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SOS Exam-AID

SOS Exam-AID. CHM 1311. Brought to you by Jeremy. Topics in this Exam-AID 1. The Fundamentals Balancing Chemical Equations Stoichiometry % Composition by mass + Determining Chemical Equations Trivial things profs might ask to trip you up (molality, mole fraction, %w/w) Molecular Geometry.

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SOS Exam-AID

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  1. SOS Exam-AID CHM 1311 Brought to you by Jeremy

  2. Topics in this Exam-AID 1 • The Fundamentals • Balancing Chemical Equations • Stoichiometry • % Composition by mass + Determining Chemical Equations • Trivial things profs might ask to trip you up (molality, mole fraction, %w/w) • Molecular Geometry

  3. Topics in this Exam-AID 2 • Thermochemistry • Enthalpy • Heats of Reaction and Heats of Formation • Calorimetry • Entropy • Free Energy, ΔG • Equilibrium and Equilibrium Constants for: • Gases • Solutions • Acids/Bases • Solubility

  4. Topics in this Exam-AID 3 • Electrochemistry • Redox Reactions • Cell potentials • Kinetics • Rate equations • Rate-Determining Steps • Quantum Numbers

  5. Balancing Chemical Equations 1 Unbalanced: Ca(OH)2 + H3PO4 = Ca3(PO4)2 + H2O • Balancing the atom which appears the fewest times on each side • Then balance the other atoms • It can help to make a table

  6. Balancing Chemical Equations 2 Balanced: 3 Ca(OH)2 + 2 H3PO4 → Ca3(PO4)2 + 6 H2O • It helps to know what the products of your reactions are! • Unbalanced: C6H5COOH + O2 → ??

  7. Balancing Chemical Equations 3 C6H5COOH + O2 → H2O + CO2 • Balance the atom that shows up the fewest times on each side C6H5COOH + O2 → 3 H2O + 7 CO2 LS RS C 7 7 H 6 6 O 4 17 C6H5COOH + 15/2 O2 → 3 H2O + 7 CO2

  8. Stoichiometry • Ratios and Recipes 2 cups Baking Mix + 1 Cup Chocolate chips → 2 (terrible) Chocolate Chip Cupcakes • 2 moles H2 + 1 mole O2 → 2 moles H2O • 2:1:2 ratio

  9. Stoichiometry 0.005 mols H2 + Y moles O2 → Z moles H2O 2 moles H2 + 1 mole O2 → 2 moles H2O • Because the ratios are the same, you can divide the equations in order to figure out the number of moles you need

  10. % Composition by Mass 1 • Given a chemical formula: • C6H12O6 • Pretend you have one mole of the molecule, and multiple moles of the component atoms • 1 mole C6H12O6 = 6 moles C, 12 moles H, 6 moles O

  11. % Composition by Mass 2 • How much do 6 moles of C weigh? 6 moles * 12.011g/mol = 72.066g • How much do 12 moles of H weigh? • How much do 6 moles of O weigh?

  12. Composition by mass 3 • How much would one mole of the substance weigh? • The molar mass of C6H12O6 is 180.16g/mol • How much do each of the components weigh? • 6 moles of Cweigh (6 * 12.011), or 72.066g/mol • Take percentages • 72.066/180.16 = 40.0%

  13. Determining Chemical Formulas • Steps • Assume one mole of molecule • Use molar ratios to determine how many moles of atoms you have • Convert from moles to mass • To determine formula, do the opposite • Assume 100g, find mass of each element • Convert from mass to moles • Use the number of moles to create molar ratios to make your formula

  14. Determining Chemical Formulas You have a substance made of : 40% Ca, 12% C, and 48% O, molar mass of 100.0869 g/mol • Step One: Assume 100g of substance, find mass of each element • 40% Ca * 100g = 40g • 12% C * 100g = 12g • 48% O * 100g = 48g

  15. Determining Chemical Formulas 2 • Step Two: Convert from mass to moles • Remember: moles = mass / molar mass • 40g Ca / 40.078g/mol = 0.998 mole Ca • 12g C / 12.011g/mol = 0.999 mole C • 48g O = ??? moles O

  16. Determining Chemical Formulas 3 • Step Three: Use the number of moles to create molar ratios • Take the smallest number of moles you have and divide into all the other numbers • 0.998 mole Ca, 0.999 mole C, 3.00 moles O 0.998 0.998 0.998 = 1 mole Ca, 1 mole C, 3 moles O • If you don't use the smallest number... • 0.998 mole Ca, 0.999 mole C, 3.00 moles O 3 3 3 • = 0.33 mole Ca? 0.33 mole C? 1 mole O?

  17. Determining Chemical Formulas 4 • Step 4: Use the simplest formula to find the true formula • Find the molar mass of the simplest formula • The molecular weight of the true compound will be provided • Make it match by multiplying the number of atoms by an integer • Why multiply?

  18. Determining Chemical Formulas 5 You have an empirical formula CH2O, mm = 30.03g/mol The formula of the true compound is 60.06g/mol What is the true formula??? 30.03 * 2 = 60.06 CH2O * 2 = C2H4O2

  19. Trivial Things Molality: = Moles of Solute / Mass of Solvent (not solution) Mole Fraction [A]: = Moles of A / Moles of A + Moles of B %m/w (or w/w): =(mass of solute / mass of solution) * 100%

  20. Wrap-Up! • We discussed • Balancing chemical equations • Stoichiometry • % composition by mass • Determining chemical formula Any questions?

  21. Molecular Geometry • Sadly, molecular geometry is mostly memorization • Especially the names of the configurations • Key Tricks to drawing molecules with proper geometry: • Figure out the electron arrangement first • Electrons repel, and a tetrahedron provides the most space for four pairs of electrons • Memorize all the names though =/

  22. Molecular Geometry

  23. Why do Reactions Happen? Thermochemistry and Kinetics! • Thermochemistry can be broken into • Enthalpy (ΔH) and Entropy (S) • The interaction between the two form Free Energy (ΔG) • Free Energy, Enthalpy and Entropy also affect equilibrium (K)

  24. Thermochemistry - Enthalpy • Change in Enthalpy (ΔH) • = Change in Internal Energy (ΔU) [+ Work done/by on the system(W)] • Most times, work = 0 , ΔH = ΔU, except for gases ΔH = ΔU + W • If so, change in internal energy is ΔH = ΔU = q • Where q is heat energy transferred

  25. All about q • q is all about heat transfer • If q is negative, heat is given off • If q is positive, heat is absorbed • Reactions that give off heat usually happen • Reactions that take up heat usually don't • How do you figure out what q is?

  26. Calorimetry • Experimentally: q = mc ΔT • c tells you how much heat the material holds per gram • Check the units of c, usually in J/g*K but sometimes in J/mol*K • If so, q =nC ΔT • If you are given a calorimetry question, c will be given to you • Except for water. It's 4.18 J/gK

  27. Hess's Law • Finding ΔHo is a pain. • Luckily, Hess's Law states you can find ΔH using any combination of reactions as long as they add up to the reaction you want C(s) + O2 → CO2(g) ΔH = ?? C(s) + ½O2(g) → CO(g) ΔH = -110.5 kJ CO(g) + ½O2(g) → CO2(g) ΔH = -283.0 kJ • Think of it like a building Shamelessly ripped from UOttawa, Pell/Mayer 2009

  28. Hess's Law • Taken one step further, you can also take the ΔH of formation of the molecules in the equation • ΔH of formation is ΔHf • ΔHf is the energy it takes to form the molecule from the constituent elements • ΔHf of all products - ΔHf of all reactants = ΔH for reaction

  29. Hess's Law CH4(g)ΔHf= -74.87 kJO2 (g)ΔHf= 0 kJH2O(g) ΔHf= -241.83 kJCO2(g) ΔHf=-393.509 kJ CH4(g) + 2O2(g) → 2H2O(g) + CO2(g) ΔH=ΔHf (H2O(g) + CO2(g)) - ΔHf (CH4(g) + O2(g))

  30. ΔH: Endothermic or Exothermic? • When ΔH is negative, the reaction is exothermic • Heat is given off to the atmosphere • Good for the reaction! • When ΔH is positive, the reaction is endothermic • Heat is absorbed from the atmosphere • Not so good.

  31. When does Work Matter? • Remember, W = ΔPV • When you have a gas that expands, it cools off and does work • If you just measured the temperature change, you would get a false reading • Therefore, ΔH = ΔU = q + w =q + PΔV

  32. Entropy • Entropy is a measure of disorder in the universe • Having more things will make them more disordered • Having a molecule that can twist and turn makes it more disordered • Gases are more disordered than solids

  33. Enthalpy, Entropy and Free Energy ΔGois free energy ΔGo= ΔHo - TΔSo • ΔSo is entropy (J/molK) • ΔS and ΔH don't change with T • ΔGo is for T = 298K, standard conditions, 1 atm • Just like ΔH, negative is good!

  34. Finding ΔGo, ΔHo,ΔSo • If ΔGoisn't given, you can calculate it from ΔHand ΔS • You can also calculate it from ΔGof from a table of ΔGoby using Hess's law

  35. Finding ΔGo, ΔHo,ΔSo ΔGo = ΔGfo (products – reactants) = -204.6 + 137.2 = -67.4 kJ ΔGo = ΔHfo (products – reactants) - T ΔSo (products - reactants) = -218.8kJ + 110.5kJ + (298) (283.5-223.1-197.7)J/molK = -108.3 kJ + 40915.4J = -108.3 kJ + 40.91 kJ = -67.4 kJ Pirated from Waterloo Chemistry Pages

  36. Finding ΔGo, ΔHo,ΔSo H2(g) + N2(g) ↔ 2NH3(g) ΔHf for NH3(g) is -46.1kJ/molΔS for H2(g) is 130.7J/molK, for N2(g) is 191.56J/molK, and for NH3(g) is 192.77J/molK What is ΔGo , assuming standard conditions?

  37. ΔGo andΔG? • ΔGo is great! And useless by itself • It is only for T = 298K • ΔG (no o) is the value for the actual condition • ΔGo lets you get ΔG! • ΔGo = ΔHo - TΔSo • ΔG= ΔH - TΔS

  38. ΔG and Reaction Favourability • If ΔG = negative, reaction likely to happen If ΔG = positive, reaction will not happen; it might happen in the reverse direction • By plugging values into ΔG= ΔH - TΔS, you can find ΔG= 0 • At ΔG= 0, the reaction is in equilibrium

  39. Decomposition in Equilibrium 2AlCl3 (s) ↔ 2Al(s) + 3 Cl2 (g) At what temperature will the decomposition of AlCl3(s) be in equilibrium? Cl2 (g) Chlorine gas 0 223.08 0 Al(s) Aluminum solid 0 28.3 0 AlCl3 (s) Aluminum Chloride -705.63 109.29 -630.0

  40. Decomposition in Equilibrium ΔH = ΔHf (products) – ΔHf (reactants) = 2 (-706.63) – 0 = -1413.26 kJ/molΔS = ΔS(products) – ΔS(reactants) = 2(-630.0) – 0 = -1260 J/molK ΔG= ΔH - TΔS 0 = -1413.26 + (T) 1.26 T = 1121 K

  41. Equilibrium Constant: K • K is the equilibrium constant; it is temperature dependent • Given balanced formula aA + bB ↔ cC + dD • K = [C]c [D]d [A]a [B]b • In the case of gases Kp = p[C]c p[D]d p[A]a p[B]b K > 1 is product favoured K< 1 is reactant favoured

  42. K • Keq /Kc= K for equation (c for concentration) • Ksp = K for dissolved substances (sp for solubility product) • Ka /Kb = K for an acid / K for a base ALL Ks are calculated the same way! Do not include solids/pure liquids in the calculation. K = [C]c [D]d [A]a [B]b • Except for gases, which * can * be calculated with pressure

  43. Kp and Gases • For gases, Kp = p[C]c p[D]d p[A]a p[B]b • Pressures are just easier to use, especially since Total Pressure = Sum of the individuals pressures of the gases • Kp = Kc (RT)n • P = RTn/V

  44. Examples of K H2SO4(aq) + H2O(l) → HSO4-(aq) + H3O+(aq) Ka = [HSO4-] [H3O+] / [H2SO4] 2SO2(g) + O2(g) → 2SO3(g) Kp = p[SO3]2 / p[SO2]2 [O2] CaCO3(s) → Ca2+(aq) + CO32-(aq) Ksp = [Ca] [CO3]

  45. K, ΔG, and Equilibrium • When ΔG = 0, the reaction was at equilibrium How can this be related to K? • ΔG = ΔGo + RT ln K • ΔGo = -RT ln K ΔGo = -600 kJ/mol T = 200K T = 1200K RlnK = 3 RlnK = 0.5

  46. Q and Lechatalier's Principle • Lechatalier's principle is like a seesaw Reactants ↔ Products • Where ΔG is experimental, while ΔGo is for standard conditions • Q is experimental, while K is for theoretical equilibrium

  47. Q If the experimental conditions = thereoretical equilibrium, Q = K I2(g) + H2(g) ↔ 2HI(g)at some temperature/ pressure For this T and P, Kc = 1 Therefore 1 = [HI]2 / [H2] [I2] You have 10 moles of HI, two moles of H, and two moles of I in the vessel. Which way will the reaction go?

  48. Q Q = [HI]2 / [H2] [I2] =[10]2 / [2] [2] Q = 100/4 = 25 • if Q = K, the reaction would be at equilibrium • Q > K, the reaction will go towards reactants • Q < K, the reaction will go towards products ΔG= ΔGo + RT ln Q

  49. Q in a Ksp Problem PbCl2 is highly insoluble in water. The Ksp of PbCl2 is1.6 * 10−5. PbCl2 is added to 0.1L of water at 298K, 1atm pressure. a) How much PbCl2 dissolves?b) What is ΔGo? 0.058g of NaCl are added to the solution. c) Find Q. d) What is ΔG right when the NaCl is added? e) How much PbCl2 is left dissolved?

  50. Q in a Ksp Problem a) How much PbCl2 dissolves?Ksp = 1.6 * 10−5 Keep in mind that ICE tables are for concentration! Therefore the concentration of the ions is x and 2x, not actual amount. PbCl2(s) ↔ Pb2+(aq) + 2Cl-(aq)I / 0 0C / +x +2xE / x 2x Ksp = (x) (2x)2 1.6* 10-5 = 4x3x = 0.0159 Therefore 0.00159 moles of PbCl2 dissolve

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