1
SEMESTER 1
FINAL EXAM REVIEW
Unit I – Chemical Foundations
I. Good things to know:
 chemistry
 scientific method: hypothesis, data, theory, variables
 matter  substances: elements, compounds; mixtures: homogeneous, heterogeneous
 graphing: x-axis, independent variable; y-axis, dependent variable
 graphing: directly proportional, inversely proportional
 commonly used lab equipment and how to use properly
 lab safety, hazard symbols
 significant figures  their importance in measurement
 scientific notation: base, power  where are the sig figs?
 accuracy vs. precision
 SI system: what are the base units? what units are appropriate in what situations?
 density; intensive properties
II. Problems
a, d, e, h
1) Which of the following would be considered proper laboratory procedure?
a. determining the odor of a substance by gently wafting the vapors
if you know the substance is not harmful
b. weighing a crucible while it is still hot
c. measuring a liquid in a graduated cylinder by taking a reading from
the top of the liquid
d. rinsing a pipet with the test solution before performing a titration
e. using a pipet bulb to draw liquid into a pipet
f. pulling your goggles up briefly to rub your eyes
g. measuring the volume of a liquid in a beaker
h. diluting sulfuric acid by measuring out the desired amount of water,
and then slowly pouring the concentrated acid to the water
2). How many sig figs in the following numbers?
1
3)
a) 0.003
3
2
b) 10.0
c) 550
5.1348  10 30
 1.231x10 60
6.012  10 34 6.938  10 3


5) 687 mg = ? kg
687mg 

5
d) 310.01
4)
13,4300.0235  0.00109
40.017240.01
6) 80.10 km = ? um
1kg
109 um
 8.010 x1010 um
 6.87 x10  4 kg 80.10km 
6
1km
10 mg
7) 4.01 x 105 J = ? kJ
4.01x105 J 
1kJ
 401kJ
103 J
8) You are in Paris and want to buy some peaches for lunch. The sign on the fruit stand says that peaches
cost 4.00 euros per kilogram. Assuming that there are 1.14 euros to the dollar, calculate the cost
of a pound of peaches in dollars.
1lb 
1kg
4
$1


 $1.59
2.2046lb 1kg 1.14 
9) If the speed of light is approximately 3.00x 10 8 m/s, what is the approximate speed in miles/hr?
3.00 x108 m 1km 0.6214mi 3600s
 3 

 6.71x108 mph
s
10 m
1km
1hr
SEMESTER 1
FINAL EXAM REVIEW
2
10) If you have a graduated cylinder which initially contains 15.7 mL of water, and when a 65.64 g object
is placed inside, shows a new volume of 22.6 mL, what is the object’s density?
Vobject  22.6mL  15.7mL  6.9mL
D
m 65.64 g

 9.5 g mL
V 6.9mL
11) The approximate mass of the earth is 5.94 x 10 21 metric tons. If the circumference of the Earth at the
equator is approximately 25,000 miles, what is the approximate density of the earth in g/mL ?
Vsphere = 4/3 πr3
C = 2πr
1 metric ton = 1000 kg
1000kg 103 g
5.94 x10 mtons 

 5.94 x10 27 g
1mton 1kg
21
25,000mi
1km
105 cm
 3979mi 

 6.4 x108 cm
2
0.6214mi 1km
3
8
3
27
3
V  4 3 r  4 3  (6.4 x10 cm)  1.1x10 cm
C  25,000mi  2r  r 
D
m 5.94 x10 27 g

 5.4 g cm3
V 1.1x10 27 cm3
Unit II – Atomic Theory & The Periodic Table
1. Explain the Photoelectric effect as it pertains to quantam theory: Why is it that purple light will expel
electrons from a piece of metal, but red light will not? Why is it that as the intensity of the light increases,
the expelled electrons do not become more energetic, there are simply more of them expelled?
According to the equation E = hv, violet light has greater energy than red light, therefore the
violet photons have sufficient energy to expel the electrons from the metal, whereas the red ones do not.
More intense light is emitting more photons per second, so more electrons per second are expelled,
but they do not contain any more energy.
2. Briefly describe the concept of DeBroglie’s Equation. How does the wavelength vary with mass?
Generally, the mass is the most important term in the equation, as the mass increases, the
measurable wavelength of the object decreases (they are inversely proportional). Another way to see it is
that more massive objects have less measurable wavelengths (they are more “matter-like”), whereas less
massive objects have more measurable wavelengths (they are more “energy-like”).
3. Briefly describe Bohr’s Atomic Theory for the hydrogen atom. What phenomenon gave rise to his
theory? How do the colors of the bright-line spectra for hydrogen relate to the Energy levels in a hydrogen
atom?
Bohr saw the electrons in an atom occurring in specifically defined energy levels, as defined by
Planck’s Quantum Theory. As an electron passes from a more excited state (higher energy level) to a
ground state, it releases a photon of light. The energy of that photon corresponds to a certain
wavelength of light, which has a certain characteristic color.
4. Explain why adding an electron to a fluorine atom is an exothermic process.
Fluorine has seven valence electrons. Adding one more electron will give it a complete valence
shell(stable octet) which is a very favorable arrangement (lower in Energy). When one more electron is
added, the atom will give off energy.
5. Explain why adding an electron to a neon atom is an endothermic process.
Neon has eight valence electrons which is a very favorable arrangement. Although it is possible
to add another electron, because the electron would need to be placed in the next higher energy level, an
input of energy would be required.
3
SEMESTER 1
FINAL EXAM REVIEW
6. Explain the trend in atomic radius from Li to Na to K.
As you go Li  Na  K, the atomic radius increases. As you go down a column on the table,
there are more energy levels containing electrons which are a greater distance from the nucleus,
therefore increasing the effective size of the atom.
7. Explain the trend in atomic radius from Na to Mg to Al.
As you go Na  Mg  Al, the atomic radius decreases. As you travel across a row on the table,
more protons are gained by the atom without any additional electron shielding. As a result, there is a
greater effective nuclear charge, therefore a greater force of attraction between the nucleus and the
valence electrons, therefore the valence shell is drawn in closer.
8. Explain the trend in first ionization energy from Na to Mg to Al.
As you go Na  Mg  Al, the 1st ionization energy increases. As you travel across a row on
the table, more protons are gained by the atom without any additional electron shielding. As a result,
there is a greater effective nuclear charge, therefore a greater force of attraction between the nucleus
and the valence electrons, therefore it requires more energy to remove the first valence electron.
9. Explain why the 2nd ionization energy for sodium is so much greater than that for magnesium.
Once the first valence electron is removed, a sodium ion now has a stable octet, therefore
making it very unfavorable to remove a second electron. Magnesium has two valence electrons, so its
first two ionization energies are comparably small.
10. What would the first two quantum numbers be for an electron in a 4f subshell? How many possible
quantum number sets are there for a 4f subshell? Give two examples. How many electrons can
be held in the entire 4th energy level?
* (4,3
* 14 options: (4,3  3,2,1,0,-1,-2,-3, ±½ on each
* (4,3,1,-½) (4,3,-2,½)
* 32 (2 in 4s, 6 in 4p, 10 in 4d, 14 in 4f)
11. Fill in the table below
Element
Potassium
Magnesium
Cadmium
Dysprosium
Family
Name
alkalai
metals
alkaline earth
metals
transition
metals
rare earths
(lanthanides)
# of
Valence
Electrons
Physical State
(metal, nonmetal, metalloid)
Electron Configuration
Paramagnetic
or Diamagnetic?
1
metal
[Ar] 4s1
paramagnetic
2
metal
[Ne] 3s2
diamagnetic
XX
metal
[Kr] 5s24d10
diamagnetic
XX
metal
[Xe]6s24f10
paramagnetic
Chlorine
halogens
7
nonmetal
[Ne]3s23p5
paramagnetic
Neon
noble gases
8
nonmetal
[He] 2s22p6
diamagnetic
4
SEMESTER 1
FINAL EXAM REVIEW
a
12. Which of the following could be the quantum numbers for the valence electron in a ground
state sodium atom?
a. (3,0,0,½) Na is in 3s row b. (3,1,1,½) c. (4,0,0,½) d. (4,1,1,½)
e. (4,2,1,½)
c
13. Which of the following is an impossible set of quantum numbers? Explain your answer.
a. (3,0,0,½) b. (3,1,1,-½)
c. (3,1,-2,½)
d. (3,2,-1,½)
e. (3,2,2,½)
absolute value of
3rd # can’t exceed 2nd
14. Which fundamental atomic theory is violated by the following list of quantum numbers representing
silicon’s 4 valence electrons?
(3,0,0,-½), (3,0,0,-½), (3,1,1,-½), (3,1,1,-½)
Pauli Exclusion Principle  No 2 electrons may have the same set of quantum numbers
15. Which fundamental atomic theory is violated by the following list of quantum numbers representing
silicon’s 4 valence electrons?
(3,0,0,-½), (3,0,0,+½), (3,1,-1,-½), (3,1,-1,+½)
Hund’s Rule  the electrons in the 3p sublevel (3,1), must be evenly distributed with the
same spin, so 3rd number must be different, with the same spin (4th) number
16. Which of the following atoms or ions is larger:
sulfide a.
sulfur or sulfide
* when an atom gains electrons
its radius increases
sodium atoms b.
K c. Na or K
* K is lower on the column than Na
S
Na+
d. S or Cl
* S is further to the left from Cl
Cl-1
e. Na+1 or Mg+2
* Na is further to the left
sodium atoms or sodium ions
* when an atom loses electrons
its radius decreases
f. F-1 or Cl-1
* Cl is lower in the column
17. Explain why a fluorine atom gains in size when it accepts an electron to become a fluoride ion.
When a fluoride atom gains an electron, the number of protons in the nucleus remains the
same, so the effective nuclear charge remains the same, but that charge is distributed over a larger
number of electrons, making the attractive force less. As a result, the electrons move further away from
the nucleus, making the ion effectively larger.
18. Give the number of protons, neutrons, and electrons in the following:
a)
131
54
Xe
b)
56
26
Fe 3
c)
127
53
I 1
54
protons
26
protons
53
protons
77
neutrons
30
neutrons
74
neutrons
54
electrons
23
electrons
54
electrons
5
SEMESTER 1
FINAL EXAM REVIEW
Unit III – Nomenclature & Equations
I. Give the name or formula for the following:
H2SO4
Ca(OH)2
XeF6
ammonia (nitrogen trihydride)
a. sulfuric acid
b. calcium hydroxide
c. xenon hexafluoride
FeCr2O7
d. iron(II) dichromate
k. NH3
nickel(II) sulfide
l. NiS
ammonium carbonate
m. (NH4)2CO3
aluminum chloride
n. AlCl3
HCN
e. hydrocyanic acid
chromium(III) hydroxide
o. Cr(OH)3
Rb2O
f. rubidium oxide
carbon tetrachloride
p. CCl4
acetic acid
r. HC2H3O2
sodium hydride
s. NaH
magnesium fluoride
t. MgF2
Sn3(PO4)4
CO
h. tin(IV) phosphate
i. carbon monoxide
BaSO4-8H2O
j. barium sulfate
octohydrate
II. Balance the following equations:
1.
1 Ba(OH)2(aq) + 2 AgC2H3O2(aq) 
2.
2 KI 
3.
3 Zn + 2 Fe(NO3)3 
4.
1 C25H52 + 38 O2 
5.
3 H3AsO3 + 1 NaBrO3 
2 K+
1 Ba(C2H3O2)2(aq) + 2 AgOH(s)
1 I2
3 Zn(NO3)2 + 2 Fe
25 CO2 + 26 H2O
1 NaBr +
3 H3AsO4
III. Give the net ionic equation for the following precipitation reactions:
1. A solution of silver nitrate is mixed with a solution of sodium chromate.
2 Ag+1 + CrO4-2  Ag2CrO4
2. A solution of lead(II) chloride is mixed with a solution of sodium hydroxide.
Pb+2 + 2 OH-1  Pb(OH)2
3. A solution of barium nitrate is mixed with a solution of potassium phosphate.
3 Ba+2 + 2 PO4-3  Ba3(PO4)2
SEMESTER 1
FINAL EXAM REVIEW
6
Unit IV – Stoichiometry
I. Good things to know
 stoichiometry, chemical formula
 ionic compound  formula unit
 hydrate, anhydrous
 molecular compound  molecule
 mole, Avagadro’s Number
 formula mass, molar mass, molecular mass
 Standard Temperature and Pressure (STP)
 percent yield
 limiting reactant, excess reactant
 Law of Definite Proportions, Law of Multiple Proportions
 empirical formula, molecular formula
 mass spectrometer
 titration, standard solution, unknown solution
II. Problems
1) Calculate the number of moles of HF molecules in:
a) 0.385 g HF
b) 3.02 x 1024 molecules of HF
0.385 g HF 
1mol HF
 0.0193mol HF
20 g HF
3.02 x10 24 mcs HF 
c) 50.0 mL of a 0.600M HF solution
(0.600M )(0.050 L)  0.0300mol HF n 
1mol HF
 5.02mol HF
6.02 x10 23 mcs HF
d) 1.50L of HF at 0.989 atm of pressure
and 23.1oC
PV (0.989atm)(1.50 L)

 0.0610mol HF
RT (0.0821)( 296.25K )
2) Answer the following questions using the following equation:
4 HF(g) + 1 SiO2(s)  1 SiF4(s) + 2
a) balance the equation
H2O(l)
b) How many grams of silicon tetrafluoride will you get if you react 5.00g of silicon dioxide with
excess HF?
5.00 g SiO 2 
1mol SiO 2 1mol SiF4 104 g SiF4


 8.65g SiF4
60.1g SiO 2 1mol SiO 2 1mol SiF4
c) How many grams of silicon tetrafluoride will you get if you react 5.00g of silicon dioxide with
5.00g of HF? How many grams of the excess reactant will be left over?
1mol HF 1mol SiF 4

 0.0625mol SiF 4  LR
20.0 g HF 1mol HF
1mol SiO 2 1mol SiF 4
5.00 g SiO 2 

 0.0832mol SiF 4
60.1g SiO 2 1mol SiO 2
5.00 g HF 
0.0625mol SiF 4 
104 g SiF 4
 6.50 g SiF 4
1mol SiF 4
0.0625mol SiF 4 
1mol SiO 2 60.1g SiO 2

 3.76 g SiO 2  5.00  3.76  1.24 g SiO 2 left
1mol SiF 4 1mol SiO 2
d) If you actually produce 5.92g of silicon tetrafluoride in problem (c), what is the percent yield?
5.92 g
x100%  91.1%
6.50 g
SEMESTER 1
FINAL EXAM REVIEW
7
3) A sample of 0.6760g of an unknown compound containing barium ions (Ba +2) is dissolved in water
and treated with an excess of Na2SO4. If the mass of the barium sulfate precipitate formed is
0.4105g, what is the percent by mass of barium in the original unknown compound?
BaSO 4  FM  233.4  % Ba 
137
x100%  58.5%
233.4
0.2415 g Ba +2
(0.588)(0.4105 g ) 
x100%  35.7%
0.6760 g cmpd
4) A hydrocarbon was found to be 20% hydrogen by weight. If one mole of the hydrocarbon has a
mass of 30 grams, what is its molecular formula?
1C

 6.66 C
12 g C
1 C
30
  6.66 
 CH 3 (FM = 15) 
 2  C2H6
1H
3H
15


12 g H 
 20.0 H 
1g H

80 g C 
5) If you add 20.0 mL of 0.100 M iron(III) nitrate to 20.0 mL of 0.100 sodium hydroxide, how many
grams of precipitate will be formed? Write the net ionic equation for the reaction.
LR
(0.100M )(0.0200 L)  0.00200mol Fe +3 
+3
-1
Fe
+
3OH
 Fe(OH) 3

(0.100M )(0.0200 L)  0.00200mol OH-1  0.002mol 0.002mol
1mol Fe(OH) 3 106.8 g Fe(OH) 3
0.00200mol OH-1 

 0.0712 g Fe(OH) 3
3mol OH-1
1mol Fe(OH) 3
6) If it requires 25.0 L of a 0.500M KI solution to precipitate all of the lead(II) ions out of a 100.0
mL sample, what is the concentration of the lead ions?
Pb +2 + 2 I -  PbI 2
(0.500M )( 25 x10 6 L)  1.25 x10 5 I - 
1 Pb + 2
 6.25x10 -6 Pb + 2  0.100 L  6.25 x10 5 M
2I
7) What is the concentration of an unknown H2SO4 solution if it requires 156.3 mL of 1.50M NaOH
standard to titrate a 100.0 mL sample of the unknown?
(1.50M )(0.1563L)  0.23445 OH- 
1 H + 1 H 2SO 4

 0.117H 2SO 4  0.100 L  1.17 M
1 OH2 H+
8) In your lab, you titrated hydrogen peroxide with potassium permanganate:
6 H+1 + 2 MnO4-1 + 5 H2O2  5 O2 + 2 Mn+2 + 8 H2O
If 36.44 mL of a 0.01652 M KMnO4 solution is required to completely oxidize 25.00 mL of
a H2O2 solution, calculate the molarity of the peroxide solution.
(0.01652M )(0.03644L)  6.02 x104 MnO 4 
-
5 H 2O2
 0.00150H 2O2  0.02500L  0.0602M
2 MnO 4
8
SEMESTER 1
FINAL EXAM REVIEW
Unit V – Bonding
I. Good things to know
 ionic bonding  Coulomb’s Law
 covalent bonds  lone pair, shared pair
 stable octet
 polar vs. nonpolar, dipole moment
 sigma bond, pi bond
 formal charge, bond length  resonance
 hybrid orbitals, VSEPR model
 organic chemistry
 4 allotropes of carbon
 saturated and unsaturated hydrocarbons
 functional groups – hydroxyl, ethers, aldehydes, ketones, organic acids, esters, amines
 isomers, polymers
 aromatic compounds
II. Problems
1) Which of the following ionic compounds would you expect to have the higher melting point? Explain.
NaCl vs. KBr
a.
NaCl. The ionic radius is smaller, therefore, according to Coulomb’s Law, the ions will exhibit
a greater attractive force for each other.
2) Put the following molecules in increasing order of their C – O bond length. Explain.
CH3OH
CO3-2
CO2
shortest  longest: CO2, CO3-2, CH3OH. In CO2, both C – O bonds are double bonds, which
are shorter than single bonds. In CO3-2, the resonance structure causes all three C – O bonds to posess a
bond order of 1 1/3, which will be longer than double bonds, and shorter than single bonds. In CH3OH,
the C – O bond is a single bond, which is the longest of the C – O bonds.
3) For each of the following, give the Lewis Structure, type of hybrid orbitals used by the molecule, shape,
polarity, bond angle, and number of sigma and pi bonds:
a. PCl3
c. NO3-1
b. CO2
O
Cl
P
N
O
Cl
O C
Cl
sp3, trigonal pyramid, P, 109.5o, 3σ
d. BrF5
O
O
sp, linear, NP, 180o, 2σ, 2π
sp2, trigonal planar, P, 120o,
3σ, 1π
e. PF5
F
F
F
F
F
F
Br
F
3 2
P
F
F
F
o
sp d , square pyramid, P, 90 , 5σ
3
sp d, trigonal bipyramid, NP, 90o/120o, 5σ
9
SEMESTER 1
FINAL EXAM REVIEW
4) Without using the table of bond energies, which of the following molecules would you expect to
require the most energy to break its C – C bonds? Explain your answer.
C2H6
C2H4
C2H2
C2H2. This molecule has a triple C-C bond which will be much stronger than the double bond
in C2H4 and the single bond in C2H6. Multiple bonds are always stronger than single bonds, therefore
they require more energy to break.
5) Give the IUPAC name for the following:
CH3
CH3
2,4-dimethyl pentane
CH3
CH2
CH
CH
CH3
a.
Br
CH3
3-bromo hexane
b.
CH2
CH
CH2
CH2
CH3
NH2
1-amino, 3,5-dibromobenzene
c. Br
Br
Unit VI – Phases & Gas Laws
I. Good Stuff to Know
 definitions of solid, liquid, gas, fluid, condensed state
 Pressure - relationship between force and area
 barometer - atmospheric pressure, air pressure, barometric pressure
 standard temperature and pressure (STP): 1 atm = 76θ Torr = 76θ mmHg = 101.3 kPa and 273K
 manometer
 vaporization – boiling (how does it relate to vapor pressure?), evaporation
 amorphous solid- example
 crystal lattice – crystals
 density of solids, liquids, gases - water (most dense at 4oC)
 intermolecular forces: dipole-dipole interactions, London Dispersion Forces, hydrogen bonding, ionic
crystals, network solid. metallic crystals
 Phase Diagrams – melting, freezing, boiling, condensation, sublimation, deposition
 Phase Diagrams - triple point, critical temperature, unique properties of water
 substances that exist as gases
 Avogadro’s Law
 Kinetic Theory  use it to explain Boyle’s Law, Charles’ Law, Avogadro’s Law, Dalton’s Law,
compressibility of gases
 root-mean-square speed, mean free path
 diffusion vs. effusion
 deviations from ideal behavior, vanderWaal’s equation
10
SEMESTER 1
FINAL EXAM REVIEW
II. Multiple Choice
For questions 1-4, select from the following answers
a. metallic bonding
b. network covalent bonding
d. ionic bonding
e. London Dispersion Forces
c. hydrogen bonding
e
1. Nonpolar substances such as methane (CH4) demonstrate this type of bonding.
b
2. This kind of bonding is exhibited by diamond and quartz, and explains their hardness and
extremely high melting points.
c
3. This type of bonding is only exhibited by ammonia, water, and hydrogen fluoride, and results
in these substances having unusually high melting and boiling points.
d
4. This type of bonding results in solids that are poor conductors of heat and electricity, but which,
when melted, are good conductors of electricity.
* When an ionic compound is melted, the ions separate and can flow freely, therefore conduct
Questions 5-8 refer to the following phase diagram:
B
5. At this point the substance represented by the phase diagram will be solely in the solid phase at
equilibrium.
D
6. This point represents the boiling point of the substance.
A
C
a
7. At this point, the substance represented by the phase diagram could be undergoing sublimation.
8. At this point the substance represented by the phase diagram will be solely in the liquid phase
at equilibrium.
9. Which of the following lists of species is in order of increasing boiling point?
a. H2 , N2 , NH3 b. N2, NH3, H2 c. NH3, H2, N2 d. NH3, N2, H2 e. H2, NH3, N2
* H2 and N2 are both nonpolar, therefore they use London Dispersion forces. Because the FM
of N2 is greater, the forces are stronger. NH3 uses hydrogen bonding.
III. Essay/Problems
1. Explain why the boiling point of argon is -186oC, but the boiling point of neon is -246oC.
Both are nonpolar, therefore both exhibit London Dispersion Forces, but because the formula
mass of argon is greater, it has more electrons, therefore the dispersion forces are stronger, making it
more difficult to separate the atoms, therefore causing the boiling point to be greater.
2) What is the pressure of the gas in the enclosed container?
P = 1.27 atm
243 mm
1.27atm  965mmHg  243  1208mmHg  1.59atm
SEMESTER 1
FINAL EXAM REVIEW
11
3) Consider three identical flasks filled with different gases:
Flask A: CO at 760 Torr and 0oC
Flask B: N2 at 250 Torr and 0oC
Flask C: H2 at 100 Torr and 0oC
all
C
a. In which flask will the molecules have the greatest average kinetic energy?
* KE=3/2 RT Since all 3 are at the same T, all have same KE
b. In which flask will the molecules have the greatest average velocity?
* KE = ½ mv2  Since KE is the same for all 3, the one with the least
mass (H2) will have the greatest velocity
4) A gas occupies a volume of 34.2 mL at a temperature of 15.0oC and a pressure of 800.0 Torr. What
will the volume of this gas be at STP?
15.0  273  288K
(800.0Torr )(34.2mL) (760Torr )V2

 V2  34.1mL
288K
273K
5) Find the formula mass of a gas which diffuses at a rate 1.16 times faster than that of sulfur dioxide gas.
1.16 
64.1
64.1
 1.3456 
 x  47.6 g mol
x
x
6) 40.0 mL of helium gas is collected over water at 20.0oC. If this gas exerts a pressure of 790.0 Torr,
what would the volume of the dry gas be at STP? (Remember this is a mixture of two gases: water vapor
and helium)
(Pwater = 17.54 mmHg at 20.0oC)
PHe  790.0  17.54  772.5mmHg  20.0  273  293K
(772.5mmHg )( 40.0mL) (760mmHg )V2

 V2  37.9mL
293K
273K
7) What is the density of oxygen gas collected at 21.0oC and 103.5 kPa?
21.0  273  294 K  FM O2  32 g mol  suppose you have 1 mol of O 2 :
PV  nRT  (103.5kPa)V  (1mol )(8.31)( 294 K )  V  23.6 L
m
32 g
D 
 1.36 g L
V 23.6 L
8) If you collected 0.506 g of a gas which you know to be composed of 30.4% nitrogen and 69.6% oxygen,
and it occupied a volume of 0.134 L at a temperature of 2θoC, and a pressure of 0.986 atm, what would the
formula mass of the gas be? Give the molecular formula of the gas as well.
1N
 2.17 N
1N
14 g N
 2.17
ef : NO 2 ( FM  46)
1O
2O
69.6g O 
 4.35 O
16 g O
30.4 g N 
n
PV (0.986atm)(0.134 L)
m
0.506 g
92

 0.00549mol  FM  
 92 g mol 
 2  mf : N 2 O 4
RT
(0.0821)( 293K )
n 0.00549mol
46
SEMESTER 1
FINAL EXAM REVIEW
Unit VII – Chemical Kinetics & Nuclear Chemistry
I. Good things to know
 Collision Theory; collision effectiveness, collision frequency
 factors that affect reaction rates: temperature, concentration, presence of a catalyst
 graphs of zero order, first order, and second order reactions
 isotopes, transmutation, fission, fusion
 proton, neutron, beta, alpha, postitron
 types of decay: which types of nuclei do which types of decay?
 nuclear binding energy, half life
 applications of nuclear chemistry, basic design of a nuclear power plant
II. Multiple Choice/Short Answer. Select the best possible answer:
Questions 1-3
A+BC
The following are possible rate laws for the hypothetical reaction given above:
a. rate = k[A] b. rate = k[B]2 c. rate = k[A][B] d. rate = k[A]2[B] e. rate = k[A]2[B]2
a
1. This is the rate law for a first order reaction
b
2. When [A] is doubled, the initial rate of reaction will not change.
e
3. When [A] and [B] are doubled, the initial rate of reaction will increase by a factor of sixteen.
d
4.
H2(g) + I2(g)  2 HI(g)
When the reaction above takes place in a sealed isothermal container, the rate law is rate = k[H 2][I2]
If a mole of H2 gas is added to the reaction chamber, which of the following will be true?
a. The rate of reaction and the rate constant will increase
b. The rate of the reaction and the rate constant will not change.
c. The rate of the reaction will increase and the rate constant will decrease.
d. The rate of reaction will increase and the rate constant will not change.
e. The rate of reaction will not change and the rate constant will increase.
b
6.
5. After 44 minutes, a sample of 44K is found to have decayed to 25% of the original amount
present. What is the half life of 44K?
a. 11 minutes b. 22 minutes c. 44 minutes d. 66 minutes e. 88 minutes
S + F2
Na + F2
Na + F
2 NaF + SF2
 SF2
 NaF + F
 NaF
 Na2S + 2 F2
a. Give the overall reaction
b. Name any intermediates:
c. Name the catalyst:
2 Na + S  Na2S
SF2 , NaF , F
F2
d. Given that the rate law is Rate = k(F2)(Na)2, which step is the rate-determining step? Step 3
* Note that the “rate law” for step 3 is technically: rate = k(Na)(F), but the F is an
intermediate which is produced by both steps 1 & 2, so the reactants from there are
included in the rate law as well.
12
13
SEMESTER 1
FINAL EXAM REVIEW
A(g) + B(g)  C(g)
7. The reaction above is endothermic, and first order with respect to A and first order with respect to B.
Reactants A and B are present in a closed container. Predict how each of the following changes to
the reaction system will affect the rate and the rate constant.
a. Write the rate law for the reaction.
rate = k(A)(B)
b. More gas B is added to the container.
rate increases, k does not change
c. An inert gas D is added to the container.
NO EFFECT
d. The temperature of the container is increased.
rate and k both increase
8. Use your knowledge of kinetics to answer the following questions:
B
Potential
Energy
A
C
Reaction Coordinate
A
a. The letter of the reactants.
activated complex
C
b. The letter of the products.
c. The term for the “stuff” at letter B.
d. The activation energy is the difference between points
A
e. The enthalpy of reaction (ΔH) is the difference between points
exothermic
and
A
f. Is this reaction exothermic or endothermic?
B
and
C
SEMESTER 1
FINAL EXAM REVIEW
14
III. Essay Section
1. Use your knowledge of kinetics to explain each of the following statements.
a. Adding a catalyst increases the rate at which a reaction takes place.
adding a catalyst will allow the reaction to go through an alternate mechanism
which has a lower activation energy, therefore the reaction rate will increase
without a change in ΔH.
b. Increasing the concentration of reactants increases the rate of a reaction.
increasing the concentration of the particles forces the reactant particles
closer together, which increases the frequency of collisions, therefore
increasing the reaction rate.
2. Use your knowledge of nuclear chemistry to explain each of the following.
a. Gamma radiation causes cancer, but infrared rays do not.
Gamma radiation has a shorter wavelength. As a result, it has more
energy (E = hv) and is more likely to collide with the atoms of the body
tissues and cause damage.
b. A nucleus weighs less than the sum of the weights of its neutrons and protons.
When protons and neutrons are combined (fusion), a mass defect
occurs, and the mass decreases slightly while energy is released.
SEMESTER 1
FINAL EXAM REVIEW
15
IV. Problems.
1. Uranium-238 decays by alpha decay. Write the balanced nuclear reaction for this process.
238
92
U
4
234
 2 He + 90Th
21
2. 11 Na goes through positron emission to become more stable. Write the balanced nuclear reaction for
this step.
21
11
Na
 1  + 10 Ne
0
21
A + B + C ABC
3. The following results were obtained in experiments designed to study the rate of the reaction above.
Initial Concentration (M)
Experiment
1
2
3
4
[A]
0.05
0.05
0.10
0.10
[B]
0.05
0.10
0.10
0.10
a. Write the rate law for the reaction.
[C]
0.05
0.05
0.05
0.10
Initial Rate of
Disappearanceof A (M/sec)
3.0 x 10-3
6.0 x 10-3
2.4 x 10-2
2.4 x 10-2
rate = k(A)x(B)y(C)z
2x
A: 2 vs. 3:
=4 x=2
B: 1 vs. 2: 2y = 2 y = 1
C: 3 vs. 4: 2z = 1 z = 0
rate = k(A)2(B)
b. Calculate the value of the rate constant, k, for the reaction. Include the units.
0.024 Ms-1 = k(0.10M)2(0.10M)
24 M-2s-1 = k
c. If another experiment is attempted with [A], [B], and [C] all 0.02M, what will be the initial
rate of disappearance of A?
rate = (24)(0.02)2(0.02) = 1.92x10-4 Ms-1
d. The following reaction mechanism was proposed for the reaction above.
A + B  AB
A + AB  A2B
A2B + C  A + ABC
(i) Show that the mechanism is consistent with the balanced reaction.
Overall Reaction: A + B + C  ABC
(ii) Show which step is the rate-determining step and explain your choice.
the second step is rate-determining because its rate law would be written
as rate = k(A)(AB), but since AB isn’t in the overall reaction, you need
to find the step that generated it, which is step 1, whose rate law is rate =
k(A)(B). Integrating the two rate laws together gets the desired result:
rate = k(A)2(B)