Transcript Chapter 5: Reaction Types

Evidence for a
Chemical Reaction

Chemical reactions give a visual signal.
Examples: Rusted steel & bleached hair,
Clues that a Chemical Reaction (rxn) has Occurred
1. Color changes
2. Solid forms
3. Bubbles form
4. Heat is produced or absorbed by monitoring the change
in temperature
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Chemical Equations
Two important types of information:
1. The identities of the reactants and products.
2. The relative numbers of each.
Physical States
Symbol
State
(s)
solid
(l)
liquid
(g)
gas
(aq)
dissolved in water
(in aqueous solution)
Examples include 2K(s) + 2H2O(l)  H2(g) + 2KOH(aq)
2H2(g) + O2(g)  2H2O(g)
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Types of Rxns
 Single displacement
 Double displacement
 Combustion
 Synthesis
 Decomposition
 Acid-base neutralization
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Single displacement
 A + BC  AC + B
 Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)
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Double displacement
 AB + CD  AC + BD
 2KI(aq) + Pb(NO3)2(aq)  2KNO3(aq) + PbI2(s)
 When solid formed

 precipitate
 Can we predict the formation of a
precipitate?

Yes: solubility rules
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Solubility
 The ability for a substance to dissolve in
a medium
 Soluble substances can form a 0.10molar solution at 25 °C.
 If unable to dissolve, said to be insoluble
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Solubility Rules
 Check it out:
 http://web.clark.edu/aaliabadi/CHEM131_fall200
6_solubility_rules.htm
 Let’s take a look
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Solubility
 What happens when you dissolve NaCl in water?
 It breaks up into individual cations and anions
 They are surrounded by water molecules
 There are strong water-ion interactions
 Give the equation for the ionization of the following
compounds:

Na2CO3

Al2(SO4)3
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Combustion
 CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
 Requires oxygen gas and a hydrocarbon
 Yields carbon dioxide and water
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Synthesis
 A + B  AB
 2Mg(s) + O2(g)  2MgO(s)
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Decomposition
 AB  A + B
 2KClO3(s)  2KCl(s) + 3O2(g)
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Acid-base neutralization (a
special type of DD)
 HX(aq) + MOH(aq)  MX(aq) + H2O(l)
 HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
 CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
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More on chemical equations
 Can be expressed three ways



Molecular
Ionic
Net-ionic
 See next three slides
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Molecular equation
 2KI(aq) + Pb(NO3)2(aq)  2KNO3(aq) + PbI2(s)
 Gives the complete chemical equation
 Includes all the bells-and-whistles
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Ionic equation
 2K+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq)  2K
+
(aq)
+ 2NO3-(aq) + PbI2(s)
 Displays all the components in their
broken-down form

Shows all ions in soln
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Net-ionic equation
 Pb2+(aq) + 2I-(aq)  PbI2(s)
 Shows only that which creates a rxn

Gets rid of spectator ions
 Ions
that still remain in solution and don’t play
a role
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Practice
Write the balanced molecular, ionic, and net-ionic equations (if
necessary), including symbols of states, for each of the
chemical reactions below. Also, name the type of rxn.
1. Solid magnesium metal reacts with liquid water to form solid
magnesium hydroxide and hydrogen gas.
2. Solid ammonium dichromate decomposes to solid chromium
(III) oxide, gaseous nitrogen, and gaseous water.
3. Gaseous ammonia reacts with gaseous oxygen to form
gaseous nitrogen monoxide and gaseous water.
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More practice
4. When solutions of strontium bromide
and magnesium sulfate are mixed, a ppt
is formed.
5. Copper metal is thrown in to a beaker
containing highly concentrated nitric
acid. This exothermic reaction yields
copper (II) nitrate, nitrogen dioxide gas,
and water.
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Oxidation-Reduction
Reactions: a subset
 Oxidation-Reduction Reaction: or redox reactions

Rxns in which one or more electrons are transferred.
 Oxidation: Loss of electrons
 Reduction: Gain of Electrons
 Example: 2Na(s) + Cl2(g)  2NaCl(s)
 Sodium is oxidized to Na+
 Chlorine is reduced to Cl Chlorine is oxidizing agent

Itself being reduced
 Sodium is reducing agent

Itself being oxidized
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Oxidation-Reduction
Reactions

Identify what is oxidized/reduced, and what is the
oxidizing/reducing agent.


2Cu(s) + O2(g)  2CuO(s)
Ba(s) + H2O(l)  Ba(OH)2(aq) + H2(g)
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Oxidation-Reduction
Reactions Between Nonmetals
 Oxidation

Loss of Electrons/gain in oxygens/loss of hydrogens
 Reduction

Gain of Electrons/gain in hydrogens/loss of oxygens
 Oxidizing Agent

Causes the elements to be oxidized: electron acceptor
 Reducing Agent

Causes the elements to be reduced: electron donor
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Example
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
 Reactants:



C = -4
H = +1
O=0
 Products:



C = +4
H = +1
O = -2
 So which are oxidized
and reduced?
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Practice
 The combustion of C2H5SH to yield
carbon dioxide, water, and sulfur dioxide
 State what is being oxidized/reduced
and what is the oxidizing/reducing agent
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Solubility
 Oil and water do not mix. Oil is non-polar and water is
polar. Oil floats on the surface of water because its
density is less than that of water.
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Solution Composition:
An Introduction
 Unsaturated solution
 A solution that has not reached the limit of solute
that will dissolve
 Saturated solution
 A solution that contains as much solute as will
dissolve at a given temperature
 Supersaturated solution
 A solution that contains more than its allowable limit
of dissolved solute
 Demo
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Solution Composition: Molarity
 Molarity

Number of moles of
solute per volume of
solution in liters.
 Example: A solution that is
1.0 M contains 1.0 mol per 1
L of solution.
moles of solute
mol
Molarity = M =
=
liters of solution
L
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Example

Calculate the concentration of a solution prepared by
dissolving 1.00 g of ethanol, C2H6O, in 2.50 liters of
water
mol
1
1.00 g  (
)(
) = 8.68  10-3 M
46.069 g
2.50 L

Practice: How many grams of NaCl are needed to make
1.25 liters of a 0.050 molar solution?
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More practice
 Give the concentrations of the ions in
each of the following solutions:
 0.10 M Na2CO3
 0.010 M Al2(SO4)3
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More
 Stock solution

Solution that is routinely used and purchased or
prepared in concentrated form
 Dilution

The process of adding more water to a solution
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Making dilutions
 The equation for solving dilution problems is given below.
C1 x V1 = C2 x V2
 Moles of solute after dilution = Moles of solute
before dilution
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Practice
 What volume of 12 M HCl must be taken to prepare 0.75
L of 0.25 M HCl?
 12M x V1 = 0.25M x 0.75L
 V1 = 0.016L of acid
 So what’s the next step?
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Stoichiometry of
Solution Reactions
 Calculate the mass of lead (II) sulfate formed when 1.25
L of 0.0500 M lead (II) nitrate and an excess of sodium
sulfate are mixed.
 Na2SO4(aq) + Pb(NO3)2(aq)  2NaNO3(aq) + PbSO4(s)
0.0500 mol
1 mol
303.3 g
1.25 L  (
) (
) (
) = 19.0 g
L
1 mol
mol
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Example
 Calculate the mass of barium oxalate
formed when 0.0255 L of 0.0305 M
barium chloride and 0.0354 L of 0.0257
M rubidium oxalate are mixed.
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