Download Lecture 9

CHEM 121
Introduction to Fundamental Chemistry
Summer Quarter 2008 SCCC
Lecture 9
http://seattlecentral.edu/faculty/lcwest/CHE121
Second midterm will be next Tuesday and will cover material
from chapters 4, 5 and 6.
Chemical reactions can be divided into two classes, redox
reactions and nonredox reactions.
Within each class there are several subclasses which we will
learn about as the course progresses.
In decomposition reactions one substance is broken to form two
or more new substances.
A→B+C
Decomposition reactions may or may not be redox reactions.
e.g.
2HgO(s) → 2Hg(l) + O2(g)
In combination reactions two or more substances combine to
form a single substance.
B+C→A
Combination reactions may or may not be redox reactions.
redox: 2Mg(s) + O2(g) → 2MgO(s)
non-redox: SO3(g) + H2O(l) → H2SO4(aq)
When ionic compounds are dissolved in water to form
aqueous solutions (aq) the compounds dissociate into their
component ions.
NaOH(aq)
→
Na+(aq)
+
OH-(aq)
Similar behaviour is observed for a few covalent compounds
particular those formed between hydrogen and the halogens.
HCl(aq)
→
H+(aq) +
Cl-(aq)
Reactions between compounds that dissociate into ions can be
written by a total ionic equation which indicates the ions
present:
e.g.
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → H2O(l) + Na+(aq) + Cl-(aq)
In the total ionic equation there are some ions that appear in the
same form on both the reactant and product side of the
equation. These are called spectator ions.
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → H2O(l) + Na+(aq) + Cl-(aq)
Eliminating the spectator ions from both sides results in the net
ionic equation:
H+(aq) + OH-(aq) → H2O(l)
The next class of reactions we will examine are replacement
reactions.
Single replacement or substitution reactions involve the reaction
between an element and a compound to form a new element
and a new compound.
A
+
BC → B
+
AC
Single replacement reactions are redox reactions.
e.g.
3C(s) +
2Fe2O3(s)
→
4Fe(s)
+
3CO2(g)
Double replacement or metathesis reactions occur between two
compounds and involve the pairwise exchange of atoms.
AB +
CD → BD +
AC
Metathesis reactions are never redox reactions.
e.g.
HCl(aq) +
NaOH(aq)
→
H2O(l) +
NaCl(aq)
When ionic compounds are dissolved in water to form aqueous
solutions (aq) the compounds dissociate into their component
ions.
NaOH(aq)
→
Na+(aq)
+
OH-(aq)
Similar behaviour is observed for a few covalent compounds
particular those formed between hydrogen and the halogens.
HCl(aq)
→
H+(aq) +
Cl-(aq)
Reactions between compounds that dissociate into ions can be
written by a total ionic equation which indicates the ions
present:
e.g.
HCl(aq) + NaOH(aq) →
H2O(l) + NaCl(aq)
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → H2O(l) + Na+(aq) + Cl-(aq)
In the total ionic equation there are some ions that appear in the
same form on both the reactant and product side of the
equation. These are called spectator ions.
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → H2O(l) + Na+(aq) + Cl-(aq)
Eliminating the spectator ions from both sides results in the net
ionic equation:
H+(aq) + OH-(aq) → H2O(l)
In some reactions energy is released into the surroundings
(heat, motion or sound).
CH4 + O2 → CO2 + 2H2O + energy
Reactions of this type are called exothermic.
In some reactions energy is stored in the products, resulting in
the surroundings losing
energy.
H2O(l) + energy → H2O(g)
Reactions of this type are called endothermic reactions.
Stoichiometry is the quantitative study of reactants and
products in chemical reactions.
Stoichiometric calculations are concerned with amounts of
reactants and products.
It is essential to use a balanced equation when performing
these calculations.
Consider the following balanced equation:
N2(g)
+
3H2(g) →
2NH3(g)
This equation indicates that 1 molecule of N2 reacts with 3
molecules of H2 to give 2 molecules of NH3.
A more precise explanation of the equation would be:
“1x molecules of N2 reacts with 3x molecules of H2 to give 2x
molecules of NH3”
Where x is any number.
For example if x = 4 according to our definition :
N2(g)
+
3H2(g) →
2NH3(g)
We would say 4 molecules of N2 reacts with 12 molecules of H2 to
give 8 molecules of NH3.
What if x = NA?
If x = NA according to our definition :
N2(g)
+
3H2(g) →
2NH3(g)
We would say 1 mole of N2 reacts with 3 moles of H2 to give 2
moles of NH3.
How many moles of NH3 can be produced by reaction of 1.5
moles of N2 with 4.5 moles of H2?
Consideration of the equation indicates that for every mole of N2
present we produce 2 moles of NH3.
N2(g)
+
3H2(g) →
2NH3(g)
Therefore if we have 1.5 moles of N2 we will produce 3 moles of
NH3. This is called the yield of the reaction. Mathematically we
would say:
2(nN2) = nNH3
What would the yield of the reaction be if we reacted 1 mole of N2
with 2 moles of H2?
N2(g)
+
3H2(g) →
2NH3(g)
We need to recognize that we don’t have enough H2 present to
consume all the N2 present. We would say N2 is in excess
(INXS).
In this scenario H2 is the limiting reagent as it determines the
yield.
N2(g)
+
3H2(g) →
2NH3(g)
Dividing both sides by three we get:
1/3N2(g)
+
H2(g)
→
2/3NH3(g)
This means:
nH2 = 2/3nNH3
2 moles of H2 will produce (2/3)x2 moles of NH3
or 1.333 moles of NH3
Stoichiometric calculations are used to determine:
• Yields
• Limiting reagent
• Excess Reagent
They require:
• A balanced equation
• Skill in performing mole to mass conversions and vice
versa.
NOTE: we must use moles not mass in these calculations.
Oxidation number is a property of a single atom.
We cannot define the oxidation number for a molecule or a
polyatomic ion.
The sum of oxidation numbers of the atoms in a polyatomic ion
or molecule can be calculated. This is not the oxidation number
of the molecule or ion.
Polyatomic ions have an overall charge, this is not the oxidation
number of the ion.
The oxidation number of an atom depends upon what atoms it is
combined with:
e.g.
In CO2 carbon has O.N. = 4 while in CH4 carbon has O.N. = -4
We use the rules described on page 141 to determine O.N.’s.
This is the only method we should be using.
The subscript in the formula of a compound indicates the number
of atoms of that type present.
e.g.
CO2 is a covalent compound containing 1 carbon atom and 2
oxygen atoms.
CH4 is a covalent compound containing 1 carbon atom and 4
hydrogen atoms.
Na2O is an ionic compound containing 2 sodium ions and 1
oxide ion.
What kind of compound is C2H2 and how many atoms of each
type does it contain?
C2H2 is a covalent compound containing 2 carbon atoms and 2
hydrogen atoms.
We can determine the oxidation number of an atom in a
polyatomic ion or molecule by subtracting the known oxidation
numbers from the known sum of the oxidation numbers.
e.g. The oxidation number of C in CO2
0 = -2(2) + O.N. of C
O - (-2(2)) = O.N. of C
O.N. of C = 4
What is the oxidation number of C in C2H2?
0 = 2(1) + 2(O.N. of C)
0 - (2(1)) = 2(O.N. of C)
-2÷2 = O.N. of C
O.N. of C = -1
The key is remembering the oxidation number is a property of a
single atom and not forgetting to divide by the number of atoms
in the molecule or ion as indicated by the subscript.
When we write an equation for a chemical reaction we indicate
the number of molecules or ions of each type by putting numbers
in front of their formulas.
e.g.
CH4
+
2O2
Æ
CO2
+
2H2O
In the reaction above 1 molecule of CH4 reacts with 2 molecules
of oxygen to produce 1 molecule of CO2 and 2 molecules of H2O.
When we calculate changes in oxidation numbers of atoms
during chemical reactions we do not need to consider the
number that appears in front of the formula for a compound.
What is the change in oxidation number of carbon in the following
reaction?
CH4
+
2O2
Æ
CO2
+
2H2O
CH4
+
2O2
Æ
CO2
+
2H2O
Products
Reactants
O.N. O = 0
O.N. O = -2
O.N. H = 1
O.N. H = 1
O.N. C = -4
O.N. C = 4
The Oxidation number of an atom depends on what other atoms
are in the molecule or ion not how many of those ions or
molecules are present.
In scientific language we may say oxidation number is an
intensive property.
“An intensive quantity is a physical quantity whose value does
not depend on the amount of the substance for which it is
measured. It is the counterpart of an extensive quantity.“
Another example of an intensive property is boiling point.
i.e. 1L of H2O has the same boiling point as 100L (1000C).
In contrast the energy required to boil water is an example of an
extensive property.
Matter exists in three common states:
ƒ Solid
ƒ Liquid
ƒ Gas
At 250C and 1 atmosphere all elements are solids, except:
ƒ The group 8 elements are all gases.
ƒ H2, N2, O2, F2 and Cl2 are gases.
ƒ Hg and Br2 are liquids.
We can define a kinetic theory of matter based on the following
postulates:
1. Matter is composed of small particles called molecules
2. The particles are in constant random motion
ƒ
They possess kinetic energy
3. There are repulsive and attractive forces between
particles.
ƒ
They posses potential energy
4. Average particle speed increases with temperature
5. No energy is lost when the particles collide
The kinetic energy of a particle is given by the equation:
1
KE = mv 2
2
Where:
m = particle mass and
v = particle velocity
According to postulate 4 of our kinetic theory particle velocity
increases with temperature. This means as temperature
increases our theory says the kinetic energy increases.
Potential energy is the sum of the attractive and repulsive
forces between particles.
Examples of these types of forces are the gravitational attractive
forces between objects and the repulsive forces between the
same poles of magnets.
Alternatively we can say forces between particles may be either
cohesive or disruptive.
Cohesive forces are of the type we described in chapter 4 and
include dipole-dipole interactions, dispersion forces, attraction
between oppositely charged ions.
Cohesive forces are largely temperature independent.
e.g. magnets and gravity function the same way at different
temperature.
Disruptive forces are those forces that make particles move
away from each other.
These forces result predominately from the particle motion.
Disruptive forces increase with temperature in agreement with
postulate 4.
We can conclude that as we increase the temperature particles
will become further apart from each other.
Gases are characterized by:
ƒ Being easily compressible
ƒ Variable shape
ƒ Shape adapts to that of the container.
ƒDensity is lowest in this phase and is variable (d = m/V)
ƒ Fill whatever container they are placed in
ƒThe largest thermal expansion of all phases.
ƒGases rapidly expand when heated.
The shape, density and volume of a gas depends on the
container!!
The particles in gases are well separated from each other, move
in rapid random motion and do not have fixed positions.
Can you think of any applications of gases that make them
useful in everyday life?
Read sections 6.1-6.7
Study Examples 6.1-6.7
Complete Exercises 6.2, 6.6, 6.13, 6.16