Download Chapter 5—Chemical Reactions

Chapter 5—Chemical
Reactions
5.1—Chemical Equations
• A chemical equation shows the chemical change taking place.
• The state of each substance is written in parentheses after the
formula: s for solids, l for liquids, g for gases, and aq for
aqueous.
Reactants
H2 (g) + O2 (g) → H2O (l)
Products
Balancing Chemical Equations
• Balanced chemical equation—the number of atoms of each
element in the reactants is equal to the number of atoms of
that same element in the products.
• Reactions must be balanced to obey the law of conservation
of mass.
• Coefficients are written to the left of each reactant or product
in order to achieve balance.
2 H2 (g) + O2 (g) → 2 H2O (l)
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Examples of Balancing Chemical Equations
• Methane, CH4,is the main ingredient in natural gas. It
combines with oxygen, O2, when it burns to form carbon
dioxide, CO2, and water, H2O. Write a balanced chemical
equation for this reaction.
• Balance the following chemical equations:
• Mg(s) + O2(g) → MgO(s)
• Na3PO4(aq) + Ca(NO3)2(aq) → NaNO3(aq) + Ca3(PO4)2(s)
• C6H14(l) + O2(g) → CO2(g) + H2O(g)
More Examples
• An aqueous solution of potassium carbonate is mixed with an
aqueous solution of iron(III) nitrate resulting in a precipitate of
iron(III) carbonate. Write a balanced chemical equation for
this reaction.
• Ammonia gas reacts with oxygen gas to form nitrogen
monoxide gas and steam. Write a balancd chemical equation
for this reaction.
5.2—Types of Reactions
• Chemical reactions are often classified into categories
according to characteristics of the reactions. The following is a
useful classification scheme:
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5.3—Redox Reactions
• Oxidation-reduction (redox)
reactions involve one
substance losing electrons
(being oxidized) and another
substance gaining electrons
(being reduced).
• LEO say GER
• Lose Electrons Oxidation, Gain
Electrons Reduction
• OIL RIG
• Oxidation Is Loss, Reduction Is
Gain
Oxidation Numbers
• Oxidation numbers (also called oxidation states) are positive
or negative numbers assigned to elements in chemical
formulas according to a specific set of rules
• Rule 1:
The O.N. of an element is 0.
Examples: Fe (0), Cl2 (0), and Ca(0)
• Rule 2:
The O.N. of a monatomic ion is equal to the charge of the
ion. This includes when the monatomic ion is in a
compound.
Examples:
Mg2+(+2), O2-(-2), Cl-(-1), Na2S (Na = +1) and Ca(NO3)2
(Ca = +2)
Oxidation Numbers (continued)
• Rule 3:
The O.N. of hydrogen is +1.
Examples: HBr (H = +1) and H2SO4(H = +1)
• Rule 4:
The O.N. of oxygen is -2 except in peroxides it is -1.
Examples: MgO & HBrO3 (O = -2), and H2O2 (O = -1)
• Rule 5:
The algebraic sum of the oxidation numbers of all atoms in
a compound or polyatomic ion equals the charge of the
compound or ion. For the compound, the charge always
equals zero.
Example: MgSO4 (O.N. of Mg = +2 by rule 2, O.N. of O = 2 by rule 4, and O.N. of S = +6 by algebra and rule 5)
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Oxidation Number Practice
• Determine the oxidation number for each atom in the
following compounds or ions.
• CaCl2
• Na3PO4
• ClO31• H2SO3
Oxidizing and Reducing Agents
• Reducing agent—the substance that is oxidized. It caused
the other substance to be reduced.
• Oxidizing agent—the substance that is reduced. It caused
the other substance to be oxidized.
Oxidizing and Reducing Agent Examples
• Determine the oxidizing and reducing agents as well as what
is being reduced and what is being oxidized in the following
reactions:
2Na(s) +2H2O(l)
H2 (g) + 2NaOH(aq)
S2O82-(aq) + 2 I-(aq)
4NO2(g) + O2(g)
I2(aq) + 2SO42-(aq)
2N2O5(g)
• Hint: First, assign oxidation numbers.
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5.4—Decomposition Reactions
• In decomposition reactions, one substance is broken down
into two or more simpler substances. Decomposition
reactions may be either redox or nonredox reactions.
• The general form of the equation for a decomposition
reaction is:
A
B + C.
• An example of a redox decomposition reaction is:
2HI(g)
H2 (g) + I2 (g)
• An example of a nonredox decomposition reaction is:
H2CO3(aq)
CO2(g) + H2O(l)
5.5—Combination Reactions
• In combination reactions two or more substances react to
form a single substance. Combination reactions may be
either redox or nonredox reactions.
• The general form of the equation for a combination reaction
(also known as addition or synthesis reactions) is:
A+B
C
• An example of a redox combination reaction is:
2Fe(s) + 3O2 (g)
3Fe2O3 (s)
• An example of a nonredox combination reaction is:
N2O5(g) + H2O(l)
2HNO3(aq)
5.6—Replacement Reactions
• Single-replacement reactions—one element replaces another
in a compound. They are always redox reactions because an
element turns into an ion and an ion turns into an element
• The general form of the equation for a single replacement
reaction (also known as substitution reactions) is:
A + BX
B + AX
In this equation, A and B represent elements and AX and
BX are compounds.
• An example of a single replacement reaction is:
Fe(s) + Cu(NO3)2(aq)
Cu(s) + Fe(NO3)2(aq)
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Double-Replacement Reactions
• Double-replacement reactions—two ionic compounds switch
partners. They are never redox reactions.
• The following general form of the equation for double
replacement reactions (also known as metathesis reactions)
shows the partner-swapping characteristic of the reactions:
AX + BY
BX + AY
• An example of a double-replacement reaction is:
AgNO3(aq) + NaCl(aq)
AgCl(s) + NaNO3(aq)
Reaction Classification
• Classify each of the following equations as redox or nonredox
reactions and as combination, decomposition, singlereplacement, or double-replacement reactions.
• NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
• 2 Mg(s) + O2(g) → 2 MgO(s)
• Na2CO3(s)→ Na2O(s) + CO2(g)
• Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
5.7—Ionic Equations
• Ionic compounds and some polar covalent compounds break
apart (dissociate) when they dissolve in water and form ions.
Reactions between these dissolved substances can be
shown using molecular, total ionic, and net ionic equations.
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Ionic Equations (Continued)
• Molecular Equations: Each compound is represented by its
formula.
NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq)
• Total Ionic Equations: All soluble ionic substances are
represented by the ions they form in solution. Substances
represented by s, l, or g (do not dissolve or do not dissociate)
are represented by their formulas.
Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) → AgCl(s) + Na+(aq) + NO3-(aq)
• Net Ionic Equations: Shows only the change that is actually
taking place. All spectator ions (those that are the same on
both sides of the equation) are not included in the net ionic
equation.
Ag+(aq) + Cl-(aq) → AgCl(s)
Ionic Equations Practice
• An aqueous solution of barium chloride is mixed with an
aqueous solution of sodium sulfide. Write balanced
molecular, total ionic, and net ionic equations for this
reaction.
• Write a total ionic and a net ionic equation for the following
reaction.
BaCO3(s) + 2HNO3(aq) → Ba(NO3)2(aq) + CO2(g) + H2O(l)
5.8—Energy and Reactions
• Exothermic reactions—chemical reactions that release
heat. Ordinary combustion of a log in a fireplace is an
example of an exothermic reaction.
• Endothermic reactions—chemical reactions that absorb
heat. Example of an endothermic process is the melting of
ordinary ice (physical not chemical change). As the ice
melts, heat is absorbed from the air surrounding the ice.
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5.9—The Mole and Chemical Equations
• Stoichiometry—the study of mass relationships in chemical
reactions.
• Stoichiometry allows us to calculate:
• The amount of a substance in a rxn given the amount of
another substance in the rxn
• Must start with a balanced chemical equation!
• The following represents the process used in stoichiometry:
grams
A
Molar mass
moles
A
Equation
coefficient
moles
B
Molar mass
grams
B
Stoichiometry Examples
• Calculate the number of moles of H2S that would need to
react with excess O2 in order to produce 115 g of SO2. Water
is the other product.
• First you will need to write a balanced chemical equation.
Then follow the process as outlined previously: grams of
A to moles of A to moles of B.
• Calculate the mass of SO2 that would be produced if 56.2
grams of O2 reacts with excess H2S.
• Calculate the mass of H2S that would be needed to react with
107 g of O2.
5.10—The Limiting Reactant
• Limiting reactant—the reactant that will be used up first,
and thus determines the amount of product that can be
produced.
• How to approach limiting reactant problems: Calculate the
amount of product that could be produced by each of the
reactants available. The reactant that gives the least
amount of product is then the limiting reactant.
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Limiting Reagent in Kitchen
A BLT sandwich requires 2 slices of bread, 2 leaves of lettuce, 4
pieces of bacon, and 2 tomato slices.
How many sandwiches could you make if you were given:
7 slices of bread
14 slices of bacon
5 leaves of lettuce
9 tomatoes
Limiting Reactant Examples
• Calculate the maximum amount (in grams) of SO2 that could
be produced by reacting 55.2 g of O2 with 50.8 g of H2S.
• If 23.4 g of copper(II) nitrate reacts with 6.25 g of sodium
hydroxide, then how much of the precipitate (solid) will form?
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5.11—Reaction Yields
• A reaction will not always produce the calculated amount of
product.
• The maximum amount of product that could be produced is
called the theoretical yield.
• The amount of product actually produced is called the actual
yield.
• These two quantities are used to calculate the percentage
yield using the following equation:
% yield =
actual yield
× 100
theoretical yield
Reaction Yields Example
• A 2.00g sample of magnesium reacts with 2.00g of pure
oxygen gas to form 2.96g of magnesium oxide. What is the
percent yield for this reaction?
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