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Transcript
3/18/2014 Review Chap 11 Phase diagrams Phase diagrams display the state of a substance at various pressures and temperatures, and the places where equilibria exist between phases. The solid - liquid line marks the melting point at each pressure The liquid – vapor line is the boiling point at that pressure The high critical temperature and critical pressure are due to strong van der Waals forces between water molecules Phase Diagram of Water Interpreting a Phase Diagram Using the phase diagram for methane CH4 What is the temperature and pressure of the critical point? The critical point is where the liquid, gaseous, and supercritical fluid phases coexist. It is marked point 3 and is approximately 80C and 50 atm. If solid methane at 1 atm is heated while the pressure is held constant, will it melt or sublime? Starting at P = 1 atm and moving horizontally, liquid region is reached at T 180C Then into the gaseous region, at T 160 C. So, solid methane melts In order for methane to sublime, the pressure must be below the triple point pressure What is the temperature and pressure of the triple point The triple point is the point where the solid, liquid, and gaseous phases coexist. It is marked point 1 and is at 180 C and 0.1 atm Is methane a solid, liquid, or gas at 1 atm and 0 C? The intersection of 0 C and 1 atm is marked point 2 It is in the gaseous region If methane at 1 atm and 0 C is compressed until a phase change occurs, in which state is the methane when the compression is complete? Moving vertically up from point 2,which is 1 atm and 0 C, the first phase change we come to is from gas to supercritical fluid. This phase change happens when the critical pressure (~50 atm) is exceeded 1 3/18/2014 Chapter 13 Solutions and Their Properties Solutions, Mixtures, & Colloids • SOLUBILITY RULES – Chapter 4 • SOLUBILITY – factors that affect – Pressure (HENRY’S LAW) – Temperature • CONCENTRATION – Molecular weight • COLLIGATIVE PROPERTIES (a) A MIXTURE is heterogeneous (b) A SOLUTION is homogeneous (c) A colloidal dispersion is between a – Molecular solutions – Ionic solutions solution and a mixture • COLLOIDS Types of Aqueous Solutions • ELECTROLYTE – A substance that dissolves in water to produce IONS Example: HCl(aq), NaOH(aq), NaCl(aq) • NON ELECTROLYTE A substance that DOES NOT produce IONS, remain as molecules, when dissolved in water. Example: sugar What is in an aqueous solution of NaCl Na+(aq) + Cl – (aq) MgI2 Mg2+(aq) + 2 I-(aq) Al(NO3)3 HClO4 (NH4)2SO4 Which of the following will be water soluble NaCl Yes – all sodium salts are water soluble AgCl No – Silver, Mercury & Lead are not water soluble Yes – “Like water” {polar compound CH3OH CH3Cl Yes – “Like water” {polar compound CCl4 No – Not “Like water” { not polar Classify each of the following as a heterogeneous mixture or homogeneous mixture (a solution). Air ……………………. Mixture (homogeneous) Tomato juice ………… Mixture (heterogeneous) Al3+(aq) + 3 NO3-(aq) Iodine crystals …….…. Pure substance H+(aq) + ClO4-(aq) Mud …….………….. Mixture (heterogeneous) 2NH4+(aq) + SO42-(aq) 18 Carat White Gold … Mixture (homogeneous) 2 3/18/2014 Predicting Solubility Patterns Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride (CCl4) or in water: C7H16 Na2SO4 HCl I2 a hydrocarbon, so it is molecular and nonpolar. therefore more soluble in the nonpolar CCl4 a compound containing a metal and nonmetals, is ionic therefore more soluble in polar water a polar diatomic molecule therefore more soluble in polar water a nonpolar diatomic molecule therefore more soluble in the nonpolar CCl4 Mole Fraction (x) , Molarity (M) and Molality (m) XA Expressing Concentration (based on weight) 1 2 weight % ppm 3 ppb weight of solute x10 2 weight of solution weight of solute x106 weight of solution weight of solute x109 weight of solution A solution is made by dissolving 4.35 g glucose (C6H12O6) in 25.0 mL of water at 25 C. Calculate the molality of glucose in the solution. Moles of A Total number of moles (25.0 mL of water)(1.00 g/mL) = 25.0 g = 0.0250 kg solvent Moles of solute Molarity Liters of SOLUTION Molality = Moles of solute Kilograms of SOLVENT A solution with a density of 0.876 g/mL contains 5.0 g of toluene (C7H8) and 225 g of benzene. Calculate the molarity of the solution Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25 C. The Henry’s law constant for CO2 in water at this temperature is 3.4 102 mol/L-atm Henry’s law: The volume of the solution is obtained from the mass of the solution= 5.0 g + 225 g = 230 g and its density Sg = k Pg SCO2 = kPCO2 = (3.4 102 mol/L-atm)(4.0 atm) SCO2 = 0.14 mol/L SCO2 = 0.14 M 3 3/18/2014 Colligative Properties van’t Hoff i Factor i represents the number of ions in solution Vapor Pressure LOWERING ∆P = i x2 P°1 i=1 sugar, ethylene glycol, etc Boiling Point ELEVATION ∆Tb = i Kb m i=2 NaCl ; KNO3 ; CsC2H3O2 ; etc Freezint Point LOWERING ∆Tf = i Kf m i=3 MgCl2 ; (NH4)2ClO4 ; etc i=4 Can you think of examples ? Osmotic Pressure π = i MRT Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL. Calculate the vapor pressure of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water at 25 C. The vapor pressure of pure water at 25 C is 23.8 torr Raoult’s law: Psolution = Xsolvent Po solvent Automotive antifreeze consists of ethylene glycol, CH2(OH)CH2(OH), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 250g of ethylene glycol in 750 g water ∆Tb = kb m and ∆Tf = kf m ∆Tb = kb m = (0.51)(5.37) = 2.7 therefore Tb = 102.7 ∆Tf = kf m = (1.86)(5.37) = 10.0 so Tf = -10.0 0C The osmotic pressure of an aqueous solution containing 3.50 mg of a protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25 C was found to be 1.54 torr. Treating the protein as a nonelectrolyte, calculate its molar mass. Osmotic pressure = MRT 4 3/18/2014 Chapter 14 • Factors that affect reaction rates • Reaction Rates • Rate Law Expression Rate Law : How rate depends on concentration of reactants rate α (concentration of reactants) Rate Constant: A constant of proportionality between the reaction rate and the concentration of reactants. – Rate constant (units) – Exponents in Rate Law (order of reaction) rate = k x (concentration of reactants) --------------------------------------------- Reaction Order The EXPONENTS in the rate law rate = k x (concentration of reactants)X Determine the rate law the order of reaction and the rate constant Reaction Order the sum of the powers to which all reactant concentrations in the rate law are raised Reaction order is determined experimentally Example: for the reaction 2–(aq) S2O8 + 3 I–(aq) 2SO42–(aq) + I3–(aq) Determine rate law, order and rate constant Rate = k [S2O82–] X [I–] Y How many unknowns in rate equation? “Run” the reaction NH4+(aq) + NO2−(aq) N2 (g) + 2 H2O (liq) at least three (3) times Comparing Experiments 1 and 2, when [NH4+] doubles the initial rate doubles Therefore X = 1 in Rate = k [NH4+]x [NO2−] y for NH4+(aq) + NO2−(aq) N2 (g) + 2 H2O (liq) The rate law is Rate = k [NH4+]x [NO2−] y Find the three (3) unknowns k, X, and Y Using the method of Initial Rates NH4+(aq) + NO2−(aq) N2 (g) + 2 H2O (liq) Rate = k [NH4+]x [NO2−] y Comparing Experiments 5 and 6 when [NO2-] doubles the initial rate doubles Therefore Y = 1 in Rate = k [NH4+]1 [NO2−] y 5 3/18/2014 NH4+(aq) + NO2−(aq) N2 (g) + 2 H2O (liq) Rate = k [NH4+]x [NO2−] y Reaction of peroxydisulfate ion (S2O82–) with iodide ion (I–) is S2O82–(aq) + 3 I–(aq) 2SO42–(aq) + I3– (aq) Rate = k [NH+] 1[NO2−] 1 Rate = k [S2O82–] X [I–] Y The equation is called the rate law, the reaction is 2nd order and k is the rate constant Which is found from any of the experiments Exp 1 2 3 4 [S2O8] 0.018 0.027 0.036 0.050 [I-] 0.036 0.036 0.054 0.072 2–] X Rate = k [S2O8 Use the following data to Determine – the rate law – the order – and rate constant Rate(M/s) 2.6 x 10-6 3.9 x 10-6 7.8 x 10-6 1.4 x 10-5 Rate1 = 2.6 x 10-6 = k [0.018]x [0.036]y Rate2 = 3.9 x 10-6 = k [0.027]x [0.036]y Rate3 = 7.8 x 10-6 = k [0.036]x [0.054]y Rate4 = 1.4 x 10-5 = k [0.050]x [0.072]y [I–] Y Divide Rate 2 by Rate 1 to find X Rate1 = 2.6 x 10-6 = k [0.018]x [0.036]y Rate2 = 3.9 x 10-6 = k [0.027]x [0.036]y Rate3 = 7.8 x 10-6 = k [0.036]x [0.054]y Rate4 = 1.4 x 10-5 = k [0.050]x [0.072]y Rate1 = 2.6 x 10-6 = k [0.018]x [0.036]y Rate2 = 3.9 x 10-6 = k [0.027]x [0.036]y Rate3 = 7.8 x 10-6 = k [0.036]x [0.054]y Rate4 = 1.4 x 10-5 = k [0.050]x [0.072]y How do you find y ? Remember that X is no longer unknown ! Rate = k [S2O82–]1 [I-]y Divide Rate 4 by Rate 1 to find Y R4 14 x 10- 6 k .050 .072 (2.78)(2) y R1 2.6 x 10 -6 k .0181 .036Y 1 5.38 R2 3.9 x 10 - 6 k .027 .036 .027 (1.5) X R1 2.6 x 10 -6 k .018X .036Y .018X X 1 .5 Y X Therefore X = 1 For the Reaction S2O82–(aq) + 3 I–(aq) 2SO42–(aq) + I3–(aq) Rate law : Rate = k [S2O82–] [I-] Order : X+Y=1+1= 2 Rate constant : 2.2 x 10-6 = k (0.018)(0.036) Y k = 1.5 x 10-3 Units ??? 6 3/18/2014 For First Order Reactions: final Extra 1. Integrated form of the rate equation 2. Half life initial 1st ln [ A]0 kt [ A] t 0 k dt [A]0 = initial concentration at time t = 0 [A] = concentration at time t ln For Order Reactions d [ A] [ A] AB [ A]0 kt [ A] When ln P is plotted as a function of time, for reaction CH3NC CH3CN get a straight line ln [A]t = -kt + ln [A]0 Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k. II. Reaction Rates & Half-Life The half-life (t1/2) of a reaction is the TIME required for the concentration of a reactant to decrease to one half its initial value Therefore CH3NC CH3CN is a 1st order process and k is the negative of the slope = 5.1 10-5 s−1. Half-life {t1/2 } is defined as the time required for one-half of a reactant to react For AB at t1/2 , the concentration of A is one-half the initial concentration of A 7 3/18/2014 Half-life is defined as the time required for one-half of a reactant to react All radioactive decay is 1st order [ A ]0 kt [ A] At t1/2 concentration is ½ initial ln [A] at t1/2 is one-half of the original [A], at 13,000 sec [CH3NC] = 0.5 [A]0 ln [ A]0 k t1/2 2 A0 1 ln 2 = k t1/2 Plants take up atmospheric CO2 by photosynthesis, and are ingested by animals, so every living thing is constantly exchanging 14C with its environment as long as it lives Once it dies this exchange stops, and the amount of 14C gradually decreases through radioactive beta decay What is the rate constant for the radioactive C14 decay? Carbon-14 has a half-life of 5,730 ± 40 year All radioactive decay is 1st order ln 2 = k t1/2 k = ln 2 / t1/2 = 0.693 / 5730 k = 1.21 x 10 – 4 yr -1 8