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 [email protected]

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 C 6 H 12 O 6 + 6O

Reactants

2

chemical equation

:  6CO 2 + 6H

Products

2 O (+ energy) (4.1.1)

Balanced

:

Reactants

: 6 C, 12 H, 6 + 12 = 18 O

Products:

6 C, 12 H, 12 + 6 = 18 O

Chemical Reactions and Equations

States of matter of reaction participants: CaCO 3 (

s

) + 2HCl(

aq

)  CO 2 (

g

) + CaCl 2 (

aq

) + H 2 O(

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: NH 3 (

aq

) + H 2 O(

l

)    for heat added NH 4 + (

aq

) + OH (

aq

) (4.1.3)

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): MnO right: 2 + C 2 H 6  Mn + CO + H 2 O There will have to be at least 2 C atoms and 6 H atoms on the MnO 2 + C 2 H 6  Mn + 2CO + 3H the left must be a multiple of 5: 2 O This gives 5 O atoms on the right, so the number of O atoms on 5MnO This means that there must be 10 O atoms on the right, so multiply CO and H 5MnO 2 2 + C + C 2 2 H H 6 6  2  Mn + 2CO + 3H O on the right by 2: Mn + 4CO + 6H 2 2 O O

Balancing Chemical Equations (Continued)

There must be 4 C atoms and 12 H atoms on the left: 5MnO 2 + 2C 2 H 6  Mn + 4CO + 6H 2 O The products must have 5 Mn atoms: 5MnO 2 + 2C 2 H 6  5Mn + 4CO + 6H 2 O 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

) + H 2 SO 4 (

aq

)  H 2 (

g

) + FeSO 4 (

aq

)

The following reaction does not occur:

(4.3.1) Cu(

s

) + H 2 SO 4 (

aq

)  H 2 (

g

) + CuSO 4 (

aq

) (4.3.2)

Alternative way to make copper sulfate,CuSO 4

2Cu(

s

) + O 2 (

g

)  2CuO(

s

) CuO(

s

) + H 2 SO 4 (

aq

)  CuSO 4 (

aq

) + H 2 O(

aq

:

(4.3.4) ) (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, FeSO 4 .

First method: Fe(

s

) + H 2 SO 4 (

aq

)  H 2 (

g

) + FeSO 4 (

aq

) Could use scrap iron and waste sulfuric acid (4.3.1) Generates elemental H 2 , which is explosive But H 2 could be used in a fuel cell Second pathway: FeO(

s

) + H 2 SO 4 (

aq

)  FeSO 4 (

aq

) + H 2 O(

aq

) (4.3.6) No dangerous H 2 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

final products.

is defined as the fraction of reactants that go into 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

) + H 2 SO 4 (

l

)  2HCl(

g

) + Na 2 SO 4 (

s

) (4.4.1) When all of the NaCl and H 2 SO 4 react, there is 100% yield.

Byproduct Na 2 SO 4 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   16.0) = 215 23.0 + 32.0 + 4 (4.4.4) Percent atom economy = 73.0  100 = 23.0% 215 (4.4.5) Alternatively, the following occurs with 100% atom economy: H 2 (

g

) + Cl 2 (

g

)  2HCl(

g

) (4.4.2)

4.5. CATALYSTS THAT MAKE REACTIONS GO

Carbon monoxide burns in air: 2CO + O 2  2CO 2 (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, energy.

aerobic respiration

, in which glucose reacts with oxygen in living organisms using enzyme catalysts and producing C 6 H 12 O 6 + O 2  6CO 2 + 6H 2 O + 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 + O 2  CO 2 CaO + SO 2  CaSO 3 Addition reactions are 100% atom economical.

Decomposition reaction

two or more products.

in which a compound decomposes to 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 Na 2 SO 4 2H 2 O(

l

)

Electrolysis

2H 2 (

g

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

) + O 2 (

g

) (4.6.3)

Decomposition Reactions (Continued)

Decomposition reaction to make sodium carbonate, Na 2 CO 3 , from sodium bicarbonate, NaHCO 3 2NaHCO 3 (

s

)  Na 2 CO 3 (

s

) + CO 2 (

g

) + H 2 O(

g

) (4.6.4) to produce sodium carbonate, Na ingredient of glass.

2 CO 3 , commonly used as an industrial chemical to treat water, in cleaning solutions, and as an

Kinds of Chemical Reactions (Continued)

Substitution

or

replacement

iron and sulfuric acid, reaction is one such as the reaction of Fe(

s

) + H 2 SO 4 (

aq

)  This reaction is also H 2 (

g

) + FeSO 4 (

aq

evolution of a gas

.

) (4.3.1) A

double replacement

or

metathesis

compounds trade ions or other groups.

reaction, in which two H 2 SO 4 (

aq

) + Ca(OH) This is also a 2 (

aq

)  CaSO 4 (

s

) + 2H 2 O(

l

)

neutralization reaction

react to produce water and a salt.

(4.6.5) in which an acid and a base

Precipitation

reactions produce

precipitates

that come out of water solution: of insoluble substance CaCl 2 (

aq

) + Na 2 CO 3 (

aq

)  CaCO 3 (

s

) + 2NaCl(

aq

) Calcium removal from water is

water softening

.

(4.6.5) • 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 - Ca

2+

O

2

- Ca

2+ (4.7.1) 2Ca + O 2  2CaO Calcium metal is

oxidized

.

Elemental oxygen is

reduced

to produce the oxide ion, O 2 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 + H 2  Fe +H 2 O (4.7.2) In this case the Fe in FeO is reduced to iron metal and the hydrogen in elemental H 2 is oxidized to H oxygen, it is acting as an 2 O.

When elemental oxygen reacts to produce chemically combined

oxidizing agent

and is

reduced

.

Oxidation-reduction in photosynthesis, 6CO 2

h

 + 6H 2 O +

h

  represents light energy C 6 H 12 O 6 + 6O 2 (4.7.3) Oxidation-reduction in respiration C 6 H 12 O 6 + 6O 2  Oxidation of fossil fuel 6CO 2 CH 4 + 2O 2  CO 2 + 2H 2 + 6H 2 O + energy O + energy (4.1.1) (4.7.4)

Electrolysis of Water in Which H

2

O is Oxidized at One Electrode and Reduced at the Other

H 2 O 2 + Battery 2H 2 O



O 2 + 4H + + 4e 2H 2 O + 2e-



H 2 + 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, C ethanol, C 2 H 6 2 H 6 .

Many raw materials are partially oxidized hydrocarbons, such as O, which can be made by: 2C 2 H 6 + O 2  2C 2 H 6 O (4.7.5) Alternate biosynthesis of ethanol by fermentation of carbohydrates: C 6 H 12 O 6  2C 2 H 6 O + 2CO 2 (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 2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O (4.8.1) In terms of moles, 2 moles of C yield 4 moles of CO 2 2 H 6 react with 7 moles of O 2 mass of C 2 H 6 and 6 moles of H is 30.0 g/mol, that of O 2 2 O.

Given the atomic masses H 1.0, C 12.0, and O 16.0 the molar 32.0 g/mol, that of CO 2 44.0 g/mol, and that of H 2 O = 18.0 g/mol.

to is

Quantitative Information from Chemical Reactions (Cont.)

For the reaction: 2C produced 2 H 6 + 7O 2  4CO 2 + 6H 2 O In terms of the minimum whole number of moles reacting and • • • • 2 moles of C 2 H 6 7 moles of O 2 with a mass of 2 with a mass of 7   30.0 g = 60.0 g of C 32.0 g = 224 g of O 2 2 H 6 4 moles of CO 2 with a mass of 4  44.0 g = 176 g of CO 2 6 moles of H 2 O with a mass of 6  18.0 g = 108 g of H 2 O The total mass of reactants is 60.0 g of C 2 H 6 + 224 g of O 2 and the total mass of products is = 284.0 g of reactants 176 g of CO 2 + 108 g of H 2 O = 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

2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O (4.9.1) At the mole level, this chemical equation states that 2 moles C 2 H 6 react with 7 moles of O 2 to produce 4 moles of CO 2 of H 2 O.

For 10 times as much material, 20 moles C 2 H 6 and 6 moles react with 70 moles of O 2 to produce 40 moles of CO 2 and 60 moles of H 2 O.

Suppose that it is given that 18.0 g of C 2 H 6 mass of O of CO 2 2 A. Mass of C 2 H 2 reacting B. Convert to moles of C 2 H 2 react. What is the that will react with this amount of C 2 H 6 ? What mass is produced? What mass of H 7 mol O 2 mol C 2 2 H 6 4 mol CO 2 mol C 2 H 2 6 To solve for the mass of O 2 6 mol H 2 mol C 2 2 O H 2 6 O is produced?

To solve this problem, the following mole ratios are used: reacting use the following steps: C. Convert to moles of O 2 D. Convert to mass of O 2

Mole Ratio Calculation (Continued)

Given the molar mass of C 2 H 6 O 2 (18.0 g/mol), and the mole ratio relating moles of O 2 of C 2 H 6 , as 30.0 g/mol, the molar mass of to moles Mass of O 2 = 18.0 g C 2 H 6  1 mol C 2 H 6 30.0 g C 2 H 6  7 mol O 2 mol C 2 2 H  6 32.0 g O 1 mol O 2 2 = 67.2 g O 2 The masses of CO 2 and H 2 O produced are calculated as follows: Total mass of reactants, 18.0 g C 2 H 6 Total mass of products, 52.8 g CO 2 + 67.2 g O 2 = 85.2 g + 32.4 g H 2 O = 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 CaCl 2 mixed with a solution of sodium sulfate, was CaCl 2 (

aq

) + Na 2 SO 4 (

aq

) have a mass of 28.3 g, the  CaSO 4 (

s

) + 2NaCl( Removed by filtration and dried, the precipitate was found to

measured yield

aq

) (4.10.2) . What was the percent yield?

Percent Yield (Continued)

The stoichiometric yield of CaSO 4 method is 30.6 g CaSO 4 calculated by the mole ratio Mass CaSO 4 = 25.0 g CaCl 2  1 mol CaCl 111 g CaCl 2  1 mol CaSO 2 4  136 g CaSO 4 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 = measured yield x 100 stoichiometric yield Percent 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 quanti tative 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

(4.11.1)

Moles of solute = mass of solute, g molar mass of solute, g/mol

(4.11.2)

M = 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 percentage of Ca(OH) 2 in the solution and from that calculate the in the sample. Assume that the solid sample is placed in water and titrated with 0.112 mol/L standard HCl (concentration designated M HCl ), 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 ratio given below reacts as follows with the mole Ca(OH) 2 + 2HCl  CaCl 2 + 2H 2 O 1 mol Ca(OH) 2 2 mol HCl At the end point the number of moles of HCl can be calculated by Mol HCl = Liters HCl x M HCl

Titrimetric Analysis of Ca(OH)

2

(Cont.)

The calculation of the percentage of Ca(OH) 2 given by the following: in the sample is

Mass Ca(OH) 2 = Liters HCl  M HCl  1 mol Ca(OH) 2  74.1 g Ca(OH) 2 2 mol HCl 1 mol Ca(OH) 2 Moles HCl reacting Con verts from moles Gives mass Ca(OH) HCl to mo les Ca(OH) 2 fro m mo les Ca(OH) 2 2 Mass Ca(OH) 2 = 0.0422 L HCl  0.112 mol HCl  1 mol Ca(OH) 1 L HCl 2 mol HCl 2  74.1 g Ca(OH) 2 1 mol Ca(OH) Mass Ca(OH) 2 = 0.175 g 2 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 (NH 3 ), then with carbon dioxide, then cooling it to precipitate solid NaHCO 3 : NaCl + NH carbonate, Na 2NaHCO 3 3 2 + CO CO 3 2 + heat  + H 2 O  The sodium bicarbonate product is heated to produce sodium , a chemical with many industrial uses: Na 2 CO 3 NaHCO + H 2 O(

g

3 (

s

) + NH ) + CO 2 (

g

) 4 Cl The CO 2 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: 3H 2 + N Ammonia is reclaimed from the reaction solution by adding lime, Ca(OH) 2 2  2NH product with water: 3 , made from heating limestone and reacting the CaO

The Solvay Process (Continued)

CaCO 3 + heat CaO + H 2 O   CaO + CO Ca(OH) 2 2 (calcination of limestone) When Ca(OH) 2 is added to the spent solution from which NaHCO 3 has precipitated, the ammonia is evolved and reclaimed: Ca(OH) 2 (

s

) + 2NH 4 Cl(

aq

)  2NH 3 (

g

) + CaCl 2 (

aq

) + 2H 2 O(

l

) Although this reaction reclaims ammonia, it generates large quantities of calcium, chloride, CaCl 2 , which has few commercial uses and tends to accumulate as waste The overall reaction for the Solvay process is CaCO Na 2 3 CO 3 + 2NaCl  Na 2 CO 3 + CaCl 2 from which the stoichiometric atom economy 48.8% (mass of product divided by total mass of reactants) In practice, the yield is less due to incomplete precipitation of NaHCO 3 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 (CaCO 3 ). A significant amount of NH 3 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 CaCO 3 (although they are abundant), generates excess, potentially waste CaCl 2 , uses relatively large amounts of energy, has a relatively low atom economy In the U.S. and some other countries Na 2 CO 3 .

NaHCO 3 .

2H 2 O (trona) is mined and the Solvay process is not used.