Upload
gerard-short
View
236
Download
4
Tags:
Embed Size (px)
Citation preview
5 - 1CH104
CHAPTER 7Chemical 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
5 - 2CH104
Examples
Color
Odor
Taste
Feel
Shape
Physical propertiesCharacteristics that can be evaluated
without changing the composition of the material.
Density
Melting / Freezing point
Boiling point
Compressibility
Form (foil, wire, powder…)
5 - 3CH104
What is Chemistry?“The study of Matter and its Changes.”
Physical Changes =
Changes in a Physical Property
Chemical Changes =
Changes in a Chemical Property
Appearance: • melting, freezing, evaporation…• stretching, molding, cutting…
Chemical Composition:
5 - 4CH104
Change in the Chemical Composition
Burning of Magnesium
Chemical Changes
Rusting of Iron
Decomposing of wood
Souring of Milk
Examples:
5 - 5CH104
Examples
Mulching leaves
Which are chemical or physical changes?Which are chemical or physical changes?
Tarnishing Silver
Fermentation
Making ice into water
Carbonated Beverage going flat
Bleaching a stain
5 - 6CH104
Mg + O2 MgO +
Energy
Chemical Reactions
Shows how the Chemical change occurs.Shows how the Chemical change occurs.
Reactants
C3H8 + O2 CO2 + H2O + Energy
Fe + O2 Fe2O3
ProductsProducts
5 - 7CH104
Chemical equations
Chemist’s shorthand to describe a reaction.
•Reactants •Products
•The state of all substances
H2 + O2 H2O + E(g) (g) (g)
•Any conditions used in the reaction
heat
•Same # & type atoms on each side• Law of Conservation of Matter
2 2
5 - 8CH104
Balancing Equations
___W10 + ___B8 ___WB___W10 + ___B8 ___WB
•Reactants •Products
Making Hot dogs:How many packages wieners & buns to buy so none is left over.
4 5 40
5 - 10CH104
Ca
H
Cl
Balancing Equations
Ca + HCl CaCl2 + H2 Ca + HCl CaCl2 + H2
•Reactants •Products
Step 1: Count atoms of each element on both sides of equation.
1
1
1
1
2
2
5 - 11CH104
Balancing Equations
Ca + HCl CaCl2 + H2 Ca + HCl CaCl2 + H2
Ca
H
Cl
Ca
H
Cl
•Reactants •Products
1
1
1
1
1
1
1
2
2
1
2
2
Step 2: Determine which atoms are not balanced.
- not balanced
- not balanced
5 - 12CH104
Balancing Equations
Ca + HCl CaCl2 + H2 Ca + HCl CaCl2 + H2
•Reactants •Products
1
1
1
1
1
1
1
2
2
1
2
2
- not balanced
- not balanced
Step 3: Balance one element at a time with coefficients in front of formulas until all balanced.
(Never change the formula!)
2
22
Ca
H
Cl
Ca
H
Cl
22
5 - 13CH104
NaPO
MgCl
Balancing Equations
Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl
•Reactants •Products
Step 1: Count atoms of each element on both sides of equation.
31412
1 2831
5 - 14CH104
NaPO
MgCl
Balancing Equations
Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl
•Reactants •Products31412
1 2831
- not balanced
- not balanced
Step 2: Determine which atoms are not balanced.
- not balanced
- not balanced
- not balanced
5 - 15CH104
NaPO
MgCl
Balancing Equations
Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl
•Reactants •Products31412
1 2831
- 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!)
5 - 16CH104
Hints:• Start with a metal in a complex
compound, or an element that only appears in one formula. (Like Mg here)
NaPO
MgCl
Balancing Equations
Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl
•Reactants •Products31412
1 2831
- not balanced
- not balanced
- not balanced
- not balanced
- not balanced66
66
63
33
22
66
1
88
2
66
5 - 17CH104
Hints:• Start with an element that only appears
in one formula on both sides of the equation.
• Leave oxygen until last.
Balancing Equations
C2H6 + O2 CO2 + H2OC2H6 + O2 CO2 + H2O
•Reactants •ProductsC
H
O
5 - 18CH104
Balancing Equations
C2H6 + O2 CO2 + H2OC2H6 + O2 CO2 + H2O
C
H
O
C
H
O
•Reactants •Products
Step 1: Count atoms of each element on both sides of equation.
2
6
2
2
6
2
1
2
3
1
2
3
5 - 19CH104
C2H6 + O2 CO2 + H2OC2H6 + O2 CO2 + H2O
Balancing Equations
•Reactants •Products
2
6
2
2
6
2
1
2
3
1
2
3
Step 2: Determine which atoms are not balanced.
- not balanced
- not balanced
- not balancedC
H
O
C
H
O
5 - 20CH104
2
6
2
2
6
2
- not balanced
- not balanced
C2H6 + O2 CO2 + H2OC2H6 + O2 CO2 + H2O
22
Balancing Equations
•Reactants •Products
1
2
3
1
2
3
- not balanced
Step 3: Balance one element at a time with coefficients in front of formulas until all balanced.
(Never change the formula!)
2
55
3
66
77
3.5
77
C
H
O
C
H
O
5 - 21CH104
C
H
O
C
H
O
C2H6 + O2 CO2 + H2OC2H6 + O2 CO2 + H2O
22
Balancing Equations
•Reactants •Products
2
6
2
2
6
2
1
2
3
1
2
3
2
55
3
66
77
3.5
77
Can’t have 3.5 O2 , so multiply equation by 2!
5 - 22CH104
3.5C2H6 + O2 CO2 + H2OC2H6 + O2 CO2 + H2O
22
Balancing Equations
C
H
O
C
H
O
•Reactants •Products
2
6
2
2
6
2
1
2
3
1
2
3
4
55
6
66
77
7
77
Can’t have 3.5 O2 , so multiply equation by 2!
2
44 44
1212 1212
1414 1414
5 - 23CH104
(NH2)2CO + H2O ______> NH3 + CO2(NH2)2CO + H2O ______> NH3 + CO2
- not balanced
- not balanced
1
6
2 NHCO
NHCO
1
3
1 2 2
Balancing Equations
5 - 24CH104
(NH2)2CO + H2O ______> NH3 + CO2(NH2)2CO + H2O ______> NH3 + CO2
Balancing Equations
1
6
2 1
3
1
2
2
6NHCO
NHCO2 2
5 - 25CH104
Example: Decomposition of urea
CH3OH + PCl5 CH3Cl + POCl3 + H2O CH3OH + PCl5 CH3Cl + POCl3 + H2O
- not balanced
- not balanced
1
4
1 CHOPCl
CHOPCl
1
5
2 1 1
5 4
- not balanced
- not balanced
5 - 26CH104
1
CH3OH + PCl5 CH3Cl + POCl3 + H2O CH3OH + PCl5 CH3Cl + POCl3 + H2O
Balancing Equations
2
2
81
4
1 CHOPCl
CHOPCl
5
2 1 1
5 4 5
2
2
82
5 - 27CH104
Types of Chemical Reactions
Combination
Decomposition
Single Replacement: Substitution
Double Replacement: Metathesis
A + BX B + AX
A + B C
C A + B
AX + BY BX + AY
5 - 28CH104
Types of Chemical Reactions
Combination
Decomposition
Single Replacement: Substitution
Double Replacement: Metathesis
2H2 + O2 2H2O
CaCO3 CaO + CO2
Al + FeCl3 Fe + AlCl3
2AgNO3 + K2SO4 Ag2SO4 + 2KNO3
5 - 29CH104
Types of Chemical Reactions
Complete:C3H8 + 5O2 3CO2 + 4H2O
Combustion
Incomplete:2C3H8 + 7O2 6CO + 8H2O
C3H8 + 2O2 3C + 4H2O
5 - 30CH104
Combination Reactions
2H2 + O2 2H2O
Formation of Acid Rain
SO3 + H2O H2SO4
Explosion of Hydrogen Balloon
Rusting of Iron
4 Fe + 3 O2 2 Fe2O3
A + B C A + B C
5 - 31CH104
Decomposition Reactions
Heating Egg Shells
CaCO3 CaO + CO2
2 H2O2 2 H2O + O2
Blood with peroxide
C A + B
5 - 32CH104
Single Replacement Reactions
Iron Deposits on an Aluminum Pan
Al + FeCl3 Fe + AlCl3
A + BX B + AX
5 - 33CH104
Activity series of metalspotassiumsodium
potassiumsodium
calciumcalcium
magnesiumaluminum
zincchromium
magnesiumaluminum
zincchromium
ironnickel
tinlead
ironnickel
tinlead
coppersilver
platinumgold
coppersilver
platinumgold
Hydrogen
Al + Fe+3 Fe + Al+3
Fe + H+ Fe+3 + H2
incr
easi
ng
rea
ctiv
ity
Element give e’s to ion
lower on list
Element give e’s to ion
lower on list
5 - 34CH104
Double Replacement Reaction
BaCl2(aq) + Na2SO4(aq) BaSO4(s) + 2NaCl(aq)
AX + BY BX + AY AX + BY BX + AY
Ba+2 Cl-1
Na+1 SO4-2
Insoluble PrecipitateFormed
5 - 35CH104
Predict the products:
AgNO3(aq) + AlCl3 (aq)
Ag+ NO3
-
Al+3 Cl-
AgCl(s) + Al(NO3)3(aq)
Double Replacement ReactionAX + BY BX + AY AX + BY BX + AY
Balance later as needed to get:
3AgNO3(aq) + AlCl3 (aq) 3AgCl(s) + Al(NO3)3(aq)
5 - 36CH104
Oxidation and reductionREDOX
Where reactants exchange electrons -
Examples:
•All types of batteries
alkaline, NiCad, car batteries•Rusting and corrosion•Metabolism•Antioxidants (Vit C, E prevent oxidation)
5 - 37CH104
Oxidation and reductionREDOX
Where reactants exchange electrons -
Oxidation = Losing electronsLEO: Lose Electrons Oxidation
LEO the lion says GER
Reduction = Gaining electronsGER: Gain Electrons Reduction
OIL : Oxidation Is Losing
RIG : Reduction Is Gaining
OIL RIG
5 - 38CH104
Oxidation and reduction
2 Na(s) + Cl2 (g) 2 NaCl2 Na(s) + Cl2 (g) 2 NaCl
1-1+
Assign Oxidation States:
0 0For simple ions,
Ox state = charge.
For element in natural form
Ox State = 0.
5 - 39CH104
Oxidation and reduction
2 Na(s) + Cl2 (g) 2 NaCl2 Na(s) + Cl2 (g) 2 NaCl
1-1+
Who’s loosing or gaining electrons?
0 0
Loses 1 e- = LEO
Gains 1 e-=GER
Na loses e- (LEO) Na gets oxidized
Cl gains e- (GER) Cl gets reduced
5 - 40CH104
Oxidation and reduction
Oxidation - when reactant loses e-(s). (LEO)
Na (s) Na+ + e-
Reduction - when reactant gains e-(s). (GER)
Cl2 (g) + 2 e- 2 Cl-
These are half reactions
5 - 41CH104
2 Na(s) + Cl2 (g) + 2 e- 2 Na+ + 2e- + 2 Cl-
Oxidation and reduction
2 half reactions make a complete reaction
Na (s) Na+ + e-
Cl2 (g) + 2 e- 2 Cl-
2 2 2
2 Na(s) + Cl2 (g) 2 Na+ + 2 Cl-
5 - 42CH104
Oxidizing agent =• The chemical that caused an oxidation.• It is reduced.
Reducing agent =•The chemical that caused a reduction.• It is oxidized.
Oxidation and reduction
5 - 43CH104
Oxidation and reduction
2 Na(s) + Cl2 (g) 2 NaCl2 Na(s) + Cl2 (g) 2 NaCl
1-1+
Who’s loosing or gaining electrons?
0 0
• Na loses e- (LEO) so Na gets oxidized
• Na caused Cl to get reduced• Na is the Reducing agent
Loses 1 e- =LEO
Gains 1 e- =GER
5 - 44CH104
Oxidation and reduction
2 Na(s) + Cl2 (g) 2 NaCl2 Na(s) + Cl2 (g) 2 NaCl
1-1+
Who’s loosing or gaining electrons?
0 0
Loses 1 e- =LEO
Gains 1 e- =GER
• Cl gains e- (GER) so Cl gets reduced • Cl caused Na to get oxidized• Cl is the Oxidizing agent
5 - 46CH104
Oxidation state
Rules• Oxidation state of element in natural form = 0.
• For simple monoatomic ions, oxidation state = charge.
Describes the charge of each element.
• For certain groups at certain times, oxidation number = group number
Examples Na+1, Cl-1, Ca2+
Examples N+5, Cl+7
Examples N2, Na, O2, H2
5 - 47CH104
Hydrogen H1+ if bonded to nonmetal
H1- if bonded to metal
Example HCl
Oxidation statesOxygen O-2 Usually
O-1 in peroxides
Example H2O
Example H2O2
Example NaH
5 - 48CH104
Assign the oxidation states for all elements in
H2O
+1 -2
+1
+2 -2
Oxidation states
5 - 49CH104
Assign the oxidation states for all elements in
H2O2
+1 -1
+1
+2 -2
-1
Oxidation states
5 - 50CH104
REDOX reactions
2 H2 + O2 2 H2O
Oxidation state of H is 0Oxidation state of H is 0
Oxidation state of O is 0Oxidation state of O is 0
Oxidation state of H is 1+Oxidation state of H is 1+
Oxidation state of O is 2-Oxidation state of O is 2-
Hydrogen is oxidized and is a reducing agent.
Oxygen is reduced and is an oxidizing agent.
Lose 1 e- = LEO
Gain 2 e- = GER
5 - 51CH104
Find the oxidation state for all elements in:
HNO3
Nitrogen must be
Hydrogen is 1+Oxygen is 2-
2- 2-
5+
+1 +5 -6 = 0
Start with whatWe know
Start with whatWe know
Oxidation states
5 - 52CH104
NaCl+1
+3
-2
NaClO
Oxidation states of Cl
NaClO4
NaClO2
NaClO3
+1
+1
+1
+1
-2-2
-2-2-2
-2-2-2-2
+1
+5
+7
-1
5 - 53CH104
Types of chemical reactions
Chemical ReactionsChemical Reactions
NonredoxNonredox
Doublereplacement
Doublereplacement
CombinationCombination
RedoxRedox
Singlereplacement
Singlereplacement
CombinationCombination
Decomposition Decomposition
5 - 54CH104
Combination reactionsB + C A
REDOX or NONREDOX types
Non-REDOX reaction.
Formation of Acid Rain
SO3 + H2O H2SO4
-2+6-2-2-2-2
-2-2-2
+1+1
+1+1
+6
5 - 55CH104
Combination reactionsB + C A
REDOX or NONREDOX types
Rusting of Iron
4 Fe + 3 O2 2 Fe2O3
0-2-2-2
0 +3+3
REDOX reaction.Fe goes from 0 to +3 LEOO goes from 0 to -2 GER
5 - 56CH104
A B + C REDOX or NONREDOX types
Decomposition reactions
Non-REDOX reaction.
Decomposition of Egg Shells
CaCO3 CaO + CO2
-2+2 -2
-2-2-2-2
+4 +4+2
5 - 57CH104
Decomposition reactions
2 H2O2 2 H2O + O2
REDOX reaction.Some O goes from -1 to 0 (O2) LEO
Some O goes from -1 to -2 (H2O) GER
Decomposition of hydrogen peroxide
+1+1
-1-1
+1+1
-2 0
5 - 58CH104
A + BX B + AX Always REDOX
Iron Deposits on an Aluminum Pan
Al + FeCl3 Fe + AlCl3
00-1-1-1
-1-1-1
+3 +3
Single Replacement Reaction
REDOX reaction.Al goes from 0 to +3 LEOFe goes from +3 to 0 GER
5 - 59CH104
Activity series of metalspotassiumsodium
potassiumsodium
calciumcalcium
magnesiumaluminum
zincchromium
magnesiumaluminum
zincchromium
ironnickel
tinlead
ironnickel
tinlead
coppersilver
platinumgold
coppersilver
platinumgold
Hydrogen
Al + Fe+3 Fe + Al+3
Fe + H+ Fe+3 + H2
incr
easi
ng
rea
ctiv
ity
Element give e’s to ion
lower on list
Element give e’s to ion
lower on list
5 - 60CH104
AX + BY BX + AY Always non-REDOX
+1+1-1-1-1
-2-2-2
+5 +1
Double Replacement Reaction
Non -REDOX reaction.
AgNO3(aq) + NaCl(aq) AgCl(s)+ NaNO3(aq)
-1 -1 +1+5
5 - 61CH104
Ionic equationsIonic substances dissociate into ions when
dissolved in water.
Ag+ + NO3
-
AgNO3(aq) + NaCl (aq)
Certain ions join together
Others remain unchanged.
Ag+ NO3
-
Na+ Cl-
AgCl(s) + NaNO3(aq)
AgCl(s)+ Na+
+ Cl- + Na+ + NO3-
5 - 62CH104
Ionic equations
Total ionic equationAg+
+ NO3- +
Na+ + Cl- AgCl(s)+ Na+ +NO3-
NO3- and Na+ are spectator ions.
Net ionic equation
Ag+ + Cl- AgCl(s)
5 - 63CH104
1 pair =
1 dozen =
1 mole =
1 pair =
1 dozen =
1 mole =
The Mole
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
2
12
6.02 x 1023
602,000,000,000,000,000,000,000.
5 - 64CH104
1 car ___ 4 wheels
The Mole & Formulas
1 mol cars_
4 mol wheels1 doz cars4 doz wheels
1 mole H2O
2 mol H
1 mole H2O
1 mol O
5 - 65CH104
= doz wheels
1 car ___ 4 wheels
The Mole & Formulas
1 mol cars_
4 mol wheels1 doz cars4 doz wheels
4 doz wheels
1doz cars
5 doz cars
120
2 mol H
1 mol H2O
5 mol H2O
1
10 = mol H
5 - 66CH104
1 mole = MW in g’s The Mole & Molar Mass
1 mol Au_
197 g Au
1 mol Au_
197 g Au
1 mole Au = 197 g Au1 mole Au = 197 g Au
197 g Au
1 mol Au
197 g Au
1 mol Au
__197 g Au _
6.02 x 1023 atoms Au
__197 g Au _
6.02 x 1023 atoms Au
5 - 67CH104
1 mole = MW in g’s The Mole & Molar Mass
1 mol S_
32 g S
1 mol S_
32 g S
1 mole S = 32 g S1 mole S = 32 g S
32 g S
1 mol S
32 g S
1 mol S
1 mol C
12 g C
1 mol C
12 g C
1 mole C = 12 g C1 mole C = 12 g C
12 g C
1 mol C
12 g C
1 mol C
5 - 68CH104
The Mole & Molar Mass 1 mol H2O_
18.0 g H2O 1 mole H2O has:
1.0 g H =
1 mol H
2 mol H
1
2.0 g H
16.0 g O =
1 mol O
1 mol O
1
16.0 g O
18.0 g H2O
1 mol H2O
18.0 g
5 - 69CH104
Molar MassFind the MW of Glucose; C6H12O6
1.0 g H =
1 mol H
12 mol H
1
12.0 g H
16.0 g O =
1 mol O
6 mol O
1
96.0 g O
180.0 g C6H12O6
1 mol C6H12O6
12.0 g C =
1 mol C
6 mol C
172.0 g C
5 - 70CH104
1 mol H2O =
18 g H2O
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?
36 g H2O
1
mol H2O2.0
5 - 71CH104
1 mol H2O
18 g H2O
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
mol H4.0 2 mol H =
1 mol H2O
5 - 72CH104
180 g Gluc =
1 mol Gluc
Mole to Mass ConversionsHow 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
1g Glucose900
5 - 73CH104
You need a balancedequation.
You need a balancedequation.
H2 + O2 -----> H2O H2 + O2 -----> H2O
The mole and chemical equations
Stoichiometry - Calculations of quantities in a chemical rxn.
2 2
5 - 74CH104
Moles in Chemical Equations
2 Na(s) + Cl2 (g) 2 NaCl
2 mol Na
1 mol Cl2
2 mol Na
2 mol NaCl
1 mol Cl2
2 mol NaCl
5 - 75CH104
Moles in Chemical Equations2 Na(s) + Cl2 (g) 2 NaCl
2 mol Na
1 mol Cl2
2 mol Na
2 mol NaCl
1 mol Cl2
2 mol NaCl
How many moles of Cl2 are needed to completely react with 40 moles of Na?How many moles of Cl2 are needed to completely react with 40 moles of Na?
1 mol Cl2 =
2 mol Na
What is Unique to the problem?
40 mol Na 1
20 mol Cl2
What should the answer look like?
5 - 76CH104
1mol Na23.0g Na
2 Na(s) + Cl2 (g) 2 NaCl
1 mol Cl2
2 mol Na
What is Unique to the problem?
40 g Na1
61.7 g Cl2
What should the answer look like?
g’s Cl2
Mass in Chemical Equations
mols Na mols Cl2
70.9 g Cl2 =
1 mol Cl2
How many g’s of Cl2 are needed to completely react with 40 g’s of Na?How many g’s of Cl2 are needed to completely react with 40 g’s of Na?
40 g’s Na
5 - 77CH104
1mol Fe2O3
159.7g Fe2O3
Fe2O3 + 3H2 2Fe + 3H2O
2 mol Fe1 mol Fe2O3
What is Unique to the problem?
20.g Fe2O3
114 g Fe
What should the answer look like?
g’s FeMass in Chemical Equations
mols Fe2O3 mols Fe
55.8 g Fe = 1 mol Fe
How many g’s of Fe can be produced from 20. g’s of Fe2O3?
How many g’s of Fe can be produced from 20. g’s of Fe2O3?
20.g’s Fe2O3
5 - 78CH104
1mol Fe2O3
159.7g Fe2O3
Fe2O3 + 3H2 2Fe + 3H2O
2 mol Fe1 mol Fe2O3
20.g Fe2O3
1
14 g Fe
Percent Yield
55.8 g Fe = 1 mol Fe
How many g Fe can be made from 20.g Fe2O3?How many g Fe can be made from 20.g Fe2O3?
What is the percent yield if I only got 12 g’s of Fe from 20. g’s of Fe2O3?
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 = Actual x 100 Theoretical
% yield = 12 g x 100 = 14 g% yield = 12 g x 100 = 14 g
85.7 %85.7 %
5 - 79CH104
Limiting reactant1 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
From 2 eggs
5 c. Bquick
2 eggs
18 p cakes =
2 c. Bquick
45 p cakes
36 p cakes18 p cakes =
1 egg
5 - 80CH104
Limiting reactant1 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
From 2 eggs
45 p cakes
36 p cakes
So: Eggs are the limiting reagent.I can make only 36 pancakes I’ll have bisquick left over.
5 - 81CH104
Limiting reactant
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
14 g Fe
Fe2O3 + 3H2 2Fe + 3H2O
1mol H2
2.02 g H2
2 mol Fe3 mol H2
2.g H2
137 g Fe 55.8 g Fe =
1 mol Fe
So: Fe2O3 is the limiting reagent.
I can make only 14 g FeI’ll have H2 left over.
5 - 82CH104
31.7 18 g H2O =
1 molH2O
13.5 18 g H2O =
1 molH2O
Limiting reactant
If I have 20.g of NH3 and 20.g O2,
how many g’s H2O can I make?From 20. g NH3
From 20. g O2
4NH3 + 5O2 4NO + 6H2O
1mol NH3
17 g NH3
6 mol H2O
4 mol NH3
20.g NH3
1g H2O
1mol O2
32 g O2
6 mol H2O
5 mol O2
20.g O2
1
g H2O
5 - 83CH104
Limiting reactant
If I have 20.g of NH3 and 20.g O2,
how many g’s H2O can I make?
From 20. g NH3
From 20. g O2
31.7 g H2O
4NH3 + 5O2 4NO + 6H2O
So: O2 is the limiting reagent.
We can make only 13.5 g H2OWe’ll have NH3 left over.
13.5 g H2O
5 - 84CH104
13 76 g CS2 =
1 mol CS2
5.9 76 g CS2 =
1 mol CS2
Limiting reactant
If I have 10.g of C and 10.g SO2,
how many g’s CS2 can I make?From 10. g C
From 10. g SO2
5C + 2SO2 1CS2 + 4CO
1mol C
12 g C
1 mol CS2
5 mol C
10.g C1
g CS2
1mol SO2
64 g SO2
1 mol CS2
2 mol SO2
10.g SO2
1
g CS2
5 - 85CH104
Limiting reactant
13 g CS2
So: SO2 is the limiting reagent.
We can make only 5.9 g CS2
We’ll have C left over.
5.9 g CS2
If I have 10.g of C and 10.g SO2,
how many g’s CS2 can I make?From 10. g C
From 10. g SO2
5C + 2SO2 1CS2 + 4CO
5 - 86CH104
Study of Energy changes in Reactions =
• used to calculate the amount of useful work produced by chemical reactions.
Thermodynamics
Enthalpy
andEntropy
= Change in Energy = DH
= Change in = DS
5 - 87CH104
law of Thermodynamics
“Energy can’t be created or destroyed in a chemical reaction”
• Energy just gets
converted from one form to another.
(you can’t get something from nothing)
5 - 88CH104
Exothermic Reactions
Mg + O2 MgO + Energy
CH4 + 2O2 CO2 + 2H2O + 211 kcal
Breaking bondscosts E(+DH)
Making bondsgives E(-DH)
Old bonds break New bonds get made
Hproducts - Hreactants = H
5 - 89CH104
Activation Energy (Eact)The minimum amount of energy required to produce a chemical reaction.
• The energy of collision must be great enough to break the old bonds and form the new ones.
5 - 90CH104
H2OH2O
more stablemore stable
Energy in Chemical Reactions
Exothermic reactionExothermic reaction
-H= heat of reaction
En
ergy
Rxn Progress
Reactants
Products
Eact= Activation Energy
(Gets hot)(Gets hot)
H2 + O2H2 + O2
2H2 + O22H2 + O2 2H2O + Energy2H2O + Energy
5 - 91CH104
Exothermic (Exergonic) RxnsE
ner
gy
Rxn Progress
2Mg + O2 2MgO + Energy
2H2 + O2 2H2O + Energy
CH4 + 2O2 CO2 + 2H2O + 213 kcal
5 - 92CH104
Exothermic ReactionsE
ner
gy
Rxn Progress
Reactants(Water)
Products(Ice)
Energy is released
Products are more stable.
-DH
5 - 93CH104
more stablemore stable
Energy in Chemical Reactions
Endothermic reactionEndothermic reaction
+H= heat of reactionEn
ergy
Rxn Progress
Reactants
Products
Eact= Activation Energy
(Gets cold)(Gets cold)
5 - 94CH104
En
ergy
Rxn Progress
Endothermic (Endergonic) RxnsEnergy is required
Products are less stable.
Products(Water)
Reactants(Ice)
+DH
5 - 95CH104
Examples of Energy diagrams
Exothermic reaction
Endothermic reaction
H > 0reactants are more stable
H < 0products are more stable
+H
-H
5 - 96CH104
Examples ofenergy diagrams
High activation energyLow heat of reaction
Low activation energyHigh heat of reaction
5 - 97CH104
1780
Energy in Reactions
How many kilojoules are released when 100. g Na reacts with chlorine?
2Na(s) + Cl2(g) 2NaCl + 819 kJ
1mol Na
23.0 g Na
819 kJ 2 mol Na
100.g Na1
= kJ