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Chapter 5 - Chemical Reactions and
Quantities
(Chapter 6 and 7 in 1st Edition)
Physical change versus Chemical change
A physical change involves a change in the appearance of
a substance but not its composition.
e.g.
a) Melting of ice
b) Breaking glass
c)
d)
In a chemical change, a substance reacts to give a new
substance with a different composition and different
properties.
e.g.
a) Burning wood
b) Metal rusting
c)
d)
Chemical Equations
Since many chemical changes can occur in nature,
chemists have devised a ‘language’ to keep track of these
changes. When describing a chemical change, chemists
write chemical equations. Chemical equations tell us
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what compounds we need and what products will form in
a chemical reaction.
e.g.
2H2(g) + O2(g)
2H2O(l)
The above equation tells us that we need two molecules of
hydrogen and one molecule of oxygen to form two
molecules of water. Also note that chemical reactions
always involve the __________ and __________ of
bonds. Since ___________ are responsible for chemical
bonds, chemical reactions are a result of electron
transfer.
Important points regarding chemical equations:
a) reactants are always on the left hand side.
b) products are always on the right hand side.
c) physical states are abbreviated as letters in
parentheses.
d) equations must always be balanced.
Balancing Chemical Equations.
Steps for balancing chemical equations:
Step 1. Use the correct formulas
Step 2. Determine if the equation is balanced
Step 3. Balance the equation one element at a time using
coefficients.
Step 4. Check to see if the equation is balanced.
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e.g. Balance the following equations:
Al + Cl2
AlCl3
P4 + O2
P4O10
C4H8 + O2
CO2 + H2O
Sb2S3 + HCl
SbCl3 + H2S
Types of Reactions.
Most chemical reactions can be described as one of the
following reactions:
Combination Reactions. In combination reactions, two
or more elements or simple compounds combine to form
one product.
e.g.
2H2 + O2
2H2O
Decomposition Reactions. In decomposition reactions, a
reactant breaks up to give two or more simpler products.
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e.g.
CaCO3
CaO + CO2
Replacement Reactions. In replacement reactions,
elements in compounds are replaced by other elements. If
only one compound has an element replaced, it is a single
replacement reaction. If two compounds have elements
replaced, then it is a double replacement reaction.
e.g.
Fe2O3 + 3C
AgNO3 + NaCl
2Fe + 3CO
AgCl + NaNO3
Combustion Reactions. In combustion reactions, fuel
(usually a carbon compound) reacts with oxygen to
produce carbon dioxide, water, and heat.
e.g.
CH4 + 2O2
CO2 + 2H2O
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Interlude: Oxidation States
Rules For Assigning Oxidation States (listed in order
of importance).
1. Free elements are assigned an oxidation state of 0.
2. The sum of the oxidation states of all the atoms in a
species must equal the net charge of the species.
3. The alkali metals (Li, Na, K, Rb, and Cs) are always
assigned an oxidation state of +1.
4. Fluorine is always assigned an oxidation state of -1.
5. The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and
Ra) and Zn and Cd are always assigned an oxidation state
of +2.
6. Hydrogen is assigned an oxidation state of +1.
7. Oxygen is assigned an oxidation state of -2.
**Warning** Oxidation number is not to be confused
with formal charge.
e.g. Assign an oxidation state to each atom in the
following:
Na
NaCl
H2O2
BF4-
CrO42-
NaH
OF2
HSO3-
KNO3
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Oxidation-Reduction Reactions. Oxidation-Reduction
(Redox) reactions are characterized by the reactants
losing and gaining electrons.
Oxidation. Oxidation occurs when an element loses
electrons.
Reduction. Reduction occurs when an element gains
electrons.
‘LEO the lion goes GER’
LEO: Lose Electrons Oxidation.
GER: Gain Electrons Reduction.
The way we tell if an atom has been oxidized or reduced
is by determining its oxidation state.
2Na + S
Na2S
The reactants, being in their elemental form and neutral of
charge, are assigned an oxidation state of zero. Following
the above rules, we can assign Na an oxidation state of +1
and S an oxidation state of -2. Therefore, Na is oxidized
and S is reduced. In redox reactions, the species that
loses electrons is called the reducing agent and the
species that gains electrons is called the oxidizing agent.
Note that the reducing agent, which loses electrons, is
oxidized and that the oxidizing agent, which gains
electrons, is reduced.
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Therefore:
Na → oxidized, reducing agent
S → reduced, oxidizing agent
e.g. For each reaction, indicate which species is oxidized
and which is reduced. Indicate the reducing agent and the
oxidizing agent.
Fe + CuSO4
Zn + 2HCl
FeSO4 + Cu
ZnCl2 + H2
An older definition describes oxidation as the loss of
hydrogen atoms or gain of oxygen atoms. Conversely,
reduction is accompanied by the loss of oxygen atoms or
gain of hydrogen atoms.
e.g.
4Fe + 3O2
2Fe2O3
CH2O + H2
CH3OH
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The Mole
In everyday life, we use counting units to deal with large
quantities. For example, we buy eggs by the dozen (12),
paper by the ream (500), and pop by the case (24). In
chemistry, we count molecules, atoms, and ions by the
mole. A mole contains 6.02 X 1023 items. This very large
number is called Avogadro’s number, after the Italian
physicist Amedeo Avogadro.
One mole of any element, molecule, or ionic compound
contains Avogadro’s number of atoms.
Therefore:
1mol of C contains 6.02 X 1023 carbon atoms.
1mol of Al contains 6.02 X 1023 aluminum atoms.
1mol of Br contains 6.02 X 1023 bromine atoms.
1mol of H2O contains 6.02 X 1023 H2O molecules.
1mol of NaCl contains 6.02 X 1023 NaCl formula
units.
The subscripts in a chemical formula also represent
moles. For example, the chemical formula for aspirin is
C9H8O4. Therefore, 1 mole of aspirin contains 9 moles of
carbon atoms, 8 moles of hydrogen atoms, and 4 moles of
oxygen atoms.
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How many C atoms are in 0.350 mol of C6H12O6?
How many Cl atoms in 2.0 mole of PCl3?
How many moles of C2H6O contain 5.0 x 10 24 atoms of
H?
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Molar Mass
The mass in grams of 1 mole of substance is called its
molar mass. The molar mass (in grams) of any substance
is always equal to its formula weight (in amu).
Determine the molar masses for the following substances:
a)
b)
c)
d)
e)
C
Ag
H2O
NaCl
NO3-
What is the mass of 1 mole of glucose, C6H12O6?
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How many moles of glucose are in 5.38 g?
How many grams of glucose are in 2.74 moles?
The concept of the mole provides a bridge between
masses and numbers of particles. Using molar mass and
Avogadro’s Number as conversion factors, we can
convert grams
moles
atoms.
For example, calculate the number of Cu atoms in a 3g
copper penny (assuming the penny is 100% Cu).
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How many glucose molecules are in 5.23 g of C6H12O6?
Percent Composition and Empirical Formula
Recall that the subscripts in a molecular formula represent
a definite proportion of the elements within a compound.
Therefore, a mole of any compound contains a definite
proportion by mass of its elements. By using the molar
mass of a compound, we can determine its percent
composition. For example, determine the percent
composition of K2CO3.
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The molecular formula of a compound represents the
true formula of a compound. A formula that gives the
lowest whole-number ratio of atoms in the compound is
called the empirical formula. For example, hydrazine
with molecular formula of N2H4, has an empirical formula
of NH2. The mole concept can be used to determine
empirical formulas.
e.g. A sample of a compound contains 3.24 g Na, 2.26 g
S, and 4.51 g O. What is its empirical formula?
e.g. Calculate the empirical formula of a compound that
has percent composition 36% Al and 64% S.
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e.g. Elemental analysis of an unknown sample was found
to contain 24.27% C, 4.075% H, and 71.65% Cl. What is
the empirical formula of the unknown sample? Given
that the unknown sample has a molar mass of 98.95
g/mol, what is its molecular formula?
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Stoichiometry
Consider the following reaction:
2H2 + O2
2H2O
The coefficients from the balanced equation tell us that
we need two molecules of hydrogen to react with one
molecule of oxygen to give two molecules of water.
From the mole concept, we can interpret the equation in
terms of moles. Thus, 2 moles of hydrogen react with 1
mole of oxygen to give 2 moles of water. The
coefficients in the above equation are called
stoichiometrically equivalent quantities. The
relationship between these quantities can be expressed as
follows:
2 mol H2 K 1 mol O2 K 2 mol H2O
These stoichiometric relations can be used to give
conversion factors for relating quantities of reactants and
products in a chemical reaction.
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e.g. Calculate the number of moles of H2O produced
from 1.57 moles of O2.
e.g. Using the equation below, calculate the mass of CO2
that is produced in burning 1.00 g of butane, C4H10.
2C4H10 + 13O2
8CO2 + 10H2O
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e.g. Nitrogen gas reacts with hydrogen gas to produce
ammonia by the following equation:
N2 + 3H2
2NH3
a) If you have 1.8 moles of H2, how many grams of NH3
can be produced?
b) How many grams of H2 are needed to react with 2.80 g
of N2?
c) How many grams of NH3 can be produced from 12 g
of H2?
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Percent Yield.
All of the equations we’ve seen so far have assumed that
100% of product was formed. In reality, this is rarely the
case. Side reactions or the equilibrium can prevent
reactions from going to completion. When chemists
perform reactions, there are two things they want to
know; a) what is the theoretical yield of the reaction and
b) what is the actual yield of the reaction.
• The theoretical yield is the calculated amount of
product if the reaction proceeds to completion.
• The actual yield is the amount of product obtained
after the reaction is complete.
If we know the values of the theoretical yield and actual
yield, then we can express these values as a percentage,
called the percent yield.
e.g. What is the percent yield if 40.0 g of CO are
produced from the reaction of 30.0 g O2?
2C + O2
2CO
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e.g. From the following equation, calculate the mass of
CO2 that can be produced if the reaction of 45.0 g of
propane (C3H8) and sufficient oxygen has a 60% yield.
C3H8 + 5O2
3CO2 + 4H2O
Energy of Chemical Reactions
(from Chapter 6, 2nd Edition)
We have seen that when chemical reactions take place,
bonds are broken and formed. The amount of energy
needed to break the bonds is called the activation energy.
If the energy of the system (ie. the reaction) is less than
the activation energy, then the bonds won’t break and the
reaction will not occur.
Consider this analogy:
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Exothermic and Endothermic Reactions.
Just as it takes energy to break bonds, energy is released
when bonds are formed. The heat of reaction is the
difference between the energy of the reactants and
products. When the energy of the products is less than
that of the reactants, heat is given off and the reaction is
said to be exothermic. When the energy of the products
is more than that of the reactants, heat is absorbed and the
reaction is said to be endothermic.
e.g.
CH4 + 2O2
H2 + I2 + 12kcal
CO2 +2H2O + 213kcal
2HI
Rate of Reaction (from Chapter 9, 2nd Edition)
Although we are always interested if a reaction will or
will not occur, another important factor to consider is the
rate of reaction. The rate or speed of the reaction is
defined as the amount of reactant used up, or amount of
product formed, in a certain period of time. Therefore, a
reaction with lower activation energy will proceed faster
than a reaction with higher activation energy.
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There are three major factors that can affect the rate of
reaction:
1) Temperature
2) Concentration
3) Catalyst
Temperature. Raising the temperature will increase the
number of collisions between molecules and also provide
the collisions with the required energy of activation.
Raising the temperature almost always increases the rate
of reaction. Conversely, lowering the temperature will
reduce the rate of reaction.
Concentration. The rate of reaction increases when the
concentration of reactants is increased because there are
more collisions between molecules.
Catalyst. Adding a catalyst to the reaction increases the
rate of reaction by lowering the activation energy. A
catalyst is not changed or used up during a reaction.
How would each of the following change the rate of the
reaction shown below?
2NO + 2H2
a) adding more NO
b) lowering the temperature
N2 +2H2O
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c) removing some H2
d) adding a catalyst
Chemical Equilibrium
Although we write reactions going from left to right, that
is, going from reactants to products, most reactions can
occur in both directions. These are called reversible
reactions and are shown using a double arrow.
e.g.
2SO2 + O2
2SO3
2SO2 + O2
2SO3
2SO2 + O2
2SO3
Eventually, the rates of the forward and reverse reactions
will become equal. At this point, the reaction is said to
have reached chemical equilibrium. When chemical
equilibrium occurs, the amount of reactants and products
produced does not change. Important: this does not
necessarily mean that the amount of reactants is equal to
the amount of products.
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LeChâtelier’s Principle.
We have already seen how a change in the reaction
conditions (ie. temperature) can change the rate of the
reaction. When we make a change to a reaction at
equilibrium, we are putting a stress on the equilibrium. In
nature, a reaction will proceed in the direction that
relieves this stress. This is known as LeChâtelier’s
Principle. Consider the following endothermic reaction:
N2 + O2
a)
b)
c)
d)
2NO
What happens if the amount of O2 is increased?
What happens if the amount of O2 is decreased?
What happens if the amount of NO is increased?
What happens if heat is added to the reaction?
Consider the exothermic reaction below. What is the
effect of each of the following changes on equilibrium?
2SO2 + O2
a) add heat
b) remove SO3
c) add O2
2SO3
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Important Concepts from Chapter 5
• Physical vs Chemical Change
• Balancing Chemical Equations
• Types of Chemical Reactions
• Oxidation Number
• The Mole
• Avogadro’s Number
• Molar Mass
• Percent Composition and Empirical Formula
• Stoichiometry
• Percent Yield
• Activation Energy
• Exothermic and Endothermic Reactions
• Reaction Rates
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• Chemical Equilibrium
• LeChâtelier’s Principle