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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 moleculesElements 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 cluesnew odor, permanent new state
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Figure 6.2: Hot and cold pack reactions
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Figure 6.5(b)&(c): The reaction of potassium with water
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Figure 6.3 (a): Chemical reactions
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Figure 6.3 (b): Chemical reactions
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Figure 6.3 (c): Chemical reactions
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Chemical Equations
• Shorthand way of describing a reaction
• Provides information about the reactionFormulas of reactants and productsStates of reactants and productsRelative numbers of reactant and product
molecules that are requiredCan be used to determine weights of reactants
used and of products that can be made
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10
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 waterwhenever something burns it combines with O2(g)
CH4(g) + O2(g) CO2(g) + H2O(l)
H
HC
H
HOO+
O
O
C + OH H
1 C + 4 H + 2 O 1 C + 2 O + 2 H + O1 C + 2 H + 3 O
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Combustion of MethaneBalanced
• 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
HC
H
H
OO
+
O
O
C +
OH H
OO
+O
H H
+
1 C + 4 H + 4 O 1 C + 4 H + 4 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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Sample – Recognizing Reactants and Products• when magnesium metal burns in air it produces a
white, powdery compound magnesium oxideburning 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 chemicalmagnesium(s) + oxygen(g) magnesium oxide(s)
• write the equation in formulasidentify diatomic elementsidentify polyatomic ionsdetermine formulas
Mg(s) + O2(g) MgO(s)
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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 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
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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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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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Exampleswhen magnesium metal burns in air it produces a white, powdery compound magnesium oxideburning in air means reacting with O2
• write the equation in wordsidentify the state of each chemical
magnesium(s) + oxygen(g) magnesium oxide(s)
• write the equation in formulasidentify diatomic elementsidentify polyatomic ionsdetermine 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 oxideburning in air means reacting with O2
• count the number of atoms of on each sidecount polyatomic groups as one “element” if on both sidessplit 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 oxideburning in air means reacting with O2
• pick an element to balanceavoid 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 oxideburning 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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Exampleswhen magnesium metal burns in air it produces a white, powdery compound magnesium oxideburning 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
2 O 2
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Examples• Under appropriate conditions at 1000°C ammonia gas reacts
with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
• write the equation in wordsidentify the state of each chemical
ammonia(g) + oxygen(g) nitrogen monoxide(g) + water(g)
• write the equation in formulasidentify diatomic elementsidentify polyatomic ionsdetermine formulas
NH3(g) + O2(g) NO(g) + H2O(g)
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Examples• 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 sidecount polyatomic groups as one “element” if on both sidessplit 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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Examples• Under appropriate conditions at 1000°C ammonia
gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
• pick an element to balanceavoid 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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Examples• 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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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 16 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• 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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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) + 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• 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) + 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