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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. RULES IN PRIORITY ORDERFOR 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

  5. Balance in basic solution: NH2OH + MnO2(s) → Mn2+ + NO(g) • 1. Assign oxidation numbers -1+1-2+1 +4 -2 +2 +2-2 NH2OH + MnO2(s) → Mn2+ + NO(g) NOTE +1 -2 +1 -2 -2 +1 2 H2O ↔ H3O+ + OH-

  6. 1. Assign oxidation numbers -1+1-2+1 +4 -2 +2 +2-2 NH2OH + MnO2(s) → Mn2+ + NO(g) +1 -2 +1 -2 -2 +1 2 H2O ↔ H3O+ + OH- • 2. Balance oxidation number changes -1+1-2+1 +4 -2 +2 +2-2 2NH2OH + 3MnO2(s) → 3Mn2+ + 2NO(g) (+3) (-2) • 3. Balance charge 2NH2OH + 3MnO2(s) → 3Mn2+ + 2NO(g) net zero net 6+ 2NH2OH + 3MnO2(s) → 3Mn2+ + 2NO(g) + 6OH- net zero net zero 4. Balance remaining atoms (O and H) 2NH2OH + 3MnO2(s) → 3Mn2+ + 2NO(g) + 6OH- total 8 O atoms total 8 O atoms So nothing is needed! 5. Check last atoms (H) 2NH2OH + 3MnO2(s) → 3Mn2+ + 2NO(g) + 6OH- total 6 H atoms total 6 H atoms

  7. 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

  8. 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

  9. 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- NH3has 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

  10. 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 .03500mLBa(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

  11. 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 .08205Latmx 303.15K x 760torr = 100 g CaCO3 3mole CaCO3 moleCO2K 700torr atm = 67.4 LCO2 .300L H3PO4 x 3.00mole H3PO4 x 3mole CO2 x RT = 36.5L L H3PO42mole H3PO4 P

  12. HESS’ LAW Heats of combustion of ethanol (C2H5OH) and acetic anhydride (C4H6O3) are 327.6 and 431.9 kcal/mole, respectively. Calculate the standard enthalpy change for: 2 C2H5OH + 2 O2(g) → C4H6O3 + 3 H2O(l) 2 C2H5OH + 6 O2(g) → 4 CO2(g) + 6 H2O(l)DH1o = 2(-327.6) 4 CO2(g) + 3 H2O → C4H6O3 + 4 O2(g)DH2o = -1(-431.9) DHrxno= DH1o + DH2o = -655.2-(-431.9) = -223.3 kcal/mole rxn If formation enthalpies rather than heats of combustion are given, then in general: DHrxno = Σ nprod∆Hof,prod - Σnreact∆Hof,react 3 H2(g) + 3/2 O2(g) → 3H2O(l)DH = 3 DHf(H2O(l)) 4 C(gr) + 3 H2(g) + 3/2 O2(g) → C4H6O3DH = DHf(C4H6O3) 2 C2H5OH → 4 C(gr) + 6 H2(g) + O2(g) DH = -2DHf(C2H5OH) 2 C2H5OH + 2 O2(g) → C4H6O3 + 3 H2O(l) DH = 3 DHf(H2O(l)) + DHf(C4H6O3) -2DHf(C2H5OH)

  13. ELECTROCHEMISTRY • 1. For the voltaic cell: Zn/Zn2+(.200M)//Fe3+(.500M),Fe2+(.100M)/Pt write anode, cathode, and cell reactions (Eo = + 0.761v and -0.771v for Zn→Zn2+ and Fe2+→Fe3+, respectively). • Anode (Oxidation half-reaction): Zn → Zn2+ + 2e- • Cathode (Reduction half-reaction): Fe3+ + e- → Fe2+ • Cell reaction: Zn + 2 Fe3+ → Zn2+ + 2 Fe2+ • 2. Calculate Eo and E for the cell. • Eocell= +0.761 – (-0.771) = +1.532v. • Ecell= 1.532 - .0591 log[Zn2+][ Fe2+]2= 1.532 –(.0591) log(.100)2(.200)= +1.594 v. • n[Fe3+]2 2 (.500)2 • 3. If each electrode compartment contains 500mL of solution and the Zn electrode weighs 5.00g, how long can the cell operate at an average current of 2.00amp? • .500mole Fe3+ x eq Fe3+ x .500L = 0.250 eq Fe3+ • L solution moleFe3+ • 5.00gZn x mole Zn x 2 eq Zn = 0.153 eq Zn • 65.4gZn moleZn • .153 eq x 96487 coul x second = 7.38 x 103 seconds • eq 2.00 coul

  14. ACID-BASE EQUILIBRIA • 1. Calculate the pH of 0.50M HF solution. Ka = 7.1 x 10-4. • HF + H2O ↔ H3O+ + F- • .50-x x x • Ka= 7.1 x 10-4= [H3O+][ F-]/[HF] = x2/(.50-x) • x = 1.85 x 10-2 pH = -log[H3O+] = 1.73 • 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

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