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Matter & Measurement
EIT Review
F2007
Dr. J.A. Mack
Matter is the physical material of the universe.
Matter is anything that occupies space (volume) and
has mass.
www.csus.edu/indiv/m/mackj/
Matter is made up of relatively few elements.
Matter consists of atoms and molecules.
Part 1
Each element is a unique atom.
Atoms combine to form molecules
1
3
5
7
Classifying Matter:
1
• Physical Properties can be determined without changing
the chemical make–up of the sample.
STATES OF MATTER
• SOLIDS — have rigid shape, fixed volume.
• Examples of physical properties are:
External shape can reflect the atomic and
molecular arrangement.
–Melting Point,Boiling Point, Density, Mass, Touch,
Taste, Temperature, Size, Color, Hardness, Conductivity.
• LIQUIDS — have no fixed shape and may
not fill a container completely.
• Examples of physical changes are:
• GASES — expand to fill their container.
–Melting, Freezing, Boiling, Condensation, Evaporation,
Dissolving, Stretching, Bending, Breaking
8
• Physical Properties can be determined without changing
the chemical make–up of the sample.
9
Chemical Properties are those that do change the
chemical make–up of the sample.
• Examples of physical properties are:
–Melting Point,Boiling Point, Density, Mass, Touch,
Taste, Temperature, Size, Color, Hardness, Conductivity.
Examples of chemical properties are:
Burning, Cooking, Rusting, Color change, Souring of
milk, Ripening of fruit, Browning of Apples,
• Examples of physical changes are:
photography, Digesting food
–Melting, Freezing, Boiling, Condensation, Evaporation,
Dissolving, Stretching, Bending, Breaking
10
Note: Chemical properties are actually chemical changes.
13
2
Units of Measure:
Temperature:
English Customary
Weights and Measures
The International System of
Units (SI)
distance
(phase change)
meter
inch
Temperature is a measure of
particle motion
foot
= 12 inches
micrometer
= 10−6 meters
yard
= 3 feet
centimeter
= 10−2 meters
mile
= 5280 feet
kilometer
= 103 meters
(all motion stops)
Units with in a system can be represented by units within the system.
Units within the metric system are related by powers of 10
14
15
What are Significant Figures?
Significant figures communicate the uncertainty in a
measurement.
Given the mass:
45. 8724g
With the uncertainty:
± 0.001g
Which numbers are certain ? (significant)
You need to memorize these values and have the ability to convert
between.
16
17
3
The error (±) tells us which digit is uncertain:
Counting Significant Figures
45. 8724g
1. All non zero numbers are significant
± 0.001
2. All zeros between non zero numbers are significant
The uncertainty occurs at the thousands place:
3. Leading zeros are NEVER significant. (Leading
zeros are the zeros to the left of your first non zero
number)
Therefore,
4. Trailing zeros are significant ONLY if a decimal
point is part of the number. (Trailing zeros are the zeros
to the right of your last non zero number).
45. 8724
These digits are certain
The blue digits are significant,
This digit is uncertain
or we say there are 5
significant figures
18
Determine the
number of Sig.
Figs. in the
following
numbers
zeros written
behind the decimal
are significant…
1256
4 sf
1056007
7 sf
0.000345
3 sf
0.00046909
5 sf
1780
3 sf
770.0
4 sf
0.08040
4 sf
19
Scientific notation:
Now count each step as a power of ten to find the exponent.
not trapped by a
decimal place.
563,490.
Five steps: the exponent is 5
5.63490 × 105
20
21
4
Multiplication and Division:
Sig. Figures in Calculations
Determine the correct number of sig. figs. in the
following calculation, express the answer in scientific
notation.
Multiplication/Division
The number of significant figures in the answer is
limited by the factor with the smallest number of
significant figures.
4 sf
2 sf
4 sf
23.50 ÷ 0.2001 × 17
Addition/Subtraction
The number of significant figures in the answer is
limited by the least precise number (the number with its
last digit at the highest place value).
NOTE: Defined numbers (numbers from tables and
references) never limit calculations.
The sf in the result is limited to the number with the least
amount of sf.
The answer must be rounded to 2 sf.
22
Sig. Figs. Addition and Subtraction
23.50 ÷ 0.2001 × 17
from the calculator:
1996.501749
How many sig. figs. are allowed in the following calculation?
10 sf
391 - 12.6 +156.1456
Your calculator knows nothing of sig. figs. !!!
in sci. not.:
1.996501749 x 103
Rounding to 2 sf:
2.0 x 103
23
To determine the correct decimal to round to, align the
numbers at the decimal place:
391
-12.6
+156.1456
no digits here
One must round the calculation to the least significant decimal.
24
25
5
391
-12.6
+156.1456
534.5456
one must round to here
Combined Operations:
When there are both addition / subtraction and
multiplication / division operations, the correct
number of sf must be determined by examination of
each step.
(answer from calculator)
Example: Complete the following math mathematical
operation and report the value with the correct # of sig.
figs.
round to here (units place)
Answer: 535
(26.05 + 32.1) ÷ (0.0032 + 7.7) = ???
26
27
3 sf
(26.05 + 32.1) ÷ (0.0032 + 7.7) = ???
26.05
(26.05 + 32.1)
=
(0.0032 + 7.7)
1st determine the correct
# of sf in the numerator
(top)
(26.05 + 32.1)
=
(0.0032 + 7.7)
2nd determine the correct #
of sf in the denominator
(bottom)
The result will be limited by the least # of sf (division rule)
28
+ 32.1
0.0032
+ 7.7
2 sf
The result
may only
have 2 sf
29
6
(26.05 + 32.1)
=
(0.0032 + 7.7)
The Representation of Matter:
58.150
In chemistry we use chemical formulas and symbols to
represent matter.
7.70032
= 7.5516
Why?
We are “macroscopic”: large in size on the order of 100’s
of cm
= 7.6
2 sf
Atoms and molecules are “microscopic”:
on the order of 10-12 cm
30
Where do we begin…
31
The modern periodic table is defined by:
The Periodic Table
Groups (families)
(columns down)
Periods (rows across)
Dmitri Mendeleev (1834 - 1907)
32
33
7
Which elements do we need to know?
Chemical Symbols and Formula:
Elements:
Start here as a minimum…
H = hydrogen
O = oxygen
C = carbon
Molecules:
H2 = hydrogen
O2 = oxygen
H2O = water
UhUh-Oh!
this is confusing…
As many as possible!
Yes it is…
Get over it
and get used to it!
CO2 = carbon dioxide
34
1
35
The Groups are labeled:
The Periods are labeled:
1A
7A 8A
2A
2
3
3A 4A 5A 6A
3B 4B 5B 6B 7B
The “A” refers to
the “main group
elements”
36
8B
1B 2B
The “B” refers to the
transition metal elements.
37
8
Periodic Table: Metallic arrangement
1
IA
1
18
VIIIA
2
IIA
13
IIIA
14
IVA
2
3
4
5
Periodic Table: Metallic arrangement
3
IIIB
4
IVB
5
VB
6
VIB
7
VIIB
8
9
VIIIB
10
11
IB
12
IIB
Metals
15
VA
16
VIA
17
VIIA
1
IA
1
s
al
et
m
on
N
18
VIIIA
2
IIA
13
IIIA
14
IVA
15
VA
16
VIA
17
VIIA
2
3
3
IIIB
4
IVB
5
VB
6
VIB
7
VIIB
8
9
VIIIB
10
11
IB
12
IIB
4
5
6
6
7
7
somewhere in
between metals
and non-metals
Metalloids
38
Atomic Number, Z
Atom: The smallest divisible unit of an element
Compound: A substance made of two or more atoms
An element’s identity is defined by
the number of protons in the
nucleus: Z
13
Al
26.981
39
Ion: A charged atom or molecule
Cation: Positive ion
Anion: Negative ion
Atomic number
NOMENCLATURE
Format for naming chemical compounds using prefixes,
suffixes, and other modifications of the names of elements
which constitute compounds.
Atom symbol
Atomic weight
40
41
9
Ion Charges:
inert or “noble” gasses:
+1
+3
-3 -2
-1
+2
Variable
Metals form Cations
non-metals form anions
Metalloids can do both
42
43
Binary Compounds: Metal & non-Metal
Compounds fall into one of two classes:
Inorganic Salts
Molecules
metal cation
non-metal
+
+
non-metal or
polyatomic anion
non-metal
(no cations or anions)
Metal of fixed oxidation (charge) state combined with a
non-metal.
non-metal takes on “ide” suffix
Examples:
The two use different formalisms for naming…
44
Cation
Anion
Formula
Name
K+
Cl−
KCl
Potassium chloride
Ca2+
O2-
CaO
Calcium Oxide
Na+
S2-
Na2S
Sodium sulfide
Al3+
S2-
Al2S3
Aluminum sulfide
45
10
Metals of variable charge (transition) with a non-metal
Examples:
Cation
Pb2+
Some common polyatomic ions:
modify transition metal name with roman numeral
Anion
Formula
Name
Cl−
PbCl2
lead (II) chloride
pronounced: lead - two - chloride
Pb4+
Cl−
PbCl4
lead (IV) chloride
Fe3+
O2−
Fe2O3
Iron (III) oxide
NH4+
H3O+
CO32–
HCO32–
NO2–
NO3–
SO42–
SO32–
PO43–
C2H3O2–
ammonium
hydronium
carbonate
hydrogencarbonate or bicarbonate
nitrite
nitrate
sulfate
sulfite
phosphate
acetate
46
Ternary Compounds: Those with three different elements
Type A: metal of fixed charge with a complex ion
Cation
Anion
Formula
47
Metal of variable charge transition) with a complex ion
Cation
Anion
Formula
Name
Fe3+
NO3−
Fe(NO3)3
Iron (III) nitrate
Fe2+
NO −2
Fe(NO2)2
Iron (II) nitrite
Name
K+
OH−
KOH
Potassium hydroxide
Ca2+
OH−
Ca(OH)2
Calcium hydroxide
Na+
SO 24−
Na2SO4
Sodium sulfate
Al3+
SO 24−
Al2(SO4)2
Aluminum sulfate
48
49
11
Examples:
Formula
NonNon-metal with a nonnon-metal
When non-metals combine, they form molecules.
They may do so in multiple forms:
CO
CO2
Because of this we need to specify the number of each
atom by way of a prefix.
BCl3
boron trichloride
SO3
sulfur trioxide
NO
nitrogen monoxide
we don’t write:
1 = mono
5 = penta
2 = di
3 = tri
6 = hexa
4 = tetra
N2O4
7 = hepta
Name:
nitrogen monooxide
or mononitrogen monoxide
dinitrogen tetraoxide
50
51
Type I Acids: Acids derived from –ide anions.
D) Writing formulas for acids and Bases
•An acid is a substance that produces H+ when dissolved
in water.
•Certain gaseous molecules become acids when dissolved
in water.
•A base produces OH− when dissolved in water.
•Bases often are Group I and Group II hydroxide salts.
52
The names for these acids follows the formula:
“hydro”
+
the root of the ide anion
+
ic “acid”
Anion:
Acid:
Name:
chloride
HCl
hydrochloric acid
fluoride
HF
hydrofluoric acid
53
12
Examples:
H+ and S2-
Anion:
it takes 2 H+ to
cancel one S2-
Acid:
Name:
(nitrate)
NO3−
HNO3
nitric acid
(sulfate)
SO24 −
H2SO4
sulfuric acid
(acetate)
C2 H 3O2−
HC2H3O2
acetic acid
vinegar
H2S
hydro sulfuric acid
54
55
Practice:
CO2
carbon dioxide
BaCO3
carbon monoxide
CO
FeO
Iron (II) oxide
P2O5
diphosphorous pentaoxide
zinc phosphide
Zn3P2
nitrogen trihydride
NH3
NiBr2
nickel (II) bromide
N2SO4
sodium sulfate
barium carbonate
(ammonia)
56
57
13
Common Names:
H2O
water
ammonia
NH3
please go to my chem. 1A web site and download a
nomenclature practice worksheet if you need more review.
CH4
methane
http://www.csus.edu/indiv/m/mackj/chem1a/chem1A_lab.html
NO
nitric oxide
The link to the worksheet is at the top of the page.
N2O
nitrous oxide
58
59
The Composition of an Atom:
Modern Atomic Theory:
•Atoms are made of protons, neutrons and electrons.
•The nucleus of the atom carries most of the mass.
The atom is mostly
empty space
• It consists of the protons and neutrons surrounded by a
cloud of electrons.
The charge on the electron is –1
The charge on the proton is +1
•protons and neutrons in the nucleus.
There is no charge on the neutron
•the number of electrons is equal to the
number of protons.
The Atomic Number or number of protons in the
nucleus defines an element.
•electrons in space around the nucleus.
60
61
14
Isotopes, Atomic Numbers, and Mass Numbers
•Atomic number (Z) = number of protons in the nucleus.
•Mass number (A) = total number of nucleons in the
nucleus (i.e., protons and neutrons).
75 protons +
neutrons
•One nucleon has a mass of 1 amu
75
34 Se
(Atomic
Atomic Mass Unit)
Dalton”
Unit a.k.a “Dalton
34 protons
•Isotopes have the same Z but different A.
•The elements are arranged by Z on the periodic table.
protons = 34
By convention, for element X, we write
A
Z
X
electrons = 34
neutrons = 75-34 = 41
62
63
Molar Masses
Avagadro’s Number
Since we can equate mass (how much matter) with moles
(how many particles) we now have a conversion
that relates the two.
Since one mole of 12C has a mass of 12g (exactly), 12g of 12C
contains 6.022142 x 1023 12C-atoms.
But carbon exists as 3 isotopes: C-12, C-13 &C-14
mols
The average atomic mass of carbon is 12.011 u.
molar mass (g/mol) = grams
The Molar Mass of a substance is the amount of matter that
contains one-mole or 6.022 × 1023 particles.
From this we conclude that 12.011g of carbon contains
6.022142 x 1023 C-atoms
Is this a valid
assumption?
×
factor
aka: Avogadro's number (NA)
Yes, since NA is so large,
the statistics hold.
The atomic masses on the Periodic Table also represent the
molar masses of each element in grams per mole (g/mol)
64
65
15
Molar Masses (Molecular Weights) of Compounds:
So if you have 12.011g of carbon…
you have 6.022×1023 carbon atoms!
The molar mass of a molecular compound is the sum of the
molar masses of its atoms.
So if you have 39.95g of argon…
you have 6.022×1023 argon atoms!
Example:
The molar mass of CO2 is:
if you have a mole of dollar bills… you are Bill Gates…
you have 6.022×1023 bucks!
1 x (12.01 g/mol)
+ 2 x (16.00 g/mol)
and if you have 6.022×1023 avocados…
= 44.01 g/mol
a “guacamole”
you have…
66
How many oxygen atoms are there in 25.1g of chromium (III) acetate?
Cr(C2H3O2)3
step 2: calculate the molar mass…
Question: What is the weight % of each element in C2H6?
229.13 g/mol
First determine the molar mass of C2H6:
1 mol of C 2 H 6 has a mass of 30.07 g
step 3: use dimensional analysis to solve the problem…
25.1g Cr(C 2 H3O 2 )3 × 1mol Cr(C 2 H3O 2 )3
229.13g Cr(C2 H 3O 2 )3
×
6.022 × 1023 O − atoms
=
1mol O
3.96 ×
1023
Percent Composition:
The relative amounts of each atom in a molecule or
compound can be represented fraction of the whole.
step 1: write the correct chemical formula…
Cr3+ & C2H3O2−
68
(2×12.01 + 6 ×1.008) g/mol
Next determine the mass of hydrogen in 1 mol of the
compound:
6mol O
×
1mol Cr(C 2 H 3O 2 )3
1 mol C2 H 6 ×
O-atoms
69
6 mol H
1.0079 g H
×
= 6.047 g H
1 mol C2 H 6
1 mol H
70
16
Determining a Formula from Percent Composition:
Now relate the mass of H in one mol of the
compound to the molar mass of the compound
Given the relative percentages of each element in a
compound,
1
× 100 = 20.11% H
6.047 g H ×
30.07g C2 H 6
10 % X,
Since there is only C as the remaining element:
% C = 100 − % H
20 % Y,
30 % Z
etc…
one can find the empirical formula of the compound.
= 79.89 % C
The empirical formula of a compound or molecule
represents the simplest ratio of each element in 1 mol
of the compound or molecule.
The compound C2H6 is 20.11% H & 79.89% C
71
Example: A compound is found to be 64.82 % carbon,
21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
Solution:
Example: A compound is found to be 64.82 % carbon,
21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
2. Convert the grams of each element to moles.
determine X, Y & Z in (CXHYOZ)
(g element X → mole X etc…)
1. Since the percentages for each element sum to 100%,
if one equates % to grams (g), the sum of the masses must
be 100g.
(i.e. one can assume 100g of the compound)
64.82 g C
21.59 g O
72
13.59 g H
73
64.82g C ×
1 mol C
12.011 g C
= 5.397 mol C
21.59g O ×
1 mol O
16.00 g O
= 1.349 mol O
13.59g H ×
1 mol H
= 13.48 mol H
1.0079 g H
74
17
3. Divide each of the individual moles by the smallest number of
moles to gain the molar ratios for each element in the compound.
These are the formula subscripts. (X2Y3 etc…)
Subscript for C
Subscript for H
**If the ratios are fractional (0.5, 1.5 or 0.333) multiply each ratio by a
whole number to get even number formula subscripts.
0.5 × 2 = 1
X = 4.001 = 4
Y = 9.992 =10
Z = 1.000 = 1
Subscript for O
13.48
1.349
5.397
= 9.992 Z =
= 1.000
X=
= 4.001 Y =
1.349
1.349
1.349
Examples:
Rounding to the nearest whole numbers:
0.25 × 4 = 1
C4H10O
The empirical formula is:
The results of this calculation tells us only about the
empirical formula of the compound.
To determine the molecular formula, we need more
information. This will be shown in a later example.
0.333 × 3 = 1
75
76
Balancing Chemical Reactions: Example
Chemical Equations:
C2H6 +
2 C’s & 6 H’s
Mass is conserved in a chemical reaction.
→
O2
2 O’s
balance last
CO2
+
H2O
1 C & 2 O’s
2 H’s & 1 O
balance H first
Total mass of
reactants
=
Total mass of
products
2 2H6
___C
+
→
O2
CO2
63 H2O
___
+
balance C next
Chemical equations must therefore be balance for mass
2C2H6 +
balance O
The number and type of atoms on either side of the equation must
be equal!
2C2H6 +
O2
→
7 O2
____
4 C’s 12 H’s 14 O’s
77
4 CO2
___
→
+
4CO2 +
6H2O
6H2O
4 C’s 12 H’s 14 O’s
78
18
Reduction and Oxidation Reactions: RedOx
Oxidation Numbers
+1
+2
+3
-3 -2 -1
Oxidation involves an atom or compound losing electrons
Variable
Reduction involves an atom or substance gaining electrons
Niether process can occur alone… that is, there must be an exchange
of electrons in the process.
Metals take on their formal charge:
Non-metals do the same
The substance that is oxidized is the reducing agent
The substance that is reduced is the oxidizing agent
Chemists use oxidation numbers to account for the transfer of electrons
in a RedOx reaction.
79
Electrochemistry: Oxidation numbers
In the compound potassium bromate (KBrO3), the
oxidation number of bromine (Br) is?
The compound is neutral so the sum of the oxidation numbers should
be zero.
5
+1
3×(-2) = -6
??
KBrO3
1+ ?? + (-6) = 0
80
Balancing REDOX reactions:
Fe + O2 → Fe2O3
oxidation states:
0
0
+3
-2
oxidation half reaction:
{Fe → Fe+3
reduction half reaction:
{O2
+ 3e-} x 4
+ 4e- → 2O2-} x 3
Balance electrons transferred then sum the half RXN’s:
4Fe + 3O2 + 12e- → 2Fe2O3 + 12e-
?? = 5
4Fe + 3O2 → 2Fe2O3
81
82
19
Quantitative calculations: Mass and moles
2HCl (aq) + Ba(OH)2 (aq) Æ 2H2O (l) + BaCl2(aq)
Consider the following reaction:
HCl (aq) + Ba(OH)2 (aq) Æ H2O (l) + BaCl2(aq)
g BaCl2 Æ mol BaCl2
Balacing:
Æ mol HCl
2 HCl (aq) + Ba(OH)2 (aq) Æ 2 H2O (l) + BaCl2(aq)
How many moles of HCl are consumed if 1.50 g of BaCl2 are
produced assuming that Ba(OH)2 is in excess?
Solution:
convert
to
g BaCl2
then convert
Æ
using the Molar mass of BaCl2
mol BaCl2
1.50 g BaCl 2 ×
1 mol BaCl 2
2 mol HCl
×
= 0.0144 mol HCl
208.24 g BaCl 2 1 mol BaCl 2
to
Æ
Molar mass BaCl2
mol HCl
Molar ratio for equation
using the Molar ratio for equation
83
84
Reaction Yields:
Stoichiometry: Conversions between masses & moles in
chemical reactions.
The theoretical yield is the maximum product yield that
can be expected based on the masses of the reactants and
the reaction stoichiometry.
Limiting Reactant:
ÖWhen one reactant is present in an amount such that it
is completely consumed before all other reactants, we
say that it limits the reaction.
The actual yield is the experimentally measured amount
of products that results upon completion of the reaction.
Ö The other reactants are said to be in excess.
excess
Ö The Theoretical Yield is determined by the stoichiometry
of the limiting reactant.
Ö The limiting reactant can only be determined through
molar ratios. It cannot be identified by mass.
The percent yield is a measure of the extent of the
reaction in terms of the actual vs. the theoretical yield.
% Yield =
85
Actual Yield (in grams or moles)
Theoretical Yield (in grams or moles)
× 100
86
20
Example: A solution of sodium phosphate is added to a solution of
aqueous barium nitrate. A white ppt is observed.
The Behavior of Solutes:
Strong Electrolytes: complete dissociation into ions
Unbalanced Equation:
1 M Na3PO4(aq) Æ 3M Na+ (a) + 1M PO43–
Na3PO4 (aq) + Ba(NO3)2 (aq) → Ba3(PO4)2 (s) + NaNO3 (aq)
Molecular:
4M overall in ions
Non–Electrolytes:
2Na3PO4 (aq) + 3Ba(NO3)2 (aq) → Ba3(PO4)2 (s) + 6NaNO3 (aq)
no dissociation into ions
1M CH3OH (aq)
methanol
Ionic:
1M overall in
molecules
/
/
6Na+ (aq) + 2PO43- (aq) + 3Ba2+ (aq) + 6NO3- (aq)
/
1M HC2H3O2(aq) Æ
H+
(aq) +
/
→ Ba3(PO4)2 (s) + 6Na+ (aq) + 6NO3- (aq)
Weak Electrolytes: partial dissociation into ions
Net Ionic:
C2H3O2–(aq)
2PO43- (aq) + 3Ba2+ (aq) → Ba3(PO4)2 (s)
between 1 & 2 M overall
87
How many grams of sodium phosphate are in 35.0 mL of a
1.51 M Na3PO4 solution?
Solutions and Concentration
Molarity:
88
Moles of solute per liter of solution.
mL solution Æ L Æ mols Na3PO4 Æ g Na3PO4
moles of solute
use M as a
conversion factor
{units: mol/L}
Molarity (M) =
L of Solution
35.0mL ×
molarity is a conversion factor that transforms units
of volume to mole and vise–versa
163.94g
L
1.51 mol Na 3 PO4
×
×
1
mol
Na 3 PO 4
10 mL
L
3
= 8.66g Na 3 PO 4
89
90
21
The pH Scale:
Dilutions:
old Molarity × old Volume
 1 
pH = log  + 
 H  
 
New Molarity =
new Volume
1
Since log   = − log(x)
x
M1 × V1
M2 =
Rearranging:
[H+] = Molarity of H+
(or anything for that matter…)
pH = -log[H+]
In a neutral solution, [H+] = [OH-] = 1.00 x 10-7 M at 25 oC
V2
pH = - log [H+] = -log (1.00 x 10-7) = - [0 + ((-7)]
M1 × V1 = M2 × V2
= 7
91
What is the pH of the 0.0515 M HCl solution that Jane made?
pH = -log[H+]
92
Reaction Quotient & Equilibrium
Constant
HCl (aq) → H+(aq) + Cl−(aq)
Strong electrolyte!
Equilibrium Established
Therefore [H+] = [HCl]
pH = − log[H + ] = − log(0.0515)
H2 + I2 → 2 HI
H 2 + I 2 U 2HI
= 1.29
93
94
22
The Reaction Quotient, Q
THE EQUILIBRIUM CONSTANT
In general, ALL reacting chemical systems are
For any type of chemical equilibrium of the type
aA + bB→cC + dD
the following is a CONSTANT (at a given T)
characterized by their REACTION QUOTIENT, Q.
aA + bB→cC + dD
conc. of products
K =
[C]c [D]d
[A]a [B]b
equilibrium constant
conc. of reactants
If Q < K, then system will shift to the right, convert reactants to products.
If Q > K, then system will shift to the left, products convert to
to reactants.
If Q = K, then system is at equilibrium.
If K is known, then we can predict concentrations of products or reactants.
95
Forms of Chemical Bonds
• There are 2 extreme forms of connecting
or bonding atoms:
96
When there exists a difference in the Electronegativity (EN)
between the two bonding atoms, (∆EN) the bond is said to be
polar.
Example:
∆EN
Bond
Br2
0
electrons from one atom to another
HCl
0.9
polar-covalent
• Covalent—some valence electrons
NaF
3.1
ionic
• Ionic—complete transfer of 1 or more
non-polar
shared between atoms
• Most bonds are somewhere in
between.
97
98
23
Electronegativity:
Rank the following bonds by increasing strength:
The measure of the tendency for a given atom to polarize the
electrons in a covalent bond is called the Electronegativity (EN)
of an atom.
Electronegativity is related to the ionization energy and the
electron affinity of an atom.
H−F, H−Cl and H−I
Each bond has hydrogen in common so we have a basis for
comparison.
From the periodic table, in terms of
electronegativity:
F > Cl > I
EN increases
Periodic Table
Bond Strength
H−I < H−Cl < H−F
inc. EN
EN
decreases
One concludes that the H-F bond is the most
polar, followed by H-Cl then H-I.
99
Multiple bonds in covalent molecules:
Bond Strength and Bond Properties:
The oxygen and nitrogen that makes up the bulk of the atmosphere also
exhibits covalent bonding in forming diatomic molecules.
: :
→
O=O
“double bond”
Covalent bond strength increases with increasing ∆EN
example:
HCl bond is stronger than the HBr bond
Covalent bond strength increases with increasing bond order
:
:
N.
O:
: :
+
: :
: :
O:
100
+
N⋅
→
:N≡N:
“triple bond”
:
:
example:
O=O bond in stronger than O–O bond
triple > double > single
Polyatomic Molecules (More than two atoms)
Bond Order:
3
2
1
Bond length decreases with increasing bond order (Strength)
: :
: :
Carbon dioxide: CO2
O=C=O
example:
101
O=O bond is shorter than O–O bond
102
24
Heat Transfer:
Surroundings
heat in
heat out
System
H–bonding in water brings about a network of interactions which explain
phenomena such as:
capillary action
surface tension
q<0
(–)
∆E < 0
q>0
(+)
∆E > 0
why ice floats
103
Thermochemistry:
Energy in
E Final
Ef > Ei
Heat and the Specific Heat Capacity:
Energy out
Ef < Ei
— When heat is absorbed or lost by a body, the temperature
must change as long as the phase (s, g or l) remains constant.
E initial
work out
energy
work in
104
— The amount of heat (q) transfer is related to the mass and
temperature by:
q = m × C × ∆T
q = heat lost or gained (J)
q in

E Final
∆Esystem > 0
(+)
∆Esystem < 0
(–)
J

C = the Specific Heat Capacity of a compound  o 
 g⋅ C 
q out
E initial
m = mass of substance (g)
105
∆T is the temperature change in degrees Celsius or Kelvin’s
106
25
Enthalpy: ∆H
Ideal Gas Law
How many kJ of energy are released when 128.5 g of methane
(CH4(g)) is combusted?
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
128.5g CH 4 ×
P×V
= constant for a set number of gas moles
n×T
∆H = -802 kJ
802 kJ
mol CH 4
×
= 6.42×103 kJ
1mol CH 4
16.04 g
P×V = n×R ×T
How many hours would this power a 100 W light bulb if one could
use all of this energy?
100 W = 100 Js–1
6.42×103
103 J × s × min × hr =
kJ ×
100J
60s 60 min
kJ
R = "gas constant" = 0.08206
L ⋅ atm
mol ⋅ K
PV = nRT
17.8 hrs
107
Problems: How many grams of krypton (Kr) will it take to exert a
pressure of 11.2 atm in an 18.5 L at 28.2 oC.
PV = nRT
PV
n=
=
RT
108
What wavelengths correspond to FM radio (93.5 MHz) signals?
λ×υ = c
c
λ=
υ
L ⋅ atm
R = 0.082057
mol ⋅ K
3.00 × 108 m
11.2 atm × 18.5 L
0.08206
L ⋅ atm
× ( 28.2 + 273.15) K
mol ⋅ K
8.3789 mols Kr ×
83.80 g
= 702 g Kr
mol
= 8.3789 mols Kr
λ=
93.5 MHz ×
s
106 Hz 1s −1
×
1 MHz Hz
= 3.21 m
(3sf)
109
110
26
What wavelengths correspond to FM radio (93.5 MHz) signals?
λ×υ = c
c
λ=
υ
3.00 × 108 m
λ=
93.5 MHz ×
s
106 Hz 1s −1
×
1 MHz Hz
= 3.21 m
111
112
We can use the periodic table to determine the electron configuration by
counting:
Each element’s outermost electrons (valence) are related to the
elements position on the periodic table.
N: 7 electrons
1
3
2
4
5
1: 1s1
All of the subshells below the valence are full.
113
2: 1s2
3: 2s1
4: 2s2
1s2
2s2
5: 2p1
6
6: 2p2
7
7: 2p3
2p3
114
27
Electron Configurations cont…
Periodic Trends
Orbital box notation:
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
1s
2s
2p
3s
↑
3p
↑
3p
↑↓
Aluminum: Al (13 electrons)
Electron Configuration
3s
↑↓
↑↓
↑↓
↑↓
2p
2s
1s2 2s2 2p6 3s2 3p1
↑↓
1s
115
116
Kinetics: Rate Law & Reaction Order
Reaction Orders:
The reaction rate law expression relates the rate of a reaction to
the concentrations of the reactants.
Each concentration is expressed with an order (exponent).
The rate constant converts the concentration expression into the
correct units of rate (Ms−1). (It also has deeper significance,
which will be discussed later)
For the general reaction:
A reaction order can be zero, or positive integer and fractional number.
a A + b B → cC + d D
Rate = k [A]x [B]y
x and y are the reactant orders
determined from experiment.
x and y are NOT the
stoichiometric coefficients.
117
Order
0
1
2
0.5
1.5
0.667
Name
zeroth
first
second
one-half
three-half
two-thirds
118
28
Recognizing a first order process:
AÆproducts
Determining the Rate constant for a first order process
Taking the log of the integrated rate law for a first order process we find:
Whenever the conc. of a
reactant falls off
exponentially, the kinetics
follow first order.
ln ([A] = [A]o e − kt )
ln[A] = ln[A]o − k × t
A plot of ln[A] versus time (t)
is a straight line with slope -k
and intercept ln[A]o
[A] = [A]o e − kt
119
A certain reaction proceeds through t first order kinetics.
The half-life of the reaction is 180 s.
What percent of the initial concentration remains after 900s?
120
A certain reaction proceeds through t first order kinetics.
The half-life of the reaction is 180 s.
What percent of the initial concentration remains after 900s?
Using the integrated rate law, substituting in the value of k and 900s we find:
Step 1: Determine the magnitude of the rate constant, k.
k = 0.00385 s-1
t 1 = ln 2 = 0.693
2
k
k
k=
[A]
= e − kt
[A]o
ln 2
ln 2
=
= 0.00385s −1
t1
180s
−1
[A]
= e −0.00385 s × 900 s
[A]o
= 0.0312
Since the ratio of [A] to [A]0 represents the fraction of [A] that remains, the
% is given by:
2
100 × 0.0312 = 3.12%
121
122
29