CHEMICAL REACTIONS: MAKING CHEMICALS SAFELY WITHOUT DAMAGING THE ENVIRONMENT

CHAPTER 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: C, 12 H, Products: C, 12 H, + 12 = 18 O 12 + = 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) ∆ for heat added (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): MnO2 + C2H6 → Mn + CO + H2O There will have to be at least C atoms and H atoms on the right: MnO2 + C2H6 → Mn + 2CO + 3H2O This gives 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 C atoms and 12 H atoms on the left: 5MnO2 + 2C2H6 → Mn + 4CO + 6H2O The products must have Mn atoms: 5MnO2 + 2C2H6 → 5Mn + 4CO + 6H2O The equation should be balanced, check the results: Reactants: Mn, 10 O, C, 12 H Products: Mn, + = 10 O, 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 Τοταλ µασσ οφ προδυχτ (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 = × (1.0 + 35.5) = 73.0 (4.4.3) Total mass product = × (1.0 + 35.5) + (2 × 23.0 + 32.0 + × 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 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 mole of the element or compound Consider 2C2H6 + 7O2 → 4CO2 + 6H2O (4.8.1) In terms of moles, moles of C2H6 react with moles of O2 to yield moles of CO2 and 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 • moles of C2H6 with a mass of × 30.0 g = 60.0 g of C2H6 • moles of O2 with a mass of × 32.0 g = 224 g of O2 • moles of CO2 with a mass of ì 44.0 g = 176 g of CO2 ã moles of H2O with a mass of × 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 moles C 2H6 react with moles of O2 to produce moles of CO2 and 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 mol mol O mol O C 2 2H H 62mol 62H 6C 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: 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 of O2 of O2 reacting Η µολ 32.0 µο 2 Mass = H × 18.0 × of × = O g 2630.0 Η γ µ µ 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, 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 measu Percent 100 x stoich 28.3 Percen 100 x 30.6 g à2ì à4ì 136 γ 4Χα Mass 4CaSO = 25.02×g111CaCl γ ΧαΧλ µολ4Χ µολ 1ΧαΧλ Μασσ 30.6 4γ ΧαΣΟ 4=ΧαΣΟ The percent yield is calculated by the following (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: Pipet quanti V o lume fo Bur et accurate for tative a spec tr measurem solution volum varying v 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: moles o M = number mass o Moles of molar m mass o M = (molar x(num m (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 → CaCl 2H2O + Ca(OH) mol 2 mol HCl At the end point the number of moles of HCl can be calculated by MolHCl = LitersHCl x MHCl 74.1 mol Ca( 2 Mass = × M × Liters × Ca(OH) HCl H Cl 2Moles H HCl fro to 0.112 74 mo 2 Mass = HCl × 0.042 × × Ca(OH) L H m 0.175 mass Ca(OH Percent = ×100 = Ca(OH) 2×100 sa Titrimetric Analysis of Ca(OH)2 (Cont.) The calculation of the percentage of Ca(OH) in the sample is given by the following: 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 NaHCO and other factors Degree to Which the Solvay Process is “Green” It is green in that 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 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