1 / 28

Chapter 4 Chemical Reactions

Chapter 4 Chemical Reactions. HW #3 Due by Monday, 5/5/2014 HW #4 due by Friday, 5/9/2014 Quiz #2: Monday, 5/12/2014 Exam #2: Wednesday, 5/14/2014. End-of-Chapter Homework: pp 167 - 177. 5, 6, 9-11 14-18 19 (a,d,g,h) 28 29 31 32 33 38 41 43

weldon
Download Presentation

Chapter 4 Chemical Reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 4Chemical Reactions HW #3 Due by Monday, 5/5/2014 HW #4 due by Friday, 5/9/2014 Quiz #2: Monday, 5/12/2014 Exam #2: Wednesday, 5/14/2014

  2. End-of-Chapter Homework: pp 167 - 177 5, 6, 9-11 14-18 19(a,d,g,h) 28 29 31 32 33 38 41 43 47 51 55 57 59 61 65b 67 69 71 73 79 81 83 89 100

  3. I. Solubility - Many Ionic compounds dissolve in water yielding ions. -These soluble ionic compounds conduct electricity due to the ions. Are called electrolytes. - Driving force is water being attracted to the ions, forming weak bonds, & releasing energy. 1) Nonelectrolyte (does not conduct electricity in water): a) Molecular & Soluble - when dissolves, yields no ions - CH3OH b) Ionic & Insoluble like AgCl or PbI2 2) Strong Electrolyte (Ionic, Soluble, Conducts well in water) NH4Cl(s) -----) NH4+(aq) + Cl-(aq) 3) Weak Electrolyte (few % ionization in water; Weak Conductor) NH3 + H2O ------) NH4+ + OH- (~ 1% ionized) HC2H3O2 --------) H+ + C2H3O2- (~ 1 % ionized)

  4. Svante Arrhenius NaI in H2O – Soluble & Ionic; Many Ions H2O – Molecular & No Ions Nonelectrolyte – No Current Electrolyte – Much Current

  5. I. Solubility Continued • Know general solubility rules : These allow you to predict both the solubility & products of many Reactions (Rxns). #1 Grp 1, NH4+, C2H3O2- &NO3- compds are soluble (except H). #2 Cl- Br- I-compds are solubleexcept : Ag+ Pb2+ Hg22+ #3 SO4-2solubleexcept : Ca2+ Sr2+ Ba2+ Ag+ Pb2+ Hg22+ #4 OH-compds are insolubleexcept#1 & Ca2+ Sr 2+ Ba 2+ #5 Most CO3-2 (carbonates), PO4-3 (phosphates), & S-2 (sulfides) are insoluble except for #1, (rule 1).

  6. II. Writing Reactions - Ways to write; Example 1 [Note: Ag ion is always Ag+1 does not need (I); Solvent is H2O] silver nitrate + sodium chloride --------) silver chloride + sodium nitrate 1) Molecular Equation (ME):Write all as if not ionized AgNO3(aq) + NaCl(aq) --------) AgCl(s) + NaNO3(aq) 2) Complete Ionic Equation (CIE):Write soluble, ionic compounds as ions; should use subscripts like (aq) Ag+ + NO3- + Na+ + Cl- ----) AgCl(s) + Na+ + NO3- 3) Net Ionic Equation (NIE):Eliminate Spectator Ions Ag+(aq) + Cl-(aq) -----) AgCl(s)

  7. III. Reaction Types A. Precipitation Reactions - Formation of an insoluble compound (precipitate, ppt) is a driving force in a chemical reaction. - Can predict a product from the solubility rules. Method:Write soluble ionic reactants as ions. Check solubility rules & see if insoluble product forms from ions. Example: Predict product when mix K2SO4(aq) & BaCl2(aq) Write CIE & NIE. 2K+ + SO42- + Ba2+ + 2Cl- ---) BaSO4(s) + 2K+ + 2Cl-(CIE) Ba2+(aq) + SO42-(aq) -----) BaSO4(s)(NIE)

  8. III. Reaction Types B. Acid-Base Rxns 1. Definitions (Note: Proton = H+) Arrhenius:Acid =substance which releases H+ in water. Base =substance which releases OH- in water. Examples: HCl(aq) -----) H+(aq) + Cl-(aq) NaOH(aq) -----) Na+(aq) + OH-(aq) Bronsted-Lowry:Acid = Proton Donor Base = Proton Acceptor Example:(air is the solvent) NH3(g) + HCl(g) -----) NH4Cl(s)

  9. III. Reaction Types B. Acid-Base Rxns 2. Properties Note: salt = any ionic compound Acids: - Sour taste - Turn Indicator dyes a specific color: Phenolphthalein colorless; Litmus red - React with bases to yield salt + H2O Bases: - Bitter Taste, Feel Slippery - Turn Indicator dyes a specific color: Phenolphthalein red; Litmus blue - React with acids to yield salt + H2O - Strong Acids/Bases ionize 100% in water - Weak Acids/Bases ionize < < 100% in water

  10. III. Reaction Types B. Acid-Base Rxns 3. Examples (Know these strong & weak acids & bases) Strong Acids:HCl HBr HI HNO3 H2SO4 HClO4 Strong Bases: NaOH KOH Ba(OH)2 (but not too sol.) Example in water: HCl(aq) -----) H+ + Cl-100% to the right Weak Acids: HF HC2H3O2 HCN H2CO3 Weak Bases: NH3[= NH4OH (in water)]NaHCO3 Examples in water: NH4OH NH4+ + OH-~1% to the right HC2H3O2H+ + C2H3O2-~1% to the right

  11. III. Reaction Types B. Acid-Base Rxns 4. Acid-Base or Neutralization Reactions - Acids react with bases to give: salt(ionic compd) + H2O - Driving force is to produce stable H2O& sometimes a gas. - Examples:NaOH + HF -----) NaF + H2O 2KOH + H2SO4 ---) K2SO4 + 2H2O • Some Neutralization Rxns produce a gas:(memorize these six) H+ + HCO3- -----) CO2(g) + H2O 2H+ + CO32------)CO2(g)+ H2O 2H+ + SO32- -----) SO2(g) + H2O 2H+ + S2- -----) H2S(g) H+ + CN- -----) HCN(g) NH4+ + OH------) NH3(g)+ H2O

  12. III. Reaction Types B. Acid-Base Rxns 5. Write three rxn formats (ME, CIE & NIE) for reaction of potassium hydroxide with hydrochloric acid: ME:KOH + HCl -----) KCl + H2O CIE:K+ + OH- + H+ + Cl- -----) K+ + Cl- + H2O NIE:H+(aq) + OH-(aq) -----) H2O(l) Note: 1) The above net ionic equation is common for many acid-base reactions. The production of stable H2O is a driving force for the reaction. 2) H+ in water is better represented by H3O+ 3) H3O+is a polyatomic ion = hydronium ion

  13. III. Reaction Types B. Acid-Base Rxns 5. ME, CEI and NIE - Write the ME, CIE & NIE for: HCl + NaHCO3(in water) ME: HCl + NaHCO3 -----) NaCl + H2O + CO2 CIE: H+ + Cl- + Na+ + HCO3- ---) Na+ + Cl- + H2O + CO2 NIE: H+(aq) + HCO3-(aq) -----) H2O(l) + CO2(g) 6. Titration - Process of adding known (amount & concentration) of Base (or Acid) in a burette to an unknown concentration of Acid (or Base). End point is determined with an indicator or a pH meter. Can now determine the amount of the unknown (quantitative analysis).

  14. III. Reaction Types B. Acid-Base Rxns Known M & Volume (Standard Solution) + indicator After titn can calculate moles & M.

  15. III. Reaction Types C. Predicting products - Review • Predicting products:write a) strong acids & b) solubleionic reactants as ions; look to see if the ions can form: (1) aprecipitate(solubility rules) (2) agas(six gas equations) (3) anonionized molecule like H2O or HC2H3O2 • Write NIE (in water) for: 1) BaI2 + AgNO3 & 2) HCl + NaCN 1) Ba+2 + I- + Ag+ + NO3- -----) AgI(s) + Ba+2 + NO3- Ag+(aq)+ I-(aq) -----) AgI(s)(Balanced NIE) 2) H+ + Cl- + Na+ + CN- -----) HCN(g) + Na+ + Cl- H+(aq) + CN-(aq) -----) HCN(g)(Balanced NIE)

  16. III. Reaction Types D. Redox Rxns 1. Definitions & Introduction: - Oxidation Reduction Reaction (Redox Rxn) is one in which electrons are transferred. - Oxidation = Loss of Electrons - Reduction = Gain of Electrons 2. Examples: (Note: electrons lost & gained must be same) 2Mg + O2 -----) 2MgO [ 2Mgo ---) 2Mg2+ + 4e- = Oxidation ] Half Rxn [ O2o + 4e- -----) 2O2- = Reduction ] Half Rxn 3Cu+2 + 2Al -----) 3Cu + 2Al+3 [ 2Alo -----) 2Al+3 + 6e- = Oxidation] Half Rxn [ 3Cu+2 + 6e- -----) 3Cuo = Reduction] Half Rxn

  17. III. Reaction Types D. Redox Rxns Note: Oxidizing agent causes oxidation Reducing agent causes reduction 3. Uses of RedoxRxns: a) Create new chemicals b) Theoretical Importance (chem 123) c) Battery – allow the e’stransferred to perform work 4. Oxidation Numbers (ON): Definition:ON = the charge that an atom or group of atoms would have if they were ionic. Uses:1)Redox Rxn(ID & what’s been oxid/red & #e-) 2) NomenclatureFeF2 = Iron(II)Fluoride; (II) = ON 3) Used in balancing redox rxns(chem 123) Rules: Know the following rules [Order gives priority]

  18. III. Reaction Types D. Redox Rxns 5. Rules for Determining ON (Know) Note: When is a conflict, 1st rule controls result. a) The sum of ON’s add up to give the charge b) The ON of a neutral element by itself is 0 c) Group 1, 2, 13 ions are +1, +2 & +3 respectively; (exception: H is -1 when combined with metal: KH , NaH) d) F is -1 e) O is -2 (exception: O is -1 when found as a peroxide, O22-) f) Cl, Br, I are -1 [when together like BrCl, most electronegative = -1]

  19. III. Reaction Types D. Redox Rxns Examples:Determine the ON of single N in each: N2 N3-1 Na3N N2O NO NF3 NO2 NO3-1 0 -1/3 -3 +1 +2 +3 +4 +5 - Which of the above forms of N can explosively react with organics? Why? - Determine ON of each atom in FeSO3(Note: SO3-2 ON = -2) [+2 +4 -2] • Which are redox: (ones with change in ON; ∆ON) 1) Mg + Cl2 -----) MgCl2 2) CH4 + 2 O2 -----) CO2 + 2 H2O 3) HCl + NaOH -----) H2O + NaCl

  20. III. Reaction Types D. Redox Rxns 6. Redox Reaction Types Combination: 2Al + 3F2 -----) 2AlF3 Decomposition: 2HgO -----) 2Hg + O2 Combustion: CH4 + 2O2 -----) CO2 + 2H2O Single Replacement: 2Na + 2HCl -----) 2NaCl + H2 [ Can predict reaction from table 4.6, Pg 153; The more active metal will replace the cation ] 7. Balancing redox rxns: Can become difficult; so, rules were generated & will cover in Chemistry 123.

  21. IV. Quantitative Aspects of Solutions A. Molarity, M - Many reactions take place in solution & we need a way to measure concentration; Mis common way. - M = moles of Solute M = m (use as eqn) L Solution L solvent = substance present in largest amount solute = substances dissolved in solvent solution = solute + solvent (everything present) B. Molarity Calculations 1) Calculate the M if 4.5 moles of HCl are dissolved in a total volume of 500 mL (0.500 L) M = m = 4.5 moles = 9.0 mHCl = 9.0 MHCl L 0.500 L L

  22. IV. Quantitative Aspects of Solutions 2) Calculate M if 0.020 g of CaCO3 are dissolved in 40 L a) convert g CaCO3to mols; b) plug into M eqn; M = m/L 0.020 g CaCO3 x 1 mole CaCO3/100. g CaCO3 = 2.0x10-4 mol CaCO3 M = moles = 2.0x10-4 mol CaCO3 = 5.0x10-6M CaCO3 L 40 L 3) How many moles of HCl are present in 2.0 L of 0.30 MHCl? M = moles/Lmoles = M x L moles= 0.30 m HCl x 2.0 L = 0.60 moles HCl L 4) How many L of 12 MHCl will provide 2.0 moles of HCl ? M = m L = m = 2.0 mole HCl = 0.17 L HCl LM 12 mole/L

  23. IV. Quant. Aspects of Solutions C. Dilution Problems - Frequently we are given a concentrated solution & we need to add solvent to get a more dilute solution. This is a dilution problem. M1 x V1 = M2 x V2 Note: 1) Initial moles = Final moles 2) Works with any conc. or vol. units 3) Use eqn ONLY for dilution problem Example: How many mL of 12 M HCl are needed to make 100. mL of 3.0 M? M1 x V1 = M2 x V2 V1 = M2 x V2= (3.0 M x 100. mL) = 25 mL M112 M - Take 25 mL of 12 M HCl and add enough water to make 100 mL.

  24. IV. Quantitative Aspects of Solutions D. Gravimetric Analysis - A type of analysis where one converts the analyte to an insoluble compound which is then isolated, dried & weighed. Example: A 1960 dime (2.50g) is dissolved in nitric acid. HCl is then added to form insoluble AgCl, which is collected, washed, dried & weighed. Calculate a) the g of Ag & b) w/w % Ag in the dime if 2.98 g of AgCl was obtained. Ag ---) Ag+1 ---) AgCl a) g Ag in dime: 2.98 g AgCl 1 mol AgCl 1 mol Ag 108 g Ag = 2.25 g Ag 143 g AgCl1 mol AgCl1 mol Ag b) % Ag in dime: w/w % Ag = g Ag x 100% = 2.25 g Ag x 100% = 90.0% g Dime 2.50 g

  25. IV. Quantitative Aspects of Solutions E. Volumetric Analysis - An analysis technique where the M & volumeof a reagent in a burette is used to calculate the mols or M of an unknown. - NOTE: This is a typical stoichiometric problem except that we calculate mols of reagent from M & Linstead of from grams. Example: 25.0 mL of HCl is titrated with 45.5 mL of 0.433 MNaOH. Calculate the a) mols of HCl present & b)M of the original HCl. 1 HCl + 1 NaOH -----) 1 NaCl + 1 H2O a) m 0.433 m NaOH 0.0455 L NaOH 1 m HCl = 0.0197 m HCl L NaOH 1 m NaOH b) M = moles HCl = 0.0197 m HCl = 0.788 M HCl L 0.0250 L

  26. E. Vol Anal, Example: Note HC2H3O2 = HA, MW = 60.05 g/m 1) Moles HA (12.1 mL NaOH = 0.0121 L) 0.500 m NaOH x 0.0121 L = 0.00605 m NaOH L 0.00605m NaOH x 1m HA/1m NaOH = 0.00605 mHA 2) M HA (10.0 mL HA = 0.0100 L HA) M= m = 0.00605 m HA = 0.605 m/L HA L 0.0100 L HA 3) w/w % HA = (g HA / g Soln) x 100% g HA = 0.00605m x 60.05 g/m HA = 0.3633 g HA w/w% HA = (0.3633g HA /10.3 g soln) x100 = 3.53%

  27. General Example • Give the NIE for reaction of Lead (II) Nitrate with hydroiodic acid. Pb(NO3)2 + HI ----) Pb+2 + NO3- + H+ + I- ----) PbI2(s) + H+ + NO3- Pb+2(aq) + 2I-(aq) ----) PbI2(s) -? Mols PbI2 from 0.10 m HI. 0.10 m I-1 m PbI2 = 0.050 m PbI2 2 m I- - ? Mols PbI2 from 2.0 L of 0.050 m/L HI? 0.050 m HI 2.0L 1 m PbI2 = 0.050 m PbI2 L 2 m HI

  28. Chapter 3 & 4 Review • Mole • % Composition • EF & MF • Stiochiometry: Reg, LR, Amt. Left, % Yield • Predicting Products: Solubility, Gas, H2O & Weak Acids • Writing Rxns: ME CIE NIE Phases • Electrolytes • Acids & Bases • Redox Oxidation & Reduction • ON • MM & Stiochiometry & Titrations; Dilution

More Related