3/18/2014 Phase diagrams display the state of a substance at Review Chap 11 Phase diagrams 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 What is the temperature and pressure of the triple point Using the phase diagram for The triple point is the point methane CH4 where the solid, liquid, and What is the temperature and gaseous phases coexist. It is pressure of the critical point? marked point 1 and is at The critical point is where 180 C and 0.1 atm the liquid, gaseous, and Is methane a solid, liquid, or supercritical fluid phases gas at 1 atm and 0 C? coexist. The intersection of 0 C and It is marked point 3 and is 1 atm is marked point 2 approximately 80C and 50 It is in the gaseous region atm. If solid methane at 1 atm is heated while the pressure If methane at 1 atm and 0 C is compressed until a is held constant, will it melt or sublime? phase change occurs, in which state is the methane Starting at P = 1 atm and when the compression is complete? moving horizontally, liquid region is reached at T 180C Moving vertically up from point Then into the gaseous region, 2,which is 1 atm and 0 C, the at T 160 C. first phase change we come to is from gas to supercritical fluid. So, solid methane melts This phase change happens when In order for methane to the critical pressure (~50 atm) is sublime, the pressure must be exceeded below the triple point pressure 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) (a) A MIXTURE is heterogeneous – Temperature • CONCENTRATION (b) A SOLUTION is homogeneous – Molecular weight • COLLIGATIVE PROPERTIES (c) A colloidal dispersion is between a – Molecular solutions solution and a mixture – Ionic solutions • COLLOIDS Types of Aqueous Solutions Which of the following will be water soluble • ELECTROLYTE – A substance that NaCl Yes – all sodium salts are water soluble dissolves in water to produce IONS Example: HCl(aq), NaOH(aq), NaCl(aq) AgCl No – Silver, Mercury & Lead are not water soluble • NON ELECTROLYTE A substance that Yes – “Like water” {polar compound CH3OH DOES NOT produce IONS, remain as CH Cl Yes – “Like water” {polar compound molecules, when dissolved in water. 3 No – Not “Like water” { not polar Example: sugar CCl4 What is in an aqueous solution of Classify each of the following as a heterogeneous mixture or homogeneous mixture (a solution). NaCl Na+(aq) + Cl – (aq) Air ……………………. Mixture (homogeneous) MgI Mg2+(aq) + 2 I-(aq) 2 Tomato juice ………… Mixture (heterogeneous) Al3+(aq) + 3 NO -(aq) Al(NO3)3 3 Iodine crystals …….…. Pure substance H+(aq) + ClO -(aq) Mud …….………….. Mixture (heterogeneous) HClO4 4 18 Carat White Gold … Mixture (homogeneous) 2NH +(aq) + SO 2-(aq) (NH4)2SO4 4 4 2 3/18/2014 Predicting Solubility Patterns Expressing Concentration (based on weight) Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon 1 weight of solute 2 tetrachloride (CCl4) or in water: weight% x10 weight of solution C H a hydrocarbon, so it is molecular and nonpolar. 7 16 therefore more soluble in the nonpolar CCl 4 2 a compound containing a metal and nonmetals, is weight of solute 6 Na2SO4 ppm x10 ionic therefore more soluble in polar water weight of solution a polar diatomic molecule therefore more soluble HCl in polar water 3 I a nonpolar diatomic molecule therefore more weight of solute 9 2 soluble in the nonpolar CCl ppb x10 4 weight of solution Mole Fraction (x) , Molarity (M) A solution is made by dissolving 4.35 g glucose and Molality (m) (C6H12O6) in 25.0 mL of water at 25 C. Calculate the molality of glucose in the solution. Moles of A X 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 Moles of solute Molality = Kilograms of SOLVENT A solution with a density of 0.876 g/mL contains 5.0 g Calculate the concentration of CO2 in a soft drink that of toluene (C7H8) and 225 g of benzene. Calculate the is bottled with a partial pressure of CO of 4.0 atm over molarity of the solution 2 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: Sg = k Pg The volume of the solution is obtained from the mass S = kP = (3.4 102 mol/L-atm)(4.0 atm) CO2 CO2 of the solution= 5.0 g + 225 g = 230 g and its density S = 0.14 mol/L CO2 S = 0.14 M CO2 3 3/18/2014 van’t Hoff i Factor Colligative Properties i represents the number of ions in solution ° i = 1 sugar, ethylene glycol, etc Vapor Pressure LOWERING ∆P = i x2 P 1 i = 2 NaCl ; KNO3 ; CsC2H3O2 ; etc Boiling Point ELEVATION ∆Tb = i Kb m i = 3 MgCl2 ; (NH4)2ClO4 ; etc Freezint Point LOWERING ∆Tf = i Kf m Osmotic Pressure π = i MRT i = 4 Can you think of examples ? Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a Automotive antifreeze consists of ethylene glycol, density of 1.26 g/mL. Calculate the vapor pressure of a CH2(OH)CH2(OH), a nonvolatile nonelectrolyte. solution made by adding 50.0 mL of glycerin to 500.0 Calculate the boiling point and freezing point of a 250g mL of water at 25 C. The vapor pressure of pure water of ethylene glycol in 750 g water at 25 C is 23.8 torr ∆T = k m and ∆T = k m o b b f f Raoult’s law: Psolution = Xsolvent P solvent ∆Tb = kb m = (0.51)(5.37) = 2.7 therefore Tb = 102.7 ∆Tf = kf m = (1.86)(5.37) = 10.0 0 so Tf = -10.0 C 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 Rate Law : How rate depends on concentration of reactants rate α (concentration of reactants) • Factors that affect reaction rates Rate Constant: • Reaction Rates A constant of proportionality between the reaction • Rate Law Expression rate and the concentration of reactants. – Rate constant (units) rate = k x (concentration of reactants) --------------------------------------------- – Exponents in Rate Law (order of reaction) Reaction Order The EXPONENTS in the rate law rate = k x (concentration of reactants)X Determine the rate law Reaction Order the order of reaction and the rate constant the sum of the powers to which all reactant + − for NH4 (aq) + NO2 (aq) N2 (g) + 2 H2O (liq) concentrations in the rate law are raised The rate law is + x − y Reaction order is determined experimentally Rate = k [NH4 ] [NO2 ] Example: for the reaction Find the three (3) unknowns 2– – 2– – S2O8 (aq) + 3 I (aq) 2SO4 (aq) + I3 (aq) k, X, and Y Determine rate law, order and rate constant Using the method 2– X – Y Rate = k [S2O8 ] [I ] of Initial Rates How many unknowns in rate equation? + − “Run” the reaction NH4 (aq) + NO2 (aq) N2 (g) + 2 H2O (liq) + − + x − y NH4 (aq) + NO2 (aq) N2 (g) + 2 H2O (liq) Rate = k [NH4 ] [NO2 ] at least three (3) times + - Comparing Experiments 1 and 2, when [NH4 ] doubles Comparing Experiments 5 and 6 when [NO2 ] doubles the initial rate doubles the initial rate doubles + x − y + 1 − y Therefore X = 1 in Rate = k [NH4 ] [NO2 ] Therefore Y = 1 in Rate = k [NH4 ] [NO2 ] 5 3/18/2014 + − 2– – NH4 (aq) + NO2 (aq) N2 (g) + 2 H2O (liq) Reaction of peroxydisulfate ion (S2O8 ) with iodide ion (I ) + x − y 2– – 2– – Rate = k [NH4 ] [NO2 ] is S2O8 (aq) + 3 I (aq) 2SO4 (aq) + I3 (aq) + 1 − 1 Rate = k [NH ] [NO2 ] 2– X – Y Rate = k [S2O8 ] [I ] The equation is called the rate law, Use the following data to Determine nd the reaction is 2 order – the rate law and k is the rate constant – the order Which is found from any of the experiments – and rate constant - -6 x y Exp [S2O8][I] Rate(M/s) Rate1 = 2.6 x 10 = k [0.018] [0.036] -6 1 0.018 0.036 2.6 x 10 -6 x y 2 0.027 0.036 3.9 x 10-6 Rate2 = 3.9 x 10 = k [0.027] [0.036] -6 -6 x y 3 0.036 0.054 7.8 x 10 Rate3 = 7.8 x 10 = k [0.036] [0.054] -5 4 0.050 0.072 1.4 x 10 -5 x y Rate4 = 1.4 x 10 = k [0.050] [0.072] Rate = k [S O 2–] X [I–] Y 2 8 Divide Rate 2 by Rate 1 to find X Rate = 2.6 x 10-6 = k [0.018]x [0.036]y 1 -6 X Y X R2 3.9 x 10 k .027 .036 .027 X -6 x y 1.5 (1.5) Rate2 = 3.9 x 10 = k [0.027] [0.036] -6 X Y X R1 2.6 x 10 k .018 .036 .018 -6 x y Rate3 = 7.8 x 10 = k [0.036] [0.054] Therefore X = 1 -5 x y Rate4 = 1.4 x 10 = k [0.050] [0.072] -6 x y Rate1 = 2.6 x 10 = k [0.018] [0.036] For the Reaction -6 x y 2– – 2– – Rate2 = 3.9 x 10 = k [0.027] [0.036] S2O8 (aq) + 3 I (aq) 2SO4 (aq) + I3 (aq) -6 x y Rate3 = 7.8 x 10 = k [0.036] [0.054] 2– - Rate law : Rate = k [S2O8 ] [I ] -5 x y Rate4 = 1.4 x 10 = k [0.050] [0.072] How do you find y ? Order : X + Y = 1 + 1 = 2 Remember that X is no longer unknown ! 2– 1 - y Rate = k [S2O8 ] [I ] Rate constant : Divide Rate 4 by Rate 1 to find Y 2.2 x 10-6 = k (0.018)(0.036) -6 1 Y R4 14 x 10 k .050 .072 y 5.38 -6 1 Y (2.78)(2) -3 R1 2.6 x 10 k .018 .036 k = 1.5 x 10 Units ??? 6 3/18/2014 For First Order Reactions: A B final d[ A] t k dt Extra 0 initial [ A] 1.