Download Chapter 6 Chemical Reactions: An Introduction

Chapter 6
Chemical Reactions:
An Introduction
Chemical Reactions
• Reactions involve
chemical changes in
matter that result in new
substances.
• Reactions involve
rearrangement and
exchange of atoms to
produce new molecules.
• Reactants → Products
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
Evidence of Chemical Reactions
(cont.)
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
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.
Conservation of Mass (cont.)
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
C
H H
H
+
O
O
C
+
O
H
H
O
1C+4H
+
2O
1C+2O +2H +O
1C+2H+3O
Combustion of Methane Balanced
• To show a 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
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 equations show the relationship between
the relative numbers of molecules of reactants
and products.
– Can be used to determine mass relationships
Symbols Used in Equations
• Symbols used after chemical formula to
indicate state:
– (g) = gas; (l) = liquid; (s) = solid
– (aq) = aqueous, dissolved in water
– e. g. NH3(aq) indicates ammonia dissolved in
water
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.
Recognizing Reactants
and Products (cont.)
• 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) (unbalanced)
Balancing by Inspection
• Count atoms of each element
– Polyatomic ions may be counted as one “element” if
they do 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 element separately and add.
CO + O2 → CO2
1 + 2 O 2
Balancing by Inspection
(cont.)
• Pick an element to balance.
• Find least common multiple (LCM) 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.
Example #1
• When magnesium metal burns in air it
produces a white, powdery compound,
magnesium oxide.
– Burning in air means reacting with O2
Example #1 (cont.)
• 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) (unbalanced)
Example #1 (cont.)
• 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
Example #1 (cont.)
• 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
Example #1 (cont.)
• Use factors as coefficients in front of
compound containing the element
– If coefficient is already there, multiply them
together
Mg(s) + O2(g) → 2 MgO(s)
1 ← Mg → 1 x 2
1x2←O→1x2
Example #1 (cont.)
• 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
Example #2
• Under appropriate conditions, at 1000°C
ammonia gas reacts with oxygen gas to
produce gaseous nitrogen monoxide and
gaseous water.
Example #2 (cont.)
• 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)
Example #2 (cont.)
• 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
Example #2 (cont.)
• Pick an element to balance
– Avoid elements 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
2x3←H→2x3
2←O→1+1
Example #2 (cont.)
• 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
Example #2 (cont.)
• 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→1x2
6←H→6
2←O→1+3
Example #2 (cont.)
• Recount
2 NH3(g) + O2(g) → 2 NO(g) + 3 H2O(g)
2←N→2
6←H→6
2←O→2+3
Example #2 (cont.)
• Repeat
– 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
Example #2 (cont.)
• Multiply all the coefficients by a number to
eliminate fractions
– x.5 2, x.33 3, x.25 4, x.67 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