Lecture

General Chemistry
402101-4

1
 ‫‪Lecture‬‬
‫االختبارات و‬
‫المواعيد المهمة‬

‫النظري ‪ 70‬درجة‬ ‫العملي ‪ 30‬درجة‬

‫‪ 40‬درجة اختبار‬ ‫‪ 10‬درجات كويز‬ ‫‪ 20‬درجة نصفي‬

‫نهائي‬ ‫األسبوع ‪ 13‬أو‬ ‫‪10‬درجات (اختبار)‬
‫أالسبوع ال‪10‬‬ ‫‪ 20‬درجة (اجراء‬
‫األسبوع ‪18‬‬ ‫‪14‬‬ ‫االسبوع ‪ 16‬أو ‪17‬‬ ‫التجارب)‬

‫‪2‬‬ ‫‪2‬‬
Lecture

Chapter
Matter
1

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What is Chemistry?

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Matter Any thing that occupies space and has space

Has a definite Separation by physical methods A combination of two

composition and or more substances in
distinct pure substance mixture which the substances
properties retain their distinct
identities

Separation by chemical methods

element compound heterogeneous homogeneous

composed of two The composition is not The composition is the

cannot be separated
different elements or uniform same throughout
into simpler substances
more chemically united
by chemical means
in fixed proportions.

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compound
NaCl
Salt water homogeneous mixture

Iron element

sugar compound

air homogeneous mixture

helium element

water compound
heterogeneous mixture
salad
compound element homogeneous mixture heterogeneous

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Element Compound

Molecules
containing
Atoms different two
Molecules atoms or
more

C, Al, Ti H2O, H2SO3

Polyatomic Diatomic

S8, O3 O2 , H2

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Matter States

The difference between the

states is the distance between
the molecules.

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Matter
properties

is a property when the Can be measured and

matter undergoes a observed without changing
chemical change or reaction Chemical Physical the composition or identity
of a substances

reactivity, color,
flammability mass,
size

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Measurable
properties of
matter

depends on how much Does not depend on how

matter is being considered extensive intensive much matter is being
considered

Mass Density
volume temperature

How can these properties be measured ?

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Measurement

Base Quantity Name of unit Symbol

Length meter m
Mass Kilogram Kg
Time Second s
Electrical current Ampere A
Temperature Kelvin K
Amount of substance Mole mol
Luminous intensity candela cd
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Prefix Symbol Multiple of Base Unit

Giga G 1,000,000,000 or 109
Mega M 1,000,000 or 106
kilo k 1,000 or 103
deci d 0.1 or 10-1
centi c 0.01 or 10-2
milli m 0.001 or 10-3
micro m 0.000001 or 10-6
nano n 10-9
pico p 10-12
Femto f 10-15

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Mass and weight

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Volume
Volume – SI derived unit for volume is cubic meter (m3)

1 cm3 = (1 x 10-2 m)3 = 1 x 10-6 m3

1 cm3 = 1 mL

1 L = 1000 mL = 1000 cm3 = 1 dm3

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Dimensional Analysis Method of Solving Problems

How many mL are in 1.63 L?

1000 mL
1.63 L x = 1630 mL
1L

1L L2
1.63 L x = 0.001630
1000 mL mL

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Density
Density is defined as the mass per unit volume.

density = mass/volume m S.I. units for density = kg/m3

d= V

g/cm3 for solids

g/ml for liquids
g/L for gases

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A piece of platinum metal with a density of 21.5 g/cm3 has a volume of

4.49 cm3. What is its mass?

m
d= V

m=dxV

m = 21.5 g/cm3 x 4.49 cm3 = 96.5 g

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Temperature
Temperature
scales

Fahrenheit 0F 9 0
=( x C )+ 32
oF 5
32 0F = 0 0C
212 0F = 100 0C
Celsius
oC
273 K = 0 0C
373 K = 100 0C
Kelvin
K T(in Kelvin) = T(in Celsius) + 273.15
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Precision and Accuracy

Student A Student B Student C
1.964 g 1.972 g 2.000 g
1.978 g 1.968 g 2.002 g
Average 1.971 g 1.970 g 2.001 g

The true mass of object= 2.000 g

Precision: how close a set of measurements are to each other

(reproducibility).
Accuracy: how close your measurements are to the true value.

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Significant Figures
• Any digit that is not zero is significant
1.234 kg 4 significant figures
• Zeros between nonzero digits are significant
606 m 3 significant figures
• Zeros to the left of the first nonzero digit are not significant
0.08 L 1 significant figure
• If a number is greater than 1, then all zeros to the right of the decimal point
are significant
2.0 mg 2 significant figures
• If a number is less than 1, then only the zeros that are at the end and in the
middle of the number are significant
0.00420 g 3 significant figures
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Exact Numbers
Numbers from definitions or numbers of objects are considered
to have an infinite number of significant figures

The average of three measured lengths; 6.64, 6.68 and 6.70?

6.64 + 6.68 + 6.70

= 6.67333 = 6.67 =7
3

Because 3 is an exact number

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Scientific Notation
N x 10n The number of atoms in 12 g of carbon:
n is a positive or 602,200,000,000,000,000,000,000
negative integer
N is a number 6.022 x 1023
between 1 and 10
The mass of a single carbon atom in grams:
0.0000000000000000000000199
1.99 x 10-23

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How many significant figures are in each of the following

measurements?

Tip: start to count the sig. fig. from the left when you see a non zero number until the end of the number.

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Significant Figures: Addition & Subtraction

If addition or subtraction:
1- must have same power before addition or subtraction
2- sig. fig. in the answer is as the smaller digits after decimal point

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Significant Figures: Multiplication & Division

If multiplication or division:
1- add exponent for multiplication or subtract exponent for division
2- write the answer with the smaller sig. fig.

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First
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lecture Questions
Question 1 Question 3
Which of the following is an example of a Convert 240 K and 468 K to the Celsius scale.
physical property? A) 513oC and 741oC
A) combustibility
B) -59oC and 351oC
B) corrosiveness
C) -18.3oC and 108oC
C) explosiveness
D) -33oC and 195oC
D) density
E) A and D
Question 4
Question 2 Calculate the volume occupied by 4.50 X 102 g of
Which of the following represents the greatest gold (density = 19.3 g/cm3).
mass? A) 23.3 cm3
A) 2.0 x 103 mg B) 8.69 x 103 cm
B) 10.0 dg C) 19.3 cm3
C) 0.0010 kg D) 450 cm3
D) 1.0 x 106 μg
E) 3.0 x 1012 pg
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Question 5 Question 7
The melting point of bromine is -7oC. What How many significant figures should you
is this melting point expressed in oF? report as the sum of 8.3801 + 2.57?
A) 45oF A) 3

B) -28oF B) 5

C) -13oF C) 7

D) 19oF D) 6

E) None of these is within 3oF of E) 4

the correct answer.
Question 6 Question 8
How many significant figures are there in How many significant figures are there in the
the measurement 3.4080 g? number 0.0203610 g?
A) 6 A) 8
B) 5 B) 7
C) 4 C) 6
D) 3 D) 5
E) 2

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lecture Questions
Question 9 Question 11
The value of 345 mm is a measure of A laboratory technician analyzed a sample
A) temperature three times for percent iron and got the
following results: 22.43% Fe, 24.98% Fe, and
B) density 21.02% Fe. The actual percent iron in the
sample was 22.81%. The analyst's
C) volume
A) precision was poor but the
D) distance average result was accurate.
E) Mass B) accuracy was poor but the
precision was good.
Question 10 C) work was only qualitative.
The measurement 0.000 004 3 m,
expressed correctly using scientific D) work was precise.
notation, is E) C and D.
A) 0.43 x 10-5 m
B) 4.3 x 10-6
C) 4.3 x 10-7
D) 4.3 x 10-5

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Atomic,
Molecular Weights Chapter
&
Moles Calculations 2
in Chemical Equations

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Atomic Mass
The mass of an atom in atomic mass units (amu)

6 Atomic number The atomic mass of

elements is relative to a
C standard atom 12 C
(6 protons, 6 neutrons)
12.01 Atomic mass

Molar Mass (Atomic weight Aw)

The mass of an element atoms per one mole (g/mol)
= Atomic Mass numerically

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Mole (mol)
The amount of a substance that contains as many
elementary particles (atoms, molecules or ions), where
each mole has number of 6.022 × 1023 particles.

1 mole= 6.022 × 1023 particles = Avogadro’s number Na

1 mol Al = 6.02 × 1023 atoms

1 mol CO2 = 6.02× 1023 molecules
23 + 23 -
1 mol NaCl = 6.02× 10 Na ions = 6.02× 10 Cl ions

The number of atoms in exactly 12 g of 12C is one mole

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Lecture

Molar Mass ( Atomic weight Aw):

mass (weight) of 1 mole of atoms in grams

1 mol C atoms = 12.01 g Aw of C = 12.01* g/mol

1 mol Cl atoms = 35.45 g Aw of Cl = 35.45* g/mol
1 mol Fe atoms = 55.85 g Aw of Fe = 55.85* g/mol

*( get from periodic table)

Think: What is the difference between the mass and weight?

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Molar Mass ( Molecular weight Mw):

The sum of atomic weights of 1 mol of the molecule

Mw of 1 mol of H2O = 2 (Aw of H) + Aw of O

= (2× 1.008) + 16
= 18.02 g/mol

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Lecture

Learning check
What are the molecular weights of the following:
C2H6
N2O4
C8H18O4N2S
Al2(CO3)3
MgSO4.7H2O

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Number of moles (n)

wt  g 
n
Mwg / mol 
Remember: No. of particles = No. of moles × Avogadro's number

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Example
Helium (He) is a valuable gas used in
industry, low temperature research, deep-sea
diving tanks, and balloons. How many moles of He
atoms are in 6.46 g of He?

1 mol of He = 4.003 g i.e. Aw = 4.003 g/mol

n = wt/Aw = 6.46/4.003 = 1.61 mol He

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Example

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Learning check
What is the number of moles in 21.5 g CaCO3?

What is the mass in grams of 0.6 mol C4H10?

How many atoms of Cu are present in 35.4 g of Cu?

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Lecture

Percent Composition of Compounds

Mass percent (weight percent) of each element in a

compound.

n  Aw ( x)
%x  100
Mw

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Lecture

Example
n  Aw ( x )
%x   100
Mw
Calculate the mass percent of each element in ethanol (C2H5OH) ?
Mass of 1 mol (molar mass) of C2H5OH = 24.02+6.048+16.00= 46.07 g/mol
2 x 12.01 g/mol
Mass percent of C = x 100 = 52.14 % (4 sf)
46.07 g/mol
6 x 1.008 g/mol
Mass percent of H = x 100 = 13.13 % (4 sf)
46.07 g/mol
1 x 16.00 g/mol
Mass percent of O = x 100 = 34.73 % (4 sf)
46.07 g/mol
Total mass = 52.14 +13.13 + 34.73 =100%
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Lecture

Percent composition

Determining the Formula of a Compound

empirical formula molecular formula

CH2O C6H12O6
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑀𝑓
Molecular formula = (Empirical formula)𝒙 , 𝑥 =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝐸𝑓

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Lecture

Determine the empirical and molecular formulas for a compound

that gives the following mass: 30.46 % N 69.54 % O
Molar mass = 92.02 g/mol

Firstly: Determination of the empirical formula:

Assume we have a 100.00 g sample

WN = 30.46 g, WO = 69.54 g

1- Calculate number of moles of each element in 100 g of a compound

nN = (30.46 g)/(14.01 g mol-1) = 2.174 mol N round to 2sf

nO = (69.54 g)/(16.00 g mol-1) = 4.346 mol O round to 2sf

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Lecture

2 -Determine the molar ratio by dividing by the smallest subscript mole

nN : nO
2.2 : 4.3
1 : 2
The empirical formula NO2
Secondly: Determination of molecular formula
1- Calculate the empirical formula mass
Empirical molar mass = (1 x 14.01) + (2 x 16.00) = 46.01 g
2- Calculate the ratio between the molar mass and the empirical molar mass:
𝑀𝑓
𝑥= = (92.02 g)/(46.01 g) =2
𝐸𝑓

Molecular formula = (NO2)𝒙 = (NO2)2 = N2O4

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Lecture
Learning check

wt  g 
n
Mwg / mol 

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Lecture

Chemical Reactions
Reactants Products

A process in which one or more substances is

changed into one or more new substances.

2H2 (g) + O2 (g) 2H2O (l)

2HgO (s) 2Hg (l) + O2 (g)

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Lecture

Chemical Equations
It is a way to represent the chemical reaction.
It shows us:
• The chemical symbols of reactants and products
• The physical states of reactants and products– (s), (l), (g), (aq)
• Balanced equation (same number of atoms on each side)

2H2 (g) + O2 (g) 2H2O (l)

Reactants (starting materials) Products ( materials formed)

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Balancing Chemical Equations

The number of atoms of each element must be the same on
both sides of the equation.
C2H6 + O2 CO2 + H2O C2H6 + 7/2O2 2CO2 + 3H2O
2C2H6 + 7O2 4CO2 + 6H2O
Reactants Products Reactants Products
2C 1C 4C 4C
6H 2H 12 H 12 H
2O 3O 14 O 14 O

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Lecture
Balance the following equations:

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Lecture

Stoichiometry

The quantitative study of reactants and

products in a chemical reaction

CH4 + 2O2  CO2 +2 H2O

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Lecture

The Mole Method:

Stoichiometric coefficients in a chemical equation can be
interpreted as the number of moles of each substance.
N2 (g) + 3 H2 (g)  2NH3 (g)

1 mole
N2 6.022 ×1023 molecules
3 mole
H2 3×6.022 ×1023 molecules
2 mole
NH3 2×6.022 ×1023 molecules

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Lecture
Mole Ratios
N2 (g) + 3 H2 (g)  2NH3 (g)

Recall that the coefficient on N2 is 1 but is not explicitly

written in the reaction Coefficients:

N2 = 1
H2 = 3
NH3 = 2

Using the coefficients we can write mole ratios

Definition: mole ratio gives the relative amounts of reactants

and products

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N2 (g) + 3 H2 (g)  2NH3 (g)

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MOLE to MOLE Stoichiometry

N2 (g) + 3 H2 (g)  2NH3 (g)

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Lecture
Example
Silicon tetrachloride (SiCl4) can be prepared by heating Si
in chlorine gas:
Si (s) + 2Cl2 (g) SiCl4 (l)
In one reaction, 0.507 mole of SiCl4 is produced. How many
moles of molecular chlorine were used in the reaction?
Cl2 SiCl4
2 mol 1 mol
?? 0.507 mol

n of Cl2 used = (0.507 molSiCl4 ×( 2 molCl2 /1 molSiCl4)

= 1.01 molCl2
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Second lecture
Lecture
Example
If 85.0 g of CH4 is consumed by a person over a certain period,
what is the mass of CO2 produced?
CH4 + 2O2  CO2 + 2H2O
Convert 85.0 g to moles:
n (CH4) = wt/Mw = (85.0/16.04) = 5.30 mol(CH4)
CH4 + 2O2  CO2 + 2H2O
1 mole of CH4 1 mole of CO2
5.30 moles ??
n (CO2) = 5.30 mol(CH4) × (1mol (CO2) /1mol (CH4) ) = 5.30 mol(CO2)
wt (CO2) = n × Mw = 5.30 mol(CO2)× (44.01g (CO2) / 1mol (CO2) )
= 233.2 g = 2.33 × 102 g (CO2)
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Lecture

You can use this version of the mole map to solve stoichiometry problems.

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Second lecture
Lecture Questions
Question 1 Question 3
Determine the number of moles of aluminum One mole of H2
in 0.2154 kg of Al. A) contains 6.0 x 1023 H atoms
A) 1.297 x 1023 mol B) contains 6.0 x 1023 H2 molecules
B) 5.811 x 103 mol C) contains 1 g of H2
C) 7.984 mol D) is equivalent to 6.02 x 1023 g of H2
D) 0.1253 mol E) None of the above
E) 7.984 x 10-3 mol
Question 4
Question 2 How many oxygen atoms are present in 5.2 g of
How many phosphorus atoms are there in O2?
2.57 g of P? A) 5.4 x 10-25 atoms
A) 4.79 x 1025 B) 9.8 x 1022 atoms
B) 1.55 x 1024 C) 2.0 x 1023 atoms
C) 5.00 x 1022 D) 3.1 x 1024 atoms
D) 8.30 x 10-2 E) 6.3 x 1024 atoms
E) 2.57

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Lecture Questions
Question 5 Question 7
How many protons and neutrons are in Determine the mass percent of iron in
sulfur-33? Fe4[Fe(CN)6] 3.
A) 2 protons, 16 neutrons A) 45% Fe
B) 26% Fe
B) 16 protons, 31 neutrons
C) 33% Fe
C) 16 protons, 17 neutrons D) 58% Fe
E) None of the above
D) 15 protons, 16 neutrons
Question 6 Question 8
What is the mass of 5.45 x 10-3 mol of What is the coefficient for CO2 when the
glucose, C6 H12O6? following chemical equation is properly
A) 0.158 g balanced using the smallest set of whole
B) 982 g numbers?
C) 3.31 x 104 g C4H10 + O2 ----> CO2 + H2O
D) 0.982 g A) 1
E) None of the above. B) 4
C) 6
D) 8
E) 12

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Second lecture
Lecture Questions
Question 9
Calculate the number of moles of H2O formed
when 0.200 mole of Ba(OH)2 is treated with 0.500
mol of HClO3 according to the chemical reaction
shown below.
Ba(OH)2+2HClO3 ----> Ba(ClO3)2 + 2 H2O
A) 1.00 mol
B) 0.600 mol
C) 0.500 mol
D) 0.400 mol
E) 0.200 mol

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Lecture

Chemical Reactions in Chapter

3
Solutions: Acids & Bases
Concentrations & pH
Calculations

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Lecture

Solutions
Solution: a homogeneous mixture of
two or more substances
Solute: a substance that is being
dissolved (smaller amount)
Solvent: a substance which dissolves
Solute a solute (larger amount)
particle

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Lecture

Concentrations
The concentration of a solution is the amount of solute
present in a given quantity of a solvent or solution.

Concentrations

Molar Percentages
concentrations

Mole
Molarity Molality Normality
fraction
v/v w/w w/v

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Lecture

Molarity
The number of moles of solute dissolved in one liter of
solution.

What is the unit of molarity?

What is the relationship between weight and molarity?

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Lecture

Example
A solution has a volume of 2.0 L and contains 36.0 g of
glucose (C6H12O6). If the molar mass of glucose is 180
g/mol, what is the molarity of the solution?

No. of mol of glucose = wt (g) / Mw (g/mol) = 36.0 g /180 g/mol

= 0.2 mol

M = n (mol) / V (L) = 0.2 mol /2.0 L = 0.1 mol/L

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Lecture

Molality
The number of moles of solute dissolved in one kilogram of solvent

Molality (m) Molarity (M)

moles of solute moles of solute
m = M =
mass of solvent (kg) liters of solution

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 Example
Third lecture
Lecture

What is the molality of a 5.86 M ethanol (C2H5OH) solution

whose density is 0.927 g/mL?
moles of solute
m =
mass of solvent (kg)
Assume 1 L of solution:
5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL x 0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg

moles of solute 5.86 moles C2H5OH

m = = = 8.92 m
mass of solvent (kg) 0.657 kg solvent

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Lecture

Learning check

What is the concentration of a solution in mol/L when 80 g

of calcium carbonate, CaCO3, is dissolved in 2 L of
solution?

How many liters of 0.25 M NaCl solution must be measured

to obtain 0.1 mol of NaCl?

A student needs to prepare 250 ml of 0.1 M of Cd(NO3)2

solution. How many grams of cadmium nitrate are
required?

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Third lecture
Lecture Questions
Question 1 Question 3
Molarity is the number of …… of solute dissolved Molarity is the number of moles of solute dissolved
Solution 1 …… of the Solution
a) Grams a) Grams
b) Milliliter b) Liter
c) Second c) Second
d) moles d) moles

Question 2 Question 4
Molality is the number of moles of ……. A solution has a volume of 2.0 L and contains 36.0 g of
dissolved in 1kg solvent glucose (C6H12O6). If the molar mass of glucose is 180
g/mol, what is the molarity of the solution
a) Solvent
a) 1.0
b) Solute
b) 1.00
c) Solution
c) 0.1
d) acid
d) 0.01

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Third lecture
Lecture Questions
Question 5 Question 7
How many liters of 0.25 M NaCl solution must be A student needs to prepare 250 ml of 0.1 M of Cd(NO3)2
measured to obtain 0.1 mol of NaCl
solution. How many grams of cadmium nitrate are required?
A) 1
(Molecular weight of Cd(NO3)2 = 236 g/mol
B) 2
A) 5.9
C) 2.5 B) 5.1
C) 5.4
D) 3.5 D) 5.6
Question 6
What is the concentration of a solution in mol/L when 80
g of calcium carbonate, Ca(CO3)2, is dissolved in 2 L of
solution? (Molecular weight of Ca(CO3)2 = 100g/mol
A) 0.4
B) 4
C) 0.004
D) 1

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Lecture

Chapter
Chemical
Equilibrium 4
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Fourth lecture
Lecture
Equilibrium
Equilibrium is a state in which there are no observable changes as time goes by
Chemical equilibrium is achieved when:
• the rates of the forward and reverse reactions are equal and
• the concentrations of the reactants and products remain constant
Chemical equilibrium Physical equilibrium
N2O4 (g) 2NO2 (g) H2O (l) H2O (g)

Physical equilibrium is between two

states of the same substance

N2O4 Equilibrium NO2

Colorless state Dark-brown

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Fourth lecture
Lecture

N2O4 (g) 2NO2 (g)

equilibrium equilibrium

equilibrium

Start with NO2 Start with N2O4 Start with NO2 & N2O4

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Lecture

Equilibrium Constant K
N2O4 (g) 2NO2 (g)

[NO2]2 2
PNO
Kc = Kp = 2
[N2O4] PN2O4
[NO2]2
K= = 4.63 x 10-3
[N2O4]

aA + bB cC + dD

[C]c[D]d
K= Law of Mass Action
[A]a[B]b

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Lecture
Equilibrium Position

K>>1 K<<1
Products are favored Reactants are favored
at equilibrium at equilibrium
(the equilibrium lie to the right) (the equilibrium lie to the left)

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Fourth lecture
Lecture
Relation between Kc and Kp
N2O4 (g) 2NO2 (g)

[NO2]2 P2
Kc = NO2
Kp =
[N2O4] P
N2O4
In most cases

Kc  Kp

aA (g) + bB (g) cC (g) + dD (g)

Kp = Kc (RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)

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Fourth lecture
Lecture
Homogeneous Equilibrium
Homogenous equilibrium applies to reactions in which all reacting species
are in the same phase.

CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)

′ [CH3COO-][H3O+]
Kc = [H2O] = constant
[CH3COOH][H2O]

[CH3COO-][H3O+]
Kc = = Kc′ [H2O]
[CH3COOH]

General practice not to include units for the equilibrium constant.

76
Fourth lecture
Lecture

The equilibrium concentrations for the reaction between carbon monoxide and
molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M,
and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.

CO (g) + Cl2 (g) COCl2 (g)

[COCl2] 0.14
Kc = = = 220
[CO][Cl2] 0.012 x 0.054

Kp = Kc(RT)Dn

Dn = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K

Kp = 220 x (0.0821 x 347)-1 = 7.7

77
Fourth lecture
Lecture

The equilibrium constant Kp for the reaction: 2NO2 (g) 2NO (g) + O2 (g)

is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm
and PNO = 0.270 atm?

78
Fourth lecture
Lecture
Heterogeneous Equilibrium
Heterogenous equilibrium applies to reactions in which reactants and products
are in different phases

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

[CaO][CO2]
Kc′ = [CaCO3] = constant
[CaCO3] [CaO] = constant

[CaCO3]
Kc = [CO2] = Kc′ x
[CaO]

The concentration of solids and pure liquids are not included in the
expression for the equilibrium constant.

79
Fourth lecture
Lecture

Consider the following equilibrium at 295 K:

NH4HS (s) NH3 (g) + H2S (g)

The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction?

80
Fourth lecture
Lecture
Reaction Quotient Qc
The reaction quotient (Qc) is calculated by substituting the initial concentrations
of the reactants and products into the equilibrium constant (Kc) expression.
• Qc > Kc system proceeds to left to reach equilibrium
• Qc = Kc the system is at equilibrium
• Qc < Kc system proceeds to right to reach equilibrium

81
Fourth lecture
Lecture

• Find the value of Q and determine which side of the reaction is favored. Given Keq=0.5
HCl (g) + NaOH (aq) ⇌ NaCl (aq)+ H2O (l)
[HCl]= 3.2 M [NaOH]= 4.3 M [NaCl]=6 M

𝐍𝐚𝐂𝐥 𝟔
𝐐𝐜 = = = 0.436
𝐇𝐂𝐥 [𝐍𝐚𝐎𝐇] (𝟑.𝟐) (𝟒.𝟑)

Qc = 0.436 … Q is less than Keq so the reaction shifts

RIGHT, favors the products.

82
Fourth lecture
Lecture

Chemical Kinetics & Chemical Equilibrium

kf ratef = kf [A][B]2
A + 2B AB2
kr rater = kr [AB2]

Equilibrium
ratef = rater

kf [A][B]2 = kr [AB2]

kf [AB2]
= Kc =
kr [A][B]2

83
Fourth lecture
Lecture
Equilibrium Constant Calculations
• If a reaction can be expressed as the sum of two or more reactions,
the equilibrium constant for the overall reaction is given by the
product of the equilibrium constants of the individual reactions.

A+B C+D Kc′ [C][D] [E][F]

Kc′ = Kc′′=
[A][B] [C][D]
C+D E+F Kc′′
A+B E+F Kc [E][F]
Kc =
[A][B]

Kc = Kc′ x Kc′′

84
 Lecture
Fourth lecture
Equilibrium Constant Calculations
• When the equation for a reversible reaction is written in the opposite
direction, the equilibrium constant becomes the reciprocal of the
original equilibrium constant.

N2O4 (g) 2NO2 (g) 2NO2 (g) N2O4 (g)

[NO2]2 [N2O4] 1
K= = 4.63 x 10-3 K′ = = = 216
[N2O4] [NO2] 2 K

85
Fourth lecture
Lecture
Equilibrium Calculations
A closed system initially containing 0.001M H2 and 0.002 M I2 is allowed to reach equilibrium
at 448oC.
Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M.
Calculate Kc at 448oC for the reaction taking place, which is:
H2(g)  I2(g)  2HI(g)
What Do We Know?

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0
Change
At equilibrium 1.87 x 10-3

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Fourth lecture
Lecture

[HI] Increases by 1.87 x 10-3 M

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At 1.87 x 10-3
equilibrium

87
Fourth lecture
Lecture
Stoichiometry tells us [H2] and [I2] decrease by
half as much …
H2(g)  I2(g)  2HI(g)

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At 1.87 x 10-3
equilibrium

88
Fourth lecture
Lecture

We can now calculate the equilibrium

concentrations of all three compounds…

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

equilibrium

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Fourth lecture
Lecture

…and, therefore, the equilibrium constant is

calculated as:

 
2
3
[HI]2 1.87x10
Kc    51

[H2 ][I2 ] 6.5x10
5
1.065x10 
3

90
Fourth lecture
Lecture
Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the system adjusts in such a
way that the stress is partially offset as the system reaches a new equilibrium position.

I. Changes in Concentration

N2 (g) + 3H2 (g) 2NH3 (g)

Equilibrium
Add
shifts left to
NH3
offset stress

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Fourth lecture
Lecture
Changes in Concentration continued

Add Add

aA + bB cC + dD

Remove Remove

Change Shifts the Equilibrium

Increase concentration of product(s) left
Decrease concentration of product(s) right
Increase concentration of reactant(s) right
Decrease concentration of reactant(s) left
92
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Fourth lecture
Lecture

II. Changes in Volume and Pressure

A (g) + B (g) C (g)

Change Shifts the Equilibrium

Increase pressure Side with fewest moles of gas
Decrease pressure Side with most moles of gas
Increase volume Side with most moles of gas
Decrease volume Side with fewest moles of gas

93
Fourth lecture
Lecture

III. Temperature Changes

Consider heat as a product in exothermic reactions

A + B = AB + Heat
Add heat  Shift to reactants
Remove heat  Shift to products

Consider heat as a reactant in endothermic reactions

A + B + heat = AB
Add heat  Shift to products
Remove heat  Shift to reactants

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Fourth lecture
Lecture

Adding a Catalyst
• does not change K
• does not shift the position of an equilibrium system
• system will reach equilibrium sooner

Catalyst lowers Ea for both forward and reverse reactions.

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Fourth lecture
Lecture
Le Châtelier’s Principle - Summary

Change Shift Equilibrium Change Equilibrium Constant

Concentration yes no

Pressure yes* no

Volume yes* no

Temperature yes yes

Catalyst no no

*Dependent on relative moles of gaseous reactants and products

96
Fourth lecture
Lecture Questions
Question 1 Question 3
Which equilibrium in gaseous phase would be The equilibrium constant (Kc) for the reaction is
unaffected by an increase in pressure: 2SO3(g) -> 2SO2(g) + O2(g)
system as described by the above equation is:
(a) N2O4 ->2NO2
(b) N2 + O2 ->2NO (a) [SO2]2/[SO3] (b) [SO2]2[O2]/[SO3]2
(c) N2 + 3H2 ->2NH3 (c) [SO3]2/[SO3]2[O2] (d) [SO2][O2]
(d) CO + ½ O2 ->O2 + CO2

Question 2 Question 4
At equilibrium, __________.
For the equilibrium ,
2NO2(g) -> N2O4(g) + 14.6 kcal
(a) the rates of the forward and reverse reactions are equal
An increase of temperature will:
(b) the rate constants of the forward and reverse reactions are equal
(c) all chemical reactions have ceased
(a) Favour the formation of N2O4
(d) the value of the equilibrium constant is 1
(b) Favour the decomposition of N2O4
(c) Not affect the equilibrium
(d) Stop the reaction

97
Fourth lecture
Lecture Questions
Question 5 Question 7
The value of Keq for the following reaction is 0.25: What is the correct equilibrium constant expression for the
SO2 (g) + NO2 (g) -> SO3 (g) + NO (g) following reaction? 2 Cu(s) + O2(g) → 2 CuO(s)
The value of Keq at the same temperature for the
reaction below is __________. (a) Keq = 1/[O2]2
2SO2 (g) + 2NO2 (g) -> 2SO3 (g) + 2NO (g) (b) Keq = [CuO]2/[Cu]2[O2]
(c) Keq = [O2]
(a) 0.062 (d) Keq = 1/[O2]
(b) 16
(c) 0.25
(d) 0.50 Question 8
What is the relationship of the equilibrium constants for the
Question 6 following two reactions?
(1) 2 NO2(g) ↔ N2O4(g);
Consider the reaction: 2 SO2(g) + O2(g) ↔ 2
K1
SO3(g). If, at equilibrium at a certain temperature, (2) N2O4(g) ↔ 2 NO2(g)
[SO2] = 1.50 M, [O2] = 0.120 M, and [SO3] = 1.25 K2
M, what is the value of the equilibrium constant?
(a) K1= 1/ K2 (b) K2= 1/ K1
(a) 5.79 (b) 6.94 (c) K1= K2 (d) both a and b are correct
(c) 8.68 (d) 0.14

98
Fourth lecture
Lecture Questions
Question 9 Question 11
Consider the following endothermic reaction: Which of these four factors can change the value of the
H2(g) + I2(g) ↔ 2 HI(g). If the temperature is increased, equilibrium constant?

(a) more HI will be produced (a) catalyst

(b) some HI will decompose, forming H2 and I2 (b) pressure
(c) the magnitude of the equilibrium constant will (c) concentration
decrease (d) temperature
(d) the pressure in the container will increase
Question 12
Question 10 Which general rule helps predict the shift in direction of an
Consider the following reaction at equilibrium: equilibrium reaction?
NO2(g) + CO(g) ↔ NO(g) + CO2(g). Suppose the volume
of the system is decreased at constant temperature, (a) Le Chatelier's principle (b) Haber process
what change will this cause in the system? (c) Equilibrium constant (d) Bosch theory

(a) A shift to produce more NO

(b) A shift to produce more CO
(c) A shift to produce more NO2
(d) No shift will occur

99
Fifth lecture
Lecture

Type of Chemical Reactions Chapter

in Aqueous Solutions
Ionic equilibrium 5
100
Fifth lecture
Lecture

Type of Chemical Reactions

in Aqueous Solutions
1) Acid-Base Reactions

2) Oxidation-Reduction Reactions

3) Precipitation Reactions

101
Fifth lecture
Lecture

I. Acid-Base Reactions
acid + base → salt + water

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

102
Fifth lecture
Lecture

II. Oxidation-Reduction Reactions

Redox reactions are electron transfer reactions
Mg + 2Ag+ →Mg2++2Ag
Half-reactions:
Mg (s) → Mg2++ 2e
2Ag+ + 2e→2Ag

Oxidation Reactions : half-reaction that involves a loss of electrons

Reduction Reactions : half-reaction that involves a gain of electrons

103
Fifth lecture
Lecture

III. Precipitation Reactions

A precipitate is an insoluble solid that separates from
the solutions
Pb(NO3)2 (aq) + 2KI (aq) → PbI2 (s) + 2KNO3 (aq)
Pb2+ (aq) +2NO3- (aq) + 2K+ (aq) + 2I- (aq) → PbI2 (s) + 2K+ (aq) +2NO3- (aq)
Pb2+ (aq) + 2I- (aq) → PbI2 (s)

104
Fifth lecture
Lecture

Acids & Bases

Definition of acids and bases

Arrhenius Brønsted-Lowry
concept concept

Lewis
concept
105
Fifth lecture
Lecture

1- Arrhenius Concept
An acid is a compound that releases H+ ions in water
A base is a compound that releases OH- in water.

HCl (aq) H+ (aq) + Cl- (aq)

NaOH (aq) Na+ (aq) + OH- (aq)

Limitations: Some bases do not contain OH-

106
Fifth lecture
Lecture

2- Brønsted-Lowry Concept
An acid is any molecule or ion that can donate a proton
H+. A base is any molecule or ion can accept a proton.

•proton-transfer reaction

107
Fifth lecture
Lecture

3- Lewis Concept
An acid as an electron pair acceptor and a base as an
electron pair donor.
:

:
:F H :F H
: :

: :
:B
F + :N H :B
F N H
:F H :F H
:

:
Another examples: hydration of AlCl3, BCl3, OH-

108
Fifth lecture
Lecture

Strength of Acids and Bases

A strong acid or base ionizes completely in water

Strong Acids Strong bases

HCl LiOH
HBr NaOH
HI KOH
HNO3 Ca(OH) 2
H2SO4 Sr(OH) 2
HClO4 Ba(OH) 2

109
Fifth lecture
Lecture

Weak Acids and Bases

A weak acid or base ionizes only to a limited extent in
water

Examples: CH3COOH, NH3

110
Fifth lecture
Lecture

Acid or Base Ionization Constant

It is a measure of the strength of acid or base.
The ionization constant has the same equilibrium expression.

CH3COOH + H2O CH3COO- + H3O+

[CH 3COO  ] [ H 3O  ]
Ka 
[CH 3COOH ]
NH3 + H2O NH4+ + HO-

[ NH 4 ] [ HO ]
Kb 
[ NH 3 ]

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Fifth lecture
Lecture

Self-ionization of water
Water acts either as an acid or a base
H 2O(l)  H 2O(l)  H 3O (aq)  OH (aq)

 
K w  [H 3O ][OH ]
Or
 
K w  [ H ][OH ]

Kw = water dissociation constant

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Fifth lecture
Lecture

Self-ionization of water
 
K w  [ H ][OH ]
K w  1.0 1014 at 25 oC
  14 7
[ H ]  [OH ]  1.0 10  1.0 10
At 25°C, you observe the following conditions.

an acidic solution, [H+] > [OH-]

a neutral solution, [H+] = [OH-]
a basic solution, [H+] < [OH-]

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Fifth lecture
Lecture
pH of Solutions
The pH of a solution is defined as the negative logarithm of the
molar hydrogen-ion concentration


pH   log[ H ]

pH  pOH  14.00
In a neutral solution, whose hydrogen-ion
concentration is 1.0 x 10-7, the pH = 7.00

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Fifth lecture
Lecture

At 25°C, you observe the following conditions

In an acidic solution, [H+] > 1.0 x 10-7 M, pH<7

In a neutral solution, [H+] = 1.0 x 10-7 M, pH=7

In a basic solution, [H+] < 1.0 x 10-7 M, pH>7

115
Fifth lecture
Lecture

Example
For a solution in which the hydrogen-ion concentration
is 1.0 x 10-3, the pH is:
3
pH   log(1.0  10 )  3.00
Note that the number of decimal places
in the pH equals the number of
significant figures in the hydrogen-ion
concentration

116
 Examples
Fifth lecture
Lecture

The hydrogen ion concentration of a fruit juice is

3.3 x 10-2 M. What is the pH of the juice? Is it
acidic or basic?
2
pH   log( 3.3 10 )  (1.48)  1.48
If a solution has pH of 5.50, calculate its [OH-]
14  pH  pOH
pOH  14.00  5.50  8.50
pOH   log[ OH  ]
log[ OH  ]  8.50

117
Fifth lecture
Lecture

pH of Strong Acids and Bases

Dissociation of a strong base:

NaOH Na+ + OH-

complete dissociation of a base
and no base in the form of NaOH will be left in solution

pOH = -log[OH-]

pH = 14 - pOH = 14 + log [OH-]

118
Fifth lecture
Lecture

Example
An ammonia solution has a hydroxide-ion concentration
of 1.9 x 10-3 M. What is the pH of the solution?
You first calculate the pOH:
3
pOH   log(1.9  10 )  2.72
Then the pH is:

pH  14.00  2.72  11 .28

119
Fifth lecture
Lecture

pH of Weak Acids and Bases

Examples: Ka (HF)=7.1 x 10-4 , Ka (HCOOH)=1.7 x 10-4

[A-][H+] x2 x2
HA A- + H+ Ka = = =
[HA] c-x c
c-x x x
c-x = concentration of an acid at equilibrium c >> x
x = concentration of products at equilibrium for diluted
c = concentration of an acid at the beginning weak acids

[H+] = x = (Ka c)1/2

pH = -log [H+] = -log (Ka c)1/2 pKa = -logKa

120
Fifth lecture
Lecture Questions
Question 1 Question 4
The solution with the lowest pH is Which of the following describes the relationship
A. 1.0M HF between [H3O+] and [OH-]?
B. 1.0M HCN A. [H3O+][OH-] = 14.00
C. 1.0M HCOOH B. [H3O+] + [OH-] = 14.00
D. 1.0M CH3COOH C. [H3O+][OH-] = 1.0 x 10-14
D. [H3O+] + [OH-] = 1.0 x 10-14
Question 2
As the [H3O+] in a solution decreases, the [OH-] Question 5
A. increases and the pH increases What is the pOH of 0.1 M NaOH?
B. increases and the pH decreases A. 1
C. decreases and the pH increases B. 0.0032
D. decreases and the pH decreases C. 0.40
D. 13.60
Question 3
The value of pKw at 25°C is Question 6
A. 1.0 x 10-14 The pH of a solution for which [OH–] = 1.0 x 10–6 is
B. 1.0 x 10-7 A. 1.00
C. 7.00 B. 8.00
D. 14.00 C. 6.00
D. –6.00

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Fifth lecture
Lecture Questions
Question 7 Question 10
The ionization of water at room temperature is Addition of HCl to water causes
represented by A. both [H3O+] and [OH-] to increase
A. H2O = 2H+ + O2- B. both [H3O+] and [OH-] to decrease
B. 2H2O = 2H2 + O2 C. [H3O+] to increase and [OH-] to decrease
C. 2H2O = H2 + 2OH- D. [H3O+] to decrease and [OH-] to increase
D. 2H2O = H3O+ + OH-
Question 11
Question 8
Which of the following statements concerning
According to the Bronsted-Lowry theory, a base is a(n)
Arrhenius acids and Arrhenius bases is correct?
A. proton donor
A. In the pure state, Arrhenius acids are
B. proton acceptor
covalent compounds.
C. electron donor
B. In the pure state, Arrhenius bases are
D. electron acceptor
ionic compounds
C. Dissociation is the process by which
Question 9 Arrhenius acids produce H+ ions in
the pH of 1.0 M acetic acid (Ka is 1.86 x10-5 at 20 °C). solution
A. 1.37 D. Arrhenius bases are also called hydroxide
B. 2.37 bases
C. 3.73
D. 4.73

122
Sixth lecture
Lecture

Chapter
Thermochemistry
6
123
Sixth lecture
Lecture

Energy
Energy is the capacity to do work.
• Thermal energy is the energy associated with
the random motion of atoms and molecules
• Chemical energy is the energy stored within the
bonds of chemical substances
• Nuclear energy is the energy stored within the
collection of neutrons and protons in the atom
• Potential energy is the energy available by virtue
of an object’s position

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Sixth lecture
Lecture

Kinds of Systems
Closed system Isolated system
Open system
allows the transfer of doesn't allow
can exchange mass transfer of either
energy (heat) but not
and energy mass or energy
mass

125
Sixth lecture
Lecture

Examples

126
Sixth lecture
Lecture

Thermodynamics

Thermodynamics is the scientific study of the interconversion

of heat and other kinds of energy

127
Sixth lecture
Lecture

Heat (q)
Heat is the transfer of thermal energy between
two bodies that are at different temperatures.

Temperature is a measure of the thermal energy

Temperature = Thermal Energy

128
Sixth lecture
Lecture

First Law of Thermodynamics

First Law: Energy of the Universe is Constant

E=q+w
q = heat. Transferred between two bodies

w = work. Force acting over a distance (F x d)

w=Fxd

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Sixth lecture
Lecture
Thermodynamic State Functions
Thermodynamic State Functions: Thermodynamic properties that
are dependent on the state of the system only regardless of the
pathway. Examples: (Energy, pressure, volume, temperature)
DE = Efinal - Einitial
DP = Pfinal - Pinitial

DV = Vfinal - Vinitial
DT = Tfinal - Tinitial
Other variables will be dependent on pathway (Examples: q and w).
These are Path Functions. The pathway from one state to the other
must be defined.

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Sixth lecture
Lecture

Thermochemistry
Thermochemistry is the study of heat change in chemical reactions.

Exothermic process is any process that gives off heat –

transfers thermal energy from the system to the surroundings.

2H2 (g) + O2 (g) 2H2O (l) + energy

H2O (g) H2O (l) + energy

Endothermic process is any process in which heat has to be

supplied to the system from the surroundings.
energy + 2HgO (s) 2Hg (l) + O2 (g)

energy + H2O (s) H2O (l)

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Sixth lecture
Lecture

Enthalpy of Chemical Reactions

Definition of Enthalpy

Thermodynamic Definition of Enthalpy (H):

H = E + PV

E = energy of the system

P = pressure of the system
V = volume of the system

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Sixth lecture
Lecture

Changes in Enthalpy (ΔH)

Consider the following expression for a chemical process:

DH = Hproducts - Hreactants

If DH >0, then qp >0. (+) The reaction is endothermic

If DH <0, then qp <0. (-) The reaction is exothermic

DH=qp
qp : heat at constant pressure

Calorimetry: the measurement of heat change

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Lecture

Kinds of Processes
(chemical reactions or physical changes)
Endothermic processes Exothermic processes

H2O (s) CH4 (g) + 2O2 (g)

Heat absorbed from Heat released from

system to surroundings system to surroundings

q=+ q=-

H2O (l) CO2 (g) + 2H2O (l)

134
12
Sixth lecture
Lecture
Heat Capacity
Heat Capacity (C) Specific heat (S)

The amount of heat required to raise The amount of heat required to

the temperature of a given quantity raise the temperature of one
of the substance by one degree gram of the substance by one
Celsius or kelvin. degree Celsius or kelvin.

J/oC J/g.oC
J/K mass J/g.K
q = C ΔT
135
Sixth lecture
Lecture

Specific heats of some common substances

Substance Specific Heat (J/g.oC)

Al 0.900
Au 0.129
C (graphite) 0.720
C (diamond) 0.502
Cu 0.385
Fe 0.444
Hg 0.139
H2O 4.184
C2H5OH 2.46

136
15
Sixth lecture
Lecture

Standard Enthalpy (Heat) of reaction (ΔHorxn)

Enthalpy change at standard conditions (25 oC, 1 atm)

N2 (g) + 3H2 (g) 2NH3 (g) ΔHo = - 92.38 kJ

Thermochemical reaction

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Sixth lecture
Lecture

Standard Heat of formation (ΔHfo)

The heat change that results when 1 mol of the

compound is formed from standard state of its
elements

The standard enthalpy of formation of any element

in its most stable form is zero.

DH0 (C, diamond) = 1.90 kJ/mol

What is ΔHfo of O2 (g), Hg(l), C(graphite)?

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Sixth lecture
Lecture

Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890.4 kJ/mol
- It shows the physical states of all products and reactants
- Balanced
- It shows Heat of reaction kJ

H2O (s) H2O (l) DH = 6.01 kJ/mol

• If you reverse a reaction, the sign of DH changes
H2O (l) H2O (s) DH = -6.01 kJ/mol
• If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.

2H2O (s) 2H2O (l) DH = 2 x 6.01 = 12.0 kJ

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Sixth lecture
Lecture

How to calculate ΔHro

1- Direct method 2- indirect method

1- Direct method: by standard heat of formation

ΔHo = ∑ n ΔHfo (products) - ∑ n ΔHfo (reactants)

n = no. of moles in the balanced thermochemical equation

140
Sixth lecture
Lecture

Example
Calculate the enthalpy of the following reaction:

2Al (s) + Fe2O3 (s) Al2O3 (s) + 2Fe(l)

ΔHfo of Fe2O3, Al2O3 and Fe(l) = - 822.2, - 1669.8 and 12.40 kJ/mol

ΔHo = ∑ n ΔHfo (products) - ∑ n ΔHfo (reactants)

ΔHo = [(ΔHfo (Al2O3))+ (2× ΔHfo (Fe))]-[(2×ΔHfo (Al))+(ΔHfo (Fe2O3))]

ΔHo = [ (-1669.8)+ (2×12.40)] – [2×(0)+(-822.2)] = -822.8 kJ

141
Sixth lecture
Lecture

2- indirect method :(Hess’s Law)

DH for a process involving the transformation of reactants
into products is not dependent on pathway. Therefore, we
can pick any pathway to calculate DH for a reaction.

When reactants are converted

to products, the change in
enthalpy is the same whether
the reaction takes place in one
step or in a series of steps.
DH is a state function

142
Sixth lecture
Lecture

Hess’ Law
Details
Once can always reverse the direction of a reaction when
making a combined reaction. When you do this, the sign
of DH changes.

N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ

2NO2(g) N2(g) + 2O2(g) DH = -68 kJ

143
Sixth lecture
Lecture

If the coefficients of a reaction are multiplied by a constant, the

value of DH is also multiplied by the same integer.

N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ

2N2(g) + 4O2(g) 4NO2(g) DH = 136 kJ

144
Sixth lecture
Lecture

Example
Calculate the enthalpy of the following reaction:
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g)
ΔHfo of Fe2O3, CO and CO2= - 822.2, - 110.5 and – 393.5 kJ/mol

1- direct method
ΔHo = ∑ n ΔHfo (products) - ∑ n ΔHfo (reactants)
ΔHo = [(2× ΔHfo (Fe))+ (3× ΔHfo (CO2))]- [(ΔHfo (Fe2O3)+(3xΔHfo (CO))]
ΔHo = [ (0)+ (3×(-393.5))] – [(-822.2)+(3×(-110.5))]

145
Sixth lecture
Lecture

2- Hess’s Law
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g)

1- CO (g) + ½O2 (g) CO2 (g) ΔHo = - 283.0 kJ

2- 2Fe (s) + 3/2O2 (g) Fe2O3 (s) ΔHo = - 822.2 kJ

3CO (g) + 3/2O2 (g) 3CO2 (g) ΔHo = 3× - 283.0 = - 849 kJ

Fe2O3 (s) 2Fe (s) + 3/2O2 (g) ΔHo = + 822.2 kJ

Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g)

23 146
Sixth lecture
Lecture Questions
Question 1 Question 3
An exothermic reaction causes the surroundings The specific heat of aluminum is 0.214 cal/g.oC.
to: Determine the energy, in calories, necessary to
A. become basic raise the temperature of a 55.5 g piece of
B. decrease in temperature aluminum from 23.0 to 48.6oC.
C. condense
D. increase in temperature A. 109 cal
E. decrease in pressure B. 273 cal
C. 577 cal
Question 2 D. 347 cal
E. 304 cal
How much heat is evolved when 320 g of SO2 is
burned according to the chemical equation
shown below? Question 4
2 SO2(g) + O2(g) ----> 2 SO3(g) ΔHorxn = -198 kJ Energy is the ability to do work and can be:

A. 5.04 x 10-2 kJ A. converted to one form to another

B. 9.9 x 102 kJ B. can be created and destroyed
C. 207 kJ C. used within a system without
D. 5.0 x 102 kJ consequences
E. None of the above D. none of the above

147
Sixth lecture
Lecture Questions
Question 5 Question 7
To which one of the following reactions, occurring The specific heat of aluminum is 0.214 cal/g.oC.
at 25oC, does the symbol ΔHof [H2SO4(l)] refer? Determine the energy, in calories, necessary to
raise the temperature of a 55.5 g piece of
A. H2(g) + S(s) + 2 O2(g) ----> H2SO4(l) aluminum from 23.0 to 48.6oC.
B. H2SO4(l) ----> H2(g) + S(s) + 2 O2(g)
C. H2(g) + S(g) + 2 O2(g) ----> H2SO4(l) A. 109 cal
D. H2SO4(l) ----> 2 H(g) + S(s) + 4 O(g) B. 273 cal
E. 2 H(g) + S(g) + 4 O(g) ----> H2SO4(l) C. 577 cal
D. 347 cal
Question 6 E. 304 cal
Given: SO2(g) + ½O2(g) ----> SO3(g) ΔHorxn = -99 kJ,
what is the enthalpy change for the following Question 8
reaction? 2 SO3(g) ----> O2(g) + 2 SO2(g) Standard enthalpy of reactions can be
calculated from standard enthalpies of
A. 99 kJ formation of reactants.
B. -99 kJ
C. 49.5 kJ A. True
D. -198 kJ B. False
E. 198 kJ

148
Sixth lecture
Lecture Questions
Question 9 Question 10
Calculate ΔHorxn for the combustion reaction of Find the standard enthalpy of formation of
CH4 shown below given the following: ethylene, C2H4(g), given the following data:
ΔHof CH4(g) = -74.8 kJ/mol; C2H4(g) + 3 O2(g) ----> 2 CO2(g) + 2 H2O(l)
ΔHof CO2(g) = -393.5 kJ/mol; ΔHorxn = -1411 kJ;
ΔHof H2O(l) = -285.5 kJ/mol. C(s) + O2(g) ----> CO2(g) ΔHof = -393.5 kJ;
CH4(g) + 2 O2(g) ----> CO2(g) + 2 H2O(l) H2(g) + ½O2(g) ----> H2O(l) ΔHof = -285.8 kJ

A. -604.2 kJ A. 731 kJ
B. 889.7 kJ B. 2.77 x 103 kJ
C. -997.7 kJ C. 1.41 x 103 kJ
D. -889.7 kJ D. 87 kJ
E. None of the above E. 52 kJ

149
Seventh lecture
Lecture

Chapter

Gases
7
150
Seventh lecture
Lecture

Elements that exist as gases at 250C and 1 atmosphere

151
Seventh lecture
Lecture

152
Seventh lecture
Lecture

Physical Characteristics of Gases

• Gases assume the volume and shape of their containers.
• Gases are the most compressible state of matter.
• Gases will mix evenly and completely when confined to the same container.
• Gases have much lower densities than liquids and solids.

153
Seventh lecture
Lecture

Force
Pressure = Area

(force = mass x acceleration)

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa

154
Seventh lecture
Lecture • Boyle’s law: The volume of a given amount of gas held at
constant temperature varies inversely with the applied pressure.
1 atm

2 atm

4 Liters
2
P a 1/V T = Constant
P x V = constant n = Constant
P1 x V1 = P2 x V2
155
Seventh lecture
Lecture

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg.

What is the pressure of the gas (in mmHg) if the volume is reduced at constant
temperature to 154 mL?

P x V = constant
P1 x V1 = P2 x V2
P1 = 726 mmHg P2 = ?
V1 = 946 mL V2 = 154 mL
P1 x V1 726 mmHg x 946 mL
P2 = = = 4460 mmHg
V2 154 mL

156
Seventh lecture
Lecture
Variation in Gas Volume with Temperature at Constant Pressure

As T increases V increases
157
Seventh lecture
Lecture

Charles’s Law: The volume of a given amount of gas held at constant

.
pressure is directly proportional to the Kelvin temperature

VaT
V = constant x T
V1/T1 = V2 /T2

Temperature must be
in Kelvin
T (K) = t (0C) + 273.15

158
Seventh lecture
Lecture

A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature
will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1 /T1 = V2 /T2

V1 = 3.20 L V2 = 1.54 L
T1 = 398.15 K T2 = ?
T1 = 125 (0C) + 273.15 (K) = 398.15 K

V2 x T1 1.54 L x 398.15 K
T2 = = = 192 K
V1 3.20 L

159
Seventh lecture
Lecture

• Gay-Lussac’s Law: The pressure of a given amount of gas held at

constant volume is directly proportional to the Kelvin temperature.
P P1 P2
= a constant or =
T T1 T2
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A
certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant
volume. What is the final pressure of argon in the lightbulb (in atm)?

P1 P2 P1 = 1.20 atm P2 = ?
=
T1 T2 T1 = 291 K T2 = 358 K

T2 358 K
P2 = P1 x = 1.20 atm x = 1.48 atm
T1 291 K
160
Seventh lecture
Lecture

RELAT- CON-
LAW LAW
IONSHIP STANT

Boyle’s P V P 1 V1 = P 2 V 2 T, n
Charles’ V T V1/T1 = V2/T2 P, n
Gay-
P T P1/T1 = P2/T2 V, n
Lussac’s

161
Seventh lecture
Lecture
Avogadro’s Law
V a number of moles (n) Constant temperature
Constant pressure
V = constant x n

V1 / n1 = V2 / n2

162
Seventh lecture
Lecture

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many
volumes of NO are obtained from one volume of ammonia at the same temperature
and pressure?

4NH3 + 5O2 4NO + 6H2O

1 mole NH3 1 mole NO

At constant T and P

1 volume NH3 1 volume NO

163
Seventh lecture
Lecture
Ideal Gas Equation
Boyle’s law: P a V (at 1constant n and T)

Charles’ law: V a T (at constant n and P)

Avogadro’s law: V a n (at constant P and T)

nT
Va
P
nT
V = constant x = R nT R is the gas constant
P P
PV = nRT

164
Seventh lecture
Lecture

The conditions 0 0C and 1 atm are called standard temperature and pressure
(STP).

Experiments show that at STP, 1 mole

of an ideal gas occupies 22.414 L.

PV = nRT
PV (1 atm)(22.414L)
R= =
nT (1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)

165
Seventh lecture
Lecture

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

T = 0 0C = 273.15 K

P = 1 atm
PV = nRT
1 mol HCl
nRT n = 49.8 g x = 1.37 mol
V= 36.45 g HCl
P
L•atm
1.37 mol x 0.0821 mol•K
x 273.15 K
V=
1 atm

V = 30.7 L

166
Seventh lecture
Lecture

Density (d) Calculations

m PM m is the mass of the gas in g

d= =
V RT M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRT
M= d is the density of the gas in g/L
P

167
Seventh lecture
Lecture

A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0C. What is the molar
mass of the gas?

dRT m = 4.65 g g
M= d= = 2.21
P V 2.10 L L

g L•atm
2.21 x 0.0821mol•K x 300.15 K
L
M=
1 atm

M = 54.5 g/mol

168
Seventh lecture
Lecture

• Dalton’s law of partial pressures: the total pressure, Ptotal, of a

mixture of gases is the sum of their individual gas partial
pressures V and T are constant

P11 P2 Ptotal = P1+ P2

169
169
Seventh lecture
Lecture

Consider a case in which two gases, A and B, are in a container of volume V.

nART
PA = nA is the number of moles of A
V
nBRT nB is the number of moles of B
PB =
V
nA nB
PT = PA + PB XA = XB =
nA + nB nA + nB

PA = XA PT PB = XB PT

ni
Pi = Xi PT mole fraction (Xi ) =
nT
170
Seventh lecture
Lecture

A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116
moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial
pressure of propane (C3H8)?

Pi = Xi PT PT = 1.37 atm

0.116
Xpropane = = 0.0132
8.24 + 0.421 + 0.116

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

171
Seventh lecture
Lecture Kinetic Molecular Theory of Gases
1. A gas is composed of molecules that are separated from each other by
distances far greater than their own dimensions. The molecules can be
considered to be points; that is, they possess mass but have negligible
volume.
2. Gas molecules are in constant motion in random directions, and they
frequently collide with one another. Collisions among molecules are
perfectly elastic.
3. Gas molecules exert neither attractive nor repulsive forces on one
another.
4. The average kinetic energy of the molecules is proportional to the
temperature of the gas in kelvins. Any two gases at the same
temperature will have the same average kinetic energy

KE = ½ mu2
172
Seventh lecture
Lecture

Compressibility of Gases
• Boyle’s Law
P a collision rate with wall
Collision rate a number density
Number density a 1/V
P a 1/V
• Gay-Lussac’s Law
P a collision rate with wall
Collision rate a average kinetic energy of gas molecules
Average kinetic energy a T
PaT

173
Seventh lecture
Lecture

• Avogadro’s Law
P a collision rate with wall
Collision rate a number density
Number density a n
Pan

• Dalton’s Law of Partial Pressures

Molecules do not attract or repel one another
P exerted by one type of molecule is unaffected by the
presence of another gas
Ptotal = SPi

174
Seventh lecture
Lecture

Gas diffusion is the gradual mixing of molecules of one gas with molecules of
another by virtue of their kinetic properties.

NH4Cl

NH3 HCl
17 g/mol 36 g/mol

175
Seventh lecture
Lecture

Gas effusion is the is the process by which gas under pressure escapes from one
compartment of a container to another by passing through a small opening.


r1 t2 M2
= =
r2 t1 M1

Nickel forms a gaseous compound of the formula Ni(CO)x What is the value of x given that under the
same conditions methane (CH4) effuses 3.3 times faster than the compound?
r1 2
r1 = 3.3 x r2 M2 = (r )2
x M1 = (3.3)2 x 16 = 174.2

M1 = 16 g/mol 58.7 + x • 28 = 174.2 x = 4.1 ~ 4

176
Seventh lecture
Lecture

Deviations from Ideal Behavior

1 mole of ideal gas

PV = nRT
PV = 1.0
n=
RT

177
Seventh lecture
Lecture

Van der Waals equation

Non ideal gas

an 2
( P+ 2
V )(V – nb) = nRT
}

}
corrected corrected
pressure volume

178
Seventh lecture
Lecture Questions
1.Which of the following is not a characteristic of substances in the gas 3.A steel tank contains carbon dioxide at a pressure of 13.0 atm
phase? when the temperature is 34oC. What will be the internal gas
A)Substances in the gas phase have much lower densities than the same pressure when the tank and its contents are heated to 100oC.
substances would have in the liquid or solid phase. A) 38.2 atm
B)A mixture of substances in the gas phase will form a homogeneous
solution, whereas the same mixture might not form a homogeneous B) 9.40 atm
solution in the liquid phase. C) 10.7 atm
C)Substances in the gas phase retain their shapes easily.
D)Substances in the gas phase are compressible. D) 15.8 atm
E) None of the above.
2. A sample of gas occupies 2.78 x 103 mL at 25oC and 760 mm Hg. What
volume will the gas sample occupy at the same temperature and 475 mm 4.Calculate the density of nitrogen gas, in grams per liter, at STP.
Hg? A) 0.625 g/L
A) 0.130 L
B) 0.800 g/L
B) 1.04 L
C) 1.25 g/L
C) 1.74 L
D) 2.50 g/L
D) 4.45 L
E) None of the above
E) None of the above

179
Seventh lecture
Lecture Questions
5. A gas evolved during the fermentation of alcohol had a volume of 19.4 L 7. A 1.325 g sample of an unknown vapor occupies 368 mL at 114oC
at 17oC and 746 mm Hg. How many moles of gas were collected? and 946 mm Hg. The empirical formula of the compound is NO2.
A) 1.25 mol What is the molecular formula of the compound?
A) NO2
B) 0.800 mol
B) N4O8
C) 10.5 mol
C) N3O6
D) 13.6 mol
D) N2O4
E) 608 mol
E) N5O10.
6. How many grams of carbon dioxide are contained in 550 mL of this gas 8.An organic compound was analyzed and found to contain 55.8%
at STP? C, 7.03% H, and 37.2% O. A 1.500 g sample of the compound was
A) 0.0245 g vaporized and found to occupy 530 cm3 at 100oC and 740 torr.
Which of the following is the correct molecular formula of the
B) 0.0280 g compound?
C) 1080 g A) C2H3O

D) 0.560 g B) C6H4O2

E) 1.1 g C) C3H2O
D) C4H6O2
E) C2H3O2

180
Seventh lecture
Lecture Questions
9.What volume of chlorine gas at 646 torr and 32oC would be produced by the 11.A steel tank contains carbon dioxide at a pressure of 13.0 atm when
reaction of 14.75 g of MnO2 according to the following chemical equation? the temperature is 34oC. What will be the internal gas pressure when
the tank and its contents are heated to 100oC.
MnO2(s) + 4 HCl(aq) ----> MnCl2(aq) + Cl2(g) + 2 H2O(l)
A) 38.2 atm
A) 5.00 L
B) 9.40 atm
B) 0.170 L
C) 10.7 atm
C) 2.33 L
D) 15.8 atm
D) 0.200 L
E) None of the above.
E) None of the above

10. A mixture of neon, argon, and xenon had a total pressure of 1560 mm Hg 12.Which of the following correctly identifies Boyle's law?
at 298 K. The mixture was found to contain 1.50 mol Ne, 2.65 mol Ar, and 1.75
mol Xe. What is the partial pressure of Xe? A) PV=k1

A) 701 mm Hg B) V=k2T

B) 658 mm Hg
13.The magnitude of one Kelvin, one Celsius degree, and one degree on
C) 396 mm Hg the absolute temperature scale is the same.
D) 463 mm Hg A) True
E) None of the above B) False

181
Seventh lecture
Lecture Questions
14.The Kelvin temperature scale is useful when comparing: 16. A sample of CO2(g) has a volume of 2L at pressure P and
temperature T. If the pressure becomes triple the original value, at the
A) various gas samples at different densities same absolute temperature, the volume of CO2 will be
B) volume of a gas sample with temperature at constant pressure A) L
C) pressure of gas samples at different volumes and constant temperature B) 2/3 L
D) various liquids at constant pressure C) 6L

D) 2L
15. Which of the following is not an assumption of the kinetic theory
of gases?
17. Using the van der Waals equation, calculate the pressure exerted
A) Elasticity refers to the molecules random interactions resulting in no net by 15.0 mol of carbon dioxide confined to a 3.00-L vessel at 56.0 oC.
energy change. Note: Values for a and b in the van der Waals equation: a = 3.59
L2.atm/mol, and b = 0.0427 L/mol.
B) Gas molecules are viewed as points due to large distances between them.
A) 23.2 atm

C) Gas molecules can attract other gas molecules. B) 2.16 atm

D) Temperature in Kelvin and average kinetic energy are proportional. C) 81.9 atm

D) 81.9 atm

182
Eighth lecture
Lecture

Chapter
Electrochemistry
8
183
Eighth lecture
Lecture

Electrochemical processes are oxidation-reduction reactions

in which:
• the energy released by a spontaneous reaction is
converted to electricity or
• electrical energy is used to cause a nonspontaneous
reaction to occur
0 0 2+ 2-
2Mg (s) + O2 (g) 2MgO (s)

2Mg 2Mg2+ + 4e- Oxidation half-reaction (lose e-)

O2 + 4e- 2O2- Reduction half-reaction (gain e-)

184
Eighth lecture
Lecture
Oxidation-Reduction Reactions
Oxidizing Agent- a substance that accepts electrons from another substance, causing the
other substance to be oxidized.

Reducing Agent- a substance that donates electrons to another substance, causing it to become
reduced.

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

Zn Zn2+ + 2e- Zn is oxidized

Zn is the reducing agent

Cu2+ + 2e- Cu Cu2+ is reduced

Cu2+ is the oxidizing agent
185
Eighth lecture
Lecture Galvanic Cells

anode cathode
oxidation reduction

spontaneous
redox reaction

186
Eighth lecture
Lecture

The difference in electrical

potential between the anode
and cathode is called:
• cell voltage
• electromotive force (emf)
• cell potential
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M and [Zn2+] = 1 M
Cell Diagram
phase boundary
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
anode salt bridge cathode
187
Eighth lecture
Lecture
Standard Reduction Potentials
Standard reduction potential (E0) is the voltage associated
with a reduction reaction at an electrode when all solutes
are 1 M and all gases are at 1 atm.

Reduction Reaction

2e- + 2H+ (1 M) H2 (1 atm)

E0 = 0 V

Standard hydrogen electrode (SHE)

188
Eighth lecture
Lecture

0 = 0.76 V
Ecell 0 )
Standard emf (Ecell
0 = E0
Ecell 0
cathode - Eanode

0 = E 0+ - E 0 2+
Ecell H /H2 Zn /Zn

0 2+
0.76 V = 0 - EZn /Zn
0 2+
EZn /Zn = -0.76 V

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

Anode (oxidation): Zn (s) Zn2+ (1 M) + 2e-

Cathode (reduction): 2e- + 2H+ (1 M) H2 (1 atm)
Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)
189
Eighth lecture
Lecture

0 = 0.34 V
Ecell
0 = E0
Ecell - E 0
cathode anode

Cu /Cu – EH +/H 2
0 = E 0 2+
Ecell 0

0 2+
0.34 = ECu /Cu - 0
0 2+
ECu /Cu = 0.34 V

Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)

Anode (oxidation): H2 (1 atm) 2H+ (1 M) + 2e-
Cathode (reduction): 2e- + Cu2+ (1 M) Cu (s)
H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)
190
Eighth lecture
Lecture
• E0 is for the reaction as
written
• The more positive E0 the
greater the tendency for the
substance to be reduced
• The half-cell reactions are
reversible
• The sign of E0 changes
when the reaction is
reversed
• Changing the stoichiometric
coefficients of a half-cell
reaction does not change
the value of E0
191
Eighth lecture
Lecture
What is the standard emf of an electrochemical cell made
of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr
electrode in a 1.0 M Cr(NO3)3 solution?

Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V Cd is the stronger oxidizer

Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V Cd will oxidize Cr

Anode (oxidation): Cr (s) Cr3+ (1 M) + 3e- x 2

Cathode (reduction): 2e- + Cd2+ (1 M) Cd (s) x3
2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)

0 = E0
Ecell 0
cathode - Eanode

0 = -0.40 – (-0.74)
Ecell
0 = 0.34 V
Ecell
192
Eighth lecture
Lecture

Spontaneity of Redox Reactions

DG0 = -nFEcell
0

193
Eighth lecture
Lecture
The Effect of Concentration on Cell Emf
DG = DG0 + RT ln Q DG = -nFE DG0 = -nFE 0

-nFE = -nFE0 + RT ln Q

Nernst equation

RT
E = E0 - ln Q
nF

At 298 K

0.0257 V 0.0592 V
E = E0 - ln Q E = E0 - log Q
n n

194
Eighth lecture
Lecture
Will the following reaction occur spontaneously at 250C if
[Fe2+] = 0.60 M and [Cd2+] = 0.010 M?
Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)

Oxidation: Cd Cd2+ + 2e-

n=2
Reduction: 2e- + Fe2+ 2Fe
2+/Fe – ECd2+/Cd
0
E0 = EFe 0

E0 = -0.44 – (-0.40) 0.0257 V

E= - E0 ln Q
n
E0 = -0.04 V
0.0257 V 0.010
E = -0.04 V - ln
2 0.60
E = 0.013
E>0 Spontaneous

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Eighth lecture
Lecture

Batteries

Dry cell

Leclanché cell

Anode: Zn (s) Zn2+ (aq) + 2e-

Cathode: 2NH+4 (aq) + 2MnO2 (s) + 2e- Mn2O3 (s) + 2NH3 (aq) + H2O (l)

Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)

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Eighth lecture
Lecture

Lead storage
battery

Anode: Pb (s) + SO2-4 (aq) PbSO4 (s) + 2e-

Cathode: PbO2 (s) + 4H+ (aq) + SO2-

4 (aq) + 2e
- PbSO4 (s) + 2H2O (l)

Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2-

4
(aq) 2PbSO4 (s) + 2H2O (l)

197
Eighth lecture
Lecture

A fuel cell is an
electrochemical cell
that requires a
continuous supply of
reactants to keep
functioning

Anode: 2H2 (g) + 4OH- (aq) 4H2O (l) + 4e-

Cathode: O2 (g) + 2H2O (l) + 4e- 4OH- (aq)

2H2 (g) + O2 (g) 2H2O (l)

198
Eighth lecture
Lecture
Corrosion
Corrosion is the term usually applied to the degradation of metals by an
electrochemical process.

199
Eighth lecture
Lecture

Cathodic Protection of an Iron Storage Tank

200
Eighth lecture
Lecture
Electrolysis is the process in which electrical energy is used
to cause a nonspontaneous chemical reaction to occur.

Electrolysis of Water Electrolysis of molten NaCl

201
Eighth lecture
Lecture

Electrolysis and Mass Changes

charge (C) = current (A) x time (s)

1 mol e- = 96,500 C

202
Eighth lecture
Lecture

How much Ca will be produced in an electrolytic cell of

molten CaCl2 if a current of 0.452 A is passed through the
cell for 1.5 hours?

Anode: 2Cl- (l) Cl2 (g) + 2e-

Cathode: Ca2+ (l) + 2e- Ca (s)

Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)

2 mole e- = 1 mole Ca

C s 1 mol e- 1 mol Ca
mol Ca = 0.452 x 1.5 hr x 3600 x x
s hr 96,500 C 2 mol e-
= 0.0126 mol Ca = 0.50 g Ca

203
Eighth lecture
Lecture Questions
1. Separating redox equations is simpler if the 4. A galvanic cell is prepared using zinc and iron. Its
equation is separated into oxidation and cell notation is as follows: Zn|Zn2+(1 M)||Fe2+(1 M),
reduction portions. Fe3+(1M)|Pt. Which of the following reactions occurs
(a) True (b) False at the cathode?
(a) Fe3+ + e- ----> Fe2+ (b) Fe2+ ----> Fe3+ + e-
2. Which of the following does not denote the
difference in electrical potential between the anode (c) Zn ----> Zn2+ + 2 e- (d) Zn2+ + 2e- ----> Zn
and cathode?
(a) cell voltage (b) electromotive force 5. A cell can be prepared from zinc and iron. What is
(c) Voltmeter (d) cell potential the Eocell for the cell that forms from the following
half reactions? Fe3+ + e- ----> Fe2+, Eo = 0.77 V; Zn2+ +
3. The same electrode may act as either the anode 2e- ----> Zn, Eo = -0.76 V
or the cathode depending on the second electrode (a) -1.53 V (b) -0.01 V
used in setting up the galvanic cell (c) 0.01V (d) 1.53 V
(a) True (b) False
6. The electrode at which oxidation occurs is
called the
(a) anode (b) cathode
(c) salt bridge

204
Eighth lecture
Lecture Questions
7. Calculate the free energy change per mole of Cu2+ formed 9. Which is a secondary cell?
in the following reaction at 25 oC. (a) Mercury cell. (b) dry Cell
Cu + 2 Ag+ ----> 2 Ag + Cu2+. (c) H2-O2 Cell (d) Lead storage battery
Cu2+ + 2 e- ----> Cu Eo = 0.34 V;
Ag+ + e- ----> Ag, Eo = 0.80 V. 10. The principal function of a fuel cell is to
(a) 0.46 kJ (b) 89 kJ (a) Produce fuel. (b) Electrolyze fuel.
(c) 44.5 kJ (d) -89 kJ (c) Produce hydrogen. (d) Produce electricity
11. Corrosion is the disintegration of metals through
8. One of the differences between a galvanic cell and an (a) combustion. (b) oxidation
electrolytic cell is that in a galvanic cell (c) reduction (d) exposure
(a) an electric current is produced by a chemical reaction.
(b) electrons flow toward the anode. 12. What conditions are required for rusting to take place?
(c) a non spontaneous reaction is forced to occur using an (a) Water only (b) oxygen only
electric current from an external source. (c) Water and oxygen (d) Water and carbon dioxide
(d) reduction occurs at the anode. 13. What mass of copper can be deposited by the
passage of 12.0 A for 25.0 min through a solution of
copper(II) sulfate? [1 C = 1 A·s; F = 96,500 C]
(a) 5.93 g (b) 3.95 g
(c) 1.97 g (d) 11.85 g

205