The Major Classes of Chemical Reactions

The Major Classes of Reactions 수용액에서의 반응 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 The Major Classes of Chemical Reactions 4.1 The Role of Water as a Solvent 4.2 Writing Equations for Aqueous Ionic Reactions 4.3 Precipitation Reactions 4.4 Acid-Base Reactions 4.5 Oxidation-Reduction (Redox) Reactions 4.6 Elements in Redox Reactions Charge distribution in H2 and H2O and Molecular Dipole Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. 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+ Enthalpy of Hydration (ΔHhyd kJ/mol) of Some Typical Ions (ΔHhyd kJ/mol) of Some Typical Ions Ion ΔHhyd Ion ΔHhyd Ion ΔHhyd H+ -1130 Al3+ -4665 Fe3+ -4430 Li+ -520 Be2+ -2494 F - 505 Na+ -406 Mg2+ -1921 Cl - 363 K+ -322 Ca2+ -1577 Br - 336 Rb+ -297 Sr2+ -1443 I - 295 Cs+ -276 Ba2+ -1305 ClO4 --238 Cr2+ -1904 Mn2+ -1841 Fe2+ -1946 Co2+ -1996 Ni2+ -2105 Cu2+ -2100 Zn2+ -2046 Cd2+ -1807 Hg2+ -1824 An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. nonelectrolyte electrolyte solution solution sugar solution NaCl solution 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 (b) 78.5 g of cesium bromide dissolved in water SOLUTION: H2O (a) (NH4)2SO4(s) → 2NH4+(aq) + SO42-(aq) + nNH4 = 5.0 mol ×2/1 = 10. mol nSO42-= 5.0 mol ×1/1 = 5.0 mol H2O (b) CsBr(s) → Cs+(aq) + Br-(aq) nCsBr = 78.5 g CsBr /(212.8 g /mol) = 0.369 mol = n Cs+ = nBr Determining Moles of Ions in Aqueous Ionic Solutions How many moles of each ion are in the following solutions? (c) 7.42×1022 formula 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 nCu2+ = nCu(NO3)2 = 0.123 mol nNO3-= 2×nCu(NO3)2 = 0.246 mol H2O (d) ZnCl2(aq) → Zn2+(aq) + 2Cl-(aq) nZnCl2 = 35 mL× (L/103mL) ×(0.84 mol/L)= 2.9×10-2 mol nZn2+ = nZnCl2 = 2.9×10-2 mol nCl-= 2×nZnCl2 = 5.8×10-2 mol 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? One mole of H+(aq) is released per mole of nitric acid (HNO3) SOLUTION: H2O HNO3(l) → H+(aq) + NO3-(aq) nH+ = nHNO3 , so MH+ = MHNO3 = 1.4M Writing Equations for Aqueous Ionic Reactions The molecular equation shows all of the reactants and products as intact, undissociated compounds. 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. Precipitation Reactions The reaction of Pb(NO3)2 and NaI. Precipitate .insoluble solid that separates from solution molecular equation ionic equation Na+(aq) and NO3 -(aq) spectator ions precipitate 2Na+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3 -(aq) →PbI2(s) + 2Na+(aq) + 2NO3 2NaI(aq) + Pb(NO3)2 (aq) →PbI2(s) + 2NaNO3(aq) net ionic equation Pb2+(aq) + 2I-(aq) →PbI2(s) PbI2 2NaI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2NaNO3(aq) double displacement reaction (metathesis) -(aq) 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. S S=Soluble I I=Insoluble D=Decomposes in water U=Compound does not exist or is unstable Cation Na+ K+ NH4 + Al3+ Mg2+ Ca2+ Sr2+ Ba2+ Cr3+ Fe3+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+ Cd2+ Hg2+ Hg2 2+ Ag+ Pb2+ As3+ Sb3+ Bi3+ CH3COO -S S S S S S S S S U S S S S S S S I I S U U I Br -S S S S S S S S S S S S S S S S I Ia Ib I D D D Cl -S S S S S S S S S S S S S S S S S Ib Ib I D S D I -S S S S S S S S I U S S S U S S Ia Ia I Ia S D I ClO3 -S S S S S S S S U U U S S S S S S S S S U U U NO3 -S S S S S S S S S S S S S S S S S D S S U U D SO4 2-S S S S S Ia I Ib S S S S S S S S D I I I U D D SO3 2-S S S U S I I I I U I I I U I I U U I I U U U CrO4 2-S S S U S S Ia Ia U S I I U S Ia I Ia Ia Ia Ia U U U C2O4 2-S S S I I I I I S S I I I I I I I I I I U I D PO4 3-S S S Ia Ia Ia Ia Ia I Ia Ia I I Ia Ia I U U Ia Ia U U I CO3 2-S S S U Ia I Ia I U U I I I I Ia I I Ia Ia Ia U U U SiO3 2-S S U I Ia Ia Ia S U U I I U U Ia I U U U Ia U U I O2-D D U Ib Ia I I S I Ia Ia I I Ia I I I Ia I I I I I OH -S S U Ia Ia Ia I S I Ia Ia I I Ia Ia I U U U Ia U U D S2-S S S D D Ia I D I I Ia I I Ia Ia I I I Ia Ia I D I Ia Soluble in Acids Ib Slightly Soluble in Acids Solubility Rules for Common Ionic Compounds in Water at 25 oC oC Soluble Compounds Exception Alkali metal(1A) & Ammonium(NH4 +) NO3 -, HCO3 -, ClO3 -, CH3COOHalides( Cl-, Br-, I-) w/ Ag+, Hg2 2+, Pb2+ , Sulfates(SO4 2-) w/ Ag+, Ca2+, Sr2+, Ba2+, Hg2 2+, Pb2+ , Insoluble Compounds Exceptions CO3 2-, PO4 3-, CrO4 2, S2Alkali metal(1A) compounds, Ammonium compounds OH-Alkali metal(1A) compounds, Ba2+ , 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. Common Acids and Bases Acids Strong hydroiodic acid, HI hydrobromic acid, HBr perchloric acid, HClO4 hydrochloric acid, HCl sulfuric acid, H2SO4 nitric acid, HNO3 Weak hydrofluoric acid, HF phosphoric acid, H3PO4 acetic acid, CH3COOH salicylic acid, C6H4(OH)COOH ascorbic acid(Vitamin C), C6H8O6 Bases Strong sodium hydroxide, NaOH potassium hydroxide, KOH calcium hydroxide, Ca(OH)2 strontium hydroxide, Sr(OH)2 barium hydroxide, Ba(OH)2 Weak ammonia, NH3 Acid-Base Reactions When HCl gas dissolves in water, O+(aq) + Cl.(aq) HCl(g) + H2O(l) → H3H+ transfer HCl(aq) + NaOH(aq), [H3O+(aq) + Cl.(aq)] + [Na+ (aq) + OH.(aq)] → H2H+ transfer O(l) + Cl.(aq) + Na+(aq) + HOH H3O+(aq) + OH.(aq)] → H2O(l) H+ transfer Johannes Brønsted & Thomas Lowry : acid: proton donor, base: acceptor 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) 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 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 mol HCl/0.05000L = 0.1016 M HCl 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. 1. Free elements have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4, ... = 0 2. In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 3. The oxidation number of oxygen is usually .2. In H2 O2 and O22-it is .1. 4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is .1. 5. Group IA metals are +1, IIA metals are +2 and fluorine is always .1. HSO4 Oxidation numbers of all O = -2 H = +1 the elements in HSO4-? 4×(-2) + 1 + ? = -1 S = +6 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. oxidation states of the main group elements 3As23534.2Se24635.1Br1345736Kr237Rb138Sr249In12350.4Sn2451.3Sb3552.2Te245653.1I135754Xe246855Cs156Ba281Tl1382.4Pb2483.3Bi3584.2Po24685.1At135786Rn287Fr188Ra2 Oxidation-Reduction (redox) reactions M + X → M+ + e-+ X → M+ + X-→ MX M . . . . M loses electron(s) M is oxidized M is the reducing agent M increases its oxidation number .X gains electron(s) .X is reduced .X is the oxidizing agent .X decreases its oxidation number X Recognizing Oxidizing and Reducing Agents Identify the oxidizing agent and reducing agent in each of the following: (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) 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. +2-2 +2-2 0 +4-2 (b) PbO(s) + CO(g) → Pb(s) + CO2(g) 0 0 +1-2 (c) 2H2(g) + O2(g) → 2H2O(g) An active metal displacing hydrogen from water: Large Sodium Explosion http://www.youtube.com/watch?v=sNdijknRxfU&feature=related Displacing one metal with another. Fe(s) + CuSO4(aq) → Fe(SO4) (aq) + Cu(s) Shiny White + Blue → Green Solution + Reddish Brown The activity series of the metals. Reduction 0+1 0 →+x M + H2O → M(OH)x + H2 Oxidation Li K can displace H Ba from water Ca Na Cu cannot displace H from Hg any source Ag Au strength as reducing agents Mg Al Mn An Cr Fe Cd can displace H from steam Co Ni Sn can displace H from acid Pb H2 Types of Oxidation-Reduction Reactions 0 0 +3-1 Combination : A + B → C 2Al + 3Br2 → 2AlBr3 +1+5-2 +1 -1 0 Decomposition : C → A + B 2KClO3 → 2KCl + 3O2 0 0 +4-2 0 0 +2-2 Combustion : A + O2 → B S + O2 → SO2 2Mg + O2 → 2MgO Displacement : A + BC → AC + B 0+1+2 0 Sr + 2H2O → Sr(OH)2 + H2 H Displacement +4 00+2 TiCl4 + 2Mg → Ti + 2MgCl2 Metal Displacement -1 0 -1 0 Cl2 + 2KBr → 2KCl + Br2 Halogen Displacement 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) (a) Combination 0 0 +2-3 3Mg(s) + N2(g) → Mg3N2 (aq) Mg is the reducing agent; N2 is the oxidizing agent. (b) Decomposition +1-1 +1-2 0 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. Types of Oxidation-Reduction Reactions Disproportionation Reaction Element is simultaneously oxidized and reduced. 0 +1 -1 Cl2 + 2OH-→ ClO-+ Cl-+ H2O 다음에서 각 원소의 산화수를 결정하라 . (a) 산화 스칸듐 (Sc2O3) (b) 염화 갈륨 (GaCl3) (c) 인산 수소 이온 (d) 삼플루오린화 아이오딘 다음 각 물질이 물에 잘 녹을 것인지 말하고 , 그 이유를 설명하라 . (a) 벤젠(C6H6) (b) 수산화 소듐 (c) 에탄올(CH3CH2OH) (d) 아세트산 포타슘 (e) 질산 리튬 (f) 글라이신(H2NCH2COOH) (g) 펜테인 (h) 에틸렌 글라이콜 (HOCH2CH2OH)