Chemical Equations &
Reactions
Chemistry 6.0
Chemical Reactions
A.
B.
Definition: a process by which 1 or more
substances, called reactants, are changed into 1
or more substances, called products, with
different physical & chemical properties.
Evidence of a Chemical Reaction
1.
2.
3.
4.
5.
C.
Color change
Formation of a precipitate, ppt
Release of a gas
Energy change – heat, light, sound
Odor change
Reactions are started by the addition of energy
Chemical Equation
A.
Form
1.
2.
Reactant + Reactant  Product + Product
Symbols: (s), (l), (g), (aq)
NR
methanol
Writing Chemical Equations

Two moles of water at room temperature are exposed
to an electric current and produces two moles of
hydrogen gas and one mole of oxygen gas.
2 H2O(l)  2 H2(g) + 1 O2(g)

When two moles of aluminum pellets are added to
three moles of a copper(II) chloride solution, 3 moles
of copper precipitate out and two moles of aluminum
chloride remain in solution.
2 Al(s) + 3 CuCl2(aq)  3 Cu(s) + 2 AlCl3(aq)
Characteristics of A Balanced
Chemical Equations
1.
2.
3.
4.
The equation must represent known facts. All
substances have been identified.
The equation must contain the correct symbols
and/or formulas for the reactants and products
Can be either a word equation or a formula
equation
The law of conservation of mass must be satisfied.
This provides the basis for balancing chemical
equations. 1st formulated by Antoine Lavoisier
TOTAL MASS REACTANTS = TOTAL MASS PRODUCTS
Number of atoms of EACH element is the SAME on
both sides of the equation.
Balancing Chemical Equations
1. Balance using coefficients after correct formulas
are written.
Coefficients are usually the smallest whole number – required
when interpreted at the molecular level
2. Balance atoms one at a time
3. Balance the atoms that are combined and appear
only once on each side.
4. Balance polyatomics that appear on both sides
5. Balance H and O atoms last
NEVER CHANGE SUBSCRIPTS!!!
**Count atoms to be sure that the
equation is balanced**
BALANCING Examples
1.
sodium + chlorine  sodium chloride
2.
CH4(g) + O2(g)  CO2(g) + H2O(l)
3.
K(s) + H2O(l)  KOH(aq) + H2(g)
HOH
4.
AgNO3(aq) + Cu(s)  Cu(NO3)2(aq) + Ag(s)
5.
C5H10(g) + O2(g)  CO2(g) + H2O(g)
Interpretation of a
Balanced Equation
2Mg(s) + O2(g)  2MgO(s)
2 atoms of solid magnesium react with
1 molecule of oxygen gas to form
2 formula units of solid magnesium oxide
OR
2 moles of solid magnesium react with
1 moles of oxygen gas to form
2 moles of solid magnesium oxide
Reaction Ratios:
Classifying Chemical Reactions
A.
Pattern for prediction based
on the kind of reactants
Combustion or Burning – complete
combustion always produces carbon
dioxide and water!
1.
Hydrocarbons
CxHy + O2  CO2 + H2O
2.
Alcohols
CxHyOH + O2  CO2 + H2O
3.
Sugars
C6H12O6 + O2  CO2 + H2O
C12H22O11 + O2  CO2 + H2O
Synthesis or Composition
1.
2/more reactants  1 product
Element + Element  Compound
A + B  AB
2 Na + Cl2  2 NaCl
4 Al + 3 O2  2 Al2O3
Synthesis
Compound + Compound  Compound
EXAMPLE 1: metal
CaO
oxide + carbon dioxide  metal carbonate
+
CO2

CaCO3
oxide + water  a base (hydroxide)
Na2O +
H2O 
2 NaOH
EXAMPLE 2: metal
H(OH)
EXAMPLE 3:
nonmetal oxide + water  an acid
SO3 +
H2O 
H2SO4
**Determine oxidation numbers for molecular compounds and oxyacids**
Decomposition
Binary Compounds
1. Binary Compound  2 elements
AB

A + B
2 H2O
2 HgO

2 H 2 + O2
 2 Hg + O2
Decomposition - Ternary Compounds
Ternary Compound  Compound + Element/Compound
EXAMPLE 1:
metal chlorate  metal chloride + oxygen
2KClO3 
EXAMPLE 2: metal
+
3O2
carbonate  metal oxide + carbon dioxide
CaCO3
EXAMPLE 3: metal
2KCl

CaO + CO2
hydroxide  metal oxide + water
Mg(OH)2  MgO + H2O
EXAMPLE 4: acids
(Except Group IA metals)
 nonmetal oxide + water
H2CO3  CO2 + H2O
EXAMPLE 5: Hydrogen
Peroxide
2H2O2  2H2O + O2
Single Replacement or
Single Displacement
Element + Compound  New Compound + New Element
1.
Metals
A
+
BC
 AC
+ B
Active metals displace less active metals or hydrogen from their
compounds in aqueous solution. Refer to the Activity Series.
a. 2Al + 3CuCl2
 2AlCl3 + 3Cu
b. metal + H2O  metal hydroxide + H2
An active metal (top of series to calcium) will react with water to
form the hydroxide of the metal and hydrogen gas.
2Na + 2HOH  2NaOH + H2
Single Replacement or
Single Displacement
2. Nonmetals
D + EF  ED + F
Cl2 + 2NaBr  2NaCl + Br2
Many nonmetals displace less active
nonmetals from combination with a metal or
other cation. Order of decreasing activity is
F2  Cl2  Br2  I2
Double Replacement/Displacement
or Metathesis:
Compound + Compound  New Compound + New Compound
AB +
CD
2AgNO3 (aq) + CaCl2 (aq)
Pb(NO3)2 (aq) + 2NaCl

AD

(aq)
2AgCl
+
(s)
CB
+ Ca(NO3)2 (aq)
 PbCl2 (s) + 2NaNO3 (aq)
The driving force for these reactions is if it produces a
1.
A precipitate (ppt): See Solubility Table
2.
Water
3.
Gas: Only HCl and NH3 are soluble in water. All other gases (CO2 and
H2S) are sufficiently insoluble to force a reaction to occur if they are found
as a product.
Double Replacement Reactions
Solutions of sodium chromate and aluminum acetate are mixed.
3Na2CrO4(aq) + 2Al(CH3COO)3(aq) +  Al2(CrO4)3(s) + 6NaCH3COO(aq)


Magnesium hydroxide and ammonium bromide
Mg(OH)2 + 2 NH4Br  MgBr2 + 2 NH4OH
If NH4OH is produced, it breaks up into ammonia and water
Mg(OH)2 + 2 NH4Br  MgBr2 + 2 NH3 + 2 H2O

Sulfuric acid and magnesium carbonate
H2SO4 + MgCO3  MgSO4 + H2CO3
If H2CO3 is a product, it breaks up into carbon dioxide and water
H2SO4 + MgCO3  MgSO4 + H2O + CO2
Thermochemistry


The study of the changes in energy that
accompany a chemical reaction and physical
changes.
Chemical Reactions involve changes in energy
that result from



Bond breaking that requires energy (absorbs) from the
surroundings.
Bond making that produces energy (releases) to the
surroundings.
Changes in energy result in an energy flow or
transfer.
Types of Reactions
1.
Exothermic Reactions: a reaction that
releases heat into their surroundings.


Heat is a product of the reaction and temperature
of the surroundings increase.
This occurs during bond formation.
surroundings
surroundings
Exothermic
Reaction
(system)
surroundings
surroundings
Types of Reactions
2.
Endothermic Reactions: a reaction that
absorbs heat from the surroundings.


Heat acts as a reactant and temperature of the
surroundings decreases.
This occurs during bond breaking.
surroundings
surroundings
Endothermic Reaction
(system)
surroundings
surroundings
Energy & Chemical Equations
Coefficients are always interpreted as moles. Physical states are written –
influences the overall energy exchanged. Very specific!

Exothermic – release energy; E product
CaCl2(s)  Ca+2 (aq) + 2Cl-1(aq) + 88.0kJ

Combustion reactions are ALWAYS exothermic:
C3H8 + 5O2 → 3CO2 + 4H2O + 2043 kJ


Endothermic– absorbs energy; E reactant
2NH4Cl(s) + Ba(OH)2·8H2O(s) + 63.9 kJ 
BaCl2(s) + 2NH3(g) + 10H2O
Rewrite for 1 mole of Cl-1:
½ CaCl2(s)  ½ Ca+2 (aq) + 1 Cl-1(aq) + 44.0kJ
Heat and Enthalpy Changes

Enthalpy (H): the heat content of a system at
constant pressure.


Unit: J
Enthalpy Change (H): is the heat absorbed or
released in a physical or chemical change at
constant pressure.




H = Hproducts ─ Hreactants
This can be measured.
H is also known as the heat of the reaction.
Difference between the stored energy of the reactants
and the products.
Enthalpy Diagrams
#1
#2
#1
#2
a. Which has a higher enthalpy?
Products or Reactants
R
P
b. Was heat absorbed or released?
R
A
c. Is this an endothermic or exothermic reaction?
Exo Endo
d. Is ΔH for this reaction positive or negative?
-
+
e. Would the ΔH be on the left or right side of the
yield sign?
R
L
f. Is the reverse reaction exothermic or
endothermic?
Endo Exo
Rewrite each equation with the heat term
in the reaction as a reactant or product –
THERMOCHEMICAL equation:
#1) C3H8 + 5O2 → 3CO2 + 4H2O + 2043 kJ
#2) C + H2O + 113kJ → CO + H2
Enthalpy Diagrams
products
reactants
BaCl2 + 2NH3 + 10H2O
H
(kJ)
CaCl2
H
(kJ)
∆H = --88.0 kJ
∆H = ++63.9 kJ
products
reactants
2NH4Cl + Ba(OH)2 8H2O
Ca+2 + 2Cl-
Course of Reaction
Course of Reaction
Endothermic
Exothermic
Reaction Progress
Collision Theory
A.
In order for a reaction to occur, the particles
must collide
A successful or effective collision occurs when
1.
2.
a)
b)
3.
4.
The collision is energetic enough
The particles collide with the correct orientation
During a collision, kinetic energy is
converted to potential energy
The minimum energy needed for a successful
collision =
activation energy (Ea)
Reaction Pathways or Potential
Energy (heat content) Diagrams
Reaction Pathways or Potential
Energy (heat content) Diagrams
Answer the following questions based on the
potential energy diagram shown here:
1.
2.
3.
4.
5.
6.
Does the graph represent an endothermic or exothermic
reaction?
Label the position of the reactants, products, and activated
complex.
Determine the heat of reaction, ΔH, (enthalpy change) for this
reaction.
Determine the activation energy, Ea for this reaction.
How much energy is released or absorbed during the
reaction?
How much energy is required for this reaction to occur?
Solution
1. The graph represents an endothermic reaction
2.
3.
4.
5.
6.
ΔH = +50 kJ.
Ea = +200 kJ
50 kJ of energy are absorbed during this
endothermic reaction (this is the value of ΔH)
200 kJ of energy are required for this reaction to
occur (Ea).
Practice





Sketch a potential energy curve that
is represented by the following values
of ΔH and Ea. You may make up
appropriate values for the y-axis
(potential energy).
ΔHforward = -20 kJ
Earev = 80 kJ
Activated Complex = 120 kJ
Is this an endothermic or exothermic
reaction?
Solution

Based on your diagram, determine:




ΔHforward = -20 kJ
Eaforward = +60 kJ
Enthalpy of reactants = 60 kJ
Enthalpy of products = 40 kJ
Enthalpy Diagram - Formative
Assessment #1





Sketch a potential energy curve that
is represented by the following values
of ΔH and Ea.
ΔHreverse = -10 kJ
Eaforward = +40 kJ
Activated Complex = 50 kJ
Is this an endothermic or exothermic
reaction?
Enthalpy Diagram - Formative
Assessment #1
Based on your diagram,
determine:
1.
2.
3.
4.
5.
Endo or Exo?
ΔHforward =
Eaforward =
ΔHreverse =
Eareverse =
Enthalpy Diagram – FA #1
Answer
Based on your diagram,
determine:
1.
2.
3.
4.
5.
Exothermic
ΔHforward = -20 kJ
Eaforward = +60 kJ
ΔHreverse = +20 kJ
Eareverse = +80 kJ
Enthalpy Diagram - Formative
Assessment #2





Sketch a potential energy curve that
is represented by the following values
of ΔH and Ea.
ΔHforward = -100 kJ
Eareverse = +150 kJ
Activated Complex = 200 kJ
Is this an endothermic or exothermic
reaction?
Enthalpy Diagram - FA#2
Answer:






Activated Complex
= 200 kJ
Eareverse = +150 kJ
ΔHforward = -100 kJ
ΔHreverse = +100 kJ
Eaforward = +50 kJ
exothermic
Enthalpy Diagram Formative Assessment #2
Based on your diagram,
determine:
1.
2.
3.
4.
5.
Endo or Exo?
ΔHforward =
Eaforward =
ΔHreverse =
Eareverse =
Enthalpy Diagram - FA#2
Answer
Based on your diagram,
determine:
1.
2.
3.
4.
5.
Endo
ΔHforward = +200 kJ
Eaforward = +300 kJ
ΔHreverse = -200 kJ
Eareverse = +100 kJ
Calculating ∆H using Bond Energy
2 H2 + O2  2 H2O
Bonds Formed = exothermic (-)
Bonds Broken = endothermic (+)
Using Bond Energy Table, determine ∆H.
∆H = -482 kJ (482 kJ released = exothermic)
Hess’s Law
The enthalpy change for a reaction is the
sum of the enthalpy changes for a series
of reactions that adds up to the overall
reaction.
2.
This is also called the Law of Heat of
Summation (Σ)
3. This allows you to determine the enthalpy
change for a reaction by indirect means
when a direct method cannot be done.
1.
Steps for using Hess’s Law
1.
2.
3.
4.
5.
Write a balanced equation.
Identify the compounds.
Locate the compounds on the Heats of Reaction
Table (or given).
Write the reaction from the table so the compound
is a reactant or product as it appears in the
balanced equation.
Write appropriate ΔH for each sub equation.
a)
b)
6.
7.
If needed, multiply the sub equation and the associated
ΔH’s (coefficients).
If you reverse the equation, change the sign of the
enthalpy change.
Add the sub equations to arrive at the desired
balanced equation.
Add ΔH’s of each sub equation to calculate the ΔH
for the desired balanced equation.
Calculate ΔH for the following example:
#1) XeF2 + F2  XeF4
Xe + F2  XeF2
Xe + 2F2  XeF4
XeF2 + F2  XeF4
ΔH = ?
ΔH = -123 kJ
ΔH = -262 kJ
ΔH = -139 kJ
#2) C + H2O → CO + H2
2CO  2C + O2
2H2 + O2  2H2O
ΔH = ?
ΔH = +222 kJ
ΔH = -484 kJ
C + H2O → CO + H2 ΔH = +131 kJ
Calculate ΔH for the following example:
#3) CO + O2 → 2 CO2
2C + O2  2CO
CO2  C + O2
CO + O2 → 2 CO2
ΔH = ?
ΔH = -222 kJ
ΔH = +394 kJ
ΔH = -394 kJ
#4) H2O2 + H2 → 2 H2O
H2O + ½ O2  H2O2
2H2 + O2  2H2O
H2O2 + H2 → 2H2O
ΔH = ?
ΔH = +94.6 kJ
ΔH = -484kJ
ΔH = -336.6kJ
Calculate ΔH for the following example:
#1) C(s)
+
H2O(g)
→ CO(g)
+
H2(g)
H2O(g) → H2(g) + 0.5O2(g)
C(s) + 0.5O2 → CO
ΔH = +242.0kJ
ΔH = -111.0kJ
C(s) + H2O(g) → CO(g) + H2(g)
ΔH= +131.0kJ
Thermochemical Equation:
C(s) + H2O(g) + 131.0kJ → CO(g) + H2(g)
Calculate ΔH for the following example:
#2) 2 CO(g) + O2(g) → 2 CO2(g)
2
CO(g)
2 C(s)
+
1
O
(g)
CO(g)
→→0.5C
(g)
+
C(s)
2
2
C(s)
(g)CO2(g)
1
C(s)++O12(g)
O2→
(g) CO
→21
ΔH==+111.0
+222.0kJ
kJ
ΔH
ΔH
ΔH==-394.0
-394.0kJ
kJ
2CO(g) + O2(g) → 2CO2(g)
ΔH = -566.0 kJ
Thermochemical Equation:
2CO(g) + O2(g) → 2CO2(g) + 566.0kJ