 mol/L

3. The equilibrium constant for the reaction : H2 (g) + I2 (g) 2HI (g) is 50 at 448oC. A) How many moles of HI will be present at equilibrium when one mole of H2 is mixed with one mole of I2 in a

0.50 L container and allowed to react at 448oC ? : 1.56 mol B) How many moles of H2 and I2 are left unreacted? : 0.22 mol C) If the conversion of H2 and I2 to HI were essentially complete, how many moles of HI would be present

: 2 mol D) What is the percent yield of the equilibrium mixture ? : 78%

4. When 0.035 mol of PCl5 is heated to 250oC in a 1.0 L vessel, an equilibrium is established in which the

concentration of Cl2 is 0.025 mol/L. Find the equilibrium constant at 250oC for:

PCl5 (g) PCl3 (g) + Cl2(g) : 6.2 x 10-2 5. Assume that the analysis of another equilibrium mixture of the system in problem #4 shows that the equilibrium

[PCl5] = 0.012 M and that the [Cl2] = 0.049 M. What will be the equilibrium [PCl3] if the reaction occurs at

250oC? : 1.5 x 10-2 M

6. How many moles of PCl5 must be heated in a 1.0 L flask at 250oC in order to produce enough Cl2 to give an equilibrium concentration of 0.10 mol/L ? : 0.26 mol

7. Calculate the concentration of Ag+ ion and Cl- ion in a saturated solution of the salt at 25oC. What is the approximate solubility of AgCl in mol/L at this temperature ? see page 741 : 1.3 x 10-5 M

8. Determine the solubility of AgI in a) mol/L and b) g/L at 25oC. : a) 9.1 x 10-9 mol/L b) 2.1 x 10-6 g/L

9. The equilibrium constants for three different reactions are: 1.5 x1012 , 0.15 , 4.3 x 10-15 In which reaction is there: A) A large ratio of product to reactant B) A small ratio of product to reactant

10. At 55oC the K for the reaction : 2NO2 (g) N2O4 (g) is 1.15 A)Write the equilibrium expression. B)Calculate the [N2O4] present in equilibrium with 0.5 M of NO2. : 0.29 M

11. Calculate the Kc for the following reaction: given: [CO2] = 1.17 x 10-3 M

[CO] = 1.33 x 10-3 [H2] = 1.17 x 10-3 M [H2O] = 1.33 x 10-3

CO2 (g) + H2 (g) CO (g) + H2O (g) : 1.29 12. When 0.5 mol of CO2 and 0.5 mol of H2 were forced into a 1.0 L reaction container, the following equilibrium

was established: CO2 (g) + H2 (g) CO (g) + H2O (g) At these conditions, Kc = 2.00 A) Find the equilibrium concentration of each reactant and product. B) How would the equilibrium concentrations differ if 0.50 mol of H2O and 0.50 mol of CO had been introduced into the reaction vessel instead of CO2 and H2 ?

Equilibrium Multiple Kc 1. At 985oC, the equilibrium constant for the reaction given below is 1.63. What is the equilibrium

constant for the reverse reaction? H2(g) + CO2(g) H2O(g) + CO(g) A) 0.815 B) 0.63 C) 2.66 D) 0.613 E) 1.00 2. Choose the correct equilibrium law for the heterogeneous reaction: 2C(s) + O2(g) 2CO(g) A) Kc = [CO]/([C] [O2]) C) Kc =( [C]2 + [O2])/[CO]2 E) Kc = ([C]2[O2])/[CO]2

B) Kc = [CO]2/[O2] D) Kc = [CO]2/[O2][C]2

3. For the chemical equilibrium: NO2(g) + NO(g) N2O3(g) + heat a decrease in the volume of the container will; A) increase the amount of NO in the container B) increase the amounts of both NO and NO2 in the container C) increase the amount of N2O3 in the container D) have no effect on the position of equilibrium E) cause heat to be absorbed by the reaction 4. What is the mass action expression for the following? 2SO2(g) + O2(g) 2SO3(g) A) 4[SO3]2/[SO2] [O2] C) [SO3]2/[O2] + [SO2]2 E) [SO2]2[O2]/[SO3]2

B) [SO3]2/[O2] [SO2]2 D) 2[SO3]/2[SO2] [O2] 5. Which of the following will cause the position of equilibrium in the reaction shown below to be shifted

to the left? PCl3(g) + Cl2(g) PCl5(g) A) a decrease in the volume of the container C) addition of PCl3 E) addition of a catalyst B) removal of Cl2 D) removal of PCl5

6. At a certain temperature Kc = 1 x 109 for the reaction Cl2 (g) + F2 (g) 2ClF (g) If 1.00 mol of Cl2 and 1.00 mol of F2 are placed in a reaction vessel and allowed to react, then at

equilibrium... A) [ClF] will be much larger than [Cl2] and [F2] D) the system will contain only Cl2 and F2 B) [ClF] will be much less than [Cl2] and [F2] E) the system will contain only ClF C) [ClF] will be nearly equal to [Cl2] and [F2] 7. The appropriate mass action expression for the reaction 2N2O (g) + O2 (g) 4NO (g) is: A) [N2O]2[O2]/[NO]4 C) 4[NO]4/[2[N2O]2[O2]) E) 4[NO]/(2[N2O] [O2])

B) [NO]4/([N2O]2 + [O2]) D) [NO]4/[N2O]2[O2]

8. At a certain temperature the equilibrium: 3NO(g) N2O(g) + NO2(g) was established. What is the value of Kc for this reaction if the equilibrium concentrations were:

[NO] = 0.060 M [N2O] = 0.015M [NO2] = 0.025M

A) 1.7 B) 0.58 C) 4.8 x 102 D ) 6.4 x 10-3 E) 3.4 x 103

9. The reaction: 2ClO (g) Cl2(g) + O2(g) has Kc = 6.4 x 108 at a certain temperature. If 1.00 mol of ClO is placed in a 1.00 L reaction vessel and allowed to react, than at equilibrium...

A) [ClO] will be much larger than [Cl2] and [O2] D) the system will contain no ClO at al

B) [ClO] will be about the same as [Cl2] and [O2] E) the system will contain only ClO (g) C) [ClO] will be much less than [Cl2] and [O2]

10. Select the Kc value that represents a chemical reaction that has almost reached completion A) 27 B) 2.45 x 10-7 C) 1.8 x 10-21 D) 1.5 x 1023 E) -9 11. For the reaction: PCl3(g) + Cl2(g) PCl5(g) The addition of more Cl2 would cause the reaction to shift to the _____ because A) right, the frequency of collisions would increase between the PCl3 and the Cl2 which would increase

the reaction rate B) right, the activation energy for the exothermic reaction would be lowered C) left, the frequency of collisions would increase between the PCl3 and the Cl2 which would increase

the reaction rate D) left, the activation energy for the exothermic reaction would be lowered E) right because more of the PCl5 would decompose 12. For the reaction: CH2O (g) H2(g) + CO (g) , the addition of He (g) would A) cause a shift in the equilibrium toward the right D) have no effect on the equilibrium B) cause the activation energy to be lowered E) decrease the temperature C) cause the equilibrium reaction to shift to the left 13. Which of the following would cause the position of equilibrium in the reaction shown below to shift to

the left? 2CO(g) + O2(g) 2CO2(g) A) removal of CO2 C) a decrease in container volume E) addition of CO B) removal of O2 D) addition of a catalyst 14. An ice skater causes a change in the physical equilibrium of the ice such that A) rate of sublimation > rate of sublimation D) rate of melting < rate of solidification B) rate of condensation < rate of vaporization E) none of the above C) rate of melting > rate of solidification 15. The concentration of a solid is determined by A) measuring its viscosity D) concentration does not change at any

temperature B) multiplying the density by the molar mass E) dividing the density by the molar mass C) dividing the mass by density

16. For the reaction: 2A2(g) + O2(g) 2A2O(g) If the Kc=27 at 340oC, doubling the [O2] would; A) increase Kc C) decrease Kc E) decrease volume B) keep Kc constant D) decrease the partial pressure

17. For the following reaction: 2N2O (g) + O2 (g) 4NO (g) + heat The amount of O2 at equilibrium will be decreased by ... A) increasing the pressure by decreasing the volume D) removing N2O B) adding a catalyst E) decreasing the temperature C) adding more NO 18. For the equilibrium in question #17 the ∆H = -199.1kJ. Which of the following changes will increase

the amount of N2O at equilibrium? A) The temperature of the system is lowered D) An inert gas is added without changing the

volume B) The volume of the container is decreased E) Some NO is removed C) Some O2 is added

19. The temperature of the following equilibrium system is increased: 2A + heat B + 3C therefore; A) the catalyst stops working C) the Kc will decrease E) the Kc will increase B) the Kc will stay the same D) Dynamic Equilibrium 20. Colourless N2O4(g) is in equilibrium with red-brown NO2(g). Which of the following statements is

not true A) Increasing the temperature of the mixture produces more colour because that is the endothermic

direction of the reaction B) Reducing the pressure makes more colour C) Adding a catalyst favours the production of N2O4, so the mixture loses colour intensity D) Increasing the volume increases the rate of reaction E) decreasing the temperature of the mixture produces less colour because that is the exothermic

direction of the reaction 21. At 350oC, the equilibrium constant for the reaction given below is 70. At equilibrium, the

concentration of HI is 1.32 M and that of H2 is 0.100 M. What is the concentration of I2? H2(g) + I2(g) ↔ 2HI(g)

A) 0.25 M B) 0.19 M C) 1.0 M D) 0.32 M E) 0.57 M 22. For the following reaction: 2NO (g) + H2 (g) N2 (g) + H2O (g)+ heat The addition of N2 to the system will cause: ... A) heat to be evolved D) the concentrations of both NO(g) and H2 (g) to

increase B) the concentration of H2 to decrease E) the concentration of H2O(g) to increase C) no change in the position of equilibrium

23. For the equilibrium: 2SO2(g) + O2(g) 2SO3(g) + 47 kcal an increase in the temperature of the system will cause: A) the concentrations of both SO2(g) and SO3(g) to increase B) the concentration of SO2(g) to increase C) the equilibrium constant for the reaction to become larger D) the concentration of SO3(g) to increase E) the concentrations of both O2(g) and SO3(g) to increase For each of the questions 24 and 25, one or more of the responses may be correct. Decide which of

the responses is(are) correct, then choose: A) if i, ii and iii are correct B) if only i and iii are correct C) if only ii and iv are correct D) if only iv is correct E) is some other response or combination of responses of those given , is correct 24. When a reaction is at equilibrium: i) the amounts of each substance present are equal ii) the concentrations of the substances remain unchanged iii) the reactions between the various substances cease iv) changing the external conditions may change the concentrations 25. When SO3(g) is heated in a closed tube it decomposes and eventually reaches an equilibrium state

with SO3(g) , SO2(g) and O2(g) all present. When this state is reached: i) no more SO3 is decomposing ii) no more SO2 and O2 are combining together iii) the mass of SO3 is equal to the combined masses of SO2 and O2

iv) the mass of SO3 present does not change 1-D 6-A 11-A 16-B 21-A 2-B 7-D 12-D 17-E 22-D 3-C 8-A 13-B 18-B 23-B 4-B 9-C 14-C 19-E 24-C 5-B 10-D 15-E 20-D 25-D

Qualitative Analysis is the procedure by which one can determine the nature (ions) , but not the amount (quantitative analysis ) of the species in a mixture Types of Reactions Used in Qualitative Analysis Precipitation reaction: using solubility rules to produce insoluble products

Ag+(aq) + Cl-(aq) ! AgCl(s)

Complex Ion Formation Ag+(aq) + 2NH3(aq) ! Ag(NH3)2

+(aq)

Redox Reaction 2I-(aq) + 2Fe3+

(aq) ! I2(aq) + 2Fe2+(aq)

Acid-Base reaction ( strong acid and strong base in H2O

H3O+(aq) + OH-

(aq) ! 2H2O(l) Use your knowledge of solubility rules to precipitate the anion/cation indicated for each of the following solutions Solution Anion Solution added

to precipitate Equation

Na2CO3 KI NaOH K2CrO4 KBr NaCl Solution Cation Solution added

to precipitate Equation

CuSO4 Pb(NO3)2 NiCl2 Other methods of identifying ions by qualitative analysis include colour changes , evolution of gas , the ability to redissolve a precipitate by addition of a complexing ligand or by change of pH ( acidity or basicity )

Selective Separation of Ions from an Aqueous Mixture

Hg1+ , Mg2+ , Ba2+

add

precipitates ion remaining

precipitates

precipitates

ion remaining

add

add

Solubility Product Experiment

Purpose: to determine the solubility and the solubility product for silver acetate Apparatus: beaker , copper wire , saturated silver acetate solution , thermometer Method: 1. Clean and mass a piece of copper wire 2. Accurately measure a volume ( between 75 and 100 mL) of the silver acetate solution

in a clean beaker. Record the volume 3. Coil the copper wire so that it will submerge in the solution. 4. Add the copper wire to the solution and allow it to sit until your next class. 5. Rinse the silver crystals of the copper wire. Carefully scrape off any of the remaining

silver crystals . 6. Dry and mass the copper wire. 7. Record the room temperature. Observation: Table Conclusions: 1. Write the net ionic equation for the dissolving of AgC2H3O2 . 2. The equation for the reaction of silver acetate with copper is as follows

Cu + AgC2H3O2 -! Ag + Cu(C2H3O2 )2 Write the net ionic equation for the above reaction 3. Calculate the mole of copper consumed and the moles of silver produced. 4. Calculate the solubility and the molar solubility of the silver acetate. 5. Calculate the Ksp for silver acetate at this temperature.

6. Discuss any possible sources of error for this experiment.

Equilibrium Ksp 1. The solubility of MgF2 in water at 25oC is 0.13 g/L . What is the value of Ksp for magnesium

fluoride? A) 3.6 x 10-8 B) 2.1 x 10-7 C) 4.4 x 10-6 D) 9.1 x 10-9 E) 8.8 x 10-3

2. Which of the following is the largest concentration of Ba2+ that will not give a precipitate of BaF2 in a solution that has [F-] = 1.5 x 10-2M? For BaF2, Ksp= 1.7 x 10-6.

A) 6.0x10-3 M B) 5.0x10-4 M C) 1.0x10-1 M D) 2.0x10-5 M E) 8.8 x 10-2 M 3. What is the solubility product expression, Ksp , for Ca(OH)2?

A) [Ca2+] + [2OH-] D) [Ca2+] [OH-]2/ [Ca(OH)2]

B) [Ca2+]2[OH-] E) [Ca2+] [OH-]2 C) [Ca2+] [2OH-] 4. The Ksp of Ag2CrO4 is 1.9 x 10-12. The solubility ( in mol/L) of this salt in 0.10 M AgNO3 solution

is: A) 7.8 x 10-5 M B) 9.5 x 10-12 M C) 1.9 x 10-10 M D) 1.3 x 10-4 M E) 1.4 x 10-6 M 5. What is the molar solubility of Zn(OH)2 in 0.10 M Zn(NO3)2? Ksp=4.5 x 10-17 for Zn(OH)2.

A) 6.7 x 10-8 M B) 1.1 x 10-8 M C) 4.5 x 10-16 M D) 1.5 x 10-17M E) 2.5 x 10-6 M 6. Solid NaOH is added gradually to a solution of 0.10 M Mg2+. At which of the following [OH-]'s will a

precipitate of Mg(OH)2 first appear? For Mg(OH)2 , Ksp = 1.2 x 10-11.

A) 1.2 x 10-3 M B) 8.4 x 10-5 M C) 6.3 x 10-6 M D) 3.5 x 10-4 M E) 1.1 x 10-5 M 7. Which of the following is the largest [Ag+] that will not give a precipitate of Ag2CrO4 in a solution

having [CrO42-] = 2.0 x 10-3 M ? For Ag2CrO4 , Ksp = 1.9 x 10-12.

A) 6.0 x 10-14 M B) 4.0 x 10-7 M C) 7.0 x 10-11 M D) 2.0 x 10-5 M E) 5.0 x 10-3 M 8. The Ksp for Pb(IO3)2 is 2.8 x 10-13. What is the molar solubility of Pb(IO3)2 in 0.100 M NaIO3

solution? A) 4.1 x 10-5 M B) 2.8 x 10-15 M C) 2.8 x 10-11 M D) 5.3 x 10-7M E) 5.3 x 10-9

M 9. The solubility of MgCO3 ( MM = 84.0 g/mol) in water at 25oC is 0.012 g/L. What is the value of Ksp

for magnesium carbonate? A) 4.3 x 10-7 B) 2.0 x 10-8 C) 1.4 x 10-4 D) 2.4 x 10-3 E) 3.6 x 10-4 10. What is the Ksp of BaF2 if its molar solubility in water is 7.5 x 10-3 M?

A) 4.2 x 10-7 B) 1.7 x 10-6 C) 8.4 x 10-7 D) 3.8 x 10-5 E) 2.2 x 10-2 11. What is the molar solubility of CaCO3 in 0.10 M Na2CO3? Ksp= 9 x 10-9 for calcium carbonate

A) 3 x 10-5 M B) 9 x 10-8 M C) 0.10 M D) 6 x 10-6M E) 3 x 10-3 M

12. What is the molar solubility of AgCl in 0.43 M CaCl2? Ksp= 1.7 x 10-10 for AgCl

A) 1.97x10-10 M B) 5.0x109 M C) 2.0x10-10 M D) 1.7x10-10 M E) 4.1x10-3 M

13. The Ksp for MgF2 is 8.0 x 10-8. What is the molar solubility of MgF2 in 0.10 M Mg(NO3)2 solution?

A) 4.0 x 10-7 M B) 2.0 x 10-7 M C) 6.3 x 10-5 M D) 4.5 x 10-4 M E) 6.3 x 10-4 M

14. A solution which is 0.50 M in each of Pb2+ , Ag+ and Hg22+ has the Cl- concentration adjusted to

1.0 x 10-9 M. Which of the following chlorides will precipitate? Ksp(PbCl2)= 1.6 x 10-5 , Ksp (AgCl) = 1.7 x 10-10 , Ksp (Hg2Cl2) = 2 x 10-18 A) AgCl and Hg2Cl2 D) AgCl only E) PbCl2 , AgCl and Hg2Cl2 B) Hg2Cl2 only C) PbCl2 only 15. Find the molar solubility of a saturated solution of CaF2 in H2O. the Ksp of the salt is 1.7x10-10.

A) 5.5 x 10-4 M B) 3.5 x 10-4 M C) 4.0 x 10-11 M D) 1.7 x 10-10 M E) 1.3 x 10-5 M 16. The molar solubility of Ag2CrO4 in water is 7.7 x 10-5 M. What is the Ksp for this salt?

A) 1.9 x 10-12 B) 3.5 x 10-15 C) 4.7 x 10-13 D) 9.5 x 10-13 E) 4.2 x 10-11

17. The Ksp of Zn(OH)2 is 4.5 x 10-17. What is the molar solubility of Zn(OH)2 in 0.10 M NaOH?

A) 3.6 x 10-6 M B) 4.5 x 10-15 M C) 2.8 x 10-6 M D) 1.1 x 10-15 M E) 2.2 x 10-6

M 18. What is the molar solubility of BaF2 in water? For BaF2, Ksp= 1.7 x 10-6..

A) 1.3 x 10-3 M B) 1.2 x 10-2 M C) 9.4 x 10-3 M D) 7.5 x 10-3 M E) 3.5 x 10-4 M 19. The value of Ksp for Zn(OH)2 is 4.5 x 10-17. What is the molar solubility of zinc hydroxide in water?

A) 6.7 x 10-9 M B) 4.5 x 10-17 M C) 1.8 x 10-6 M D) 3.6 x 10-6 M E) 2.2 x 10-6

M 20. At a given temperature, the solubility of Ag2CrO4 in water is 8 x 10-5 mol/L. Calculate the value of

Ksp for Ag2CrO4.

A) 5 x 10-13 B) 2 x 10-12 C) 2 x 10-4 D) 8 x 10-5 E) 6 x 10-9 21. What is the molar solubility of Fe(OH)2 in 0.10 M NaOH? Ksp= 2 x 10-15 for Fe(OH)2

A) 4 x 10-8 M B) 4 x 10-7 M C) 5 x 10-14 M D) 2 x 10-17 M E) 2 x 10-13 M 22. What is the concentration of F- ion in a saturated solution of CaF2? Ksp= 1.7 x 10-10 for calcium

fluoride A) 7.0 x 10-4 M B) 3.5 x 10-4 M C) 1.1 x 10-3 M D) 1.3 x 10-5 M E) 5.5 x 10-4 M 23. The Ksp for Pb(OH)2 is 4.0 x 10-15 . What is the molar solubility of Pb(OH)2 in water ?

A) 1.0 x 10-5 M B) 4.0 x 10-5 M C) 1.6 x 10-5 M D) 2.5 x 10-6 M E) 4.0 x 10-7 M

1 A 7 D 13 D 19 E 2 A 8 C 14 D 20 B 3 E 9 B 15 B 21 E 4 C 10 B 16 A 22 B 5 B 11 B 17 B 23 A 6 E 12 C 18 D 24

Solubility Equilibria : Equilibrium involving the dissolving of Ionic Compounds

Solution: homogeneous combination of solute and solvent Solute: the substance being dispersed Solvent: the substance doing the dispersion

Dissociation: the process whereby polar water molecules pulls apart the anions and cations of an ionic compound. ____________________________________________________ ________________________

Solubility: Solubility Terms Soluble: solute dissolves Insoluble: solute does not dissolve Miscible: two liquids mix to form a solution

Immiscible: two liquids that do not mix (heterogeneous combination )

Solubility Curve: Graph showing the changes in solubility with change in temperature Molar Solubility:

Solubility Equilibria solute(s) ↔ cation(aq) + anion(aq) AgCl(s) ↔ Ag1+ + Cl1-

Ion Product/ Ion Product Constant Ksp = [Ag1+ ] [Cl1- ]

Na1+

Cl1- Na1+

Cl1- H

H O

H

H O

H

H O

H

H O

Solubility g/100 g H2O

temperature

supersaturated saturated

unsaturated

Na1+

Cl1-H

H O

H

H O

H

H O

H

H O

Ksp Problems A) Solving For Ksp 1. (ex17.4) Silver bromide , AgBr , is a light sensitive compound in nearly all photographic

film. At 25oC, 1.0 L of water can dissolve 7.1x10-7 mol of AgBr. Calculate the Ksp. molar solubility = n/V = 7.1x10-7mol/1L =7.1x10-7mol/L AgBr Ag1+ Br1- [I] 0 0 [C] molar

solubility 7.1x10-7 7.1x10-7

[E] Ksp=[Ag1+ ] [Br1- ] = (7.1x10-7 ) 2 = 5.0x10-13 2. (ex17.5) The molar solubility of silver chromate, Ag2CrO4 , in pure water is 6.7x10-5 mol/L at

25oC . What is Ksp for silver chromate? Ag2CrO4 2Ag1+ CrO42- [I] 0 0 [C] molar solubility 1.34x10-4 6.7x10-5 [E]

3. (17.6) Common Ion Effect At 25oC, the molar solubility of PbCl2 in 0.10 M NaCl solution is

1.7x10-3 mol/L. Calculate the Ksp. PbCl2 Pb2+ 2Cl1- [I] 0 0.10 [C] molar solubility [E]

Practice Questions p744 5 , 6 , 7 B) Calculating Molar Solubility Given Ksp 1. (17.7) What is the molar solubility of AgCl in pure water at 25oC ?

AgCl Ag1+ Cl1- [I] 0 0 [C] molar solubility [E]

Ksp=[Ag1+ ] [Cl1- ] 2. (17.8) Calculate the molar solubility of lead iodide, PbI2 , in water at 25oC .

PbI2 Pb2+ 2I1- [I] 0 0 [C] molar solubility [E]

Practice Page 745 #8 and 9

Solubility Rules For Ionic Compounds Soluble Compounds 1. All compounds of the alkali metals ( Group 1A ) are soluble 2. All salts containing NH4+ , NO3-, ClO4- , ClO3- and C2H3O2- are soluble . (except AgC2H3O2) 3. All chlorides, bromides , and iodides (Cl- , Br- , I- ) are soluble. except with Ag+ , Pb2+ , Hg22+ ) 4. All sulphates (SO42- ) are soluble except those of Pb2+ , Ca2+ , Sr2+ , Hg22+ , and Ba2+ . Insoluble Compounds 5. All hydroxides ( OH- ) and all metal oxides ( O2- ) are insoluble except those of Group IA and of Ca2+ , Sr2+ ,

Ba2+ Note: When metal oxides do dissolve, they react with water to form hydroxides. The oxide ion O2- ) does

not exist in water. Na2O + H2O ---> 2NaOH(aq) 6. All compounds that contain PO43- , CO32- , SO32- , and S2- are insoluble, Except those of Group 1A and NH4+

. Additional Rule All common hydrogen compounds are soluble

Solubility Rules I = insoluble SS = slightly soluble S = soluble D = decomposes N = not isolated

Anions/

Cation

C2H3O2-

Br- CO32- Cl- CrO4

2

-

OH- I- NO3- PO4

3- SO42- S2-

Al3+ SS S N S N I S S I S D

NH4+ S S S S S S S S S S S

Ba2+ S S I S I S S S S I D

Ca2+ S S I S S SS S S I SS D

Cu2+ S S I S I I D S I S I

Fe2+ S S I S N I S S I S I

Fe3+ S S N S I I N S I SS D

Pb2+ S SS I SS I I SS S I I I

Mg2+ S S I S S I S S I S D

Hg+ SS I I I SS N I S I SS I

Hg2+ S S I S SS I I S I D I

K+ S S S S S S S S S S S

Ag+ SS I I I SS N I S I SS I

Na+ S S S S S S S S S S S

Zn2+ S I I S S I S S I S I

Dissolving: Energy Concerns Energy must be added to overcome the attractive forces holding the ions together in a crystal ( solute ). The magnitude of this Lattice Energy is reflected in properties such as melting point , boiling point and heat of formation. Energy needs to be added to separate the particles of the liquid ( solvent ) from one another. The liquids are either polar covalent molecular ( dipole-dipole attraction ) or non-polar covalent molecular (London Dispersion Forces ) Energy is released as the ions and the solvent molecules interact ( bond ) . This �Solvation Energy� is greater for smaller more highly charged ions and more polar solvents. Consider the following Data

Substance M.P. oC B.P. oC ∆H formation Heat of Solution Solubility g/100 g H2O

NaF 980 1700 -136.3 -0.48 (12 oC) 4.22 (18 oC ) NaCl 801 1413 -98.4 -1.28 (18 oC) 35.7 (0 oC ) NaBr 755 1390 -86.3 -0.19 (18 oC) 79.5 NaI 651 1300 -69.5 +1.41 (18 oC)

Questions 1. From considerations of ionic radii and Coulomb�s Law, which would you expect to have

the greater Lattice Energy, NaF or NaCl? Is your answer supported by the values for their melting points?

2. Given the melting point of CaO as 2580 oC, arrange the compounds CaO , SrO , MgO and

BaO in the expected order of decreasing melting point. 3. Would you expect NaCl or NaF to have the greater Solvation energy when interacting

with water? Explain your answer.

4. For the equation: ∆Hsolution = Lattice energy

(energy put in to pull ions apart)

+ Energy put in to separate solvent molecules

+Solvation Energy (energy released when ions and solvent molecules interact

Consider your answers to questions 1 and 3 for NaCl and NaF. Which factor appears to predominate in explaining their relative heats of solution?

5. A) Define enthalpy and entropy B) Considering the tendencies toward minimum energy and maximum randomness,

explain the fact that, NaCl is more soluble than NaF.

6. What would be the effect on the solubility of NaCl if CCl4 were used as the solvent rather than water? Explain your answer.

7. Predict a value for the solubility of NaI in water.

Entropy/Enthalpy Questions 1. Label the following equations to show direction of minimum energy and maximum

randomness. Draw a potential energy diagram for each reaction. Equation Cd(s) + ½ O2(g) ↔ CdO(s) + 265 kJ H2(g) + ½ O2(g) ↔ H2O(g) + 241.8 kJ N2O4(g) ↔2NO2(g) ∆H = 58.9 kJ

2. For the following equilibrium: 2C4H8(g) ↔ C8H16(g) + heat A) Which side of the reaction is more random ? B) Write the equilibrium constant expression for this reaction.

C) Explain how �Kc� will affected by the driving forces: Entropy , Enthalpy 3. A) For the reaction: CuSO4(s) + heat ↔ Cu2+(aq) + SO42-(aq) If this solution is a good conductor of electricity, which is the main driving force?

B) PbI2(s) + 58.5 kJ ↔ Pb2+(aq) + 2I1-(aq) Since lead(II) iodide is only slightly soluble which is the main driving force ? C) NaOH(s) ↔ Na1+ + OH1-(aq) + 43.1 kJ Explain why this solute would be highly soluble.

4. For the reactions: CO2(g) ↔ CO2(aq) + heat solubility 0.0087 mol/L SO2(g) ↔ SO2(aq) + heat solubility 0.36 mol/L If randomness tendencies are equal, which of the above reactions would have the larger

∆H ? Explain why both solutes would be less soluble at higher temperatures. 5. Explain why ammonium nitrate easily dissolves and the temperature of the resulting

solution drops. 6. For the following reaction indicate the direction of minimum energy and maximum

randomness. Y(s) + 2W(g) ↔ 2Z(s) ∆H = 836 kJ

Solubility Problems 1. A solution in equilibrium with a precipitate of AgCl was found to contain 3.4x10-6 kmol of Ag+ per cubic metre and 5.0x10-5 kmol of

Cl- per cubic metre. Calculate the Ksp for AgCl.. (1.7x10-10 ) 2. What is the equilibrium [Cd2+ ] ions in a saturated aqueous solution of CdS made by shaking CdS(s) with water? Ksp = 6.0x10-27 .

(7.7x10-14 ) 3. How many grams of PbSO4 will dissolve in 5.0 L of water? Ksp =1.3x10-8 ( m = 0.17 g ) 4. Ksp for ferrous oxalate, FeC2O4 , is 2.1x10-7 at 25oC. What is the concentration of ferrous ion in a saturated solution of ferrous oxalate?

( 0.00046 mol/L) 5. Ksp for barium chromate, BaCrO4 , is 1.6x10-10 at 18oC. How many grams of BaCrO4 are present in 10 L of saturated BaCrO4 solution?

( 0.032 g ) 6. An experiment showed that a maximum of 1.49 g of AgBrO3 can dissolve in 1.0 L of water at 20oC. Calculate the Ksp.

( 4.0x10-5 ) 7. Ksp for Ni(OH)2 is 1.6x10-16 . Calculate the solubility of Ni(OH)2 in g/L . (0.00032 g/L) 8. A solution in equilibrium with a precipitate of Ag2S was found to contain 1.6x10-16 mol/L of S2- and 2.5x10-18 mol/L of Ag+. Calculate

the solubility product for Ag2S. ( 1.0x10-51 ) 9. A solution in equilibrium with a precipitate of Pb3(PO4)2 was found to contain 2.9x10-3 M PO43- and 1.2x10-9 M Pb2+ . Calculate the

solubility product for Pb3(PO4)2 . ( 1.5x10-32 ) 10. A speck of chalk ( CaCO3 ) dust with a mass of 0.0010 g just barely dissolved in 100 mL of water. Calculate the Ksp for CaCO3 .

( 1.0x10-8 ) 11. The molar concentration of Co2+ in solution in equilibrium with CoS(s) was found to be three times as great as the concentration of S2-.

Calculate the [Co2+ ] .Ksp= 3.0x10-26 for CoS ( 3.0x10-13 M ) Precipitation 1. Ksp= 5.0x10-22 for CoS. Will a precipitate form when 1.0 L of 3.0x 10-10 mol/L Co2+ is added to1.0 L of 2.0x10-11 mol/L S2- .

( Y ) 2. Ksp (FeS) = 4.0x10-17.. Will a precipitate form when 1.0 L of 5.0x10-9 M Fe2+ is added to 1.0 L of

4.0x10-9 M S2- ? (N ) 3. What [S2- ] must be present to just start precipitation of CuS from 0.20 M solution of CuCl2 ?

Ksp= 4.0x10-36 for CuS ( 2.0x10-35 M ) 4. Ksp = 1.06x10-8 for PbSO4 at 18oC . Predict whether a precipitate will form when 10.0 mL of 1.0x10-3 mol/L Pb(C2H3O2)2 is added to

40mL of 1.5x10-4 mol/L Na2SO4 . ( Y ) 5. Predict whether a precipitate forms when 20 mL of 1.0x10-4 M Zn(NO3)2 solution is added to 60 mL of 2.4x10-5 M Ca(OH)2 solution.

Ksp = 1.8x10-14 for Zn(OH)2 ` ( Y ) 6. When 100 mL of 2.5x10-5 M ferrous chloride is added to 150 mL of 6.7x10-5 M NaOH , a precipitate of Fe(OH)2 just starts to form. What

is the Ksp for Fe(OH)2 ? ( 1.6x10-14 ) 7. Ksp (Ag2CO3 ) = 6.2x10-12 . Assuming that a precipitate can be detected as soon as it begins to form, what is the minimum [CO32- ] that

can be detected in a solution having [Ag+] = 0.050 mol/L . Express your answer in grams per litre . (1.5x10-7 g/L)

8. Ksp (AgBr) = 1.0x10-13, and Ksp (Ag2CrO4 )= 1.0x10-12 . a solution is 0.10 mol/L in Br- and 0.010 mol/L in CrO42- . If Ag+ ions are slowly

added: A) What precipitate forms first ? ( AgBr ) B) What is the concentration of the negative ion involved in the first precipitate that forms when the second negative ions start to form

a precipitate ? ( [Br- ] = 1.0x10-8 M ) 9. The Ksp of SrCO3 is 1.6x10-9 . To a solution containing 0.175 g of Sr2+ per litre was added solid Na2CO3 until the [CO32- } was 2.50x10-6

M. What mass of SrCO3 was precipitated? (8.1x10-2 g/L) Common Ion Effect 1. Ksp(AgBr)=1.0x10-13 . How many moles of AgBr will dissolve in 5.0 L of 0.10 M KBr solution?

(5.0x10-12 mol) 2. Ksp (AgCl)=1.6x10-10 . How many moles of AgCl will dissolve in 5.0 L of a 0.10 M NaCl solution?

(8.0x10-9 mol) 3. Ksp=1.1x10-8 for PbSO4. If 5.0x10-6 mol of PbSO4 can dissolve in 1.0 L of K2SO4 solution, what is the molar concentration of the K2SO4

solution? (2.2x10-3 mol/L) 4. Powdered AgOH is slowly added to 2.0 L of a 0.020 M NaOH solution until no more AgOH will dissolve. If Ksp = 1.4x10-8 for AgOH: A) What is [Ag+] in the saturated solution? ([Ag+] = 7.0x10-7 mol/L) B) What mass of AgOH had to be added to the solution to produce this [Ag+]. (1.75x10-4 g) 5. A solution contains 0.020 mol of (NH4)2S. How many grams of MnS must be added to 1.0 L of the solution to just start forming a

precipitate of MnS? Ksp=1.4x10-15 for MnS. (6.1x10-12g) 6. What is the solubility (g/L) of Mg(OH)2 in 0.050 M MgSO4. Ksp=1.2x10-12 for Mg(OH)2? (1.43x10-4 g)

Rules for assigning oxidation states 1 Any element in the free state has an oxidation state of zero

ex. Ago silver atom , O2o oxygen molecule

2 For a monoatomic ionic species, the oxidation state is equal to the net charge of the species. Ex. O-2 , F-1 , Na+1 , Mg+2

3 the sum of all oxidation states of a polyatomic chemical species is equal to the net charge of the species ex. PbSO4 ! Pb+2 + S+6 +4O-2 = +2 + 6 - 8 = 0 net charge on compound ClO3

-1 ! Cl+5 + 3O-2 ! +5 - 6 = -1 4 The following are rules that apply to specific atoms in their chemical compounds. If a

conflict appears, that which is listed first has preference and the other oxidation number would be determined from the first

Elements Oxidation

Number Halogens: F , Cl , Br , I -1

Group I : Li , Na , K , Rb , Cs +1 Group II : Be , Mg , Ca , Sr , Ba +2

H +1 O -2

Assign Oxidation Numbers to every element in each of the following formulae KMnO4

CaSO4

Fe2O3

K2CrO4

KClO3

NaClO2

HClO

NO2

-1

NH4

+ K2Cr2O7

MnCl4

CrO2

Na2O2

ClO4

-1

Balancing Equations by the Ion-Electron Method ( Redox Method ) Rules

1. Assign oxidation states to all elements 2. Identify the species oxidized and the species reduced 3. Write partial ( half ) reactions for LEO and GER ( balance for atoms ) 4. Balance the half reactions for atoms and charge 5. Adjust ( balance) �spectator� species coefficients if needed

Example: #1 Equation : FeCl3 + SnCl2 ! SnCl4 + FeCl2

identify oxidation (LEO) and reduction (GER) balance electrons LEO

GER balance equation for redox FeCl3 + SnCl2 ! SnCl4 + FeCl2

Example#2 Equation : Cu + HNO3 ! NO2 + Cu(NO3)2 + H2O Example#3 Equation : HCl + KMnO4 ! KCl + Cl2 + MnCl2 + H2O Assignment: :Balance the following redox reactions by the ion electron method

1 S + O2 !. SO2 2 CuO + C ! Cu + CO2 3 Zn + H2SO4 ! ZnSO4 + H2 4 Fe + CuSO4 ! FeSO4 + Cu 5 HBr + H2SO4 ! SO2 + Br2 + H2O 6 C + HNO3 ! NO2 + H2O + CO2 7 MnO4

- + C2O42- + H+ ! CO2 + Mn2+ + H2O

8 H2O2 + PbS ! PbSO4 + H2O 9 Al + FeO.Fe2O3 ! Fe + Al2O3 10 H2SO3 + KMnO4 ! K2SO4 + MnSO4 + H2SO4 + H2O 11 H2S + HNO3 ! S + NO + H2O 12 P + HNO3 + H2O ! H3PO4 + NO 13 Al2O3 + C ! Al4C3 + CO 14 As + NaOH ! Na3AsO3 + H2 15 KI + HNO3 ! KNO3 + H2O + NO + I2 16 P + KOH + H2O ! KH2PO2 + PH3 17 Cu + HNO3 ! Cu(NO3)2 + NO + H2O 18 K2Cr2O7 + HI + HCl ! KCl + CrCl3 + I2 + H2O 19 KMnO4 + FeSO4 + H2SO4 ! K2SO4 + MnSO4 + Fe2(SO4) 3 +H2O 20 Cu + HNO3 ! Cu(NO3)2 + H2O + NO2

Redox: Extra Goodies

1. KMnO SnCl HCl KCl MnCl SnCl H O4 2 2 4 2+ + → + + + A) What substance is the reducing agent ? B) What species is reduced ? C) Balance the equation by the ion-electron method (writing half-reactions). 2. Zn HNO Zn(NO ) NH NO H O3 3 2 4 3 2+ → + + A) What is the reducing agent ? B) What is the oxidizing agent ? C) Use half-reactions ( ion -electron method ) to balance this equation. 3. Use REDOX ( Ion -electron Method ) to balance the following equations:

A) Cu +HNO3 Cu(NO3 )2 NO2 H2O

B) Al+H+ Al 3 H2C) HNO3 H2S NO S H2O

D) PbS+O2 PbO SO2E) K2CrO4 H2S HCl KCl CrCl3 S H2O

F) FeSO4 H2SO4 HNO3 Fe2(SO4 )3 NO H2O

G) K2Cr2O7 SnCl2 HCl KCl CrCl3 SnCl4 H2O

H) CrO2- Na2O2 H2O CrO4

2 Na OH

I) SO32 - MnO4 SO4

2 MnSO4 O2

→ + +

→ + +

+ → + +

→ +

+ + → + + +

+ + → + +

+ + → + + +

+ + → − + + + −

+ − → − + + −

4. A) From the following reactions, which is the strongest reducing agent ?

HCl H2SO4 N.R.

2HBr H2SO4 SO2 2H2O Br28HI H2SO4 H2S 4H2O 4I2

+ →

+ → + +

+ → + +

B) Compare the strength of Cl2,Br2, I2 as reducing agents. C) Can you rationalize your answer for part(B) from atomic structure. 5. For all of the following reactions determine: the oxidized and reduced elements , the oxidizing and

reducing agents. A) Smelting of Ore I) Copper Ore: CuO(s) + C(s) ! Cu(s) + CO2(g) ii) Iron ore: FeS2(s) + C(s) ! Fe(s) + CS2(g) B) Removing of Tarnish from Silver (Write the half reactions). 2Al(s) + 3Ag2S(s) + 6H2O(l) ! 6Ag(s) + 2Al(OH)3(aq) + 3H2S(aq) C) Lead-Acid Storage Battery Pb(s) + PbO2(s) + 2H2SO4(aq) <--> 2PbSO4(s) + 2H2O(l) + energy D) The Catalytic Converter: ( complete brackets with �oxidized� or �reduced� ) 2CO(g) + O2(g) -Pt/Pd-> 2CO2(g) ( ) NO(g) -Pt/Ni-> N2(g) + O2(g) + energy ( ) E) Iron Corrosion: Fe(s) + O2(g) +2H3O+(aq) (10-7 M) ! Fe(OH)2 + 2H2O(l) F) Statue of Liberty: 2Fe(s) + 3Cu+2 ! 2Fe+3 + 3Cu G) Three Mile Island Disaster (1979 ) Zr(s) + 2H2O(l) ! ZrO2(aq) + 2H2(g)

H) Funny Tasting Coke: 2Al(s) + 6H+(aq) ! 2Al+3(aq) + 2H2(g) REDOX - Activity Series For Metals

Least Active Most Active Au,Hg,Ag,Cu,H,Pb,Sn,Co,Cd,Fe,Cr,Zn,Mn,Al,Na,Ca,Sr,Ba,K,Rb,Cs

increased ease of oxidation LEO ! ← increased ease of reduction GER

reducing agents oxidizing agents

Reaction Types 1. Active metals react with acids to displace hydrogen gas 2. A more active metal ( substance oxidized ) will displace an ion of a

metal ( substance reduced ) with a lower activity 3. Very active metals will displace hydrogen from water Identify the REDOX half reactions in the following equations

Zn(s) HCl(aq) ZnCl2(aq) H2(g)

Zn(s) CuSO4(aq) ZnSO4(aq) Cu(s)

Na(s) H2O(l) NaOH(aq) H2(g)

Fe(s) Pb(NO3)2(aq) Fe(NO3 )2(aq) Pb(s)

+ → +

+ → +

+ → +

+ → +

Which of the following pairs of reactants will actually react?

Fe s CuSO aq

Fe s Al SO aq

Cu s H SO aq

Al s HNO aq

( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )

+ →

+ →

+ →

+ →

4

2 4 3

2 4

3

Ion-electron Method For Acidic Solutions

Cr O Fe Cr Fe2 72 2 3 3− + + ++ → +

Rules Write the two half reactions for REDOX Balance all atoms other than H and O Balance O by adding H2O to the side needing O

Balance the H by adding H+ to the side needing H atoms Balance charges on each side of equation by adding electrons Balance electron loss equal to electron gain Add the two reactions and cancel spectators

Ion-electron Method for Basic Solutions

SO MnO SO MnO32

4 42

2− − −+ → +

Method: First complete the ACIDIC Method 1. Divide equation into two half-reactions 2. Balance atoms other than H and O 3. Balance O by adding H2O

4. Balance H by adding H+ 5. Balance net charge by adding e- 6. Balance LEO=GER and add half-reactions 7. Cancel spectators 3SO 2MnO 2H 3SO 2MnO H O3

24 4

22

− − + −+ + → + + 2 Conversion to BASIC equation depends on the 1:1 ration of H+:OH- in water 8. Add the same number of OH- as there are H+ to both sides of the equation 9. Combine OH- and H+ to form H2O 10. Cancel any H2O that are spectators

Electrochemistry Problems

1. For the silver-copper galvanic cell: 2Ag Cu Cu 2Ag E v(aq) (s) (aq)

2(s) cell

o+ ++ → + =0 46.

If ECuo = 0 34. v, what is the value for the reduction potential of Ag+ ?

2. A galvanic cell was constructed using electrodes made of lead and lead dioxide (

PbO2), with sulphuric acid as the electrolyte. What is the cell reaction and what is the standard potential for the cell?

PbO 4H SO 2e PbSO H O E 1.69v2(s) (aq) 4(aq)2

4(aq) 2 PbOo

2+ + + ↔ + =+ − −

PbSO 2e Pb SO E 0.36v4(s) (s) 4(aq)2

PbSOo

4+ ↔ + = −− −

3. What would be the cell reaction and the standard cell potential of a galvanic cell

employing the following half-reactions?

Al 3e Al E 1.66v(aq)3

(s)o+ −+ → = −

Cu 2e Cu E v(aq )2

(s)o+ −+ → = 0 34.

Which half cell is the anode? 4. What spontaneous reaction will occur if Cl2 and Br2 are added to a solution

containing Cl- and Br- ? 5. Referring to table 17.1, predict the reaction that will occur when Ni and Fe are

added to a solution that contains both Ni+2 and Fe+2. 6. Determine whether the following reactions are spontaneous as written. If they are

not, give the reaction that is spontaneous.

A) Cu 2H Cu H(s) (aq) (aq)2

2(g)+ ↔ ++ +

B) 3Cu NO 8H 3Cu 2NO 4H O(s) 3(aq) (aq ) (aq )2

(g) 2 (l)+ + ↔ + +− + +

Electrochemical Cells 1. General Purpose Dry Cell

metal cap ( + )

graphite rod (cathode)

manganese dioxide and powdered carbon

steel disc (- )

paste ( MnO , NH Cl , ZnCl )2 4 2

zinc container ( anode )

insulator

Anode ( Oxidation ) Zn Zn 2e(s) (aq)

2→ ++ −

Cathode (Reduction)

2MnO 2NH 2eMn O H O 2NH

2(s) 4(aq )

2 3(s) 2 (l ) 3(aq)

+ + →

+ +

+ −

2. Lead-Acid Battery ( series arrangement of cells )

Anode: Pb SO PbSO 2e

Cathode: PbO 4H SO 2e PbSO 2H O(s) 4(aq)

24(s)

2(s) (aq) 4(aq)2

4(s) 2 (l)

+ → +

+ + + → +

− −

+ − −

Overall Reaction: 3. Hydrogen-Oxygen Fuel Cell

Anode(oxidation): 2H 4OH 4H O 4e

Cathode(reduction): O 2H O 4e 4OHRedoxReaction

2 2

2 2

+ → +

+ + →

− −

− −

Electrolytic Reactions 1. Downs Cell- electrolysis of molten sodium chloride

graphite anode +

steel cathodeencircles anode

moltensodium chloride

+calcium chloride

chlorine gas

sodium chloride

liquid sodium

Anode(oxidation): 2Cl Cl 2e

Cathode(reduction): Na e Na

-2

+

→ +

+ →

−

−

2. Diaphragm Cell or Mercury Cell

Overall Reaction : 2NaCl 2H O 2NaOH H Cl

Anode (oxidation): 2Cl

Cathode (reduction): 2H O 2e

(aq ) 2 (l) (aq ) 2(g) 2(aq)

(aq )-

2 ( l )

+ → + +

→

+ →−

3. Electroplating

Ag

Ag(CN)2

D.C. Powercathode anode

- +

Anode (oxidation): Ag 2CN Ag(CN) e

Cathode (reduction): Ag(CN) _ e Ag 2CN(s) (aq) 2(aq)

2(aq)-

(s) (aq )

+ → +

+ → +

− − −

− −

Corrosion of Iron ( Rusting)

cathode cathode

anodeelectronselectrons

water droplet

iron cationsrust rust

Anode (oxidation): Fe Fe 2e

Cathode (reduction): O H O 2e 2OH(s) (aq)

2

12 2(g) 2 ( l) (aq )

→ +

+ + →

+ −

− −

Ion Migration (collision) Fe OH Fe(OH) Fe O rust(aq )

2(aq ) 2(s)

oxidized2 3(s)

+ −+ → → Cathotic Protection: Sacrificial Corrosion

.Mg Mg 2e E 2.36V

Fe Fe 2e E 0.44V

2 o

2 o

→ + =

→ + =

+ −

+ −

Cell Comparison Electrolytic Electrochemical

D.C.

power

cathode anode-ve +ve

electrolyte

electrodes

e

AB

+-

V

A B

A+ B-

cathode anodee

+ -salt bridge

KNO3

V

V

Electrochemical Cell Current Depends On

Voltage Depends On

Electrolysis Reactions #1 Electrolysis of H2SO4(aq)

Electrolyte: H2SO4(aq) Ions Present: Ions migrating to anode: Ions migrating to cathode: Cathode observation: Anode observation:

Electrode Half Cell Reaction equations Electron Balance LEO=GER Anode LEO

Cathode GER

Overall Redox Reaction

Powecathodeanode

Electrolysis Reactions #2 Electrolysis of NaCl(aq)

Electrolyte: NaCl(aq) Ions Present: Ions migrating to anode: Ions migrating to cathode: Cathode observation: Anode observation:

Electrode Half Cell Reaction equations Electron Balance LEO=GER Anode LEO

Cathode GER

Overall Redox Reaction

Powecathodeanode

Electrolysis Reactions #3 Electrolysis of HCl(aq)

Electrolyte: HCl (aq) Ions Present: Ions migrating to anode: Ions migrating to cathode: Cathode observation: Anode observation:

Electrode Half Cell Reaction equations Electron Balance LEO=GER Anode LEO

Cathode GER

Overall Redox Reaction

Powecathodeanode

Electrolysis Reactions #4 Electrolysis of CuSO4(aq)

Electrolyte: CuSO4(aq) Ions Present: Ions migrating to anode: Ions migrating to cathode: Cathode observation: Anode observation:

Electrode Half Cell Reaction equations Electron Balance LEO=GER Anode LEO

Cathode GER

Overall Redox Reaction

Powecathodeanode

Organic Chemistry Organic chemistry is the study of carbon compounds, their properties and their interactions. ( except CO2 , CO , CO32-, CN-) A) Sources: i) living organisms - plants , animals, etc ii) non-living material ( coal, oil , wood, natural gas) B) Classification of Organic Compounds

Organic Compounds

Hydrocarbons

Aliphatic Aromatic

CarbohydratesAlcohols

Carboxylicacids Aldehydes

Ketones

AlkanesAlkenes

Alkynes C)Bonding: covalent bonding with carbon as the main player

C 1s 2s 2p 2p 2p2 1 11 1

Carbon Hybrid

(sp3 hybrid)

Hydrocarbons- Compounds composed of carbon and hydrogen atoms 1) Alkanes: hydrocarbons with the general formula CnH2n+2

-saturated hydrocarbon ( all single covalent bonds- paraffins ) ending in "ane" Homologous Series of Alkanes

Name Molecular Formula

Properties ( MP, BP, State ) condensed structural formula # isomers

methane CH4 MP = -182.48oC BP= -164oC CH4 1

ethane C2H6 MP = -183.3oC BP= -88.63oC CH3CH3 1

propane MP = -189.69oC BP= -42.07oC CH3CH2CH3 1

butane MP = -138.35oC BP= -0.5oC CH3(CH2)2CH3 2

pentane MP = -129.72oC BP= 36.07oC CH3(CH2)3CH3 3

hexane MP = -95oC BP= 68.95oC CH3(CH2)4CH3 5

heptane MP = -90.61oC BP=98.42oC 9

octane MP = -56.79oC BP= 125.66oC 18

nonane MP = -51oC BP= 150.79oC 35

decane

undecane

dodecane

tridecane The trend in boiling point is due to _________________________________________ Isomers: different structure forms of the same molecule and therefore different properties Examples of isomers for butane: (2)

n-butane 2-methylpropane

Rules for naming alkanes; 1. Select the longest carbon chain. ( parent chain ) 2. Identify the attached ( branched ) alkyl groups: CH3- methyl ,CH3CH2- ethyl, CH3CH2CH2- propyl Identify halogens: fluoro , chloro , bromo , iodo 3. Number the carbon chain form the end that yields the lowest total of the numerical coefficients (

check numbering from both directions ) 4. Name the attached groups in alphabetical order: chloro, ethyl, methyl . Use the prefix (di, tri, tetra )

to indicate the total number of groups ) 5. Number the position on the carbon chain that the branched group is attached 6. Commas are used to separate location of groups and hyphens are used to separate numbers from

words. ( nothing between names )

Drawing and Naming Isomers A) Isomers for Pentane (3) B) Isomers of Hexane ( 5) C) Isomers of Heptene ( 9 )

H H

H H H H H

C C C C

H H H H

HHH

HH

H

HH

H

H

C

CCC

Practice Naming Alkanes Write the name for each of the following

CH3CH3

Draw the skeletal formula for each of the following ( and write the molecular formula )

3-ethyl-3-methylhexane n-tridecane 5-(2-methylpropyl)decane

3-ethylheptane 2,2,3-trimethyloctane 5-butyl-3-ethyl-6-methylundecane

3-chlorododecane 1-iodo-3-propyloctane 3-ethyl-2-methyl-4-propyldecane

4-propylnonane 2,3,4-trimethylpentane

C C C

C C C C C C C

C

I

C CCCC

C CCCC

C

I

C C C C C C C

C C

CC CC C C C C

C

C

C

C

C C

C

C

C

C

CC

C

C

C

CCC

C C C C

C C C C

C C C C

C

C

Br

Br

C CC

C C C

C C

Reactions of Alkanes (write balanced equations for each reaction ) 1. Combustion ( react with oxygen ) A) Complete combustion - produce water and carbon dioxide C3H8 + O2 !

C6H14 + O2 !

B) Incomplete combustion - produce carbon , carbon monoxide, CO2 , H2O

C2H6 + O2 ! C + CO2 + H2O

C2H6 + O2 ! CO + CO2 + H2O

2. Substitution ( replacing hydrogen with another atom ) this is difficult because the C-H bond is very strong and therefore requires large amounts of energy

to break a hydrogen free. CH3CH2CH3 + HBr → CH3CHBrCH3 + H2

3. Alkanes are not reactive with acids or bases or chemicals that steal electrons. ( oxidizing agents )

Alkenes (unsaturated hydrocarbons) -the presence of C=C (double) bonds results in less hydrogens bonded to the carbons -functional group C=C general formula CnH2n

-this group tends to lower the FOA for other molecules Name Skeletal formula, Structural

and Molecular Formula Properties ( MP, BP, State )

ethene C=C

H

H

H

H C2H4

MP= -169.15oC BP = -103.71oC

propene ( 1-propene )

MP = -185.25oC BP = -47.4oC

1-butene 2-butene (cis) 2-butene (trans)

MP = -185.35oC BP = -6.3oC

MP = -139.91oC BP = 3.7oC

MP = -105.55oC BP = 0.88oC

C5H10 MP = BP = 121.5oC

C8H16 MP = -101.73oC BP = 121.3oC

Naming Alkenes: 1.Select the longest C-chain containing the double bond. 2.Number the spaces between the carbon atoms so that the double bond is at the lowest possible number. 3. Name the branches as with alkanes. State the numbered space where the double bond appears in front of the

alkene name. Two double Bonds The location of both double bonds must be indicated and the suffix is changed to "diene ". 1,2-butadiene 1,3-butadiene 1,3,5,7-octatetrene ( Multiple Double Bonds )

Draw the skeletal ( structural ) formulae for each of the following:

1-pentene 2-hexene 3-methyl-2-pentene

3-ethyl-5-methyl-2,4-hexadiene 2,5-dimethyl-3-nonene 6-chloro-3-ethyl-1-heptene

Name the following :

C C C C C C C C C C C C C C C C

C C C C C C C C C C C C C C C C C

C

C C C C C

C

C

C C C C C

C

C C

C

C C

Draw and name the 5 isomers of C5H10 which are alkenes:

Alkenes: Some Uses and Types of Reaction Uses: Ethene is used as a bleach, an anesthetic, a fruit colouring agent ( causes ripening of fruit in sealed box) , production of antifreeze, ethanol (alcohol) , ether , medicines and plastics Polythene ( polyethylene ) is joined units of ethene that make garbage bags and zip lock bags

Alkene Reactions 1. Addition Type:

a. Additions of Compounds will follow Markownikoff�s Rule ( The positive part of the reagent that attacks the alkene across the double bond will bond onto the �C� at the double bond that has the most �H� atoms already bonded.

Ex : CH3CH2CH=CH2 + HBr ! CH2CH2 + HCl !

propene + hydrogen chloride !

b. Halogenations: reaction of halogen across the double bond Ex. CH3CH=CH2 + Br2 !

2-methyl-2-butene + bromine ! Since Br2 becomes decolourized by this reaction it can be used as a test for saturation. Demo: cyclohexane + Br2 ! cyclohexene + Br2 !

c. Hydrogenation: reaction of hydrogen across the double bond Ex. CH2CH2 + H2 (Pt catalyst) !

2-methyl-2-butene + hydrogen Pt →

This reaction is used for converting unsaturated hydrocarbons into saturated hydrocarbons

unsaturated (liquids or oils ) + hydrogen saturated (solids or fats)→ Liver converts saturated into cholesterol. Unsaturated reverse cholesterol build up d. Oxidations : [O] -atomic uncombined oxygen comes from KMnO4 , K2CrO4 , K2Cr2O7 and other

solutions. The simplified reaction is:

3-methyl-1-butene + [O] ! + H2O ! 1-butene + [O] ! + H2O !

e. Hydrolysis: a reaction that involves adding water across the double bond

4-methyl-2-pentene + H2O !

C C [O] H2O

Unstable epoxy 1,2-ethonediol(ethylene glycol)

C C C CH

H H

H H

H

H H

H

H

H

H

O

OH OH

C C H - OH H2SO4

catalyst

C C

H OH+ ethanol

Alkyne Verification Experiment Purpose: to produce an alkyne and to observe both its chemical and its physical properties Apparatus: test tubes, stoppers, pneumatic trough, splints, Bunsen burner Method: 1. Fill the pneumatic trough with water to the overflow line. Submerge 4 test tubes in the water. 2. Add a piece of calcium carbide (CaC2) to the water and collect the ethyne (acetylene ) gas by the

downward displacement of water. Collect 3 full test tubes of the ethyne and one test tube that is 1/12 full of ethyne. Observe

3. To one of the full test tubes , add a small amount of bromine water, shake and observe. 4. To the second full test tube add KMnO4 (aq) , shake and observe.

5. Conduct an ignition test on the two remaining test tubes. Observe Observations:

Reactant description: CaC2

H2O

Product Description C2H2

Ca(OH)2

Reaction Descriptions: A) Bromine water: B) Potassium permanganate solution C) Ignition test full: 1/12 full:

Conclusions: 1. Write a balanced equation for the production of ethyne. 2. Write formula equations (4) for each of the ethyne reactions. Explain the purpose of each of the

reactions. 3. Explain the intramolecular and the intermolecular bonding that occurs in ethyne. 4. Draw and name the isomers of C5H8 ( some may not be alkynes)

5. Construct a model of an acetylene lamp that could have been used as a lantern for early automobiles. Explain how your model works to make the light and direct the light down the road.

6. Write the formula equations for each of the following: A) hydrogenation of ethyne, B) hydrolysis of ethyne , C) chlorination of propyne, D) bromination of 2-pentyne.

7. Explain the method you would use to identify the contents of three vials, containing propane, propene, and propyne.

8. Calculate the volume of ethyne produced at 27.0oC and 95.0kPa, when 20.0g of calcium carbide reacts with excess water. Calculate the volume of air (21% O2) required for the complete combustion of the

ethyne.

Reactions of Alcohols Type 1 Oxidation (gentle) A strong oxidizer (source of [O] ) is necessary for the reaction with different types of alcohols. A) Oxidation of a Primary alcohol ( 1-butanol ) Write the structural equation for 1-butanol with [O] from KMnO4(aq) + [O] ! KMnO4(aq) Note: 1-butanol (an alcohol ) was oxidized to 1-butanal ( an aldehyde. -atomic oxygen [O] takes the H from the OH group and a second H from the carbon that holds the

OH group B) Oxidation of a Secondary Alcohol ( 2-butanol , sec. butyl alcohol ) Write the structural equation for the reaction with KMnO4 2-butanol was oxidized to ________________ ( ______________________) C) Oxidation of a Tertiary Alcohol 2-methyl-2-propanol Write the structural equation for the reaction with KMnO4 Tertiary alcohols do_______react because _____________________________ Type 2 Strong Oxidations An aldehyde made by gentle oxidation is reacted with another [O] ethanal + [O] ---> ethanoic acid Summary: Primary alcohols + [O] ! ____________ + _____________ Secondary alcohols + [O] !____________ + ______________ Tertiary alcohols + [O] ! Aldehyde + [O] !

Alcohol Worksheet 1. Draw the following and indicate whether they are primary, secondary, or tertiary. Alcohols are named by numbering the longest chain containing the OH group from the end that will

give the position of the alcohol group the lowest possible number. An "ol" ending indicates an alcohol.

3-methyl-2-pentanol 2-methyl-2-butanol

2,5,5-trimethyl-1-hexanol 4-methyl-2-pentanol

2. A) Why is methanol called "wood alcohol" and ethanol called "grain alcohol"? B) List the physical properties of methanol and ethanol. formula, MP , BP, density, colour, odour, state 3. Explain using bonding concepts which should have the lower boiling point: A) 1-butanol or 1-chlorobutane B) 1-butanol or 2-butanol 4. Write structural equations for the following: [O] source KMnO4(aq) A) ethanol + [O] ! B) 2-propanol + [O] ! C) 1-propanol + [O] ! D) 3-pentanol + [O] ! E) 2-methyl-2-propanol + [O] ---> F) ____________ + [O] ! pentanal + water G) ___________ + [O] ! butanone + water H) ethanol + [O] acidified K2CrO4 ! ________ + H2O Cr3+(aq) yellow colour pale green colour Explain how this reaction could be used in a breathalyzer.

Aldehydes and Ketones Substance Functional Group General Formula

Aldehyde

Ketone 1. Draw the structural formulae for each of the following

2-methyl-3-pentanone 2-methyl-3-ethylhexanal 4,5-dimethyl-2-hexanone 4-methyl-2-bromopentanal

2. Write the structural equations for the preparation of: A) 2,3-dimethylbutanal B) 5,5,8-trimethyl-4-nonone C) Ethanal, starting with calcium carbide. Use 4 separate equations to achieve this final substance) 3. Complete this oxidation reaction equation: Copper oxide + 2-propanol ! + + 4. Draw 4 isomers of C4H8O ( aldehydes and ketones )

Carboxylic Acids substances made by the oxidation of an ____________________

Functional Group General Formula carboxylic acid

Name Molecular formula Structural Formula methanoic acid ethanoic acid (acetic acid )

propanoic acid

Esters Compounds derived from a reaction between a carboxylic acid and an alcohol.

Functional Group General Formula Ester linkage (key) Production: alcohol + carboxylic acid ! ester + water

Complete the following as structural equations: 1. propanol + ethanoic acid ! 2. propanoic acid + ethanol ! List (A) Characteristics of esters. (B) Uses of Ester ( include specific flavours )

Amines, Amino acids, Amides ( peptides ) & Polyamides Amines: 1. Production CH3CH2I + 2NH3 ! CH3CH2NH2 + NH4I ethyl iodide + ammonia ! ethylamine + ammonium iodide 2. General formula and functional group: R-NH2 -NH2

3. Names and Structural Formulae methylamine 1-propylamine 1-octylamine

4. Uses: dyes ( congo red and methyl orange ) plastics ( Melmac- melamine ) 5. Primary amine occurs when _________________________________ Secondary amine occurs when______________________________ Tertiary amine occurs when _________________________________ Amino Acids - amine links to a carboxylic acid 1. Amino acid formation:

glycine

alanine

C C C OH

HO

O

+ CCC O

O

O H2 +

N-C-C-O-H H

H O H-C-C-C-O-H

O

H H N H

H

2. Amino acids are essential both in plants and animals ( usually for synesizing polyamides

(polypeptides) called Proteins. eg. skin, hair, muscles nails etc

3. Amine + acid ! amide (peptide) linkage + water 4. Condensation Polymerization- of AMINO ACID monomers

-C-C-N C-C-C- -C-C-N-C-C-C- H2O

HH

H

O O

HO + +

C-C-N O

HO H

H H2O + 4