550 likes | 848 Views
Principles of Chemistry I Test 2 Review. UPDATED => Now with corrections to slides 28 and 43! Sorry for the typos!. Bo Marshall’s Understanding…. Rules…. Recommendations. Review Notes Memorize required tables, etc. Practice problems from the book Re-do online homework
E N D
Principles of Chemistry ITest 2 Review UPDATED => Now with corrections to slides 28 and 43! Sorry for the typos! Bo Marshall’s Understanding…
Recommendations • Review Notes • Memorize required tables, etc. • Practice problems from the book • Re-do online homework • Practice test (do this after everything else) • Review stuff you got wrong on the practice test • Read parts of book to review things that you are having difficulty with
Things to Remember from Test 1 • Poly and Monoatomic Ions • Ex. Cl-, SO42-, etc. • Grams to Moles to Atoms/Molecules • Chemistry Island • Along with other things I’m sure…
Solution Stoichiometry • Molarity = moles solute/liters soln. = M • The equation is… • MIVI=MFVF • M = Molarity, V = Volume • When making molar solns., Weigh solid and add liquid to make final volume needed for desired molarity.
Solution Stoichiometry • What volume, in L, of 10 M HCl is needed to make 2.00L of a 2.00 M HCl solution by dilution with water?
Solution Stoichiometry • M1=10 M • V1= Unkn • M2= 2.00 M • V2 = 2.00 L
Solution Stoichiometry • How many sodium ions are present in 325 mL of 0.850 M Na2SO4?
Solution Stoichiometry • What are we looking for? Na Ions • What info are we given? Lets start there…
Chapter 4 - Chemical Rxns • Solubility and electrolyte solutions • Strong electrolytes - dissociate into many ions, weak electrolytes - dissociate into few ions • Examples in Class of Electrolytes with the lightbulb: • NaCl in water => Strong electrolyte • Acetic Acid (CH3COOH) => Weak electrolyte • Ammonia (NH3) => Weak electrolyte • So what is a non-electrolyte?
Chapter 4 - Chemical Rxns • Why stuff dissolves in water… • It is polar = has separation of + and - charges • Therefore, it dissolves ionic compounds • The neg. part of water attracts cations (-, +) • The pos. part of water attracts anions (+, -) • Covalent compounds do NOT dissociate into ions when dissolved into water - except acids and bases, but well get back to that…
Why did the white bear dissolve in water? • It was polar!
Chapter 4 - Chemical Rxns • Table 4.1 Solubility Rules - You Need to Memorize these… • Soluble: • - Anything from Group 1A • - Any NH4+ (Ammonium) • - NO3- (Nitrates) • - CH3COO- (acetates) • - ClO4- (perchlorates) • - Cl-, Br-, I- except with: Ag+, Pb2+, Cu+, Hg22+ • - F- except with: Ag+, Group 2A • - SO42- (sulfates) except with: Ag+, Pb2+,Ca2+, Sr2+, Ba2+
Chapter 4 - Chemical Rxns • Table 4.1 Solubility Rules - You Need to Memorize these… • Insoluble: • - Any OH- (hydroxide) except with: Group 1A or NH4+ (b/c they are always soluble) or Group 2A starting w/ Ca2+ • - CO32- (Carbonates) except with: Group 1A or NH4+ (b/c they are always soluble) • - PO43- (Phosphates) except with: Group 1A or NH4+ (b/c they are always soluble) • - S2-(sulfides) except with: Group 1A or NH4+ (b/c they are always soluble)and Group 2A
Chapter 4 - Chemical Rxns • Precipitation Rxns • To know what Rxn will form a precipitate (ppt), you must know the solubility rules • Molecular Eqn • Total Ionic Eqn • Net Ionic Eqn • Spectator Ions…
Chapter 4 - Chemical Rxns • So what will react here? (Use Solubility Rules): • 2NaCl (aq) + Hg2(NO3)2 (aq) -> • 2NaNO3 (aq) + Hg2Cl2 (s) • Molecular Eqn • 2NaCl (aq) + Hg2(NO3)2 (aq) -> 2NaNO3 (aq) + Hg2Cl2 (s) • Total Ionic Eqn (All “aq” except the ppt) • 2Na+ + 2Cl- + Hg22+ + 2NO3 -> 2Na+ + 2NO3 + Hg2Cl2 (s) • Net Ionic Eqn • 2Cl- + Hg22+ -> Hg2Cl2 (s) • Spectator Ions • 2Na+ + 2NO3
Do lots of ppt rxns to get good at knowing what is soluble and what isn’t… • A chemistry professor couldn't resist interjecting a little philosophy into a class lecture. He interrupted his discussion on balancing chemical equations, saying, "Remember, if you're not part of the solution, you're part of the precipitate!"
Chapter 4 - Chemical Rxns • Acid Base Rxns - Very Important for Principles 2 and Organic Chem! • Forms H2O and a salt • OH- = Base • H+ (or H3O+) = Acid • Salt comes from the Cation/Anion of the acid and base • CH3COOH -> CH3COO- and H+ • NaOH -> Na+ and OH- • Salt is CH3COONa, water from H+ and OH-
Chapter 4 - Chemical Rxns • Strong Acids To know • Strong Means it completely dissociates into H+ and the anion • HCl, HBr, HI, (H+ and a halogen, except F) • HClO3, HClO4, HNO3, H2SO4 • Strong Bases to Know • Strong = Dissociates into OH- and the cation • LiOH, NaOH, KOH, RbOH, CsOH (all Gr I) • Ca(OH)2, Sr(OH)2, Ba(OH)2 (Gr II) • Note that all the strong acids and bases are all soluble by the solubility rules…
Chapter 4 - Chemical Rxns • Weak Acids You Need to Know • (NOT IN THE BOOK TABLE, THIS IS WHY YOU GO TO CLASS AND PAY ATTENTION!!!) • HF • CH3COOH • H3PO4 • Weak bases • NH3 • Amines
Chapter 4 - Chemical Rxns • NaOH + HCl -> H2O + NaCl (Molecular Eqn) • Na+ + OH- + H+ + Cl- -> Na+ + Cl- + H2O (Total Ionic) • H+ + OH- -> H2O (Net Ionic, the same for all STRONG acid/base Rxns) • Weak Acid/Base Rxns (Equilibrium Rxns) • NH3 + HCl <-> NH4+ + Cl- (Molecular) • NH3 + H+ + Cl- <-> NH4+ + Cl- (Total Ionic) • Cl- is the only spectator ion…
Chapter 4 - Chemical Rxns • Which of the following is an acid/base rxn? • Mg(s) + 2H2O(g) -> Mg(OH)2(s) + H2 (g) • HNO3 (aq) + LiOH (aq) -> LiNO3 (aq) + H2O • Which produces water and a salt? • What is the salt? • Li+ and NO3-
Chapter 4 - Chemical Rxns • Redox Reactions • Electron Transfer • Loss of e- = oxidation • Gain of e- = reduction • LEO is GER • The substance that is oxidized is the reducing agent • The substance that is reduced is the oxidizing agent • You can’t have one without the other!!!
Chapter 4 - Chemical Rxns • To keep track of who lost or gained electrons, look at the oxidation numbers and see which elements/molecules had changes in oxidation number.
Chapter 4 - Chemical Rxns • What is the O.N. for C60? • Trick question he hinted at in class, this is a compound made up of a bunch of carbon in its elemental form. O.N. for elemental form of an atom = 0 • Similar to Cl2 (g)…found this way in elemental form
Chapter 4 - Chemical Rxns • What is the oxidation state of phosphorus in Na5P3O10? • Na = +1(5) = +5 • O = -2(10)= -20 • -20+5 = -15 • So need to get +15 from 3 P atoms to equal zero overall • +15/3 = +5
Chapter 4 - Chemical Rxns http://www.youtube.com/watch?v=a6RR4kPsnlE&feature=related
Chapter 4 - Chemical Rxns • Activity Series - To see what will react with what in redox rxns… • Above on the left will react with below on the right • The above left (solid) will lose electrons (oxidize) to form the Cation and e-’s • Below right will go backwards - will gain the e-’s that the solid gave up to go from the cation to the solid (reduced) • Therefore, top = good reducing agents, bottom = good oxidizing agents
Chapter 4 - Chemical Rxns Activity Series of Metals Li (s) → Li+ (aq) + e– K (s) → K+ (aq) + e– Ba (s) → Ba2+ (aq) + 2 e– Ca (s) → Ca2+ (aq) + 2 e– Na (s) → Na+ (aq) + e– Mg (s) → Mg2+(aq) + 2 e– Al (s) → Al3+ (aq) + 3 e– Mn(s) → Mn2+(aq) + 2 e– Zn (s) → Zn2+ (aq) + 2 e– Cr (s) → Cr3+ (aq) + 3 e– Fe (s) → Fe2+ (aq) + 2 e– Co (s) → Co2+ (aq) + 2 e– Ni (s) → Ni2+ (aq) + 2 e– Sn (s) → Sn2+ (aq) + 2 e– Pb (s) → Pb2+ (aq) + 2 e– H2 (g) → 2 H+ (aq) + 2 e– Cu (s) → Cu2+ (aq) + 2 e– Ag (s) → Ag+ (aq) + e– Hg (l) → Hg2+ (aq) + 2 e– Pt (s) → Pt2+ (aq) + 2 e– Au (s) → Au3+(aq) + 3 e–
Chapter 4 - Chemical Rxns • Predict the products • 2Al(s) + 3MnSO4(aq) -> • 2Al(s) + 3Mn2+ + 3SO42- -> • Look at activity series, Al is above and left from Mn2+
Chapter 4 - Chemical Rxns • Al(s) -> Al3+ + 3 e- • Mn2+ will become Mn(s) from the electrons provided by Al • Products: 3Mn(s) + 2Al3+ + 3SO42- • 2Al3+ = 6+, 3SO42- = 6- => Al2(SO4)3 • So total eqn for Rxn • 2Al(s) + 3MnSO4(aq) -> 3Mn(s) + Al2(SO4)3
Chapter 4 - Chemical Rxns • 2Al(s) + 3MnSO4(aq) -> 3Mn(s) + Al2(SO4)3 (aq) • So what was reduced? The one that gained electrons. Mn2+ gained electrons to become Mn(s), so its ON changed from 2+ to 0, therefore it was reduced. That makes it the oxidizing agent • What was oxidized (you can’t have one w/o the other)? Al(s) became Al3+, so it lost electrons. Its ON changed from 0 to 3+, so it was oxidized. That makes it the reducing agent.
Chapter 4 - Chemical Rxns • Equilibrium Rxns • For Rxns that “go to completion” • For Equilibrium Rnxs • Forward and reverse rxns are equal MICROscopically. Don’t look like they are changing MACROscopically • Weak acids/bases are in equilibrium • NH3(g) + H2O (l) NH4+ (aq)+ OH- (aq)
Chapter 5 - Thermochemistry • System, Surroundings • ∆E = E products- E reactants • Heat = q => energy transfer from temp difference • work = w => energy transfer when an object is moved by force • Energy of a system => ∆Esys= q + w • Remember - cannot measure the energy of a system, only the change in energy (∆E )
Chapter 5 - Thermochemistry • joule = J = (kg*m2/s2) • 1 calorie = 4.184 J (On eqn sheet!) • 1000 chemical calories = 1 Nutritional Calorie
Chapter 5 - Thermochemistry • State Functions - dependent only on present state of the system, does not depend on the path taken to get there • Temp = T • Pressure = P • Volume = V • Energy = ∆E • All Capital letters!!! (so q and w aren’t state functions)
Chapter 5 - Thermochemistry • Used to define heat of a rxn at a CONSTANT PRESSURE • Enthalpy = ∆H = ∆E + ∆(PV) (eqn sheet) • Assuming constant pressure (∆H = qp) • Again, can only measure change in enthalpy… • ∆H Negative => Exothermic (Rxn gives off heat) • ∆H Positive => Endothermic (Rxn “uses up” heat)
Chapter 5 - Thermochemistry • ∆Hrxn= Enthalpy Change for a chemical rxn • For the Rxn as written, so if you use 1/2 as many moles, the ∆Hrxn = (1/2) original • 2Al + Fe2O3 -> Al2O3 + 2Fe , ∆Hrxn=-851 kJ/mol • Al + (1/2) Fe2O3 -> (1/2) Al2O3 + Fe, ∆Hrxn=-425.5 kJ/mol
Chapter 5 - Thermochemistry • ∆Hcomb = Heat of Combustion • May have fractions in final chemical eqn • ∆Hf = Heat of formation • Usually exothermic, otherwise the compound formed would be unstable • (NI3) http://www.youtube.com/watch?v=ZLfhi7MuNXg • ∆Hfus = Heat of fusion - solid to liquid • ∆Hvap = Heat of Vaporization - liquid to gas • ∆Hfus and ∆Hvap are always ENDOTHERMIC, they require energy b/c • energy solid < energy liquid < energy gas • Due to more movement of particles
Chapter 5 - Thermochemistry • What causes Rxns to have enthalpy changes? • Difference in BOND STRENGTHS of products and reactants • ∆H = ∑nD(bonds broken) - ∑nD(bonds formed) • On Eqn Sheet • Bond Energies found on table (finally something you don’t have to memorize)
Chapter 5 - Thermochemistry N N + 3H H -> 2 N trip bond = 945 kJ/mol H-H Bond = 432 kJ/mol N-H bond = 391 kJ/mol ∆H = ∑nD(bonds broken) - ∑nD(bonds formed) ∆H= [945 + 3(432)] - [6(391)] = -105kJ/mol
Chapter 5 - Thermochemistry • Specific Heat Capacity = c = temp needed to raise 1 g of a substance 1 K • Cwater = 4.184J/g*K (on eqn sheet) • Molar Heat capacity = C = temp needed to raise 1 mol of a substance 1K
Chapter 5 - Thermochemistry • q = mc∆T • Know how to solve for each given the others • Remember that ∆T is Tfinal - Tinitial • Calorimeter • Measures q for the surroundings (water), so the q for the system (rxn) equal in magnitude, opposite in sign • Should probably be familiar with the constant volume calorimeter too…
Chapter 5 - Thermochemistry • Hess’s Law • Add up∆H for each step of the rxn to get the overall ∆H. • Reverse eqn => Change sign of ∆H • Multiply coeff. By a constant => multiply ∆H by the same • Calculate ∆H for the rxn • WO3(s) + 3H2(g) -> W(s) + 3H2O(g) • Given • 2W(s) + 3O2(g) -> 2WO3(s) => ∆H = -1685.4 kJ • 2H2(g) + O2 (g) -> 2H2O(g) => ∆H = -477.84 kJ • Lets do this on the board, too much work to type…
Chapter 5 - Thermochemistry • Standard States (∆H°) • Gas = 1atm = 1bar • Liquids = Most stable form, P=1bar, must specify temp • Aq. Solns = 1M
Chapter 5 - Thermochemistry • ∆HF° = Std Heat of formation • Similar to adding equations with Hess’s law. • Find ∆HF° for the products and the reactants in the appropriate table (Appendix C in book) and then… • ∆H = ∑n ∆HF (products) - ∑n ∆HF (reactants)
Practice • In a titration experiment, 1.20 M NaOH is added to 225 mL of 3.0 M H2SO4 and reacts according to the equation below. How much NaOH has been added at the equivalence point. • H2SO4 (aq) + 2NaOH (aq) -> 2H2O (l) + Na2SO4 (aq)
Practice • H2SO4 (aq) + 2NaOH (aq) -> 2H2O (l) + Na2SO4 (aq) • How much base is needed to react with the acid that is present? That is what is asked in a titration rxn. • Must compare acid and base on mol to mol ratio • Which is the acid and which is the base?