Chemical Reactions
Natural Approach to Chemistry
Chapter 10
1
Learning Objectives
Sections 10.1-3
Characterize types of
chemical reactions
(redox,acid-base,
synthesis, and single and
double replacement)
Apply mole concept &
conservation of mass to
calculate quantities
Distinguish between
endothermic and
exothermic processes
Sections 10.4
Evaluate costs & benefits of
resources
Discuss technological effects
& impacts on environmental quality
Discuss production and use
of natural resources
Create and interpret
potential energy diagrams
2
Be sure you know these terms:
Parts of a chemical reaction
Synthesis
Decomposition
Single displacement
Double displacement
Polymerization
Precipitate
Polymer, polymerization
Exo- and endothermic
Enthalpy of
reaction/formation
Energy barrier
Photosynthesis
Chemical engineering
Biodegradable
Hazardous substances
Sustainable chemistry
Green chemistry
3
Chemical Reaction Assignments
• 10.1 322/1-7,31,37,38,52
• 10.3 322/8-15,39-41,64,65
• 10.4 3.22/16-29,43-47,66-71
4
A Chemical Equation
• Represents with symbols and formulas, the
identities and relative molecular or molar
amounts of the reactants and products in a
chemical reaction.
5
Word Equation / Formula Equation
Methane + oxygen --> carbon dioxide + water
CH4(g)
+ 02(g)
Reactants
-->
CO2(g)
+ H20(g)
Products
The above formula equation is not balanced.
6
Chemical Reaction Indications
1. Production of energy as heat and/or light.
2. Production of a gas.
3. Formation of a precipitate – a solid produced
as a result of a chemical reaction in solution.
A precipitate separates from the solution.
4. A color change.
7
Chemical Equation Requirements
1. The equation must represent known facts.
2. The formulas for the reactants and products
must be written correctly. DO NOT change
subscripts.
3. The law of conservation of mass must be
satisfied. The number of atoms of each element
must be the same on each side of the yield sign.
After a formula is written correctly, place
coefficients in front of a formula to show
conservation of mass.
8
Diatomic Molecules
Element
Symbol
Molecular Formula
Physical State at
Room Temp.
Hydrogen
H
H2
Gas
Nitrogen
N
N2
Gas
Oxygen
O
O2
Gas
Fluorine
F
F2
Gas
Chlorine
Cl
Cl2
Gas
Bromine
Br
Br2
Liquid
Iodine
I
I2
Solid
When writing a chemical equation including any of the above elements, they are
shown as diatomic molecules as in column 3 above.
9
Coefficients
• When placed in front of a correctly written
chemical formula, a coefficient multiplies the
number of atoms of each element indicated in
the formula
• 2O2 means 4 O
• 2H20 means 4 H and 2 O
10
11
Practice
Write word and balanced chemical equations
for: solid calcium reacts with solid sulfur to
produce solid calcium sulfide. Include symbols
for physical states.
Ca(s) + S(s) --> CaS(s)
Balanced: 1 Ca on each side
1 S on each side
12
Write word and balanced chemical equation for:
Hydrogen gas reacts with fluorine gas to produce
hydrogen fluoride gas.
H2(g) + F2(g) --> HF(g)
H2(g) + F2(g) --> 2HF(g)
Solid aluminum metal reacts with aqueous zinc
chloride to produce solid zinc metal and aqueous
aluminum chloride.
Al(s) + ZnCl2(aq) --> Zn(s) + AlCl3(aq)
Al(s) + 3 ZnCl2(aq) --> Zn(s) + 2 AlCl3(aq)
2 Al(s) + 3 ZnCl2(aq) -->3 Zn(s) + 2 AlCl3(aq)
13
Translate these chemical equations into sentences:
CS2(l) + 3O2(g) --> CO2(g) + 2SO2(g)
Liquid carbon disulfide reacts with oxygen gas to
produce carbon dioxide gas and sulfur dioxide gas.
NaCl(aq) + AgNO3(aq) --> NaNO3(aq) + AgCl(s)
Aqueous solutions of sodium chloride and silver
nitrate react to produce aqueous sodium nitrate
and a precipitate of silver chloride.
14
Write & Balance:
Hydrazine, N2H4, reacts violently with oxygen to
produce gaseous nitrogen and water.
N2H4(l) + O2(g) --> N2(g) + H2O(l)
Is it balanced?
No. There are 4H on reactant side and 2H on
product side and O is 2/1.
N2H4(l) + O2(g) --> N2(g) + 2H2O(l)
N: 2/2
H:4/4
O:2/2
15
Chemical Equation Indications
1. Coefficients indicate relative amounts of
reactants and products (proportions,
molecules, moles, grams, ratios).
H2(g) + Cl2(g) --> 2HCl(g)
1 molecule H2 : 1 molecule Cl2 : 2 molecules HCl
1 mole H2 : 1 mole Cl2 : 2 moles HCl
16
2. Coefficients can be used to determine
relative masses of reactants and products
H2(g) + Cl2(g) --> 2HCl(g)
1mol H2 x 2.02g H2 = 2.02 g H2
mol
1 mol Cl2 x 70.90 g Cl2 = 70.90 g Cl2
mol
2 mol HCl x 36.46 g HCl = 72.92 g HCl
mol
17
• 3. The reverse reaction for a chemical
equation has the same relative amounts of
substances as the forward reactions.
18
Balancing Chemical Equations
1. Write equation.
2. Write correct chemical formulae for each
compound.
3. Balance according to the Law of Conservation of
Mass by adjusting coefficients.
4. Start with the element appearing in the fewest
substances. Balance free elements last.
5. Count atoms to be sure the equation is
balanced.
19
Practice - Write word, formula, and balanced
chemical equations for this reaction.
1. Magnesium and hydrochloric acid react to
produce magnesium chloride and hydrogen.
Word Equation: Magnesium + hydrochloric acid
--> magnesium chloride + hydrogen
Formula Equation: Mg(s) + HCl(aq) --> MgCl2(s) +
H2(g)
Adjust coeffs: Mg(s) + 2HCl(aq) --> MgCl2(s) +
H2(g)
Count atoms: Mg: 1/1 H:2/2 Cl:2/2
20
Solid sodium combines with chlorine gas to produce
solid sodium chloride.
Sodium(s) + chlorine(g) --> sodium chloride
Na(s) + Cl2(g) --> NaCl(s)
Balance: Na(s) + Cl2(g) --> 2NaCl(s)
2Na(s) + Cl2(g) --> 2NaCl(s)
21
Types of Chemical Reactions
1.
2.
3.
4.
5.
Synthesis
A + X --> AX
Decomposition
AX --> A + X
Single-displacement
A + BX --> AX + B
Double-displacement AX + BY --> AY + BX
Combustion – a substance combines with
oxygen releasing energy as light and heat.
6. Acid + Base --> Salt + Water
7. Reduction/oxidation (Redox) – covered in a
later chapter
22
Activity Series – elements organized
according to how they react
Most active metals:
Li react w/cold
Rb H20 & acids
K replacing H2.
Ba React w/O2
Sr forming oxides
Ca
Na
Mg react w/steam
Al (not cold H20)
Mn and acids, replaZn cing H2. React
Cr with O2 forming
Fe oxides
Cd
Co
Ni
Sn
Pb
H2
Sb
Bi
Cu
Hg
Ag
Pt
Au
Do not react w/H20.
React w/acids, replacing H2. React w/O2
forming oxides.
React w/O2, forming
oxides
An element can replace any
element placed below it
BUT
It cannot replace any
element above it.
Zn can replace Cu but Au
cannot replace Mg
Fairly unreactive,
forming oxides only
indirectly.
23
Nonmetal activity series:
Most active
F
Cl
Br
I
24
Sample Problems – activity series
50oC
Zn(s) + H2O(l) --->
No reaction, water must be 100oC (steam) at least.
Sn(s) + O2(g) -->
Yes, any metal more active than Ag will react w/O2
to form an oxide. (Sn is above Ag)
2Sn(s) + O2(g) --> 2SnO
Cd(s) + Pb(NO3)2(aq) -->
Yes, Cd is above Pb. Products: Cd(NO3)2 + Pb
Cu(s) + HCl(aq) -->
No, Cu is below H
25
1. Synthesis Reaction
A + X --> AX
Samples:
Fe(s) + S(s) --> FeS(s)
2Mg(s) + O2(g) --> 2MgO(s)
H20 + SO3 --> H2SO4
26
Synthesis with Oxides (see handout for
more information!!)
CaO(s) + H2O(l) --> Ca(OH)2(s)
Pollution:
SO2(g) + H2O(l) --> H2SO3(aq)
2H2SO3(aq) + H20(l) --> 2H2SO4(aq)
Oxides:
CaO(s) +SO2(g) --> CaSO3(s)
27
2. Decomposition Reaction
AX --> A + X
electricity
H20(l) --------> 2H2(g) + O2(g) (electrolysis)
D
2HgO(s) ---> 2Hg(l) + O2(g)
28
Decomposition of Metal Oxide
D
CaCO3 --->CaO + CO2
Decomp of Metal Hydroxide
D
Ca(OH)2 --->
CaO + H2O
Decomp of Metal Chlorate
D
2KClO3 ---->
MnO 2KCl + 3O2
Decomp of Acids
H2CO3 --> CO2 + H2O (occurs at room temp)
D
H2SO4 ---> SO3 + H2O
2
29
3. Single Displacement Reaction
A + BX --> AX + B
Or Y + BX --> BY + X
1. Fe + CuSO4 --> FeSO4 + Cu
2. Cu + 2AgN03 --> Cu(NO3)2 + 2Ag
3. CI2 + 2KI --> 2KCl + I2
How is 3. different from 1. or 2.?
In 1.& 2. metals are being displaced.
In 3. a halogen is being displaced.
30
Hydrogen displaced by a metal:
Mg + 2HCl --> H2 + MgCl2
31
4. Double-Displacement Reaction
AX + BY --> AY + BX
A,X,B, and Y in reactants are ions.
AY and BX are ionic or molecular compounds.
1. Formation of a Precipitate
2KI(aq) + Pb(NO3)2(aq) --> PbI2(s) + 2KNO3(aq)
32
Formation of a Gas
FeS(s) + 2HCl(aq) –> FeCl2(aq) + H2S(g)
Formation of Water
HCl(aq) + NaOH(aq) --> NaCl(aq) + H20(l)
33
Combustion Reactions
2H2(g) + O2(g) --> 2H2O(g)
C3H8(g) + 5O2(g) --> 3CO2(g) + 4H20(g)
Other products are heat and light.
34
Predicting Activity
(use the activity series handout)
Zn(s) + H20(l) --> ?
No, not hot enough. Steam needed (100oC)
Ca(s) + H2O(l) --> ?
Yes, Ca is above H on the chart.
The products are: Ca(OH)2 + H2 (g)
Pt(s) + O2(g) --> ?
No.
35
Cd(s) + 2HBr(aq) -->
Yes, Cd is above H. Products: CdBr + H2(g)
Mg(s) + steam -->
Yes, Products: Mg(OH)2 + H2(g)
36
Activity Series
Activity of metals:
Li
Rb
K
Ba
Sr
Ca
Na
Mg
Al
Mn
Zn
Cr
Fe
Cd
Co
Ni
Sn
Pb
H2
Sb
Bi
Cu
Hg
Activity of halogen nonmetals
<--Most active metal
react w/cold
H20 & acids
replacing H2.
React w/O2
forming oxides
F2
Cl2
Br2
I2
<-- Most active nonmetal
react w/steam
(not cold H20)
and acids, replacing H2. React
with O2 forming
oxides
Do not react w/H20.
React w/acids, replacing H2. React w/O2
forming oxides.
React w/O2, forming
oxides
Ag Fairly unreactive,
Pt forming oxides only
Au indirectly.
37
Solubility Chart
38
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
Photosynthesis is an endothermic reaction: energy is absorbed
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
Photosynthesis is an endothermic reaction: energy is absorbed
Cellular respiration is an exothermic reaction: energy is released
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
Chapter 9.3
Properties of Solutions
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
Chapter 9.3
Properties of Solutions
Energy is absorbed from
the surroundings so the
pack feels cold
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
Chapter 9.3
Properties of Solutions
Energy is released into
the surroundings so the
pack feels hot
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
Chapter 9.3
Properties of Solutions
Change in enthalpy
enthalpy: the amount of energy that is released or absorbed during a
chemical reaction
Reaction
Energy
Enthalpy change (∆H,
J/mole)
Exothermic
is released
is a negative number
∆H < 0
Endothermic
is absorbed
is a positive number
∆H > 0
Chemical equation for the combustion of carbon:
C(s) + O2(g)
Reactants
CO2(g)
Products
∆H = –393.5 kJ
Energy
thermochemical equation: the equation that gives the
chemical reaction and the energy information of the reaction.
Enthalpy calculations
Chemical equation for the combustion of carbon:
C(s) + O2(g)
CO2(g)
∆H = –393.5 kJ
The reverse chemical reaction involves the same amount of energy, but the
energy flow is reversed (“in” instead of “out”):
CO2(g)
C(s) + O2(g)
∆H = +393.5 kJ
Enthalpy calculations
Chemical equation for the combustion of carbon:
C(s) + O2(g)
1 mole
1 mole
CO2(g)
∆H = –393.5 kJ
1 mole
The combustion of twice as much carbon releases twice as much energy:
2C(s) + 2O2(g)
2 moles
2 moles
2CO2(g)
2 moles
∆H = –787.0 kJ
Enthalpy calculations
Chemical equation for the formation of rust:
2Fe(s) + 3/2O2(g)
2 moles
3/2 moles
Fe2O3(s)
1 mole
∆H = –824.2 kJ
Enthalpy calculations
Chemical equation for the formation of rust:
2Fe(s) + 3/2O2(g)
2 moles
3/2 moles
Fe2O3(s)
∆H = –824.2 kJ
1 mole
Rewrite the chemical equation using coefficients
with the smallest whole numbers possible
Enthalpy calculations
Chemical equation for the formation of rust:
2Fe(s) + 3/2O2(g)
2 moles
3/2 moles
Fe2O3(s)
∆H = –824.2 kJ
1 mole
x2
4Fe(s) + 3O2(g)
4 moles
3 moles
2Fe2O3(s)
∆H = ?
2 moles
What is the enthalpy change for this reaction?
Enthalpy calculations
Chemical equation for the formation of rust:
2Fe(s) + 3/2O2(g)
2 moles
3/2 moles
Fe2O3(s)
∆H = –824.2 kJ
1 mole
x2
x2
4Fe(s) + 3O2(g)
4 moles
3 moles
2Fe2O3(s)
2 moles
∆H = –1,648.4 kJ
Enthalpy of formation
Chemical equation for the combustion of carbon:
C(s) + O2(g)
CO2(g)
∆H = –393.5 kJ
This is also the chemical equation for the formation of CO2.
∆Hreaction = ∆Hformation of CO2 = –393.5 kJ
∆Hf (CO2) = –393.5 kJ/mole
The formation of 1 mole of CO2 releases 393.5 kJ of energy
Enthalpy of formation
Enthalpies of formation of some common substances
Knowing these values and the following equation,
you can calculate unknown enthalpy values:
DHreaction  DHf  products   DHf  reactants 
Enthalpy calculations
The complete combustion of glucose (C6H12O6) releases 2,808 kJ per
mole of glucose. Calculate the enthalpy of formation of glucose.
Asked:
Given:
∆Hf(glucose) = ?
C6H12O6(s) + 6O2(g)
DHf O2 , g   0
6CO2(g) + 6H2O(g)
∆H = –2,808 kJ
kJ
mole
kJ
DHf CO2 , g    393.5
mole
kJ
DHf  H 2O, g    241.8
mole
From the table of
enthalpies of formation
Relationships:
DHreaction  DHf  products   DHf  reactants 
Enthalpy calculations
The complete combustion of glucose (C6H12O6) releases 2,808 kJ
per mole of glucose. Calculate the enthalpy of formation of
glucose.
C6H12O6(s) + 6O2(g)
6CO2(g) + 6H2O(g)
∆H = –2,808 kJ
6CO2(g) + 6H2O(g)
C6H12O6(s) + 6O2(g)
∆H = +2,808 kJ
reactants
products
Formation of glucose
Relationships:
DHreaction  DHf  products   DHf  reactants 
10.4 Chemical Reactions and Energy
Enthalpy calculations
Formation of glucose
6CO2(g) + 6H2O(g)
reactants
C6H12O6(s) + 6O2(g)
∆H = +2,808 kJ
products
DHreaction  DHf  products   DHf  reactants    2,808 kJ
 DHf  glucose   6DHf O2    6DHf CO2   6DHf  H2O     2,808 kJ
 DHf  glucose   6  0 kJ    6  393.5 kJ   6  285.5 kJ     2,808 kJ
 DHf Remember
glucose   to
0 kmultiply
J    2,361
by thekJcoefficients!
   1,713 kJ    2,808 kJ
DHf  glucose    4,074 kJ    2,808 kJ
DHf  glucose   2,808 kJ   4,074 kJ 
DHf  glucose   1,266 kJ
Enthalpy calculations
Formation of glucose
6CO2(g) + 6H2O(g)
reactants
C6H12O6(s) + 6O2(g)
∆H = +2,808 kJ
products
DHreaction  DHf  products   DHf  reactants    2,808 kJ
 DHf  glucose   6DHf O2    6DHf CO2   6DHf  H2O     2,808 kJ
 DHf  glucose   6  0 kJ    6  393.5 kJ   6  241.8 kJ     2,808 kJ
 DHf  glucose   0 kJ    2,361 kJ    1,451 kJ     2,808 kJ
DHf  glucose    3,812 kJ    2,808 kJ
DHf  glucose   2,808 kJ   3,812 kJ 
DHf  glucose   1, 004 kJ
Enthalpy calculations
The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of
glucose. Calculate the enthalpy of formation of glucose.
Formation of glucose
6CO2(g) + 6H2O(g)
C6H12O6(s) + 6O2(g)
1 mole
Asked: ∆Hf(glucose) = ?
Answer: ∆Hf(glucose) = –1,004 kJ/mole
∆H = +2,808 kJ
A
+
B
A
B
∆H = X
A
+
A
B
B
A
A
∆H = X
B
+
B
∆H = –X
The reverse reaction changes the sign of ∆H
A
+
B
A
∆H = X
B
x3
A
A
A
+
B
B
B
A
A
A
B
B
B
∆H = 3X
If three times more substances are involved, ∆H is
three times greater
A
+
B
A
B
∆H = X
∆H(reaction) = ∆Hf (products) – ∆Hf (reactants)
∆H(reaction) = ∆Hf
A
B
– ∆Hf + ∆H
A f
B
Energy profile
Thermochemical equation:
Reactants
also have
stored energy
Energy
have stored
energy
Products
We can graph the
change in energy as
the reaction takes
place
Progress of reaction
∆H = … kJ
Energy flow
during the reaction
Energy profile
Thermochemical equation:
Reactants
have stored
energy
Products
also have
stored energy
∆H = … kJ
Energy flow
during the reaction
Energy profile
Activation energy
The reaction cannot start without
this initial input of energy
Energy profile
Combustion of carbon:
C(s) + O2(g)
CO2(g)
Wood does not spontaneously
light itself up on fire
Energy profile
Reaction of sodium in water:
Na(s) + H2O(l)
2NaOH + H2(g)
Sodium reacts with water
immediately (and violently)
upon contact
Hess’s law
Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction
enthalpies of the various steps into which a reaction can be divided (2).
R
A
∆H1
A
B
∆H2
B
P
∆H3
R
P
∆H4
Hess’s law
Hess’s law:
∆H4 = ∆H1 + ∆H2 + ∆H3
1
2
R
A
∆H1
A
B
∆H2
B
P
∆H3
R
P
∆H4
Hess’s law
Given that the enthalpy of combustion for graphite (Cgr) and
diamond (Cd) are –393.5 kJ/mole and –395.4 kJ/mole, respectively,
calculate the enthalpy of formation of diamond from graphite.
Asked:
Given:
Cgr(s)
Cd(s)
Cgr(s) + O2(g)
Cd(s) + O2(g)
Relationships:
Strategy:
∆H = ?
CO2(g) ∆H = –393.5 kJ/mole
CO2(g) ∆H = –395.4 kJ/mole
Hess’s law
Create a path that leads from Cgr to Cd.
Hess’s law
Asked:
Cgr(s)
Cd(s)
∆H = ?
Given:
Cgr(s) + O2(g)
CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g)
CO2(g) ∆H = –395.4 kJ/mole
Cd(s) is a product in: Cgr(s)
Cd(s)
Cd(s) is a reactant in: Cd(s) + O2(g)
CO2(g) ∆H = –395.4 kJ/mole
Write the reverse reaction so that Cd(s) is a product, and adjust DH:
Cd(s) + O2(g)
CO2(g)
∆H = –395.4 kJ/mole
CO2(g)
Cd(s) + O2(g)
∆H = +395.4 kJ/mole
Hess’s law
Asked:
Given:
Cgr(s)
Cd(s)
Cgr(s) + O2(g)
Cd(s) + O2(g)
∆H = ?
CO2(g) ∆H = –393.5 kJ/mole
CO2(g) ∆H = –395.4 kJ/mole
Write the sum of the two equations:
Cgr(s) + O2(g)
CO2(g)
CO2(g)
Cd(s) + O2(g)
∆H = –393.5 kJ/mole
∆H = +395.4 kJ/mole
Hess’s law
Asked:
Given:
Cgr(s)
Cd(s)
Cgr(s) + O2(g)
Cd(s) + O2(g)
∆H = ?
CO2(g) ∆H = –393.5 kJ/mole
CO2(g) ∆H = –395.4 kJ/mole
Write the sum of the two equations:
Cgr(s) + O2(g)
CO2(g)
CO2(g)
Cd(s) + O2(g)
∆H = –393.5 kJ/mole
∆H = +395.4 kJ/mole
Hess’s law
Asked:
Given:
Cgr(s)
Cd(s)
Cgr(s) + O2(g)
Cd(s) + O2(g)
∆H = ?
CO2(g) ∆H = –393.5 kJ/mole
CO2(g) ∆H = –395.4 kJ/mole
Write the sum of the two equations:
Cgr(s) + O2(g)
CO2(g)
CO2(g)
Cd(s) + O2(g)
∆H = –393.5 kJ/mole
∆H = +395.4 kJ/mole
Hess’s law
Asked:
Cgr(s)
Cd(s)
Given:
Cgr(s) + O2(g)
∆H = ?
CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g)
CO2(g) ∆H = –395.4 kJ/mole
Write the sum of the two equations:
Cgr(s) + O2(g)
CO2(g)
Cgr(s)
CO2(g)
Cd(s) + O2(g)
Cd(s)
∆H = –393.5 kJ/mole
∆H = +395.4 kJ/mole
∆H = (–393.5 + 395.4) kJ/mole
∆H = +1.9 kJ/mole
Energy profile of a reaction
Hess’s law
MAKE SURE THAT YOU KNOW THE
TERMS FROM SLIDE #3!!
80