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CHEMICAL EQUATIONS
REACTANTS OF A CHEMICAL EQUATION
•  Chemical equations are a convenient way to represent
chemical reactions. Chemical equations are written in
terms of reactants and products.
•  A symbol is written in parentheses to the right of each
reactant and product to indicate the state or form in which
the substance exists. Gases are indicated by (g),
liquids by (l), solids by (s), and substances dissolved in
water by (aq).
Chapter 5:
Chemical Reactions
•  The reactants in a chemical equation are the substances
written on the left side of the arrow that points toward the
products. When two or more reactants are involved in an
equation, they are separated by a plus sign (+).
Re ac tan ts
2 H2 (g) + O2 (g) ! 2 H2O (l)
Pr oducts
PRODUCTS OF A CHEMICAL EQUATION
•  The products in a chemical equation are the substances
written on the right side of the arrow. When two or more
products are involved in an equation, they are separated
by a plus sign (+).
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solid sugar =
C12H22O11 (s) and liquid water = H2O (l)
sugar dissolved in water =
C12H22O11 (aq)
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•  The equation is balanced by inserting appropriate
coefficients on the left of each reactant and product to give
the following balanced equation:
EXAMPLES OF UNBALANCED AND BALANCED
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. This information written in
the form of an unbalanced equation is:
BALANCED CHEMICAL EQUATIONS
•  A balanced chemical equation is one in which the number
of atoms of each element in the reactants is equal to the
number of atoms of that same element in the products.
•  A reaction can be balanced by applying the law of
conservation of matter.
•  Coefficients (in red below) are written to the left of each
reactant or product in order to achieve balance.
CH4(g) + O2 (g)
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CH4(g) + 2O2 (g)
CO2 (g) + 2H2O
•  When atoms are counted, 1C, 4H and 4O atoms are on
the left and the same number are on the right.
CO2 (g) + H2O(g)
•  The equation is unbalanced as it is written because the
number of H atoms on the left is 4 and the number on the
right is 2. Also, the number of O atoms on the left is 2 and
the number on the right is 3.
2 H2 (g) + O2 (g) ! 2 H2O (l)
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REDOX REACTIONS
TYPES OF CHEMICAL REACTIONS
OXIDATION NUMBERS
•  The word redox is a combination of two words, reduction and
oxidation. These two words have multiple meanings when
applied to chemical 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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•  Oxidation numbers (also called oxidation states) are positive
or negative numbers assigned to elements in chemical
formulas according to a set of rules. The term oxidation
number is abbreviated O.N.
• Rule 1:
The O.N. of any uncombined element is 0.
For example: Fe (0), Cl2 (0), and Ca(0)
• Rule 2:
The O.N. of a simple ion is equal to the charge on the ion.
For example: Mg2+(+2), O2-(-2), and Cl-(-1).
•  The concept of oxidation numbers provides a convenient
way to work with redox reactions.
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• Rule 3:
The O.N. of group IA and IIA elements when they are in
compounds are always +1 and +2, respectively.
For example: Na2S (Na = +1) and Ca(NO3)2 (Ca = +2)
• Rule 6:
The algebraic sum of the oxidation numbers of all atoms in a
complete compound equals zero.
For example:MgSO4 (O.N. of Mg = +2 by rule 2, O.N. of O = -2
by rule 5, and O.N. of S = +6 by algebra and rule 6)
• Rule 4:
The O.N. of hydrogen is always +1.
For example: HBr (H = +1) and H2SO4(H = +1)
A closer look at H2O2 :
2 (O.N. of H) + 2 (O.N. of O ) = 0
2 (+1)
+ 2 (O.N. of O ) = 0
2 (O.N. of O ) = ! 2
(O.N. of O ) = ! 1
A closer look at MgO :
(O.N. of Mg) + (O.N. of O ) = 0
+2
+ (O.N. of O ) = 0
(O.N. of O ) = ! 2
A closer look at HCO3 ! :
A closer look at MgSO4 :
(O.N. of H) + (O.N. of C ) + 3 (O.N. of O ) = ! 1
+1
+ (O.N. of C ) + 3 (!2 )
= !1
+1
+ (O.N. of C ) +
!6
= !1
(O.N. of C ) = + 4
(O.N. of Mg) + (O.N. of S ) + 4 (O.N. of O ) = 0
+2
+ (O.N. of S ) + 4 (!2 )
=0
+2
+ (O.N. of S ) +
!8
=0
(O.N. of S ) = + 6
• Rule 5:
The O.N. of oxygen is -2 except in peroxides where it is -1.
For example: MgO (O = -2), HBrO3 (O = -2), and
H2O2 (O = -1)
Because there is only one H and one C, the total positive
oxidation number is +5. The three O atoms, with an O.N. of -2
each, give a total negative O.N. of -6. Thus, the total positive
and the total negative O.N. values add up to -1, which is the
charge on the bicarbonate ion.
Because there is only one Mg and one S, the total positive
oxidation number is +8. The four O atoms, with an O.N. of -2
each, give a total negative O.N. of -8. Thus, the total positive
and the total negative O.N. values add up to zero.
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OXIDIZING AND REDUCING AGENTS
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2Na(s) +2H2O(l)
EXAMPLE
•  In a redox reaction, the substance that contains an
element that is oxidized during the reaction is called the
reducing agent.
•  In a redox reaction, the substance that contains an
element that is reduced during the reaction is called the
oxidizing agent.
• Rule 7:
The algebraic sum of the O.N. of all the atoms in a polyatomic ion
is equal to the charge on the ion.
For example: HCO3- (O.N. of H is +1 by rule 4, O.N. of O is -2 by
rule 5, and O.N. of C is +4 by algebra and rule 7)
•  Determine the oxidizing and reducing agents in the
reaction:
2Na(s) +2H2O(l)
H2 (g) + 2NaOH(aq)
H2 (g) + 2NaOH(aq)
•  Now, determine the oxidizing and reducing agents.
•  Solution: First, assign oxidation numbers.
•  Na is oxidized; therefore, Na is the reducing agent.
•  H is reduced; therefore, H2O is the oxidizing agent.
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•  A pictorial representation of a decomposition reaction:
DECOMPOSITION REACTIONS
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COMBINATION REACTIONS
•  In decomposition reactions one substance is broken down
into two or more simpler substances. Decomposition
reactions may be either redox or nonredox 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 decomposition
reaction is: A
B + C.
•  The general form of the equation for a combination
reaction is: A + B
C
•  An example of a redox decomposition reaction is:
2HI(g)
H2 (g) + I2 (g)
•  An example of a redox combination reaction is:
S(s) + O2 (g)
SO2 (g)
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•  An example of a nonredox decomposition reaction is:
H2CO3(aq)
CO2(g) + H2O(l)
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•  An example of a nonredox combination reaction is:
N2O5(g) + H2O(l)
2HNO3(aq)
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•  A pictorial representation of a single replacement reaction:
SINGLE-REPLACEMENT REACTIONS
•  A pictorial representation of a combination reaction:
•  Single-replacement reactions are always redox reactions
because they occur when one element reacts with a
compound, displaces one of the elements from the
compound, and becomes part of a new compound.
•  The general form of the equation for a single replacement
reaction is:
A + BX
B + AX
In this equation, A and B represent elements and AX and
BX are compounds.
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Cu + AgNO 3→Ag + Cu(NO3) 2
•  An example of a single replacement reaction is:
Zn(s) + CuSO4(aq)
Cu(s) + ZnSO4(aq)
2 Al + 3 Br2 → 2 AlBr3
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•  A pictorial representation of a double-replacement
reaction:
DOUBLE-REPLACEMENT REACTIONS
•  Double-replacement reactions are never redox reactions.
These reactions often take place between substances
dissolved in water. In typical reactions, two dissolved
compounds react and exchange partners to form two new
compounds.
•  The following general form of the equation for double
replacement reactions shows the partner-swapping
characteristic of the reactions:
AX + BY
BX + AY
IONIC EQUATIONS
•  Many reactions take place between compounds or
elements that are dissolved in water. Ionic compounds
and some polar covalent compounds break apart
(dissociate) when they dissolve in water and form ions.
NaCl + AgNO3→NaNO 3 + AgCl
•  The equations for reactions that occur between dissolved
materials can be written in three ways called molecular
equations, total ionic equations and net ionic
equations.
•  An example of a double-replacement reaction is:
Ba(NO3)2(aq) + Na2S(aq)
BaS(s) +2NaNO3(aq)
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MOLECULAR EQUATIONS
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•  Write the following molecular equation in total ionic and net
ionic forms. Soluble substances are indicated by (aq) after
their formulas and insoluble solids are indicated by (s) after
their formulas.
BaCl2 (aq) + Na2S(aq)
BaS(s) + 2NaCl(aq)
•  In a net ionic equation, only unionized or insoluble
materials and ions that undergo changes as the reaction
proceeds are represented.
•  Any ions that appeared on both the left and right side of
the total ionic equation are called spectator ions and are
not included in the net ionic equation.
TOTAL IONIC EQUATIONS
•  In a total ionic equation, all soluble ionic substances are
represented by the ions they form in solution. Substances
that do not dissolve or that dissolve but do not dissociate
into ions are represented by their formulas.
•  In total ionic form, all substances except the insoluble BaS
will be written in the form of the ions they form:
Ba2+(aq) + 2Cl-(aq)
BaS(s)
+ 2Na+(aq) + S2-(aq)
+ 2Na+(aq) + 2Cl-(aq)
NaCl (aq) + AgNO3 (aq) " AgCl (s) + NaNO3 (aq)
+
Na + (aq) + Ag (aq)
!
+ Cl ! (aq) + NO3 (aq)
NaCl (aq) = Na + (aq) + Cl ! (aq)
Cl ! (aq) + Ag + (aq)
" AgCl (s)
+ Na + (aq)
+ NO3 ! (aq)
" AgCl (s)
•  In net ionic form, all spectator ions are dropped. Both the
Na+ and Cl- ions are spectator ions because they appear
on both sides of the equation. The net ionic equation is:
Ba2+(aq) + S2-(aq)
BaS(s)
Na 2S (aq) = 2 Na + (aq) + S2! (aq)
Na 3PO4 (aq) = 3 Na + (aq) + PO43 ! (aq)
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Coached Problem
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EXAMPLE
NET IONIC EQUATIONS
•  In a molecular equation, each compound is represented by
its formula.
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•  Chemical reactions that release heat as a product are
called exothermic reactions. Ordinary combustion of a
log in a fireplace is an example of an exothermic reaction.
•  While it is a physical process and not chemical, a familiar
example of an endothermic process is the melting of
ordinary ice. As the ice melts, heat is absorbed from the
air surrounding the ice.
ENERGY AND REACTIONS
•  In addition to changes in chemical composition, all
chemical reactions are also accompanied by changes in
energy. That is, all reactions either absorb or give up
energy as they proceed.
•  The energy involved in chemical reactions can take
numerous forms such as the electrical energy released by
the chemical reactions of an ordinary cell phone battery.
Often, all or most of the energy takes the form of heat.
THE MOLE AND CHEMICAL EQUATIONS
•  The mole concept can be applied to balanced chemical
equations and used to calculate mass relationships in
chemical reactions.
•  Balanced equations can be interpreted in terms of the
mole concept and the results used to provide factors for
use in factor-unit solutions to numerical problems.
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Chemistry Interactive
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Again, the four steps of the previously introduced factor-unit
method are:
EXAMPLE
•  Consider the following balanced reaction equation:
2H2S(g) + 3O2(g)
2SO2 (g) + 2H2O(l)
•  Example: Calculate the number of moles of H2S that
would need to react with excess O2 in order to produce
115 g of SO2.
•  Solution: Note that the factor used was obtained from the
two statements given earlier.
• Step 1: Write down the known or given quantity. Include
both the numerical value and units of the quantity.
•  The mole concept can be used to write useful statements
that will be the source of factors needed to solve numerical
problems. The following are two of the possible
statements:
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• Step 2: Leave some working space and set the known
quantity equal to the units of the unknown quantity.
2 mol H2S + 3 mol O2
2 mol SO2 + 2 mol H2O
• Step 3: Multiply the known quantity by one or more factors,
such that the units of the factor cancel the units of the known
quantity and generate the units of the unknown quantity.
68.2 g H2S + 96.0 g O2
128.2 g SO2 + 36.0 g H2O
• Step 4: After you generate the desired units, do the
necessary arithmetic to produce the final answer.
•  Note that the g SO2 units in the denominator of the factor
cancel the g SO2 units of the known quantity, and the mol
H2S units of the numerator of the factor generate the
needed mol H2S units of the answer.
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Coached Problem
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•  The limiting reactant present in a mixture of reactants is
the reactant that will run out first, and thus, it determines
the amount of product that can be produced.
•  A useful approach to solving limiting reactant problems is
to calculate the amount of product that could be produced
by each of the quantities of reactant that are available.
The reactant that gives the least amount of product is then
the limiting reactant.
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•  Calculation for SO2 produced from 55.2 g O2 and excess
H2S:
•  Example: Calculate the maximum amount of SO2 that
could be produced by reacting 55.2 g of O2 with 50.8 g of
H2S.
THE LIMITING REACTANT
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Coached Problem
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•  Solution:
• The amounts of SO2 that could be produced from
55.2 g of O2 reacting with excess H2S as well as from
50.8 g of H2S reacting with excess O2 will be
calculated.
•  Calculation for SO2 produced from 50.8 g H2S and
excess O2:
• The reactant giving the least amount of SO2 will be
the limiting reactant.
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Coached Problem
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Coached Problem
• The amount of SO2 produced by the limiting reactant
is the amount the reaction would produce.
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•  According to the calculations, the 55.2 g of O2 gives the
smallest amount of SO2, so O2 is the limiting reactant and the
reaction will produce 73.7 g of SO2.
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•  Example: Suppose the mixture of reactants calculated
earlier to give 73.7 g SO2 was done in the laboratory and
only 42.7 g of SO2 was collected. What is the percentage
yield of the reaction?
REACTION YIELDS
•  The amount of product calculated in the last three examples
are not the amounts that would be produced if the reactions
were actually done in the laboratory. In each case, less
product would be obtained than was calculated. There are
numerous causes. Some materials are lost during transfers
from one container to another and side reactions take place
that are different from the one that is intended to take place.
•  The amount of product calculated in the examples 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:
•  Solution:
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Coached Problem
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Coached Problem
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