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Yellow Stone National Park, Jon Sullivan, June, 2003

Chapter 4. The Major Classes of Reactions. (c) 문화재청. Yellow Stone National Park, Jon Sullivan, June, 2003. 1. 수용액의 일반적 성질 - 전해질과 비전해질 2. 침전반응 - 이온 방정식 , 용해도 규칙 3. 산 - 염기 반응 4. 산화 - 환원 반응 - 산화수 5. 용액의 농도 6. 용액의 화학양론 - 적정. 수용액에서의 반응.

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Yellow Stone National Park, Jon Sullivan, June, 2003

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  1. Chapter 4 The Major Classes of Reactions (c) 문화재청 Yellow Stone National Park, Jon Sullivan, June, 2003

  2. 1. 수용액의 일반적 성질 - 전해질과 비전해질 2. 침전반응 - 이온 방정식, 용해도 규칙 3. 산-염기 반응 4. 산화-환원 반응 - 산화수 5. 용액의 농도 6. 용액의 화학양론 - 적정 수용액에서의 반응

  3. The Major Classes of Chemical Reactions 4.1 The Role of Water as a Solvent 4.2 Writing Equations for Aqueous Ionic Reactions 4.3Precipitation Reactions 4.4 Acid-Base Reactions 4.5 Oxidation-Reduction (Redox) Reactions 4.6 Elements in Redox Reactions

  4. Charge distribution in H2and H2O and Molecular Dipole

  5. Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. 4.1

  6. Hydration of Ions (i) Hydration represents for the dissolution of a substance in water to get adsorb water molecule. Hydration of ions is the exothermic process. M(g)+ + Aq→ M+(aq); ΔH = –Hydration Energy. (ii) Smaller the cation, greater is the degree of hydration. Hydration energy is in the order of, Li+ > Na+ > K+ > Rb+ > Cs+ (iii) Li+ being smallest in size has maximum degree of hydration, moves very slowly under the influence of electric field and, therefore, is the poorest conductor current among alkali metals ions.          Relative ionic radii    Cs+ > Rb+ > K+ > Na+ > Li+          Relative hydrated ionic radii  Li+ > Na+ > K+ > Rb+ > Cs+          Relative conducting power Cs+ > Rb+ > K+ > Na + > Li+

  7. An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. nonelectrolyte solution electrolyte solution sugar solution NaCl solution

  8. Determining Moles of Ions in Aqueous Ionic Solutions How many moles of each ion are in the following solutions? (a) 5.0 mol of ammonium sulfate dissolved in water H2O (b)CsBr(s) →Cs+(aq) + Br-(aq) (b) 78.5 g of cesium bromide dissolved in water SOLUTION: nSO42-= 5.0 mol×1/1 = 5.0 mol nNH4+= 5.0 mol×2/1 = 10. mol H2O (a) (NH4)2SO4(s) →2NH4+(aq) + SO42-(aq) = nBr- • nCsBr = 78.5 g CsBr /(212.8 g /mol) = 0.369 mol = nCs+

  9. Determining Moles of Ions in Aqueous Ionic Solutions How many moles of each ion are in the following solutions? (c)7.42×1022formula units of copper(II) nitrate dissolved in water (d) 35 mL of 0.84 M zinc chloride SOLUTION: H2O (c) Cu(NO3)2(s) → Cu2+(aq) + 2NO3-(aq) nCu(NO3)2=7.42×1022 formula units /(6.022×1023/mol)= 0.123 mol nNO3-= 2×nCu(NO3)2 = 0.246 mol nCu2+ = nCu(NO3)2 = 0.123 mol H2O nZnCl2= 35 mL× (L/103mL) ×(0.84 mol/L)= 2.9×10-2 mol (d) ZnCl2(aq) →Zn2+(aq) + 2Cl-(aq) nZn2+ = nZnCl2 = 2.9×10-2 mol nCl- = 2×nZnCl2 = 5.8×10-2 mol

  10. Determining the Molarity of H+ Ions in Aqueous Solutions of Acids Nitric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H+ ion. What is the molarity of H+(aq) in 1.4M nitric acid? SOLUTION: One mole of H+(aq) is released per mole of nitric acid (HNO3) H2O HNO3(l) →H+(aq) + NO3-(aq) nH+= nHNO3 , so MH+= MHNO3 = 1.4M

  11. Writing Equations for Aqueous Ionic Reactions The molecular equation shows all of the reactants and products as intact, undissociatedcompounds. The total ionic equation shows all of the soluble ionic substances dissociated into ions. The net ionic equation eliminates the spectator ions and shows the actual chemical change taking place.

  12. precipitate PbI2 Precipitation Reactions The reaction of Pb(NO3)2 and NaI. Precipitate – insoluble solid that separates from solution 2NaI(aq) + Pb(NO3)2 (aq) →PbI2(s) + 2NaNO3(aq) molecular equation 2Na+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq)→PbI2(s) + 2Na+(aq) + 2NO3-(aq) ionic equation Pb2+(aq) + 2I-(aq) → PbI2(s) Na+(aq)and NO3-(aq) spectatorions net ionic equation 2NaI(aq) + Pb(NO3)2(aq)→PbI2(s) + 2NaNO3(aq) double displacement reaction (metathesis) 4.2

  13. Predicting Whether a Precipitate Will Form 1. Note the ions present in the reactants. 2. Consider the possible cation-anion combinations. 3. Decide whether any of the ion combinations is insoluble. solubility rules is very useful tool.

  14. Predicting Whether a Precipitation Reaction Occurs; Writing Ionic Equations Predict whether a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions. (a) sodium sulfate(aq) + strontium nitrate(aq) (b) ammonium perchlorate(aq) + sodium bromide(aq) SOLUTION: (a) Na2SO4(aq) + Sr(NO3)2 (aq) →2NaNO3(aq) + SrSO4(s) 2Na+(aq) +SO42-(aq)+ Sr2+(aq)+2NO3-(aq) → 2Na+(aq) +2NO3-(aq)+ SrSO4(s) Sr2+(aq) + SO42-(aq) →SrSO4(s) (b) NH4ClO4(aq) + NaBr(aq) →NH4Br(aq) + NaClO4(aq) All reactants and products are soluble so no reaction occurs.

  15. Common Acids and Bases Acids Bases Strong Strong sodium hydroxide, NaOH hydroiodic acid, HI hydrobromic acid, HBr potassium hydroxide, KOH perchloric acid, HClO4 calcium hydroxide, Ca(OH)2 strontium hydroxide, Sr(OH)2 hydrochloric acid, HCl sulfuric acid, H2SO4 barium hydroxide, Ba(OH)2 nitric acid, HNO3 Weak Weak hydrofluoric acid, HF ammonia, NH3 phosphoric acid, H3PO4 acetic acid, CH3COOH salicylic acid, C6H4(OH)COOH ascorbic acid(Vitamin C), C6H8O6

  16. Acid-Base Reactions When HCl gas dissolves in water, HCl(g) + H2O(l) → H3O+(aq) + Cl–(aq) H+ transfer H+ transfer H+ transfer HCl(aq) + NaOH(aq), [H3O+(aq) + Cl–(aq)] + [Na+ (aq) + OH–(aq)] → H2O(l) + Cl–(aq) + Na+(aq) + HOH H3O+(aq) + OH–(aq)] → H2O(l) Johannes Brønsted& Thomas Lowry : acid: proton donor, base: acceptor

  17. Writing Ionic Equations for Acid-Base Reactions Write balanced molecular, total ionic, and net ionic equations for each of the following acid-base reactions and identify the spectator ions. (a) strontium hydroxide(aq) + perchloric acid(aq) → (b) barium hydroxide(aq) + sulfuric acid(aq) → SOLUTION: (a)Sr(OH)2(aq)+2HClO4(aq) →2H2O(l)+Sr(ClO4)2(aq) Sr2+(aq) + 2OH-(aq)+ 2H+(aq) + 2ClO4-(aq) 2H2O(l)+Sr2+(aq)+2ClO4-(aq) 2OH-(aq)+ 2H+(aq) → 2H2O(l) (b) Ba(OH)2(aq) + H2SO4(aq) →2H2O(l) + BaSO4(aq) Ba2+(aq) + 2OH-(aq) + 2H+(aq) + SO42-(aq) 2H2O(l) + Ba2+(aq) + SO42-(aq) 2OH-(aq)+ 2H+(aq) →2H2O(l)

  18. An acid-base titration In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. H+(aq) + OH-(aq) → H2O(l) Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point End point – the point at which the indicator changes

  19. Finding the Concentration of Acid from an Acid-Base Titration You perform an acid-base titration to standardize an HCl solution by placing 50.00 mL of HCl in a flask with a few drops of indicator solution. You put 0.1524 M NaOH into the buret, and the initial reading is 0.55 mL. At the end point, the buret reading is 33.87 mL. What is the concentration of the HCl solution? SOLUTION: NaOH(aq) + HCl(aq) →NaCl(aq) + H2O(l) (33.87-0.55) mL×(L/103 mL) = 0.03332 L nNaOH 0.03332 L × 0.1524 M= 5.078×10-3 mol • At the equivalent point, nNaOH = nHCl MHCl= nHCl/VHCl 5.078×10-3 molHCl/0.05000L = 0.1016 M HCl

  20. Oxidation number(산화수) The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred to more electronegative atoms. • Free elements have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4, ... = 0 • In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 • The oxidation number of oxygen isusually–2. In H2O2 and O22- it is –1.

  21. O = -2 H = +1 4×(-2) + 1 + ? = -1 S = +6 • The oxidation number of hydrogen is +1except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. • Group IA metals are +1, IIA metals are +2 and fluorine is always –1. HSO4- Oxidation numbers of all the elements in HSO4-?

  22. Determining the Oxidation Number of an Element Determine the oxidation number (O.N.) of each element in these compounds: (a) zinc chloride (b) sulfur trioxide (c) nitric acid SOLUTION: (a) ZnCl2. The O.N. for zinc is +2 and that for chloride is -1. (b) SO3. Each oxygen is an oxide with an O.N. of -2. Therefore the O.N. of sulfur must be +6. (c) HNO3. H has an O.N. of +1 and each oxygen is -2. Therefore the N must have an O.N. of +5.

  23. oxidation states of the main group elements

  24. Oxidation-Reduction (redox) reactions • M + X → M+ + e- + X → M+ + X- → MX M X gains electron(s) X is reduced X is the oxidizing agent X decreases its oxidation number X • M loses electron(s) • M is oxidized • M is the reducing agent • M increases its oxidation number

  25. Recognizing Oxidizing and Reducing Agents Identify the oxidizing agent and reducing agent in each of the following: +2 -2 +2 -2 0 +4 -2 (a) 2Al(s) + 3H2SO4(aq) →Al2(SO4)3(aq) + 3H2(g) (b)PbO(s) + CO(g) →Pb(s) + CO2(g) (c) 2H2(g) + O2(g) →2H2O(g) SOLUTION: 0 +1 +6 -2 +3 +6 -2 0 (a) 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g) 0 0 +1 -2 The O.N. of Al increases; it is oxidized; it is the reducing agent. The O.N. of H decreases; it is reduced; H2SO4 is the oxidizing agent. (b)PbO(s) + CO(g) →Pb(s) + CO2(g) (c) 2H2(g) + O2(g) →2H2O(g)

  26. An active metal displacing hydrogen from water: Large Sodium Explosion http://www.youtube.com/watch?v=sNdijknRxfU&feature=related

  27. Displacing one metal with another. Fe(s) + CuSO4(aq) → Fe(SO4)(aq) + Cu(s) Shiny White + Blue→Green Solution+ Reddish Brown

  28. Li K Ba Ca Na can displace H from water strength as reducing agents H2 The activity series of the metals. Reduction 0 +1 → +x 0 M + H2O → M(OH)x + H2 • Mg • Al • Mn • An • Cr • Fe • Cd Co Ni Sn Pb Oxidation can displace H from steam can displace H from acid • Cu • Hg • Ag • Au cannot displace H from any source

  29. +1 0 0 +2 +4 0 0 +2 -1 0 -1 0 Types of Oxidation-Reduction Reactions Combination : A + B → C 2Al + 3Br2→ 2AlBr3 Decomposition : C → A + B 2KClO3→ 2KCl + 3O2 -1 0 0 +3 0 +4 -2 0 Combustion : A + O2→ B 2Mg + O2→ 2MgO S + O2→ SO2 Displacement : A + BC → AC + B Sr + 2H2O →Sr(OH)2 + H2 H Displacement -2 +2 0 0 +1 -2 +1 0 +5 -1 TiCl4 + 2Mg → Ti + 2MgCl2 Metal Displacement Halogen Displacement Cl2 + 2KBr → 2KCl + Br2

  30. Type of Redox Reaction Classify each of the following redox reactions as a combination, decomposition, or displacement reaction, write a balanced molecular equation for each, as well as total and net ionic equations for part (c), and identify the oxidizing and reducing agents: (a) magnesium(s) + nitrogen(g) →magnesium nitride (aq) (b) hydrogen peroxide(l) → water(l) + oxygen gas (c) aluminum(s) + lead(II) nitrate(aq) → aluminum nitrate(aq) + lead(s)

  31. 0 0 +2 -3 (a) Combination Mg(s) + N2(g) →Mg3N2 (aq) 3 Mg is the reducing agent; N2 is the oxidizing agent. +1 -1 +1 -2 0 (b) Decomposition 2 H2O2(l) →2 H2O(l) + O2(g) H2O2 is the oxidizing and reducing agent. (c) Displacement 0 +2 +5 -2 +3 +5 -2 0 Al(s) + Pb(NO3)2(aq) →Al(NO3)3(aq) + Pb(s) 2Al(s) + 3Pb(NO3)2(aq) →2Al(NO3)3(aq) + 3Pb(s) Pb(NO3)2 is the oxidizing and Al is the reducing agent.

  32. Types of Oxidation-Reduction Reactions Disproportionation Reaction Element is simultaneously oxidized and reduced. 0 +1 -1 Cl2 + 2OH-→ClO- + Cl- + H2O

  33. 다음에서 각 원소의 산화수를 결정하라. (a) 산화 스칸듐(Sc2O3) (b) 염화갈륨(GaCl3) (c) 인산 수소 이온 (d) 삼플루오린화 아이오딘

  34. 다음 각 물질이 물에 잘 녹을 것인지 말하고, 그 이유를 설명하라. 벤젠(C6H6) (b) 수산화 소듐 (c) 에탄올(CH3CH2OH) (d) 아세트산 포타슘 (e) 질산 리튬 (f) 글라이신(H2NCH2COOH) (g) 펜테인 (h) 에틸렌 글라이콜(HOCH2CH2OH)

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