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FE CHEMISTRY REVIEW STEVE DANIEL sdaniel@mines

FE CHEMISTRY REVIEW STEVE DANIEL sdaniel@mines.edu. CHEMICAL PERIODICITY Arrangement of elements in the Periodic Table allows prediction of relative properties of atoms based on the [overly] simplistic ideas: 1. Atoms in the same row have outermost electrons in the same “shell”.

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FE CHEMISTRY REVIEW STEVE DANIEL sdaniel@mines

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  1. FE CHEMISTRY REVIEW STEVE DANIEL sdaniel@mines.edu

  2. CHEMICAL PERIODICITY Arrangement of elements in the Periodic Table allows prediction of relative properties of atoms based on the [overly] simplistic ideas: 1. Atoms in the same row have outermost electrons in the same “shell”. 2. Strength of attraction of the outermost electrons depends on the effective nuclear charge and which “shell” the outermost electrons occupy (nominal distance from the nucleus). 3. Effective nuclear charge (number of protons in the nucleus partially decreased by electronic repulsions) increases left to right in the Periodic table. Relative atomic radii, ionic radii, ionization potentials, electron affinities can be rationalized. For example, arrange in order of: 1. increasing radius: P, N, F 2. increasing radius: Ar, S2-, Ca2+ 3. increasing first ionization potential: Mg, K, S 4. increasing electron affinity: S, Si, Ga

  3. 1. increasing radius: P, N, F [F < N < P] 2. increasing radius: Ar, S2-, Ca2+ [Ca2+< Ar < S2-] 3. increasing first ionization potential: Mg, K, S [K < Mg < S] 4. increasing electron affinity: S, Si, Ga [Ga < Si < S]

  4. EMPIRICAL AND MOLECULAR FORMULAE Atomic weight = mass of one mole of atoms; molecular weight = mass of one mole of molecules. A compound is 40.0% carbon, 53.3% oxygen, and 6.70% hydrogen and has a molecular weight of 60.0. What are its empirical and molecular formulae? 40.0g C x mole C x 16.0g O = 1.00 mole C 53.3g O 12.0gC mole O mole O 6.70g H x mole H x 16.0g O = 1.99 mole H 53.3g O 1.01g H mole O mole O Therefore empirical formula is COH2with formula weight 12.0 + 16.0 + 2.02 = 30.0 or half the molecular weight. So molecular formula is C2O2H4.

  5. RULES IN PRIORITY ORDER FOR OXIDATION NUMBERS 1. S(oxidation numbers) = charge 2. Group IA (Li,Na,K,etc) assign +1 3. Group 2A (Be,Mg,Ca,etc) assign +2 4. B, Al assign +3 5. Hydrogen assign +1 6. Oxygen assign -2

  6. Balance in basic solution: BaO2(s) + Cr3+ → CrO42- + Ba2+ 1. Assign oxidation numbers +2 -1 +3 +6 -2 +2 BaO2(s) + Cr3+ → CrO42- + Ba2+ NOTE +1 -2 +1 -2 -2 +1 2 H2O ↔ H3O+ + OH-

  7. 1. Assign oxidation numbers +2 -1 +3 +6 -2 +2 +1 -2 +1 -2 -2 +1 BaO2 + Cr3+ → CrO42- + Ba2+ and 2H2O ↔ H3O+ + OH- 2. Balance oxidation number changes +2 -1 +3 +6 -2 +2 3BaO2 + 2Cr3+ → 2CrO42- + 3Ba2+ (-2) (+3) 3. Balance charge 3BaO2 + 2Cr3+ → 2CrO42-+ 3Ba2+ net 6+ net 2+ 4OH- + 3BaO2 + 2Cr3+ → 2CrO42- + 3Ba2+ net 2+ net 2+ 4. Balance remaining atoms (O and H) 4OH- + 3BaO2 + 2Cr3+ → 2CrO42- + 3Ba2+ total 10 O atoms total 8 O atoms 4OH- + 3BaO2 + 2Cr3+ → 2CrO42- + 3Ba2+ + 2H2O 5. Check last atoms (H) 4OH- + 3BaO2 + 2Cr3+ → 2CrO42- + 3Ba2+ + 2H2O total 4 H atoms total 4 H atoms

  8. Balance in acidic solution: Cr2O72- + N2H5+ → NO2(g) + Cr3+ 1. Assign oxidation numbers +6 -2 -2 +1 +4 -2 +3 Cr2O72- + N2H5+ → NO2(g) + Cr3+ 2. Balance oxidation number changes +6 -2 -2 +1 +4 -2 +3 2Cr2O72- + N2H5+ → 2 NO2(g) + 4Cr3+ (-6) (+12)  3. Balance charge 2Cr2O72- + N2H5+ → 2 NO2(g) + 4Cr3+ net 3- net 12+ 15H3O++ 2Cr2O72- + N2H5+ → 2 NO2(g) + 4Cr3+ net 12+ net 12+  4. Balance remaining atoms (O and H) 15H3O++ 2Cr2O72- + N2H5+ → 2 NO2(g) + 4Cr3+ total 50 H atoms total 0 H atoms  15H3O++ 2Cr2O72- + N2H5+ → 2 NO2(g) + 4Cr3++ 25H2O  5. Check last atoms (O) 15H3O++ 2Cr2O72- + N2H5+ → 2 NO2(g) + 4Cr3++ 25H2O total 29 O atoms total 29 O atoms

  9. STOICHIOMETRY A solid sample is 15.0% Na2Cr2O7. How many grams of N2H5Cl is needed to react with 5.00 g of this sample. 15H3O++ 2Cr2O72- + N2H5+ → 2 NO2(g) + 4Cr3++ 25H2O 5.00g sample x .150g Na2Cr2O7 x mole Na2Cr2O7 x g sample 262.0g Na2Cr2O7 x mole N2H5Cl x 68.5g N2H5Cl = 0.0980 g N2H5Cl 2 mole Na2Cr2O7 mole N2H5Cl 2. How many liters NO2(g) at 30.0oC and 620.0torr are formed when the 5.00 g of sample is reacted? IDEAL GAS LAW PV = nRT V = nRT/P 5.00g sample x .150g Na2Cr2O7 x mole Na2Cr2O7 x g sample 262.0g Na2Cr2O7   x 2mole NO2 x .08205Latm x 303.15K x 760.0torr = .0873 L 2mole Cr2O7 moleNO2K 620.0torr atm

  10. CONCENTRATION UNITS • Molarity = M = moles solute/L solution •  Molality = m = moles solute/kg solvent •  Normality = N = equivalents solute/L solution •  REDOX • # equivalents/mole = ox.# change per formula •  2Cr2O72- + N2H5+ → 2 NO2(g) + 4Cr3+ • (-6) (+12) • So Na2Cr2O7 has 6 eq/mole and N2H5Cl has 12eq/mole here •  ACID-BASE • # eq/mole = # H+ gained or lost per formula •   NH3 + H3PO3 → NH4+ + HPO32- •  NH3 has 1eq/mole and H3PO3 has 2eq/mole in this reaction • How many mLs 12.0M HCl must be mixed (assuming additive volumes) with 25.0 mL 1.50M HCl to yield 3.00M HCl? • Total volume = V + 0.250 L • Total moles = (.0250L)(1.50mole/L) + V(12.0mole/L) = .0375 +12.0V • (.0375+12.0V)/(V+.0250) = 3.00 mole/L • V = 0.00417 L or 4.17 mls

  11. 4. How many mLs of 3.00N N2H5Cl solution would be required to react with 5.00g of the sample?  5.00g sample x .150g Na2Cr2O7 x mole Na2Cr2O7 x g sample 262.0g Na2Cr2O7   x mole N2H5Cl x 12 eq N2H5Cl x L N2H5Cl x 1000mL = 5.73 mL 2mole Cr2O72- mole N2H5Cl 3.00eq N2H5Cl L 5. Titration of 35.00 mL of a Ba(OH)2 solution requires 27.63 mL of 3.00 M N2H5Cl. What is the molarity of Ba(OH)2 solution? What is its normality? Ba(OH)2 + 2 N2H5Cl → BaCl2 + 2 H2O + 2 N2H4 .02763L N2H5Cl x 3.00 mole N2H5Cl x mole Ba(OH)2 = 1.18 mole Ba(OH)2 = 1.18M .03500LBa(OH)2 L N2H5Cl 2 mole N2H5Cl L 1.18 mole Ba(OH)2 x 2eq Ba(OH)2= 2.36N L Ba(OH)2 mole Ba(OH)2

  12. 6. When 250 g CaCO3 and 300 mL 3.00M H3PO4 are mixed, how many L CO2(g) at 30.0oC and 700torr result? 3CaCO3 + 2H3PO4 → 3CO2(g) + 3H2O + Ca3(PO4)2  250g CaCO3 x mole CaCO3 x 3mole CO2x .08205Latm x 303.15K x 760torr = 100 g CaCO3 3mole CaCO3 moleCO2 K 700torr atm = 67.4 LCO2 .300L H3PO4 x 3.00mole H3PO4 x 3mole CO2 x RT = 36.5L L H3PO4 2mole H3PO4 P

  13. HESS’ LAW 1. Standard heats of formation for C2H5OH(l) is -277.7, for CO(g) is -393.5 and for H2O(l) is -285.8 (all kJ/mole), respectively. Calculate the standard heat of combustion of C2H5OH(l) 2 C(gr) + 2 O2(g) → 2 CO2(g) ∆Ho = 2(-393.5) 3 H2(g) + 3/2 O2(g) →3 H2O(l) ∆Ho = 3(-285.8) C2H5OH(l) → 2 C(gr) + 3 H2(g) +1/2 O2(g)∆Ho = -1(-277.7) C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) ∆Ho = -1366.7 kJ/mole In general: DHrxno = Σ nprod∆Hof,prod - Σnreact∆Hof,react 2.  For the reaction: 2 C2H5OH(l) + O2(g) → 2 CH3CHO(l) + 2 H2O(l) ∆Ho = -348.6 kJ. Calculate the standard heat of combustion of CH3CHO(l). 2 C2H5OH(l) + 6 O2(g) → 4 CO2(g) + 6 H2O(l) ∆Ho = 2(-1366.7) 2 CH3CHO(l) + 2 H2O(l) → 2 C2H5OH(l) + O2(g)∆Ho = -(-348.6) 2 CH3CHO(l) + 5 O2(g) → 4 CO2(g) + 4 H2O(l) ∆Ho = -2384.8 kJ So ∆Hocomb = -2384.8kJ/(2 mole CH3CHO) = -1192.4 kJ/mole

  14. ELECTROCHEMISTRY • Write the anode, cathode, and cell reactions for this • voltaic cell. Eo = +0.34 v. for Cu2+ + 2e-→Cu and +1.54 v. • For MnO4- +5e- + 8H3O+ → Mn2+ + 4H2O . • Since voltaic cell potential must be positive: • 5x(Cu → Cu2+ + 2e- ) anode • 2x(MnO4- + 5e- +8H3O+ → Mn2+ + 4H2O) cathode • 5Cu +2MnO4- + 16H3O+→2Mn2+ + 5Cu2+ + 8H2O cell • Eocell = Eoanode + Eocathode = -(+0.34) + 1.54 = 1.17 v. • NOTE: Potentials are not multiplied by the coefficients used in balancing. • What is the cell potential at the initial concentration given? • E = Eo –(.0592/n)log Q here n = 10, Q = [Mn2+]2[Cu2+]5/[MnO4-]2[H3O+]16 • Cu metal does not have a variable concentration and that of water is assumed to be • nearly constant, so these do not appear in Q. • E = 1.17 –(.0592/10)log(.10)2(.10)5/(.10)2(.10)16 = 1.24 v.

  15. 3.How long could the cell operate at 2.00 amp if the Cu electrode initially weighs 5.00g and the volume of the MnO4- solution is 500.0mL? (5.00g Cu)(mole Cu/63.54g Cu)(2 eq Cu/mole Cu) = 0.157 eq. Cu (.500L)(.10 mole MnO4-/L)(5 eq MnO4-/mole MnO4-) = 0.250 eq MnO4- Therefore Cu is the limiting reactant and 0.157 mole of electrons will flow (.157 eq)(96487 coul/eq)(sec/2.00coul) = 7.57 x 103 sec ACID-BASE EQUILIBRIA 1. Calculate the pH of 0.50 M HF solution. Ka = 7.1 x 10-4 HF + H2O ↔ H3O+ + F- Ka = [H3O+][F-]/[HF] = 7.1 x 10-4 0.50-x x x 7.1 x 10-4 = (x)(x)/(.50-x) x= 1.85 x 10-2 pH = -log[H3O+] = -log(1.85 x 10-2) = 1.73

  16. 2. If 0.20 mole NaF is dissolved in 300.0mL 0.50M HF, what is the solution pH? • HF + H2O ↔ H3O+ + F- • .50-x x (.20/.300)+x • 7.1 x 10-4= [H3O+][ F-]/[HF] = x(.667+x)/(.50-x) • x = [H3O+] = 5.3 X 10-4 pH = 3.28 • Calculate the pH of 0.50 M NaF solution. • F- + H2O ↔ HF + OH- 2 H2O ↔ H3O+ + OH- Kw = 1.0 x 10-14 = [H3O+][OH-] • Kh = [HF][OH-] x [H3O+][OH-] = [HF] x [H3O+][OH-] = 1.0 x 10-14= 1.4 x 10-11 • [F-] [H3O+][OH-] [H3O+][F-] 7.1 x 10-4 • F- + H2O ↔ HF + OH- 1.4 x 10-11 = [HF][OH-] = (x)(x) • .50-x x x [F-] .50-x • x= 2.6 x 10-6 =[OH-] [H3O+] = 1.0 x 10-14/2.6 x 10-6 = 3.8 x 10-9 • pH = -log(3.8 x 10-9) = 8.42

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