Download Chapter 6 Chemical Rxns Powerpoint

Introductory Chemistry:
A Foundation
THIRD EDITION
by Steven S. Zumdahl
University of Illinois
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Mifflin Company. All rights reserved.
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Chemical Reactions:
An Introduction
Chapter 6
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Do-Now
In our demonstration 1. What were the two reactants that were
combined?
2. What was the new product produced that we
observed as a class?
3. Was this an example of a physical change or a
chemical change?
Why?
Chemical Reactions
Examples of Chemical Reactions
• Rusting of iron
• Burning (combustion) of wood
• Cooking
• Photosynthesis
o 6CO2 + 6H2O + Energy --> C6H12O6 + 6O2
Chemical reactions are like sentences.
Symbols  elements
Formula  compound
Equations  chemical reactions
Chemical reaction
• a reaction that occurs when the atoms of one of
more substances are rearranged to form
different substances
• a chemical change
Evidence of a chemical reaction
- Temperature change
- Color change
- Odor produced
- Formation of bubbles
- Appearance of a solid
Chemical Equations
2H2(g) +
O2(g)
2H2O (l)
•Reactants: elements or compounds to the left of the arrow
that combine together in a chemical reaction
•Products: elements or compounds to the right of the
arrow that are produced in a chemical reaction
•Coefficient: whole number before a chemical formula
+
: represents “and”
: direction reaction progresses, yields or produces
C(s) + O2(g)
CO2 (g)
physical states of reactants and products are
indicated by:
(s)
(g)
(l)
= solid: C(s)
= gas: CO2(g)
= liquid: H2O(l)
(aq)
= aqueous, dissolved in water:
NaCl(aq) is a salt water solution
(cr)
= crystalline
= precipitate forms
Word and Formula Equations
• A word equation is an equation represented by
words.
Example: solid sodium plus chlorine gas reacts to
produce sodium chloride
• Skeleton equation represents the reactants and
products of a chemical reaction by their formulas.
Example: Na(s) + Cl2(g)
NaCl(s)
Balancing Chemical Equations
Law of conservation of mass - in a chemical reaction
matter is neither created nor destroyed
•Chemical equation - the number of atoms of each
reactant must equal the number of atoms of each
product
•Equations must be balanced using coefficients
balanced chemical equation – shows that each side of
the equation has the same number of atoms of each
element and mass is conserved
2 Na(s) +
Cl2(g)
2 NaCl(s)
Steps for Balancing Equations
1) Assemble the correct formulas for all the
reactants and products, using “+” and “→”
2) Identify diatomic elements
3) Count the number of atoms of each type appearing
on the reactant side and the product side
4) Balance the elements one at a time by adding
coefficients in front of a formula to make number
of atoms equal on both sides of equation
5) Check to make sure equation is balanced and
coefficients are in lowest possible ratio
Practice Balancing Equations
1)
Fe +
2)
KClO3
3)
S
Ca(OH)2 + HCl
Fe2S3
KCl +
O2
CaCl2 +
H2O
Sample problems – Balancing Equations
Word Equation
Water and carbon yield hydrogen gas and carbon
monoxide
Skeleton Equation
Balanced Equation
Sample problems – Balancing Equations
Word Equation
Magnesium chloride and potassium react to
form potassium chloride and magnesium
Skeleton Equation
Balanced Equation
Combustion of Methane
Word Equation
• methane gas burns to produce carbon dioxide gas
and liquid water
– whenever something burns it combines with O2(g)
Skeleton Equation
CH4(g) + O2(g)  CO2(g) + H2O(l)
O
H
H
C
H
H
+
O
O
C
O
+
O
H
H
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Combustion of Methane
Balanced
• to show the reaction obeys the Law of
Conservation of Mass it must be balanced
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
H
H
C
H
H
O
+
O
C
+
O
O
O
1C + 4H + 4O
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O
O
+
H
H
+
O
H
H
1C + 4H + 4O
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Water and iron react to form iron III oxide and hydrogen
Propane (C3H8) burns to form carbon dioxide and water
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Chemical Reactions
• Reactions involve chemical changes in
matter resulting in new substances
• Reactions involve rearrangement and
exchange of atoms to produce new
molecules
– Elements are not transmuted during a reaction
Reactants  Products
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Evidence of Chemical Reactions
• a chemical change occurs when new
substances are made
• visual clues (permanent)
– color change, precipitate formation, gas
bubbles, flames, heat release, cooling, light
• other clues
– new odor, permanent new state
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Evidence of Chemical Reactions:
Color Change
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Evidence of Chemical Reactions
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Chemical Equations
• Shorthand way of describing a reaction
• Provides information about the reaction
– Formulas of reactants and products
– States of reactants and products
– Relative numbers of reactant and product
molecules that are required
– Can be used to determine weights of reactants
used and of products that can be made
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Conservation of Mass
• Matter cannot be created or destroyed
• In a chemical reaction, all the atoms present
at the beginning are still present at the end
• Therefore the total mass cannot change
• Therefore the total mass of the reactants
will be the same as the total mass of the
products
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Writing Equations
• Use proper formulas for each reactant and product
• proper equation should be balanced
– obey Law of Conservation of Mass
– all elements on reactants side also on product side
– equal numbers of atoms of each element on reactant side as on
product side
• balanced equation shows the relationship between the
relative numbers of molecules of reactants and products
– can be used to determine mass relationships
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Symbols Used in Equations
• symbols used after chemical formula to
indicate physical state
– (g) = gas; (l) = liquid; (s) = solid
– (aq) = aqueous, dissolved in water
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Sample – Recognizing Reactants and Products
•
¬
-
when magnesium metal burns in air it produces a
white, powdery compound magnesium oxide
– burning in air means reacting with O2
– Metals are solids, except for Hg which is liquid
write the equation in words
– identify the state of each chemical
magnesium(s) + oxygen(g) magnesium oxide(s)
write the equation in formulas
– identify diatomic elements
– identify polyatomic ions
– determine formulas
Mg(s) + O2(g)  MgO(s)
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Balancing Chemical Equations
Count atoms of each element
a polyatomic ions may be counted as one “element” if it
does not change in the reaction
Al + FeSO4 Al2(SO4)3 + Fe
1 SO4 3
b if an element appears in more than one compound on
the same side, count each separately and add
CO + O2  CO2
1 + 2 O 2
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Balancing Chemical Equations
Pick an element to balance
a avoid elements from 1b
Find Least Common Multiple and factors
needed to make both sides equal
Use factors as coefficients in equation
a if already a coefficient then multiply by new
factor
° Recount and Repeat until balanced
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Example 1:
Magnesium metal burns in air
• when magnesium metal burns in air it
produces a white, powdery compound
magnesium oxide
– burning in air means reacting with O2
write the equation in words
magnesium(s) + oxygen(g) magnesium oxide(s)
write the equation in formulas - determine formulas
Mg(s) + O2(g)  MgO(s)
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Example 1:
Magnesium metal burns in air
count the number of atoms of on each side
Mg(s) + O2(g)  MgO(s)
1  Mg 1
2O1
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Example 1:
Magnesium metal burns in air
pick an element to balance
- avoid element in multiple compounds
Mg(s) + O2(g)  MgO(s)
1  Mg 1
1x2O1x2
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Example 1:
Magnesium metal burns in air
Mg(s) + O2(g)  2 MgO(s)
1  Mg 1
1x2O1x2
Use factors as coefficients in front of compound
containing the element
if coefficient already there, multiply them together
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Example 1:
Magnesium metal burns in air
Recount
Mg(s) + O2(g)  2 MgO(s)
1  Mg 2
2O2
Repeat
2 Mg(s) + O2(g)  2 MgO(s)
2 x 1  Mg 2
2O2
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Example 2
•
•
Under appropriate conditions at 1000°C ammonia
gas reacts with oxygen gas to produce gaseous
nitrogen monoxide and gaseous water
write the equation in words
ammonia(g) + oxygen(g) nitrogen monoxide(g) + water(g)
write the equation in formulas
NH3(g) + O2(g)  NO(g) + H2O(g)
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Examples
count the number of atoms of on each side
NH3(g) + O2(g)  NO(g) + H2O(g)
1  N 1
3H2
2O1+1
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Examples
pick an element to balance - avoid element in
multiple compounds
NH3(g) + O2(g)  NO(g) + H2O(g)
1  N 1
2x3H2x3
2O1+1
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Examples
Use factors as coefficients in front of compound
containing the element
2 NH3(g) + O2(g)  NO(g) + 3 H2O(g)
1  N 1
2x3H2x3
2O1+1
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Examples
Recount
2 NH3(g) + O2(g)  NO(g) + 3 H2O(g)
2  N 1
6H6
2O1+3
Repeat
2 NH3(g) + O2(g) 2 NO(g) + 3 H2O(g)
2  N 1 x 2
6H6
2O1+3
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Examples
´
Recount
2 NH3(g) + O2(g)  2 NO(g) + 3 H2O(g)
2  N 2
6H6
2O2+3
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Examples
Repeat
– A trick of the trade, when you are forced to attack an
element that is in 3 or more compounds – find where it is
uncombined. You can find a factor to make it any amount
you want, even if that factor is a fraction!
– We want to make the O on the left equal 5, therefore we
will multiply it by 2.5
2 NH3(g) + 2.5 O2(g) 2 NO(g) + 3 H2O(g)
2  N 2
6H6
2.5 x 2  O  2 + 3
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Examples
Multiply all the coefficients by a number to
eliminate fractions:
2 x [2 NH3(g) + 2.5 O2(g) 2 NO(g) + 3 H2O(g)]
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
4  N 4
12  H  12
10  O  10
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