1. Chemical Reactions: An Introduction Chapter 6
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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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3. 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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4. Figure 6.2: Hot and cold pack reactions
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5. Figure 6.5(b)&(c): The reaction of potassium with water
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6. Figure 6.3 (a): Chemical reactions
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7. Figure 6.3 (b): Chemical reactions
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8. Figure 6.3 (c): 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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11. 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) H C O +
1 C + 4 H O
1 C + 2 O H + O
1 C + 2 H + 3 O
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12. 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 C O +
1 C H O
1 C H O
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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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14. Symbols Used in Equations
symbols used after chemical formula to indicate state
(g) = gas; (l) = liquid; (s) = solid
(aq) = aqueous, dissolved in water
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15. 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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16. Balancing by Inspection
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 SO
if an element appears in more than one compound on the same side, count each separately and add
CO + O2  CO2
O 2
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Figure 7.6: The thermite reaction gives off so much heat that the iron formed is molten
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18. Balancing by Inspection
Pick an element to balance
avoid elements from 1b
Find Least Common Multiple and factors needed to make both sides equal
Use factors as coefficients in equation
if already a coefficient then multiply by new factor
Recount and Repeat until balanced
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19. Examples
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
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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Examples
when magnesium metal burns in air it produces a white, powdery compound magnesium oxide
burning in air means reacting with O2
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
Mg(s) + O2(g)  MgO(s)
1  Mg 1
2  O  1
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Examples
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
1 x 2  O  1 x 2
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Examples
when magnesium metal burns in air it produces a white, powdery compound magnesium oxide
burning in air means reacting with O2
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
1 x 2  O  1 x 2
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Examples
when magnesium metal burns in air it produces a white, powdery compound magnesium oxide
burning in air means reacting with O2
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
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24. Examples NH3(g) + O2(g)  NO(g) + H2O(g)
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
identify the state of each chemical
ammonia(g) + oxygen(g) nitrogen monoxide(g) + water(g)
write the equation in formulas
identify diatomic elements
identify polyatomic ions
determine formulas
NH3(g) + O2(g)  NO(g) + H2O(g)
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25. Examples NH3(g) + O2(g)  NO(g) + H2O(g) 1  N 1 3  H  2
Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
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
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26. Examples NH3(g) + O2(g)  NO(g) + H2O(g) 1  N 1 2 x 3  H  2 x 3
Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
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
NH3(g) + O2(g)  NO(g) + H2O(g)
1  N 1
2 x 3  H  2 x 3
2  O  1 + 1
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27. Examples 2 NH3(g) + O2(g)  NO(g) + 3 H2O(g) 1  N 1
Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
use factors as coefficients in front of compound containing the element
2 NH3(g) + O2(g)  NO(g) + 3 H2O(g)
1  N 1
2 x 3  H  2 x 3
2  O  1 + 1
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28. Examples
Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
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
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29. Examples
Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
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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30. Examples
Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
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) 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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31. Examples
Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
Multiply all the coefficients by a number to eliminate fractions
2 x [2 NH3(g) 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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