CHAPTER 4CHEMICAL REACTIONS: MAKING CHEMICALS SAFELY

WITHOUT DAMAGING THE ENVIRONMENT

From Green Chemistry and the Ten Commandments of Sustainability, Stanley E. Manahan, ChemChar Research,

Inc., 2006manahans@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

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.

4.4. YIELD AND ATOM ECONOMY IN CHEMICAL REACTIONS

Percent atom economy = Mass of desired product ×100 (4.4.2) Total mass of product

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 ×100=23.0% (4.4.5) 215

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) 2H2(g) + O2(g) (4.6.3)

Can be 100% atom economical, but may be less than 100% because of side reactions.

Electrolysis

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:Ca + O O + Ca O 2- Ca2+ O 2- Ca2+ (4.7.1)

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

+ -Battery2H2O →O2+4H++4e-

2H2 +2O e-→H2+2OH-

O2H2

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 4 mol CO2 6 mol H2O2 mol C2H6 2 mol C2H6 2 mol C2H6

To solve for the mass of O2 reacting use the following steps: A. Mass of

C2H2 reacting

B. Convert to moles of C2H2

C. Convert to moles of O2

D. Convert to mass 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 C2H6×7 mol O2×32.0 g O2=67.2 g O230.0 g C2H62 mol C2H61 mol O2The 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

(4.10.4)

Mass CaSO4 = 25.0 g CaCl2 ×1 mol CaCl2×1 mol CaSO4×136 g CaSO4111 g CaCl21 mol CaCl21 mol CaSO4Mass CaSO4=30.6 g CaSO4 Percent yield = measured yield x 100stoichiometric yieldPercent yield = 28.3 g x 100 = 92.5%30.6 gThe percent yield is calculated by the following

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 accuratemeasurement ofvarying volumesPipet for quanti-tative transfer ofsolutionVolumetric flask containinga specific, accurately knownvolume

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 solutenumber of liters of solutionMoles of solute = mass of solute, gmolar mass of solute, g/molM = mass of solute(molar mass of solute) x (number of liters of solution)

(4.11.1)

(4.11.2)

(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 + 2H2O1 mol Ca(OH)2

2 mol HClAt 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 = LitersHCl ×MHCl × 1 mol Ca(OH)2 × 74.1 g Ca(OH)22 mol HCl 1 mol Ca(OH)2Moles HCl reacting Converts from moles Gives mass Ca(OH)2HCl to moles Ca(OH)2 from moles Ca(OH)2MassCa(OH)2 = 0.0422 L HCl × 0.112 mol HCl × 1 mol Ca(OH)2 × 74.1 g Ca(OH)21 L HCl 2 mol HCl 1 mol Ca(OH)2PercentCa(OH)2 = mass Ca(OH)2 × 100 = 0.175 g × 100 =13.9%mass sample 1.26 gMassCa(OH)2 = 0.175 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) + NH4ClThe 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.