Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Chapter 5
Chemical Quantities and Reactions
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Chemistry for Health Sciences
Prof. Ralph Duarte
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
The Mole
Counting Units
Counting units
A counting term states a specific number of items.
1 dozen eggs = 12 eggs
1 case soda = 24 cans
1 ream = 500 sheets of paper
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Avogadro's Number
Small particles such as atoms, formula units, molecules, and ions are counted using the mole.
1 mole = 6.022 x 1023 items
Avogadro’s number
602 200 000 000 000 000 000 000 = 6.022 x 1023
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Mole of Atoms
1 mole of an element = 6.022 x 1023 atoms of that element
1 mole of carbon = 6.022 x 1023 atoms of carbon
1 mole of sodium = 6.022 x 1023 atoms of sodium
4
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Number of Particles in One-Mole Samples
5
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Avogadro's Number
Avogadro’s number, 6.022 x 1023, can be written as an
equality and as two conversion factors.
Equality:
1 mole = 6.022 x 1023 particles
That means we can use the following Conversion Factors as needed:
6.022 x 1023 particles or 1 mole 1 mole 6.022 x 1023 particles
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Converting Moles to Particles
(Atoms, formula units or molecules)
Step 1 State the data and question:
Data: 0.50 mole of CO2
Question: molecules of CO2
7
Problem:
How many CO2 molecules are in 0.50 mole of CO2?
Avogadro’s number is used to convert moles of a substance to particles.
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Converting Moles to Particles
(Atoms, formula units or molecules) 8
Question = Data x “question” units
“data” units
Include conversion factor here
Problem:
How many CO2 molecules are in 0.50 mole of CO2?
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Converting Moles to Particles
(Atoms, formula units or molecules) 9
molecules
mole
Problem:
How many CO2 molecules are in 0.50 mole of CO2?
Use Avogadro’s number to write conversion factors:
1 mole of CO2 = 6.022 x 1023 molecules of CO2
6.022 x 1023 CO2 molecules or 1 mole CO2
1 mole CO2 6.022 x 1023 CO2 molecules
CO2 molecules = 0.50 mole CO2 x
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Converting Moles to Particles
(Atoms, formula units or molecules) 10
6.022 x 1023 CO2 molecules
1 mole CO2
Problem:
How many CO2 molecules are in 0.50 mole of CO2?
Use Avogadro’s number to write conversion factors:
1 mole of CO2 = 6.022 x 1023 molecules of CO2
6.022 x 1023 CO2 molecules or 1 mole CO2
1 mole CO2 6.022 x 1023 CO2 molecules
CO2 molecules = 0.50 mole CO2 x
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Converting Moles to Particles
(Atoms, formula units or molecules) 11
CO2 molecules = 0.50 mole CO2 x 6.022 x 1023 CO2 molecules
1 mole CO2
Problem:
How many CO2 molecules are in 0.50 mole of CO2?
= 3.0 x 1023 molecules of CO2
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
The number of atoms in 2.0 mole of Al atoms is:
A. 2.0 Al atoms
B. 3.0 x 1023 Al atoms
C. 1.2 x 1024 Al atoms
12
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
The number of atoms in 2.0 moles of Al atoms is:
= 2.0 moles Al x 6.022 x 1023 Al atoms
1 mole Al = 12.044 x 1023 Al atoms = 1.2 x 1024 Al atoms
13
The number of atoms in 2.0 mole of Al atoms is:
A. 2.0 Al atoms
B. 3.0 x 1023 Al atoms
C. 1.2 x 1024 Al atoms
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
Calculate the number of moles of S in 1.8 x 1024 atoms of S
14
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Set up the problem to calculate the number of moles of S:
= 1.8 x 1024 S atoms x 1 mole S
6.022 x 1023 S atoms = 0.29890 moles x 1024 x 10-23
= 3.0 moles of S atoms
15
Calculate the number of moles of S in 1.8 x 1024 atoms of S
State the needed and given quantities:
Given: 1.8 x 1024 atoms of S
Needed: moles of S
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
How many molecules of aspirin are in 0.150 mole of aspirin, C9H8O4?
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Converting Moles to Particles
(Atoms, formula units or molecules) 17
molecules
mole
Use Avogadro’s number to write conversion factors:
1 mole of C9H8O4 = 6.022 x 1023 molecules of C9H8O4
6.022 x 1023 molecules or 1 mole
1 mole 6.022 x 1023 molecules
molecules = 0.150 mole x
How many molecules of aspirin are in 0.150 mole of aspirin, C9H8O4?
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Set up the problem to calculate the number of particles.
= 0.150 mole C9H8O4 x 6.022 x 1023 molecules C9H8O4
1 mole C9H8O4
= 0.903 x 1023 molecules C9H8O4
= 9.03 x 10-1 x 1023 molecules C9H8O4
= 9.03 x 1022 molecules C9H8O4
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How many molecules of aspirin are in 0.150 mole of aspirin, C9H8O4?
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
One-Mole Quantities
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1 mole S = 32.07 g 1 mole Fe = 55.85 g 1 mole NaCl = 58.44 g 1 mole K2Cr2O7 = 294.2 g 1 mole C12H22O11 = 342.3 g
1 mole of S
= 6.022 x 1023 atoms of S
1 mole of Fe
= 6.022 x 1023 atoms of Fe
1 mole of NaCl = 6.022 x 1023 Formula Units
of NaCl
1 mole of K2Cr2O7 = 6.022 x 1023 Formula Units
of K2Cr2O7
1 mole of C12H22O11 = 6.022 x 1023 molecules
of C12H22O11
Potassium
Dichromate Sucrose
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Molar Mass The molar mass • is the mass of 1 mole of an element
• is the atomic mass expressed in grams
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
Give the molar mass for each element:
A. 1 mole of K atoms = ________
B. 1 mole of Sn atoms = ________
C. 1 mole of Ca atoms = ________
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Give the molar mass for each element:
A. 1 mole of K atoms = 39.10 g of K
B. 1 mole of Sn atoms = 118.7 g of Sn
C. 1 mole of Ca atoms = 40.08 g of Ca
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Molar Mass of the Compound
Example: Ethanol, C2H6O
To calculate the molar mass of Ethanol, C2H6O:
Step 1 Obtain the molar mass of each element.
molar mass of C = 12.01 g of C
molar mass of H = 1.008 g of H
molar mass of O = 16.00 g of O
Step 2 Multiply each molar mass by the number of subscripts in the formula
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Step 3 Calculate the molar mass by adding the masses of the elements:
2 x molar mass of C = 24.02 g of C
6 x molar mass of H = 6.05 g of H
1 x molar mass of O = 16.00 g of O
Molar mass of C2H6O = 46.07 g of C2H6O
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
What is the molar mass of each compound?
A. K2O
B. Al(OH)3
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
What is the molar mass of each compound?
Calculate the molar mass by adding the molar masses of the elements:
K2O 2 x molar mass of K = 78.20 g of C
1 x molar mass of O = 16.00 g of O
Molar mass of K2O = 94.20 g of K2O
Al(OH)3 1 x molar mass of Al = 26.98 g of Al
3 x molar mass of O = 48.00 g of O
3 x molar mass of H = 3.024 g of H Molar mass of Al(OH)3 = 78.00 g of Al(OH)3
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Calculations Using Molar Mass
Molar mass conversion factors
• are fractions (ratios) written from the molar mass.
• relate grams and moles of an element or compound.
• for methane, CH4, used in gas stoves and gas heaters, is
1 mole of CH4 = 16.0 g of CH4 (molar mass equality)
Conversion factors:
16.0 g CH4 or 1 mole CH4
1 mole CH4 16.0 g CH4
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Converting Mass to Moles of a Compound
A box of table salt, NaCl, contains 737 g of NaCl.
How many moles of NaCl are in the box?
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Converting Mass to Moles of
Compound NaCl
Step 1 State the data and question. Data: 737 g of NaCl
Question: moles of NaCl
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A box of table salt, NaCl, contains 737 g of NaCl.
How many moles of NaCl are in the box?
Step 2 Set up the problem to convert grams to moles.
moles NaCl = 737 g NaCl x 1 mole NaCl
58.44 g NaCl
= 12.6 moles NaCl
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Map: Mass – Moles – Particles
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
Allyl sulfide, C6H10S, is a compound that has
the odor of garlic. How many moles of
C6H10S are in 225 g?
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Step 1 State the data and question. Data: 225 g of C6H10S
Question: moles of C6H10S
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Step 2 Determine the molar mass and write conversion factors.
Conversion Factors 1 mole of C6H10S = molar mass of C6H10S = 114.2 g of C6H10S 114.2 g C6H10S or 1 mole C6H10S 1 mole C6H10S 114.2 g C6H10S
Step 3 Set up the problem to convert grams to moles. mole of C6H10S = 225 g C6H10S x 1 mole C6H10S 114.2 g C6H10S
= 1.97 moles of C6H10S
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Chemical Changes
A chemical change
• occurs when a substance is converted into one or more
substances with different formulas and different properties
• may be observed by the formation of bubbles, a change in
color, or production of a solid
32
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
Which of the following represent a chemical change?
A. burning a candle
B. ice melting
C. toast burns
D. water turns to steam
33
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Which of the following represent a chemical change?
A. burning a candle chemical change
B. ice melting (this is a physical change)
C. toast burns chemical change
D. water turns to steam (this is a physical change)
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Writing a Chemical Equation
35
To write a chemical equation • an arrow indicates reactants form products
• reactants are written on the left side of the arrow
• products are written on the right side of the arrow
• multiple reactants or products are separated by a + sign
• the delta {Δ} sign indicates that heat is used to start the reaction
• physical states are represented in parenthesis as (s), (l), (g) or (aq) aqueous
Example:
2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g)
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Identifying a Balanced Equation
In a balanced chemical equation, • no atoms are lost or gained
• the number of reacting atoms is equal to the number of product atoms
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Balancing a Chemical Equation
Example: Formation of Al2S3 37
Step 1 Write an equation using the correct formulas of the reactants and products.
Al(s) + S(s) Al2S3(s)
Step 2 Count the atoms of each element in the reactants and products.
Reactants Products
1 atom Al 2 atoms Al Not balanced
1 atoms S 3 atoms S Not balanced
Step 3 Use coefficients to balance each element. Starting with the most
complex formula, change coefficients to balance equation.
2Al(s) + 3S(s) Al2S3(s)
Step 4 Check the final equation to confirm it is balanced.
Make sure coefficients are the lowest ratio.
Reactants Products
2 atom Al 2 atoms Al Balanced !!!
3 atoms S 3 atoms S Balanced !!!
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
State the number of atoms of each element on the reactant side and the
product side and use coefficients to balance each of the equations:
A. P4(s) + Br2(l) PBr3(g)
B. Al(s) + Fe2O3(s) Fe(s) + Al2O3(s)
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
State the number of atoms of each element on the reactant side and the
product side and use coefficients to balance each of the equations:
A. P4(s) + Br2(l) PBr3(g)
B. Al(s) + Fe2O3(s) Fe(s) + Al2O3(s)
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Before Reactants Products
P atoms 4 1
Br atoms 2 3
Before Reactants Products
Al atoms 1 2
Fe atoms 2 1
O atoms 3 3
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
State the number of atoms of each element on the reactant side and the
product side and use coefficients to balance each of the equations:
A. P4(s) + 6Br2(l) 4PBr3(g)
B. 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s)
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After Reactants Products
P atoms 4 4
Br atoms 12 12
After Reactants Products
Al atoms 2 2
Fe atoms 2 2
O atoms 3 3
Before Reactants Products
P atoms 4 1
Br atoms 2 3
Before Reactants Products
Al atoms 1 2
Fe atoms 2 1
O atoms 3 3
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
Use coefficients to balance each element.
Fe2O3(s) + H2(g) Fe(s) + H2O(l)
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Use Coefficients to balance each element.
Fe2O3(s) + 3H2(g) 2Fe(s) + 3H2O(l)
Always check the final equation to confirm it is balanced:
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Reactants Products
Atoms Fe 2 2
Atoms O 3 3
Atoms H 6 6
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Balancing Equations with Polyatomic Ions
Example:
Na3PO4(aq) + MgCl2(aq) Mg3(PO4)2(s) + NaCl(aq)
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Balancing Equations with Polyatomic Ions
When balancing equations with polyatomic ions, count each polyatomic ion as a unit.
Example:
Na3PO4(aq) + MgCl2(aq) Mg3(PO4)2(s) + NaCl(aq)
44
before balance Reactants Products
PO43− ions 1 2
Na+ ions 3 1
Mg2+ ions 1 3
Cl− ions 2 1
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Balancing Equations with Polyatomic Ions
When balancing equations with polyatomic ions, balance each polyatomic ion as a unit.
Example:
2Na3PO4(aq) + 3MgCl2(aq) Mg3(PO4)2(s) + 6NaCl(aq)
45
After balance Reactants Products
PO43− ions 2 2
Na+ ions 6 6
Mg2+ ions 3 3
Cl− ions 6 6
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
Balance and list the coefficients from reactants to products.
__Fe2O3(s) + __C(s) __Fe(s) + __CO2(g)
a) 2, 3, 2, 3 b) 2, 3, 4, 3 c) 1, 1, 2, 3
__Al(s) + __FeO(s) __Fe(s) + __Al2O3(s)
a) 2, 3, 3, 1 b) 2, 1, 1, 1 c) 3, 3, 3, 1
__Al(s) + __H2SO4(aq) __Al2(SO4)3(aq) + __H2(g)
a) 3, 2, 1, 2 b) 2, 3, 1, 3 c) 2, 3, 2, 3
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Balance and list the coefficients from reactants to products.
2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g)
Correct answer b) 2, 3, 4, 3
2Al(s) + 3FeO(s) 3Fe(s) + Al2O3(s)
Correct answer a) 2, 3, 3, 1
2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g)
Correct answer b) 2, 3, 1, 3
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Types of Reactions
Chemical reactions can be classified as
• combination reactions
• decomposition reactions
• single replacement reactions
• double replacement reactions
• neutralization reactions
• combustion reactions
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Combination Reactions
In a combination reaction,
• two or more elements form one product
• or simple compounds combine to form one product
2Mg(s) + O2(g) 2MgO(s)
2Na(s) + Cl2(g) 2NaCl(s)
SO3(g) + H2O(l) H2SO4(aq)
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Decomposition Reaction
In a decomposition reaction, one substance splits into
two or more simpler substances.
2HgO(s) 2Hg(l) + O2(g)
2KClO3(s) 2KCl(s) + 3O2(g)
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Δ
Δ
Most decomposition reactions are endothermic: heat is needed for the reaction to proceed
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Single Replacement Reaction
In a single replacement reaction, one element takes the
place of a different element in another reacting a compound.
Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Double Replacement Reaction
In a double replacement, two elements in the reactants
exchange places
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Neutralization Reactions
In a neutralization reaction:
An acid react with a base forming a salt and water
base + acid salt + water
Examples of neutralization reactions in “antacids”:
Mg(OH)2(aq) + 2HCl(aq) MgCl2(aq) + 2H2O(l)
Al(OH)3(aq) + 3HCl(aq) AlCl3(aq) + 3H2O(l)
CaCO3(aq) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)
NaHCO3(aq) + 2HCl(aq) NaCl(aq) + H2O(l) + CO2(g)
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Milk of Magnesia
Maalox
Pepto-Bismol
Alka-Seltzer
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Combustion Reaction
In a combustion reaction,
• a carbon-containing compound burns in oxygen gas to form
carbon dioxide (CO2) and water (H2O)
• energy is released as a product in the form of heat
carbon compound + Oxygen CO2(g) + 2H2O(g) + energy
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + energy
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) + energy
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Combustion reactions are exothermic: heat is generated or released during the chemical reaction
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
Classify each of the following reactions as combination, decomposition, single replacement, double replacement, or combustion.
2Al(s) + 3H2SO4(aq) Al2(SO4)3(s) + 3H2(g)
Na2SO4(aq) + 2AgNO3(aq) Ag2SO4(s) + 2NaNO3(aq)
4Fe(s) + 3O2(g) 2Fe2O3(s)
C2H4(g) + 2O2(g) 2CO2(g) + 2H2O(g)
Mg(OH)2(aq) + 2HCl(aq) MgCl2(aq) + 2H2O(l)
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Classify each of the following reactions as combination, decomposition, single replacement, double replacement, or combustion.
2Al(s) + 3H2SO4(aq) Al2(SO4)3(s) + 3H2(g)
Na2SO4(aq) + 2AgNO3(aq) Ag2SO4(s) + 2NaNO3(aq)
4Fe(s) + 3O2(g) 2Fe2O3(s)
C2H4(g) + 2O2(g) 2CO2(g) + 2H2O(g)
Mg(OH)2(aq) + 2HCl(aq) MgCl2(aq) + 2H2O(l)
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Single Replacement
Double Replacement
Combination
Combustion
Neutralization
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
Identify each reaction as combination, decomposition,
combustion, single replacement, or double replacement.
A. 3Ba(s) + N2(g) Ba3N2(s)
B. 2Ag(s) + H2S(aq) Ag2S(s) + H2(g)
C. 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) + Heat
D. PbCl2(aq) + K2SO4(aq) 2KCl(aq) + PbSO4(s)
E. K2CO3(s) K2O(aq) + CO2(g)
F. Al(OH)3(aq) + 3HCl(aq) AlCl3(aq) + 3H2O(l)
57
Δ
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Identify each reaction as combination, decomposition,
combustion, single replacement, or double replacement.
A. 3Ba(s) + N2(g) Ba3N2(s)
B. 2Ag(s) + H2S(aq) Ag2S(s) + H2(g)
C. 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) + Heat
D. PbCl2(aq) + K2SO4(aq) 2KCl(aq) + PbSO4(s)
E. K2CO3(s) K2O(aq) + CO2(g)
F. Al(OH)3(aq) + 3HCl(aq) AlCl3(aq) + 3H2O(l)
58
Δ
Combination
Single Replacement
Combustion
Double Replacement
Decomposition
Neutralization
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Oxidation−Reduction Reactions
Oxidation–reduction reaction
• provides us with energy from food
• provides electrical energy in batteries
• occurs when iron rusts:
4Fe(s) + 3O2(g) 2Fe2O3(s)
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Oxidation−Reduction
In an oxidation–reduction reaction, electrons are transferred
from one substance to another.
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Oxidation and Reduction
2Cu(s) + O2(g) 2CuO(s)
Oxidation is loss of electrons.
2Cu(s) 2Cu2+(s) + 4e−
Reduction is gain of electrons.
O2(g) + 4e− 2O2−(s)
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The green patina on copper is
due to oxidation
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Zn replaces Cu2+
62
CuSO4(aq) + Zn(s) ZnSO4(aq) + Cu(s)
2+ 0 2+ 0
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Zn Transfers Electrons to Cu2+
OIL Oxidation is losing electrons.
RIG Reduction is gaining electrons.
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Learning Check
Identify each of the following as oxidation or reduction:
Sn(s) Sn4+(aq) + 4e−
Fe3+(aq) + 1e− Fe2+
(aq)
Cl2(g) + 2e− 2Cl−(aq)
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Identify each of the following as oxidation or reduction:
Sn(s) Sn4+(aq) + 4e−
Fe3+(aq) + 1e− Fe2+
(aq)
Cl2(g) + 2e− 2Cl−(aq)
65
Oxidation
Reduction
Reduction
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
In light-sensitive sunglasses, UV light initiates an
oxidation–reduction reaction.
UV light
2Ag+ + 2Cl− 2Ag + Cl2
A. Which reactant is oxidized?
B. Which reactant is reduced?
66
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Solution
In light-sensitive sunglasses, UV light initiates an
oxidation-reduction reaction.
uv light
2Ag+ + 2Cl− 2Ag + Cl2
A. Which reactant is oxidized? Chlorine
2Cl− Cl2 + 2e−
B. Which reactant is reduced? Silver
2Ag+ + 2e− 2Ag
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Law of Conservation of Mass
The law of conservation of mass indicates that in an
ordinary chemical reaction,
• matter cannot be created or destroyed
• no change in total mass occurs
• the mass of products is equal to mass of reactants
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Conservation of Mass
69
2(107.9 g) + 32.07 g = 247.87 g
Matter cannot be created or
destroyed, therefore the mass of
products is equal to mass of reactants
2 moles Ag + 1 mole S = 1 mole Ag2S
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Reading Equations in Moles
Consider the following equation:
2Fe(s) + 3S(s) Fe2S3(s)
An equation can be read in “moles” by placing the word “moles of” between each coefficient and formula.
2 moles of Fe + 3 moles of S 1 mole of Fe2S3
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2 (55.85 g) + 3 (32.07 g) 207.91 g
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Writing Conversion Factors from Equations
A mole–mole factor is a ratio of the moles for any two substances in an equation.
2Fe(s) + 3S(s) Fe2S3(s)
Fe and S 2 moles Fe or 3 moles S
3 moles S 2 moles Fe
Fe and Fe2S3 2 moles Fe or 1 moles Fe2S3
1 mole Fe2S3 2 moles Fe
S and Fe2S3 3 moles S or 1 mole Fe2S3
1 mole Fe2S3 3 moles S
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Calculations with Mole Factors
Example:
How many moles of Fe2O3 can form from 6.0 moles of O2?
4Fe(s) + 3O2(g) 2Fe2O3(s)
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Calculations with Mole Factors
How many moles of Fe2O3 can form from 6.0 moles of O2?
4Fe(s) + 3O2(g) 2Fe2O3(s)
Relationship: 3 moles of O2 = 2 moles of Fe2O3
Use a mole–mole factor to determine the moles of Fe2O3.
moles of Fe2O3 = 6.0 moles O2 x 2 moles Fe2O3
3 moles O2
= 4.0 moles of Fe2O3
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
How many moles of Fe are needed for the reaction of
12.0 moles of O2?
4Fe(s) + 3O2(g) 2Fe2O3(s)
A. 3.00 moles of Fe
B. 9.00 moles of Fe
C. 16.0 moles of Fe
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
How many moles of Fe are needed for the reaction of 12.0 moles of O2?
4Fe(s) + 3O2(g) 2Fe2O3(s)
The possible mole factors for the solution are:
4 moles Fe or 3 moles O2
3 moles O2 4 moles Fe
75
Set up the problem using the mole–mole factor that cancels given moles.
moles Fe = 12.0 moles O2 x 4 moles Fe = 16.0 moles Fe
3 moles O2
The answer is C, 16.0 moles Fe
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Converting from Moles to Grams
76
Determine the mass (g) of NH3 that can be produced from 32 grams of N2. N2(g) + 3H2(g) 2NH3(g)
Hint: In this case, question and data refers to grams (instead of moles)
therefore the moles need to be converted to grams !!!!
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Converting from Moles to Grams
77
Determine the mass (g) of NH3 that can be produced from 32 grams of N2. N2(g) + 3H2(g) 2NH3(g)
N2(g) + 3H2(g) 2NH3(g)
28.02g 3(2.016g) 2(17.034g)
Converting to grams:
1 mole N2(g) + 3 moles H2(g) 2 moles NH3(g)
28.02g N2(g) + 6.048g H2(g) 34.068g NH3(g)
Therefore:
grams of NH3 = 32 grams of N2 x 34.068 grams of NH3
28.02 grams of N2
grams of NH3 = 38.9071 g = 39 g of NH3
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
How many grams of O2 are needed to produce 45.8 grams of
Fe2O3 in the following reaction?
4Fe(s) + 3O2(g) 2Fe2O3(s)
A. 38.4 g of O2
B. 13.8 g of O2
C. 1.38 g of O2
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Step 1 Use molar mass to convert moles to grams:
79
How many grams of O2 are needed to produce 45.8 grams of
Fe2O3 in the following reaction?
4Fe(s) + 3O2(g) 2Fe2O3(s)
4Fe(s) + 3O2(g) 2Fe2O3(s)
4 moles Fe(s) + 3 moles O2(g) 2 moles Fe2O3(s)
4(55.85g) 3(32.00g) 2(159.7g)
223.4g 96.00g 319.4g
Now you can resolve the problem using a gram-to-gram
relationship for the above given chemical equation
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
80
How many grams of O2 are needed to produce 45.8 grams of
Fe2O3 in the following reaction?
4Fe(s) + 3O2(g) 2Fe2O3(s)
4 moles Fe(s) + 3 moles O2(g) 2 moles Fe2O3(s)
Therefore: grams of O2 = 45.8 grams Fe2O3 x 96.00 grams O2
319.4 grams Fe2O3
grams of O2 = 13.7658g
grams of O2 = 13.8g O2 (the correct answer is B)
4(55.85g) 3(32.00g) 2(159.7g)
223.4g 96.00g 319.4g
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Energy in Chemical Reactions
Exothermic Reactions
In an exothermic reaction, • heat is released
• the energy of the products is less
than the energy of the reactants
• heat is a product
C(s) + 2H2(g) CH4(g) + 18 kcal
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
In an endothermic reaction,
• heat is absorbed
• the energy of the products is greater
than the energy of the reactants
• heat is a reactant (added)
82
Energy in Chemical Reactions
Endothermic Reactions
N2(g) + O2(g) + 43.3 kcal 2NO(g)
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
Identify each reaction as exothermic or endothermic.
A. N2(g) + 3H2(g) 2NH3(g) + 22 kcal
B. CaCO3(s) + 133 kcal CaO(s) + CO2(g)
C. 2SO2(g) + O2(g) 2SO3(g) + heat
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Identify each reaction as exothermic or endothermic.
A. N2(g) + 3H2(g) 2NH3(g) + 22 kcal Exothermic
B. CaCO3(s) + 133 kcal CaO(s) + CO2(g)
Endothermic
C. 2SO2(g) + O2(g) 2SO3(g) + heat
Exothermic
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Reaction Rate
The reaction rate is the speed at which reactant is used up
The reaction rate is the speed at which product forms
The reaction rate
• increases when temperature rises because reacting molecules move
faster, providing more colliding molecules with energy of activation
• increases when agitation is used (if reactants are solids or liquids)
because reacting molecules move faster (more collisions)
• increases with increase in concentration of reactants
• increases when a catalyst is used (catalysts are chemical compounds
that lowers the energy of activation, therefore facilitating the
formation of product molecules (reaction products are formed faster)
85
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Catalyst
A catalyst
• increases the rate of a reaction
• lowers the energy of activation
• is not used up during the reaction
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Factors That Increase Reaction Rate
87
Use agitation More collisions
among molecules
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
State the effect of each on the rate of reaction as increases,
decreases, or has no effect:
A. increasing the temperature
B. removing some of the reactants
C. adding a catalyst
D. placing the reaction flask in ice
E. increasing the concentration of one of the reactants
F. using a glass rod to mix the liquid reactants
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Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
State the effect of each on the rate of reaction as
increases, decreases, or has no effect:
A. increasing the temperature increases
B. removing some of the reactants decreases
C. adding a catalyst increases
D. placing the reaction flask in ice decreases
E. increasing the concentration of one of the reactant increases
F. using a glass rod to mix the liquid reactants increases
89
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Learning Check
Indicate the effect of each factor listed on the rate of the
following reaction as increases, decreases, or no change.
2CO(g) + O2(g) 2CO2(g)
A. raising the temperature
B. adding O2
C. adding a catalyst
D. lowering the temperature
90
Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.
Solution
Indicate the effect of each factor listed on the rate of the
following reaction as increases, decreases, or no change.
2CO(g) + O2(g) 2CO2(g)
A. raising the temperature increases
B. adding O2 increases
C. adding a catalyst increases
D. lowering the temperature decreases
91