Download Introductory Chemistry: A Foundation FOURTH EDITION by Steven

Chemical Reactions:
An Introduction
Chapter 6
1
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
2
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
3
Figure 6.1: Bubbles of
hydrogen and oxygen gas
form when an electric
current is used to
decompose water
4
Figure 6.2: Hot and cold pack reactions
5
Figure 6.3 (a): Chemical reactions
6
Figure 6.3 (b):
Chemical
reactions
7
Figure 6.3 (c):
Chemical
reactions
8
Figure 6.3 (d):
Chemical
reactions
9
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
10
Figure 6.4: The
reaction between
methane and
oxygen to give
water and carbon
dioxide
11
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
12
Combustion of Methane
• methane gas burns to produce carbon dioxide gas
and liquid water
– whenever something burns it combines with O2(g)
CH4(g) + O2(g)  CO2(g) + H2O(l)
O
H
H
C
H
H
+
O
O
C
+
O
H
H
O
1C+4H
+
2O
1C+2O +2H+O
1C+2H+3O
13
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
1C + 4H + 4O
O
O
O
O
+
H
H
+
O
H
H
1C + 4H + 4O
14
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
15
Symbols Used in Equations
• symbols used after chemical formula to
indicate state
– (g) = gas; (l) = liquid; (s) = solid
– (aq) = aqueous, dissolved in water
16
Balancing by Inspection
1. Count atoms of each element
• 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
• 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
17
Balancing by Inspection
2. Pick an element to balance
3. Find Least Common Multiple and factors
needed to make both sides equal
4. Use factors as coefficients in equation
• if already a coefficient then multiply by new
factor
5. Recount and Repeat until balanced
18
Examples
•
when magnesium metal burns in air it produces a
white, powdery compound magnesium oxide
– burning in air means reacting with O2
1. write the equation in words
– identify the state of each chemical
magnesium(s) + oxygen(g) magnesium oxide(s)
2. write the equation in formulas
– identify diatomic elements
– identify polyatomic ions
– determine formulas
Mg(s) + O2(g)  MgO(s)
19
Examples
•
when magnesium metal burns in air it produces a
white, powdery compound magnesium oxide
– burning in air means reacting with O2
3. count the number of atoms on each side
– count polyatomic groups as one “element” if on both sides
– split count of element if in more than one compound on
one side
Mg(s) + O2(g)  MgO(s)
1  Mg 1
2O1
20
Examples
•
4.
5.
when magnesium metal burns in air it produces a
white, powdery compound magnesium oxide
– burning in air means reacting with O2
pick an element to balance
– avoid element in multiple compounds
find least common multiple of both sides & multiply each
side by factor so it equals LCM
Mg(s) + O2(g)  MgO(s)
1  Mg 1
1x2O1x2
21
Examples
•
when magnesium metal burns in air it
produces a white, powdery compound
magnesium oxide
– burning in air means reacting with O2
6. use factors as coefficients in front of compound
containing the element
 if coefficient already there, multiply them together
Mg(s) + O2(g)  2 MgO(s)
1  Mg 1
1x2O1x2
22
Examples
•
when magnesium metal burns in air it produces a
white, powdery compound magnesium oxide
– burning in air means reacting with O2
7. Recount
Mg(s) + O2(g)  2 MgO(s)
1  Mg 2
2O2
8. Repeat
2 Mg(s) + O2(g)  2 MgO(s)
2 x 1  Mg 2
2O2
23
Examples
•
Under appropriate conditions at 1000°C ammonia gas reacts
with oxygen gas to produce gaseous nitrogen monoxide and
gaseous water
1. write the equation in words
– identify the state of each chemical
ammonia(g) + oxygen(g) nitrogen monoxide(g) + water(g)
2. write the equation in formulas
– identify diatomic elements
– identify polyatomic ions
– determine formulas
NH3(g) + O2(g)  NO(g) + H2O(g)
24
Examples
•
Under appropriate conditions at 1000°C ammonia gas reacts
with oxygen gas to produce gaseous nitrogen monoxide and
gaseous water
3. count the number of atoms of on each side
– count polyatomic groups as one “element” if on both sides
– split count of element if in more than one compound on
one side
NH3(g) + O2(g)  NO(g) + H2O(g)
1  N 1
3H2
2O1+1
25
Examples
•
Under appropriate conditions at 1000°C ammonia
gas reacts with oxygen gas to produce gaseous
nitrogen monoxide and gaseous water
4. pick an element to balance
– avoid element in multiple compounds
5. find least common multiple of both sides & multiply
each side by factor so it equals LCM
NH3(g) + O2(g)  NO(g) + H2O(g)
1  N 1
2x3H2x3
2O1+1
26
Examples
•
Under appropriate conditions at 1000°C ammonia
gas reacts with oxygen gas to produce gaseous
nitrogen monoxide and gaseous water
6. 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
27
Examples
•
Under appropriate conditions at 1000°C ammonia gas reacts
with oxygen gas to produce gaseous nitrogen monoxide and
gaseous water
7. Recount
2 NH3(g) + O2(g)  NO(g) + 3 H2O(g)
2  N 1
6H6
2O1+3
8. Repeat
2 NH3(g) + O2(g) 2 NO(g) + 3 H2O(g)
2  N 1 x 2
6H6
2O1+3
28
Examples
•
Under appropriate conditions at 1000°C ammonia gas
reacts with oxygen gas to produce gaseous nitrogen
monoxide and gaseous water
9. Recount
2 NH3(g) + O2(g)  2 NO(g) + 3 H2O(g)
2  N 2
6H6
2O2+3
29
Examples
•
Under appropriate conditions at 1000°C ammonia gas reacts
with oxygen gas to produce gaseous nitrogen monoxide and
gaseous water
10. 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
30
Examples
•
Under appropriate conditions at 1000°C ammonia gas
reacts with oxygen gas to produce gaseous nitrogen
monoxide and gaseous water
11. Multiply all the coefficients by a number to eliminate
fractions
– 0.5 x 2, 0.33 x 3, 0.25 x 4, 0.67 x 3
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
31