CHAPTER 4
CHEMICAL REACTIONS: MAKING CHEMICALS SAFELY
WITHOUT DAMAGING THE ENVIRONMENT
From Green Chemistry and the Ten Commandments of
Sustainability, Stanley E. Manahan, ChemChar Research,
Inc., 2006
manahans@missouri.edu
4.1. DESCRIBING WHAT HAPPENS WITH CHEMICAL
EQUATIONS
In our own bodies:
A chemical reaction occurs: Glucose sugar reacts with oxygen
to give carbon dioxide, water, and energy
Represented by a chemical equation:
C6H12O6 + 6O2  6CO2 + 6H2O (+ energy) (4.1.1)
Reactants
Products
Balanced:
Reactants: 6 C, 12 H,
Products: 6 C, 12 H,
6 + 12 = 18 O
12 + 6 = 18 O
Chemical Reactions and Equations
States of matter of reaction participants:
CaCO3(s) + 2HCl(aq) 
CO2(g) + CaCl2(aq) + H2O(l)
(4.1.2)
(s) for solid, (aq) for a substance in solution, (g) for gas, and (l)
for liquid
 denotes a reversible reaction

Example:
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
(4.1.3)
 for heat added
4.2. BALANCING CHEMICAL EQUATIONS
A balanced chemical equation shows the same number of each
kind of atom on both sides of the equation.
Balancing a chemical equation (different example from one
shown in book):
MnO2 + C2H6  Mn + CO + H2O
There will have to be at least 2 C atoms and 6 H atoms on the
right:
MnO2 + C2H6  Mn + 2CO + 3H2O
This gives 5 O atoms on the right, so the number of O atoms on
the left must be a multiple of 5:
5MnO2 + C2H6  Mn + 2CO + 3H2O
This means that there must be 10 O atoms on the right, so
multiply CO and H2O on the right by 2:
5MnO2 + C2H6  Mn + 4CO + 6H2O
Balancing Chemical Equations (Continued)
There must be 4 C atoms and 12 H atoms on the left:
5MnO2 + 2C2H6  Mn + 4CO + 6H2O
The products must have 5 Mn atoms:
5MnO2 + 2C2H6  5Mn + 4CO + 6H2O
The equation should be balanced, check the results:
Reactants: 5 Mn, 10 O, 4 C, 12 H
Products: 5 Mn, 4 + 6 = 10 O, 4 C, 12 H
4.3. JUST BECAUSE YOU CAN WRITE IT DOES NOT
MEAN THAT IT WILL HAPPEN
The following reaction occurs:
Fe(s) + H2SO4(aq)  H2(g) + FeSO4(aq)
(4.3.1)
The following reaction does not occur:
Cu(s) + H2SO4(aq)  H2(g) + CuSO4(aq)
(4.3.2)
Alternative way to make copper sulfate,CuSO4 :
2Cu(s) + O2(g)  2CuO(s)
(4.3.4)
CuO(s) + H2SO4(aq)  CuSO4(aq) + H2O(aq) (4.3.5)
Alternate Reaction Pathways
Alternative reactions pathways for maximum safety, minimum
byproduct, and utilization of readily available materials.
Two ways to prepare iron (II) sulfate, FeSO4.
First method:
Fe(s) + H2SO4(aq)  H2(g) + FeSO4(aq)
(4.3.1)
Could use scrap iron and waste sulfuric acid
Generates elemental H2, which is explosive
But H2 could be used in a fuel cell
Second pathway:
FeO(s) + H2SO4(aq)  FeSO4(aq) + H2O(aq) (4.3.6)
No dangerous H2
Could also use scrap iron and waste sulfuric acid
4.4. YIELD AND ATOM ECONOMY IN CHEMICAL
REACTIONS
Yield is the percentage of the degree to which a chemical reaction
or synthesis goes to completion.
Atom economy is defined as the fraction of reactants that go into
final products.
Consider yield and atom economy for the preparation of HCl gas
By reaction of sodium chloride with sulfuric acid accompanied
by heating to drive off HCl gas:
2NaCl(s) + H2SO4(l)  2HCl(g) + Na2SO4(s) (4.4.1)
When all of the NaCl and H2SO4 react, there is 100% yield.
Byproduct Na2SO4 gives less than 100% atom economy.
Percent atom economy = Mass of desired product  100
Total mass of product
(4.4.2)
Atom Economy (Continued)
Given the atomic masses H 1.0, Cl 35.5, Na 23.0, and O 16.0
gives the following:
Mass of desired product = 2 (1.0 + 35.5) = 73.0
(4.4.3)
Total mass product = 2 (1.0 + 35.5) + (2  23.0 + 32.0 + 4
16.0) = 215
(4.4.4)
Percent atom economy = 73.0
215
 100 = 23.0%
(4.4.5)
Alternatively, the following occurs with 100% atom economy:
H2(g) + Cl2(g) 2HCl(g)
(4.4.2)
4.5. CATALYSTS THAT MAKE REACTIONS GO
Carbon monoxide burns in air:
2CO + O2  2CO2
(4.5.1)
CO is generated by automobile engines and is an undesirable air
pollutant.
CO is eliminated by reaction with oxygen over an automotive
exhaust catalytic converter.
The metals on the surface of the catalytic converter act as a catalyst
to enable the above reaction to occur efficiently.
A catalyst speeds up a chemical reaction without itself being
consumed.
Enzymes as Catalysts
Enzymes are specialized proteins that act as biological catalysts.
For example, aerobic respiration, in which glucose reacts with
oxygen in living organisms using enzyme catalysts and producing
energy.
C6H12O6 + O2  6CO2 + 6H2O + energy
(4.5.2)
Enzymes participate in many life processes including
•Protein synthesis • Repair damaged DNA •Detoxification
Chemical kinetics deals with rates of chemical reactions.
4.6. KINDS OF CHEMICAL REACTIONS
Combination reaction or addition reaction in which two
substances come together to form a new substance
C + O2  CO2
CaO + SO2  CaSO3
Addition reactions are 100% atom economical.
Decomposition reaction in which a compound decomposes to
two or more products.
Example is electrolysis of water to produce elemental hydrogen
and oxygen by passing an electrical current through water made
electrically conducting with a dissolved salt, such as Na2SO4
2H2O(l) Electrolysis2H2(g) + O2(g) (4.6.3)
Can be 100% atom economical, but may be less than 100%
because of side reactions.
Decomposition Reactions (Continued)
Decomposition reaction to make sodium carbonate, Na2CO3,
from sodium bicarbonate, NaHCO3
2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g) (4.6.4)
to produce sodium carbonate, Na2CO3, commonly used as an
industrial chemical to treat water, in cleaning solutions, and as an
ingredient of glass.
Kinds of Chemical Reactions (Continued)
Substitution or replacement reaction is one such as the reaction of
iron and sulfuric acid,
Fe(s) + H2SO4(aq)  H2(g) + FeSO4(aq)
(4.3.1)
This reaction is also evolution of a gas.
A double replacement or metathesis reaction, in which two
compounds trade ions or other groups.
H2SO4(aq) + Ca(OH)2(aq)  CaSO4(s) + 2H2O(l)
(4.6.5)
This is also a neutralization reaction in which an acid and a base
react to produce water and a salt.
Precipitation reactions produce precipitates of insoluble substance
that come out of water solution:
CaCl2(aq) + Na2CO3(aq)  CaCO3(s) + 2NaCl(aq)
(4.6.5)
Calcium removal from water is water softening.
• Calcium can cause scale in pipes
• Calcium precipitates soap in a useless solid form
4.7. OXIDATION-REDUCTION REACTIONS AND
GREEN CHEMISTRY
Oxidation-reduction reactions, frequently called redox reactions
Use of oxidation to describe the reaction of a substance with
oxygen:
(4.7.1)
Ca + O O + Ca
O2- Ca2+ O 2- Ca2+
2Ca + O2  2CaO
Calcium metal is oxidized.
Elemental oxygen is reduced to produce the oxide ion, O2- in CaO.
When a chemical species loses electrons in a chemical reaction it is
oxidized and when a species gains electrons it is reduced.
Whenever a chemical species combines with elemental hydrogen, it
is reduced.
Oxidation-Reduction Reactions (Continued)
FeO + H2  Fe +H2O
(4.7.2)
In this case the Fe in FeO is reduced to iron metal and the hydrogen
in elemental H2 is oxidized to H2O.
When elemental oxygen reacts to produce chemically combined
oxygen, it is acting as an oxidizing agent and is reduced.
Oxidation-reduction in photosynthesis,
6CO2 + 6H2O + h  C6H12O6 + 6O2
(4.7.3)
h represents light energy
Oxidation-reduction in respiration
C6H12O6 + 6O2  6CO2 + 6H2O + energy
(4.1.1)
Oxidation of fossil fuel
CH4 + 2O2  CO2 + 2H2O + energy
(4.7.4)
Electrolysis of Water in Which H2O is Oxidized at One
Electrode and Reduced at the Other
H2
O2
-
+
Battery
2H2O  O2 + 4H+ + 4e-
2H2O + 2e -  H2 + 2OH -
Oxidation-Reduction Reactions in Green Chemistry
Oxidation of fossil fuels and other materials in producing energy
Hydrogen and carbon in hydrocarbons are in reduced form, such
as in ethane, C2H6.
Many raw materials are partially oxidized hydrocarbons, such as
ethanol, C2H6O, which can be made by:
2C2H6 + O2  2C2H6O
(4.7.5)
Alternate biosynthesis of ethanol by fermentation of
carbohydrates:
C6H12O6  2C2H6O + 2CO2
(4.7.6)
4.8. QUANTITATIVE INFORMATION FROM CHEMICAL
REACTIONS
Formula mass: The sum of the atomic masses of all the atoms in
a formula unit of a compound.
Molar mass: Where X is the formula mass, the molar mass is X
grams of an element or compound, that is, the mass in grams of 1
mole of the element or compound.
Consider
2C2H6 + 7O2  4CO2 + 6H2O (4.8.1)
In terms of moles, 2 moles of C2H6 react with 7 moles of O2 to
yield 4 moles of CO2 and 6 moles of H2O.
Given the atomic masses H 1.0, C 12.0, and O 16.0 the molar
mass of C2H6 is 30.0 g/mol, that of O2 32.0 g/mol, that of CO2 is
44.0 g/mol, and that of H2O = 18.0 g/mol.
Quantitative Information from Chemical Reactions (Cont.)
For the reaction: 2C2H6 + 7O2  4CO2 + 6H2O
In terms of the minimum whole number of moles reacting and
produced
• 2 moles of C2H6 with a mass of 2  30.0 g = 60.0 g of C2H6
• 7 moles of O2 with a mass of 7  32.0 g = 224 g of O2
• 4 moles of CO2 with a mass of 4  44.0 g = 176 g of CO2
• 6 moles of H2O with a mass of 6  18.0 g = 108 g of H2O
The total mass of reactants is
60.0 g of C2H6 + 224 g of O2 = 284.0 g of reactants
and the total mass of products is
176 g of CO2 + 108 g of H2O = 284 g of products
4.9. Stoichiometry by the Mole Ratio Method
The calculation of quantities of materials involved in chemical
reactions is addressed by stoichiometry.
Based upon the law of conservation of mass which states that the
total mass of reactants in a chemical reaction equals the total
mass of products.
Holds true because matter is neither created nor destroyed in
chemical reactions.
The mole ratio method of stoichiometric calculations is based
upon the fact that the relative numbers of moles of reactants and
products remain the same regardless of the total quantity of
reaction.
Example of the Mole Ratio Method
2C2H6 + 7O2  4CO2 + 6H2O
(4.9.1)
At the mole level, this chemical equation states that 2 moles C2H6
react with 7 moles of O2 to produce 4 moles of CO2 and 6 moles
of H2O.
For 10 times as much material, 20 moles C2H6 react with 70
moles of O2 to produce 40 moles of CO2 and 60 moles of H2O.
Suppose that it is given that 18.0 g of C2H6 react. What is the
mass of O2 that will react with this amount of C2H6? What mass
of CO2 is produced? What mass of H2O is produced?
To solve this problem, the following mole ratios are used:
7 mol O2
2 mol C 2H6
4 mol CO2 6 mol H 2O
2 mol C 2H6 2 mol C 2H6
To solve for the mass of O2 reacting use the following steps:
A. Mass of B. Convert to
C. Convert
D. Convert
C2H2
moles of C2H2
to moles
to mass
reacting
of O2
of O2
Mole Ratio Calculation (Continued)
Given the molar mass of C2H6 as 30.0 g/mol, the molar mass of
O2 (18.0 g/mol), and the mole ratio relating moles of O2 to moles
of C2H6,
Mass of O2 = 18.0 g C2H6 
1 mol C 2H6
7 mol O2
32.0 g O 2


= 67.2 g O2
30.0 g C 2H6 2 mol C2H6 1 mol O 2
The masses of CO2 and H2O produced are calculated as follows:
Total mass of reactants, 18.0 g C2H6 + 67.2 g O2 = 85.2 g
Total mass of products, 52.8 g CO2 + 32.4 g H2O = 85.2 g
4.10. LIMITING REACTANT AND PERCENT YIELD
One of the reactants is almost always a limiting reactant.
Example: Reaction of 100 g of elemental zinc (atomic mass 65.4)
and 100 g of elemental sulfur (atomic mass 32.0) are mixed and
heated undergoing the following reaction:
Zn + S  ZnS
(4.9.1)
What mass of ZnS, formula mass 97.4 g/mol, is produced?
If 100 g of zinc react completely, the mass of S reacting and the
mass of ZnS produced would be given by the following calculations:
Only 48.9 g of the 100 g of S react, so zinc is the limiting reactant.
The mass of Zn required to react with 100 g of sulfur would be 204
g of Zn, but only 100 g of Zn is available.
Percent Yield
The mass of product calculated from the mass of limiting reactant
in a chemical reaction is called the stoichiometric yield of a
chemical reaction.
By measuring the actual mass of a product produced in a
chemical reaction and comparing it to the mass predicted from
the stoichiometric yield it is possible to calculate the percent
yield.
Suppose that a water solution containing 25.0 g of CaCl2 was
mixed with a solution of sodium sulfate,
CaCl2(aq) + Na2SO4(aq)  CaSO4(s) + 2NaCl(aq) (4.10.2)
Removed by filtration and dried, the precipitate was found to
have a mass of 28.3 g, the measured yield. What was the percent
yield?
Percent Yield (Continued)
The stoichiometric yield of CaSO4 calculated by the mole ratio
method is 30.6 g CaSO4
Mass CaSO 4 = 25.0 g CaCl 2 
1 mol CaCl 2
1 mol CaSO 4 136 g CaSO 4


111 g CaCl 2
1 mol CaCl 2
1 mol CaSO 4
Mass CaSO 4 = 30.6 g CaSO 4
The percent yield is calculated by the following
Percent yield =
Percent yield =
measured yield
x 100
stoichiometric yield
28.3 g x 100 = 92.5%
30.6 g
(4.10.4)
4.11. Titrations: Measuring Moles by Volume of
Solution
If the molar concentration of a solution is known, the number of
moles may be measured by the volume of the solution (see
measuring glassware below:
Buret for accurate
measurement of
varying volumes
Pipet for quantitative transfer of
solution
Volumetric flask containing
a specific, accurately known
volume
Titration
Titration uses a buret to measure the volume of a solution with a
known concentration of a reagent required to react exactly with
another substance in solution
Reagent added from the buret until a measured end point is reached
indicating that the reaction is complete
• Volume used with stoichiometry to measure amount of substance
The pertinent equations relating to solution concentration and
stoichiometry are the following:
M=
moles of solute
number of liters of solution
Moles of solute =
M=
mass of solute, g
molar mass of solute, g/mol
(4.11.1)
(4.11.2)
mass of solute
(molar mass of solute) x (number of liters of solution)
(4.11.3)
Example of Analysis by Titration (Titrimetric Analysis)
Consider a sample consisting of basic lime, Ca(OH)2, molar mass
74.1 g/mol, and dirt with a total sample mass of 1.26 g. Using
titration with a standard acid solution it is possible to determine the
mass of basic Ca(OH)2 in the solution and from that calculate the
percentage of Ca(OH)2 in the sample. Assume that the solid sample
is placed in water and titrated with 0.112 mol/L standard HCl
(concentration designated MHCl), a volume of 42.2 mL (0.0422 L)
of the acid being required to reach the end point. The dirt does not
react with HCl, but the Ca(OH)2 reacts as follows with the mole
ratio given below
Ca(OH)2 + 2HCl  CaCl2 + 2H2O
1 mol Ca(OH)2
2 mol HCl
At the end point the number of moles of HCl can be calculated by
MolHCl = LitersHCl x MHCl
Titrimetric Analysis of Ca(OH)2 (Cont.)
The calculation of the percentage of Ca(OH)2 in the sample is
given by the following:
MassCa(OH) 2 = Liters HCl MHCl  1 mol Ca(OH) 2  74.1 g Ca(OH) 2
2 mol HCl
1 mol Ca(OH) 2
Moles HCl reacting Converts from moles Gives mass Ca(OH)
HCl to moles Ca(OH) 2 from moles Ca(OH)
MassCa(OH) 2 = 0.0422 L HCl 
2
2
0.112 mol HCl 1 mol Ca(OH) 2 74.1 g Ca(OH)2


1 L HCl
2 mol HCl
1 mol Ca(OH) 2
MassCa(OH) 2 = 0.175 g
Percent Ca(OH) 2 = mass Ca(OH) 2  100 = 0.175 g  100 =13.9%
mass sample
1.26 g
4.12. INDUSTRIAL CHEMICAL REACTIONS: THE
SOLVAY PROCESS
The Solvay process consists of saturating a sodium chloride
solution (brine) with ammonia gas (NH3), then with carbon
dioxide, then cooling it to precipitate solid NaHCO3:
NaCl + NH3 + CO2 + H2O  NaHCO3(s) + NH4Cl
The sodium bicarbonate product is heated to produce sodium
carbonate, Na2CO3, a chemical with many industrial uses:
2NaHCO3 + heat  Na2CO3 + H2O(g) + CO2(g)
The CO2 from this reaction is recirculated back to the first
reaction above.
Ammonia is made by the following reaction, which requires heat,
high pressures and a catalyst:
3H2 + N2  2NH3
Ammonia is reclaimed from the reaction solution by adding lime,
Ca(OH)2, made from heating limestone and reacting the CaO
product with water:
The Solvay Process (Continued)
CaCO3 + heat  CaO + CO2 (calcination of limestone)
CaO + H2O  Ca(OH)2
When Ca(OH)2 is added to the spent solution from which
NaHCO3 has precipitated, the ammonia is evolved and reclaimed:
Ca(OH)2 (s) + 2NH4Cl(aq)  2NH3(g) + CaCl2(aq) + 2H2O(l)
Although this reaction reclaims ammonia, it generates large
quantities of calcium, chloride, CaCl2, which has few commercial
uses and tends to accumulate as waste
The overall reaction for the Solvay process is
CaCO3 + 2NaCl  Na2CO3 + CaCl2
from which the stoichiometric atom economy 48.8% (mass of
Na2CO3 product divided by total mass of reactants)
In practice, the yield is less due to incomplete precipitation of
NaHCO3 and other factors
Degree to Which the Solvay Process is “Green”
It is green in that
1. It uses inexpensive, abundantly available raw materials in the
form of NaCl brine and limestone (CaCO3). A significant
amount of NH3 is required to initiate the process with
relatively small quantities to keep it going.
2. It maximizes recycle of two major reactants, ammonia and
carbon dioxide. The calcination of limestone provides ample
carbon dioxide to make up for inevitable losses from the
process, but some additional ammonia has to be added to
compensate for any leakage.
The Solvay process is not green because it requires extraction of
non-renewable NaCl and CaCO3 (although they are abundant),
generates excess, potentially waste CaCl2, uses relatively large
amounts of energy, has a relatively low atom economy
In the U.S. and some other countries Na2CO3.NaHCO3.2H2O
(trona) is mined and the Solvay process is not used.