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ChE102 Finals Review December 12, 2010

ChE102 Finals Review December 12, 2010. About me. No clue? NO PROBLEM!. irenexliu@gmail.com. ChE102 Overview. Tips for studying ChE102 Chp 12 & 13: SLG and phase transitions Chp 15 & 18: Equilibrium Chp 20: Electrochemistry Chp 14: Chemical Kinetics. Tips for ChE102.

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ChE102 Finals Review December 12, 2010

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  1. ChE102 Finals Review December 12, 2010

  2. About me No clue? NO PROBLEM! irenexliu@gmail.com

  3. ChE102 Overview • Tips for studying ChE102 • Chp 12 & 13: SLG and phase transitions • Chp 15 & 18: Equilibrium • Chp 20: Electrochemistry • Chp 14: Chemical Kinetics

  4. Tips for ChE102 • Heavily calculation focused - do lots of HARD problems • Problems in package & past finals • Watch out for units!!! (i.e. Ideal gas constant) • Time yourself

  5. Chapter 1-6 Units, Stoichiometry, Solutions, Gases,

  6. Chapter 12 & 13 Solid, Liquid, Gas Phase Transitions

  7. 12 & 13 Overview Topics include: • Phase equilibria, vapour liquid equilibria (VLE) • Phase Diagrams and transitions • Henry’s Law, Raoult’s law • Colligative Properties • Vapour Pressure depression. • Osmotic pressure • Boiling point elevation • Freezing point depression

  8. Phase Transitions • Enthalpy all should be measured at fixed T and P.

  9. Vapour Liquid Equilibrium • VLE: Rate of evaporation = Rate of Condensation • Vapour pressure: pressure given off material when system reaches equilibrium (Po) • Po is a function of temperature and specific substance • Stronger the interactions the lower the vapour pressure for a given temperature. • Po does not depend on the size of the system (as long as material has enough to fill volume above liquid)

  10. Vapour Liquid Equilibrium hence18g of water, would sustain a vapour pressure of 0.031 atm in any vessel up to a volume of 788L, at T = 25oC. • Example: Assume water vapour behaves as an ideal gas. What is the maximum volume that would maintain the vapour pressure for 1 mole of water at 25 oC? • PV=nRT, (H2O at 25 oC, Po= 0.031 atm) • Note: Units are in atm, K, L, so R= 0.082 L∙ atm /mol∙ k • Volume = 788 L

  11. Vapour Liquid Equilibrium • This means that: • ↑ water in 788L, vapour pressure CONSTANT • ↓ water in 788L, vapour pressure ↓(not enough to fill) • ↑ volume, vapour pressure ↓ • ↓ Volume, vapour pressure CONSTANT

  12. Vapour Liquid Equilibrium * Note The C- C equation can also be applied to other phase transitions, provided that the proper heat of phase transition is used. • Vapour Pressure and Temperature • ↑ Temperature = ↑ Kinetic Energy = ↓Intermolecular forces • Hence, new equilibrium at HIGHER Po • Clausius-Clapeyron equation: • Simple ratio for P means P2 and P1 units just have to match • T must be in Kelvin, Hvap assumed to be constant, R is 8.314 J/ mol ∙ K

  13. Phase Change • Special temperature/Pressure Definition • Boiling Point: the temperature when Po= Pressure of surrounding atm • Normal boiling point @ 1 atm • Dew point: the temperature when Po = Partial pressure of water vapour in air • Summary Video: • http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/031_ChangesState.MOV

  14. Phase Diagrams • Phase diagrams show regions of uniform phases for different P and T • And regions where phase equilibria exist. • As P increases for constant T • Pure gas • Solid gas interface (sublimation) (normal sublimation T because P = 1atm). • Pure solid • As T increases for constant P D solid liquid interface (melting) • Pure liquid • As P decreases with constant T • Liquid gas interface (boiling) • Gas Carbon dioxide E D C F Triple point B G A Petrucci Fig 13-19

  15. Phase Diagram - water • Points of note • Many different solid structures • Polymorphism • Slope for solid/liquid equilibrium (ice I) implies as P increases melting point goes down. Water

  16. Chp 13: Solutions

  17. Henry’s Law Pa: partial pressure of a gas “a” (solute) Ka:(Henry’s law constant) depends on the solute/solvent pair and on temperature. xa is the mole fraction of the dissolved gas in solution • Henry’s Law: relates partial pressure of gas to its molar fraction in the solution • ↑ pressure of gas = ↑ concentration in liquid (fixed T) • ↑ Ka = ↓ solubility

  18. Raoult’s Law Vapour pressure of pure A Mole fraction A • Relates the partial pressure of a substance in solution to its mole fraction in solution and its vapour pressure in pure form • Only for “ideal solutions” (liquid/liquid and liquid/solid solutions) • assumes there are no interactions between molecules of the different components.

  19. Application of Raoult’s Law • Dalton’s Law of Partial Pressures now considered as partial vapour pressure: • The mole fraction of a solution component A in the vapour phase is a function of: • Its mole fraction in the liquid. • The vapour pressures of the solution components: A and B Composition of Vapour Phase Above Liquid/Liquid solution

  20. Liquid-Vapour Equilibrium - The vapour composition above a solution is enriched in the more volatile component, relative to its mole fraction in the liquid - mixtures with different liquid compositions will have different boiling points. - normal boiling point P = 1 atm. - Po for each component is a function of temperature.

  21. Liquid- vapour equilibrium PROBLEM 1: At 90.0oC, the vapour pressures of benzene (C6H6) and toluene (C7H8) are 1.34 atm and 0.534 atm, respectively. Benzene and toluene form 1 mole of a nearly ideal solution at 90.0oC. • What is the mole fraction of toluene in a benzene-toluene solution that begins to boil at 90.0oC under 1 atm? • What is the mass percentage of benzene in the vapour over this solution at 90.0oC and 1 atm? (Solution: 0.422, 74.5%)

  22. Colligative Properties • Colligative properties depend on the number of dissolved particles. • equations describing them must be adjusted to take dissociation into account. • insertion of an additional factor, the van’t Hoff factor “i”.

  23. Colligative Properties P = -ixBPoA iXb = the mole fraction of dissolved B, or (# of moles of B dissolved / Kg of solvent) = iCRT iC = (# moles of dissolved B/ L of solvent) • In most cases, equations are only valid for Ideal solutions • i.e. dilute solutions • Vapour-pressure lowering: Po of a solution is lower than that of pure solvent since Xliq< 1. • for a solid solute B in a solvent A the vapour pressure of solution (PoS). • Osmotic Pressure effect: • c is the concentration of the original solution in moles/L

  24. Colligative Properties Boiling point Elevation: Freezing Point Depression: Xb is molality of solution (same as previous) Kb, Kf are constants Kb ,Kf depend only on the solvent Tb = iXbKb - Tf = iXb Kf

  25. Colligative Properties PROBLEM 2: The vapour pressure of pure liquid CS2 is 0.3914 atm at 20ºC. When 40.0 g of rhombic sulphur is dissolved in 1.00 kg of CS2, the vapour pressure of CS2 falls to 0.3868 atm. Determine the molecular formula for the sulphur molecules dissolved in CS2. (solution: S8)

  26. Colligative Properties PROBLEM 3: An aqueous solution is 15.0% by weight with respect to NaBr. Assuming that the solution behaves ideally and that dissociation is complete, what would be the expected boiling point of the solution at a total pressure of 1 atm? Data: (solution: 101.76oC )

  27. Colligative Properties PROBLEM 4: A 4.0 g mixture consisting of sucrose (C12H22O11) and zinc nitrate (Zn(NO3)2) is dissolved in 150.0 g of water. If the resulting solution freezes at –0.768ºC, what is the mass fraction of sucrose in the mixture? Assume that Zn(NO3)2 dissociates completely in water. Data: (solution: 0.0275)

  28. Colligative Properties PROBLEM 5: A solution of 0.07265 g of a human hormone in 100 mL of solution has an osmotic pressure of 12.60 mmHg at 21.6 oC. What is the molecular mass of the hormone in g /mol? (solution: 1058 g/mol)

  29. Chapter 15 & 18 Equilibrium

  30. Equilibrium Topics include: The Equilibrium Process Equilibrium Constants Kc, Kp Manipulating K for Complex Reactions. External Effects on Equilibriums Ksp and common ion effect

  31. Equilibrium • Vapour pressure, Chemical equilibrium, solubility equilibrium • Chemical equilibrium: forward rate = backward rate • Equilibrium constant (Kc) is constant for a given temperature • Kc>1: products in higher amount • Kc<1: reactants in higher amount • Magnitude of Kc does not indicate if a reaction happens OR how fast it will react

  32. Equilibrium A B 2A 2 B B A Manipulating Kc:

  33. Equilibrium A B (1) B C (2) Overall Reaction A C (3) Kc for sequential reactions:

  34. Equilibrium – Gases & Solids • Kp = Kc (RT) ∆ngas • For gas: can also be defined in terms of partial pressure • in general: • For solids: only concentration of gas/dissolved reactants is considered

  35. Problem PROBLEM 6: 2008 final (1) (2)

  36. Equilibrium - Q Reaction Quotient (Q): predict direction of change in Equilib. Q is defined for ANY concentration of reactants/products K is defined ONLY when chemical equilibrium is reached

  37. Equilibrium - Q • Therefore, if • Q = Kc  chemical equilibrium Q < Kc  reactants are in excess. • System must change to reduce reactants and increase products Q > Kc  products are in excess. • System must change to reduce products and increase products.

  38. Equilibrium • Effects on equilibrium: Le Châtelier Principle: • Change in concentration of reactants/products • Change in volume • Change in pressure • Adding inert gas (no change if constant V, change if constant P) ** the above will only change Q and disturb equilibrium ** ** Temperature will change K **

  39. Equilibrium • Kc1 is the equilibrium constant at the temperature T1. • Kc2 is the equilibrium constant at the temperature T2. • T1 and T2 in Kelvin. • ∆ H is the enthalpy of reaction (assumed constant). Van’t Hoff Equation: describes relationship between temperature and equilibrium constant, K

  40. Equilibrium PROBLEM 7: Consider the homogeneous gas phase reaction: Is this a fast or a slow reaction? Effect on equilibrium if the total pressure of the system is doubled at constant temperature? if the absolute temperature of the system is doubled at constant pressure? If an inert gas is added at constant volume? (fixed T) If an inert gas is added at constant pressure? (fixed T)

  41. Equilibrium PROBLEM 8: Consider the homogeneous gas phase reaction: at 600.0 K, Kc = 45.0 L/mol. At equilibrium at 600.0 K, the total pressure of the system is 27.08 atm and the mole fraction of R is yR = 0.818181. This mixture is compressed at 600.0 K to one-half the initial volume, and the system is allowed to proceed to a new equilibrium state. -> What is the concentration of R (in mol/L) at the new equilibrium state? (Solution: 0.928 mol/L)

  42. Equilibrium- Ksp Equilibrium between precipitation and dissolution Ksp: relates to the solubility (s) of the solute in the solvent of sparingly soluble salt Solubility: maximum amount (in g or mol) of solute that dissolves in the solvent at given T

  43. Equilibrium- Ksp • Common ion effect: similar to chemical equilibrium when excess of one reagent is added • Q>Ksp => salt will precipitate until Ksp is reached • Q<ksp => everything is dissolved

  44. Ksp Problem PROBLEM 9: 2008 final

  45. Chapter 20 Electrochemistry

  46. Electrochemistry Topics include: Redox Reactions and definitions Oxidation state, balancing oxidation reduction reactions Galvanic cells Quantifying electron transfer & Faraday’s law Half-cell potential Nernst equation Cell potentials at equilibrium Concentration cells Electrolysis

  47. Electrochemistry • Introductory animation: • http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/022_REDOXREACTS1.MOV • OIL RIG: Oxidation is loss, Reduction is gain • Oxidizing agent causes something to be oxidized (reduced itself) • Reducing agent causes something to be reduced (oxidized itself)

  48. Oxidation States Certain elements may have multiple OS e.g. N • Oxidation states or “Oxidation number”: relates to electronic arrangement of atom/ion • http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/010_PERIODICOXIDA.MOV

  49. Oxidation Number RULES! • For pure elements. • OS = Zero, • For alkali metal ions (Li, Na, K, Rb, Cs, Fr). • OS = +1 • For alkaline earth metal ions (Be, Mg, Ca, Sr, Ba, Ra). • OS = +2 • for halide ions (F, Cl, Br, I, At), (X-) • OS = -1

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