CHAPTER 7
Chemical Reactions & Quantities
Reactions & Equations
Balancing Chemical Reactions
Types of Reactions
Oxidation-Reduction
The Mole and Chemical Equations
Mass Calculations
Percent Yield & Limiting Reactants
Energy Changes in Reactions
CH104
5-1
Physical properties
Characteristics that can be evaluated
without changing the composition of
the material.
Examples
CH104
Color
Density
Odor
Melting / Freezing point
Taste
Boiling point
Feel
Compressibility
Shape
Form (foil, wire, powder…)
5-2
What is Chemistry?
“The study of Matter and its Changes.”
Physical Changes =
Changes in a Physical Property
Appearance:
•melting, freezing, evaporation…
•stretching, molding, cutting…
Chemical Changes =
Changes in a Chemical Property
Chemical Composition:
CH104
5-3
Chemical Changes
Change in the Chemical Composition
Examples:
Burning of Magnesium
Rusting of Iron
Decomposing of wood
Souring of Milk
CH104
5-4
Examples
Which are chemical or physical changes?
Mulching leaves
Tarnishing Silver
Fermentation
Carbonated Beverage going flat
Making ice into water
Bleaching a stain
CH104
5-5
Chemical Reactions
Shows how the Chemical change occurs.
Reactants
C3H8 + O2
CH104
Mg +
O2
Fe +
O2
Products


CO2 + H2O + Energy
MgO
+
Energy
Fe2O3
5-6
Chemical equations
Chemist’s shorthand to describe a reaction.
2 H2 (g) + O2 (g) heat
•Reactants
2 H2O(g) + E
•Products
•The state of all substances
•Any conditions used in the reaction
•Same # & type atoms on each side
• Law of Conservation of Matter
CH104
5-7
Balancing Equations
Making Hot dogs:
How many packages wieners & buns
to buy so none is left over.
4 10 + ___B
5 8
___W
•Reactants
CH104
40
___WB
•Products
5-8
Balancing Equations
Ca
+
HCl
CaCl2 +
•Reactants
H2
•Products
1 Ca
1
1
H
2
1
Cl
2
Step 1: Count atoms of each element
on both sides of equation.
CH104
5 - 10
Balancing Equations
Ca
+
HCl
CaCl2 +
•Reactants
H2
•Products
1 Ca
1
1
H
2
- not balanced
1
Cl
2
- not balanced
Step 2: Determine which atoms
are not balanced.
CH104
5 - 11
Balancing Equations
Ca
+ 2 HCl
CaCl2 +
•Reactants
H2
•Products
1 Ca
1
2 1
H
2
- not balanced
2 1
Cl
2
- not balanced
Step 3: Balance one element at a time
with coefficients in front of formulas
until all balanced.
(Never change the formula!)
CH104
5 - 12
Balancing Equations
Na3PO4 +
MgCl2
Mg3(PO4)2 +
•Reactants
NaCl
•Products
3
1
4
1
2
Na
P
O
Mg
Cl
1
2
8
3
1
Step 1: Count atoms of each element
on both sides of equation.
CH104
5 - 13
Balancing Equations
Na3PO4 +
MgCl2
Mg3(PO4)2 +
•Reactants
NaCl
•Products
3
1
4
1
2
Na
P
O
Mg
Cl
1
2
8
3
1
- not balanced
- not balanced
- not balanced
- not balanced
- not balanced
Step 2: Determine which atoms
are not balanced.
CH104
5 - 14
Balancing Equations
Na3PO4 +
MgCl2
Mg3(PO4)2 +
•Reactants
NaCl
•Products
3
1
4
1
2
Na
P
O
Mg
Cl
1
2
8
3
1
- not balanced
- not balanced
- not balanced
- not balanced
- not balanced
Step 3: Balance elements with #’s
in front of formulas until all balanced.
(Never change the formulas!)
CH104
5 - 15
Balancing Equations
2 Na3PO4 + 3 MgCl2
•Reactants
6
2
8
3
6
Hints:
3
1
4
1
2
1 Mg3(PO4)2 + 6 NaCl
Na
P
O
Mg
Cl
•Products
1 -6not balanced
2 - not balanced
8 - not balanced
3 - not balanced
1 -6not balanced
•Start with a metal in a complex
compound, or an element that only
appears in one formula. (Like Mg here)
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5 - 16
Balancing Equations
C2H6
+
•Reactants
O2
CO2 +
C
H2O
•Products
H
Hints:
O
•Start with an element that only appears
in one formula on both sides of the
equation.
•Leave oxygen until last.
CH104
5 - 17
Balancing Equations
C2H6
+
O2
CO2 +
•Reactants
H2O
•Products
2
C
1
6
H
2
2
O
3
Step 1: Count atoms of each element
on both sides of equation.
CH104
5 - 18
Balancing Equations
C2H6
+
O2
CO2 +
•Reactants
H2O
•Products
2
C
1
- not balanced
6
H
2
- not balanced
2
O
3
- not balanced
Step 2: Determine which atoms
are not balanced.
CH104
5 - 19
Balancing Equations
C2H6
+ 3.5 O2
2 CO2 + 3 H2O
•Reactants
•Products
2
C
6
H
1 2- not balanced
2 6- not balanced
7 2
O
3 5- not7 balanced
Step 3: Balance one element at a time
with coefficients in front of formulas
until all balanced.
(Never change the formula!)
CH104
5 - 20
Balancing Equations
C2H6
+ 3.5 O2
2 CO2 + 3 H2O
•Reactants
•Products
2
C
6
H
1 2
2 6
7 2
O
3 5 7
Can’t have 3.5 O2 ,
so multiply equation by 2!
CH104
5 - 21
Balancing Equations
2 C2H6
7 O2
+ 3.5
4 CO2 + 6 H2O
•Reactants
14
•Products
4 2
C
12 6
H
1 2 4
2 6 12
7 2
O
3 5 7 14
Can’t have 3.5 O2 ,
so multiply equation by 2!
CH104
5 - 22
Balancing Equations
(NH2)2CO +
H2O
2
6
1
2
CH104
______
>
N
H
C
O
1
3
NH3 +
CO2
- not balanced
- not balanced
1
2
5 - 23
Balancing Equations
(NH2)2CO +
H2O
2
6
1
2
CH104
______
> 2 NH3 +
N
H
C
O
CO2
1 2
3 6
1
2
5 - 24
Example: Decomposition of urea
CH3OH + PCl5  CH3Cl + POCl3 + H2O
1
4
1
1
5
CH104
C
H
O
P
Cl
1
5
2
1
4
- not balanced
- not balanced
- not balanced
- not balanced
5 - 25
Balancing Equations
2CH3OH + PCl5 2CH3Cl + POCl3 + H2O
2 1
8 4
2 1
1
5
CH104
C
H
O
P
Cl
1 2
5 8
2
1
4 5
5 - 26
Types of Chemical Reactions
Combination
Decomposition
A + B  C
C A + B
Single Replacement: Substitution
A + BX  B
+
AX
Double Replacement: Metathesis
AX + BY  BX
CH104
+
AY
5 - 27
Types of Chemical Reactions
Combination
2H2 + O2  2H2O
Decomposition CaCO  CaO + CO
3
2
Single Replacement: Substitution
Al + FeCl3  Fe + AlCl3
Double Replacement: Metathesis
2AgNO3 + K2SO4  Ag2SO4 + 2KNO3
CH104
5 - 28
Types of Chemical Reactions
Combustion
Complete:
C3H8 + 5O2  3CO2 + 4H2O
Incomplete:
2C3H8 + 7O2  6CO + 8H2O
C3H8 + 2O2  3C + 4H2O
CH104
5 - 29
Combination Reactions
A + B  C
Explosion of Hydrogen Balloon
2H2 + O2  2H2O
Rusting of Iron
4 Fe + 3 O2  2 Fe2O3
Formation of Acid Rain
SO3 + H2O  H2SO4
CH104
5 - 30
Decomposition Reactions
C A + B
Heating Egg Shells
CaCO3  CaO + CO2
Blood with peroxide
2 H2O2
CH104
2 H2O + O2
5 - 31
Single Replacement
Reactions
A + BX  B
+
AX
Iron Deposits on an Aluminum Pan
Al + FeCl3 
CH104
Fe + AlCl3
5 - 32
increasing reactivity
CH104
potassium
sodium
calcium
magnesium
aluminum
zinc
chromium
iron
nickel
tin
lead
Hydrogen
copper
silver
platinum
gold
Activity series of metals
Al + Fe+3  Fe + Al+3
Fe + H+  Fe+3 + H2
Element
give e’s
to ion
lower on
list
5 - 33
Double Replacement Reaction
AX + BY  BX
+
AY
BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)
Ba+2
Cl-1
Insoluble Precipitate
Formed
Na+1 SO4-2
CH104
5 - 34
Double Replacement Reaction
AX + BY  BX
Predict the products:
AgNO3(aq) + AlCl3 (aq)
Ag+
Al+3
AY
AgCl(s) + Al(NO3)3(aq)
NO3-
Cl-
Balance later as needed to get:
3AgNO3(aq) + AlCl3 (aq)
CH104
+
3AgCl(s) + Al(NO3)3(aq)
5 - 35
Oxidation and reduction
REDOX
Where reactants exchange
electrons -
Examples:
•All types of batteries
alkaline, NiCad, car batteries
•Rusting and corrosion
•Metabolism
•Antioxidants (Vit C, E prevent oxidation)
CH104
5 - 36
Oxidation and reduction
REDOX
Where reactants exchange
electrons -
Oxidation = Losing electrons
LEO: Lose Electrons Oxidation
OIL : Oxidation Is Losing
Reduction = Gaining electrons
GER: Gain Electrons Reduction
RIG : Reduction Is Gaining
LEO the lion says GER
OIL RIG
CH104
5 - 37
Oxidation and reduction
Assign Oxidation States:
2 Na(s) + Cl2 (g)
0
0
For element in natural form
Ox State = 0.
CH104

2 NaCl
1+
1-
For simple ions,
Ox state = charge.
5 - 38
Oxidation and reduction
Who’s loosing or gaining electrons?
Loses
1 e- = LEO

2 Na(s) + Cl2 (g)
0
Gains
0 1 e-=GER
2 NaCl
1+
Na loses e- (LEO)
Na gets oxidized
Cl gains e- (GER)
Cl gets reduced
CH104
1-
5 - 39
Oxidation and reduction
Oxidation - when reactant loses e-(s).
(LEO)
Na+
Na (s)
+ e
Reduction - when reactant gains e-(s).
(GER)
Cl2 (g)
+2e
2 Cl
These are half reactions
CH104
5 - 40
Oxidation and reduction
2 half reactions make a complete reaction
+
Na

Na
+
e
2
2
2
(s)
Cl2 (g) + 2 e-

2 Cl-
2 Na(s) + Cl2 (g) + 2 e-  2 Na+ + 2e- + 2 Cl-
2 Na(s) + Cl2 (g) 
CH104
2 Na+ + 2 Cl-
5 - 41
Oxidation and reduction
Oxidizing agent =
• The chemical that caused an oxidation.
• It is reduced.
Reducing agent =
•The chemical that caused a reduction.
•It is oxidized.
CH104
5 - 42
Oxidation and reduction
Who’s loosing or gaining electrons?
Loses
1 e- =LEO
2 Na(s) + Cl2 (g)
0
Gains
0 1 e- =GER

2 NaCl
1+
1-
• Na loses e- (LEO) so Na gets oxidized
• Na caused Cl to get reduced
• Na is the Reducing agent
CH104
5 - 43
Oxidation and reduction
Who’s loosing or gaining electrons?
Loses
1 e- =LEO
2 Na(s) + Cl2 (g)
0
Gains
0 1 e- =GER

2 NaCl
1+
1-
• Cl gains e- (GER) so Cl gets reduced
• Cl caused Na to get oxidized
• Cl is the Oxidizing agent
CH104
5 - 44
Oxidation state
Describes the charge of each element.
Rules
• Oxidation state of element in natural form = 0.
Examples N2, Na, O2, H2
•
For simple monoatomic ions,
oxidation state = charge.
Examples Na+1, Cl-1, Ca2+
•
For certain groups at certain times,
oxidation number = group number
Examples N+5, Cl+7
CH104
5 - 46
Oxidation states
Oxygen
O-2 Usually
Example H2O
O-1 in peroxides
Example H2O2
Hydrogen
H1+ if bonded to nonmetal
Example HCl
H1- if bonded to metal
Example NaH
CH104
5 - 47
Oxidation states
Assign the oxidation states for all elements in
H2O
+1
-2
+1
+2
CH104
-2
5 - 48
Oxidation states
Assign the oxidation states for all elements in
H2O2
+1
+1
+2
CH104
-1
-1
-2
5 - 49
REDOX reactions
Lose 1 e- = LEO
Oxidation state of H is 0
2 H2 + O 2
Oxidation state of O is 0
Oxidation state of H is 1+
2 H2O
Oxidation state of O is 2-
Gain 2 e- = GER
Hydrogen is oxidized and is a reducing agent.
Oxygen is reduced and is an oxidizing agent.
CH104
5 - 50
Oxidation states
Find the oxidation state for all elements in:
HNO3
Hydrogen is
Oxygen is
1+
Nitrogen must be 5+
+1 +5
CH104
Start with what
We know
222-6 = 0
5 - 51
Oxidation states of Cl
NaCl
+1
NaClO3 -2
-1
+1
NaClO
+1
+1
+1
+3
CH104
NaClO4 -2
-2
+1
NaClO2
-2
-2
+5
-2
-2
+7
-2
-2
-2
5 - 52
Types of chemical reactions
Chemical Reactions
Nonredox
CH104
Redox
Combination
Combination
Decomposition
Decomposition
Double
replacement
Single
replacement
5 - 53
Combination reactions
B + C
A
REDOX or NONREDOX types
Formation of Acid Rain
SO3 +
+6
-2
-2
-2
H2O 
+1 -2
+1
H2SO4
+1 +6 -2
-2
+1
-2
-2
Non-REDOX reaction.
CH104
5 - 54
Combination reactions
B + C
A
REDOX or NONREDOX types
Rusting of Iron
4 Fe + 3 O2  2 Fe2O3
0
0
+3
+3
REDOX reaction.
Fe goes from 0 to +3 LEO
O goes from 0 to -2 GER
CH104
-2
-2
-2
5 - 55
Decomposition reactions
A
B + C
REDOX or NONREDOX types
Decomposition of Egg Shells
CaCO3 
+2
+4
-2
-2
-2
CaO +
+2
-2
CO2
+4
-2
-2
Non-REDOX reaction.
CH104
5 - 56
Decomposition reactions
Decomposition of hydrogen peroxide
2 H2O2
2 H2O + O2
+1
+1
+1
+1
-1
-1
-2
0
REDOX reaction.
Some O goes from -1 to 0 (O2) LEO
Some O goes from -1 to -2 (H2O) GER
CH104
5 - 57
Single Replacement Reaction
A + BX  B + AX
Always REDOX
Iron Deposits on an Aluminum Pan
Al + FeCl3 
Fe + AlCl3
0
0
+3
-1
-1
-1
+3
-1
-1
-1
REDOX reaction.
Al goes from 0 to +3 LEO
Fe goes from +3 to 0 GER
CH104
5 - 58
increasing reactivity
CH104
potassium
sodium
calcium
magnesium
aluminum
zinc
chromium
iron
nickel
tin
lead
Hydrogen
copper
silver
platinum
gold
Activity series of metals
Al + Fe+3  Fe + Al+3
Fe + H+  Fe+3 + H2
Element
give e’s
to ion
lower on
list
5 - 59
Double Replacement Reaction
AX + BY  BX + AY
Always non-REDOX
AgNO3(aq) + NaCl(aq)  AgCl(s)+ NaNO3(aq)
-2
-1
+1
-1 +1 +5 -1
+1 +5 -2 +1 -1
-2
-1
Non -REDOX reaction.
CH104
5 - 60
Ionic equations
Ionic substances dissociate into ions when
dissolved in water.
AgNO3(aq) + NaCl (aq)
AgCl + NaNO
(s)
Ag+
3(aq)
NO3Certain ions join together
Na+
Cl-
Others remain unchanged.
Ag+ + NO3- + Na+ + ClCH104
AgCl(s) + Na+ + NO35 - 61
Ionic equations
Total ionic equation
Ag+ + NO3- + Na+ + Cl-
AgCl(s)+ Na+ +NO3-
NO3- and Na+ are spectator ions.
Net ionic equation
Ag+ + ClCH104
AgCl(s)
5 - 62
The Mole
1 pair = 2
1 dozen = 12
1 mole = 6.02 x 1023
602,000,000,000,000,000,000,000.
1 mol eggs___
6.02 x 1023eggs
1 mol Au_______
6.02 x 1023 Au atoms
_____1 mole H2O_____
6.02 x 1023 H2O molecules
CH104
5 - 63
The Mole & Formulas
1 car ___
4 wheels
CH104
1 doz cars
4 doz wheels
1 mol cars_
4 mol wheels
1 mole H2O
1 mole H2O
2 mol H
1 mol O
5 - 64
The Mole & Formulas
1 car ___
4 wheels
5 doz cars
1
CH104
1 doz cars
4 doz wheels
1 mol cars_
4 mol wheels
4 doz wheels = 20 doz wheels
1doz cars
5 mol H2O
2 mol H
1
1 mol H2O
= 10 mol H
5 - 65
The Mole & Molar Mass
1 mole = MW in g’s
1 mole Au = 197 g Au
1 mol Au_
197 g Au
197 g Au
1 mol Au
__197 g Au
_
6.02 x 1023 atoms Au
CH104
5 - 66
The Mole & Molar Mass
1 mole = MW in g’s
1 mole S = 32 g S
1 mol S_
32 g S
32 g S
1 mol S
1 mole C = 12 g C
CH104
1 mol C
12 g C
12 g C
1 mol C
5 - 67
The Mole & Molar Mass
1 mol H2O_
1 mole H2O has:
18.0 g H2O
2 mol H 1.0 g H = 2.0 g H
18.0 g H2O
1
1 mol H
1 mol H2O
1 mol O 16.0 g O = 16.0 g O
1
1 mol O
18.0 g
CH104
5 - 68
Molar Mass
Find the MW of Glucose; C6H12O6
6 mol C 12.0 g C = 72.0 g C
1
1 mol C
12 mol H 1.0 g H = 12.0 g H
1
1 mol H
6 mol O 16.0 g O = 96.0 g O
1
CH104
1 mol O 180.0 g C6H12O6
1 mol C6H12O6
5 - 69
Mass to Mole Conversions
How many moles of water are in
36 g H2O?
What should the answer look like?
What is Unique to the problem?
CH104
36 g H2O
1 mol H2O =
1
18 g H2O
2.0 mol H2O
5 - 70
Mass to Mole Conversions
How many moles of H are in
36 g H2O?
What should the answer look like?
What is Unique to the problem?
36 g H2O
1
CH104
1 mol H2O
2 mol H = 4.0 mol H
18 g H2O 1 mol H2O
5 - 71
Mole to Mass Conversions
How many g’s of Glucose (C6H12O6)
are in 5 mol Glucose?
What should the answer look like?
What is Unique to the problem?
5 mol Gluc
1
CH104
180 g Gluc =
900 g Glucose
1 mol Gluc
5 - 72
The mole and chemical
equations
Stoichiometry
- Calculations of quantities in a chemical rxn.
You need a balanced
equation.
2H2 + O2 -----> 2H2O
CH104
5 - 73
Moles in Chemical Equations
2 Na(s) + Cl2 (g)
2 mol Na
1 mol Cl2
CH104

2 mol Na
2 mol NaCl
2 NaCl
1 mol Cl2
2 mol NaCl
5 - 74
Moles in Chemical Equations
2 Na(s) + Cl2 (g)

2 NaCl
How many moles of Cl2 are needed to
completely react with 40 moles of Na?
What should the answer look like?
What is Unique to the problem?
40 mol Na
1
2 mol Na
1 mol Cl2
CH104
1 mol Cl2 =
2 mol Na
2 mol Na
2 mol NaCl
20 mol Cl2
1 mol Cl2
2 mol NaCl
5 - 75
Mass in Chemical Equations
40 g’s Na
g’s Cl2
mols Na
mols Cl2
2 Na(s) + Cl2 (g)

2 NaCl
How many g’s of Cl2 are needed to
completely react with 40 g’s of Na?
What should the answer look like?
What is Unique to the problem?
40 g Na 1mol Na 1 mol Cl2 70.9 g Cl2 = 61.7 g Cl2
1
CH104
23.0g Na 2 mol Na 1 mol Cl2
5 - 76
Mass in Chemical Equations
20.g’s Fe2O3
mols Fe2O3
Fe2O3 + 3H2
g’s Fe
mols Fe
 2Fe + 3H2O
How many g’s of Fe can be produced
from 20. g’s of Fe2O3?
What should the answer look like?
What is Unique to the problem?
20.g Fe2O3 1mol Fe2O3 2 mol Fe
1
CH104
55.8 g Fe = 14 g Fe
159.7g Fe2O3 1 mol Fe2O3 1 mol Fe
5 - 77
Percent Yield
Fe2O3 + 3H2
 2Fe + 3H2O
How many g Fe can be made from 20.g Fe2O3?
1mol Fe2O3
2 mol Fe
55.8 g Fe = 14 g Fe
20.g Fe2O3
159.7g Fe2O3 1 mol Fe2O3 1 mol Fe
1
What is the percent yield if I only got 12
g’s of Fe from 20. g’s of Fe2O3?
% yield = Actual x 100
Theoretical
% yield = 12 g
14 g
CH104
x 100 = 85.7 %
5 - 78
Limiting reactant
1 c milk + 2 c bisquick + 1 Tbs oil +1 egg = 18
pancakes
If I have 5 c. Bisquick and 2 eggs,
how many pancakes can I make?
From 5 c. Bisquick
5 c. Bquick
18 p cakes =
45 p cakes
2 c. Bquick
From 2 eggs
2 eggs
18 p cakes =
36 p cakes
1 egg
CH104
5 - 79
Limiting reactant
1 c milk + 2 c bisquick + 1 Tbs oil +1 egg = 18
pancakes
If I have 5 c. Bisquick and 2 eggs,
how many pancakes can I make?
From 5 c. Bisquick
45 p cakes
From 2 eggs
36 p cakes
So: Eggs are the limiting reagent.
I can make only 36 pancakes
I’ll have bisquick left over.
CH104
5 - 80
Limiting reactant
Fe2O3 + 3H2
 2Fe + 3H2O
If I have 20.g of Fe2O3 and 2.0g H2,
how many g’s Fe can I make?
From 20. g Fe2O3
From 2.0 g H2
2.g H2 1mol H2
1
2 mol Fe 55.8 g Fe =
2.02 g H2 3 mol H2
So: Fe2O3 is the limiting
I can make only 14 g Fe
I’ll have H2 left over.
CH104
14 g Fe
37 g Fe
1 mol Fe
reagent.
5 - 81
Limiting reactant
4NH3 + 5O2
 4NO + 6H2O
If I have 20.g of NH3 and 20.g O2,
how many g’s H2O can I make?
From 20. g NH3
20.g NH3 1mol NH3 6 mol H2O 18 g H2O =31.7 g H2O
1
17 g NH3 4 mol NH3 1 molH2O
From 20. g O2
20.g O2 1mol O2 6 mol H2O 18 g H2O =13.5 g H2O
1
CH104
32 g O2
5 mol O2
1 molH2O
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Limiting reactant
4NH3 + 5O2
 4NO + 6H2O
If I have 20.g of NH3 and 20.g O2,
how many g’s H2O can I make?
From 20. g NH3
31.7 g H2O
From 20. g O2
13.5 g H2O
So: O2 is the limiting reagent.
We can make only 13.5 g H2O
We’ll have NH3 left over.
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Limiting reactant
5C + 2SO2
 1CS2 + 4CO
If I have 10.g of C and 10.g SO2,
how many g’s CS2 can I make?
From 10. g C
10.g C
1
1mol C
1 mol CS2 76 g CS2 = 13 g CS2
12 g C
5 mol C
1 mol CS2
From 10. g SO2
10.g SO2 1mol SO2 1 mol CS2 76 g CS2 = 5.9 g CS2
1
64 g SO 2 mol SO2 1 mol CS2
2
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Limiting reactant
5C + 2SO2
 1CS2 + 4CO
If I have 10.g of C and 10.g SO2,
how many g’s CS2 can I make?
From 10. g C
13 g CS2
From 10. g SO2
5.9 g CS2
So: SO2 is the limiting reagent.
We can make only 5.9 g CS2
We’ll have C left over.
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Study of Energy changes in
Reactions =
Thermodynamics
Enthalpy
and
Entropy
= Change in
= Change in
Energy =
=
H
S
•used to calculate the amount of useful work produced by
chemical reactions.
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law of Thermodynamics
“Energy can’t be created or
destroyed in a chemical reaction”
(you can’t get something from nothing)
•Energy just gets
converted from
one form to another.
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Exothermic Reactions
Mg + O2
MgO + Energy
CH4 + 2O2
CO2 + 2H2O + 211 kcal
Old bonds break
Breaking bonds
costs E
(+H)
New bonds get made
Making bonds
gives E
(-H)
Hproducts - Hreactants = H
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Activation Energy (Eact)
The minimum amount of energy
required to produce a chemical reaction.
•
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The energy of collision must be great
enough to break the old bonds and form
the new ones.
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Energy in Chemical Reactions
Energy
2H2 + O2
2H2O + Energy
Exothermic reaction
Reactants
H2 + O2
Eact= Activation Energy
-H= heat of reaction
(Gets hot)
H2O
Products
more stable
Rxn Progress
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Exothermic (Exergonic) Rxns
Energy
2H2 + O2
2Mg + O2
CH4 + 2O2
 2H2O + Energy
 2MgO + Energy
 CO2 + 2H2O + 213 kcal
Rxn Progress
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Exothermic Reactions
Energy is released
Energy
Products are more stable.
Reactants
(Water)
-H
Products
(Ice)
Rxn Progress
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Energy in Chemical Reactions
Endothermic reaction
Eact= Activation Energy
Energy
Products
+H= heat of reaction
(Gets cold)
Reactants
more stable
Rxn Progress
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Endothermic (Endergonic) Rxns
Energy is required
Products are less stable.
Energy
Products
(Water)
+H
Reactants
(Ice)
Rxn Progress
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Examples of Energy
diagrams
Exothermic reaction
H < 0
products are
more stable
-H
Endothermic reaction
+H
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H > 0
reactants are
more stable
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Examples of
energy diagrams
High activation energy
Low heat of reaction
Low activation energy
High heat of reaction
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Energy in Reactions
2Na(s) + Cl2(g)  2NaCl + 819 kJ
How many kilojoules are released when
100. g Na reacts with chlorine?
100.g Na 1mol Na
1
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819 kJ
= 1780 kJ
23.0 g Na 2 mol Na
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