3/18/2014. Review Chap 11 Phase diagrams

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1 Review Chap 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 CH 4 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. 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 and is at 80 C and 0. atm Is methane a solid, liquid, or gas at atm and 0 C? The intersection of 0 C and atm is marked point It is in the gaseous region If solid methane at atm is heated while the pressure is held constant, will it melt or sublime? Starting at P = atm and moving horizontally, liquid region is reached at T 80C Then into the gaseous region, at T 60 C. So, solid methane melts In order for methane to sublime, the pressure must be below the triple point pressure If methane at 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,which is 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

2 Chapter 3 Solutions and Their Properties SOLUBILITY RULES Chapter 4 SOLUBILITY factors that affect Pressure (HENRY S LAW) Temperature CONCENTRATION Molecular weight COLLIGATIVE PROPERTIES Molecular solutions Ionic solutions COLLOIDS Solutions, Mixtures, & Colloids (a) A MITURE is heterogeneous (b) A SOLUTION is homogeneous (c) A colloidal dispersion is between a solution and a mixture Types of Aqueous Solutions Which of the following will be water soluble 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 NaCl AgCl CH 3 OH CH 3 Cl CCl 4 Yes all sodium salts are water soluble No Silver, Mercury & Lead are not water soluble Yes Like water {polar compound Yes Like water {polar compound No Not Like water { not polar What is in an aqueous solution of NaCl Na + (aq) + Cl (aq) MgI Mg + (aq) + I - (aq) Al(NO 3 ) 3 Al 3+ (aq) + 3 NO 3- (aq) HClO 4 H + (aq) + ClO 4- (aq) (NH 4 ) SO 4 NH 4+ (aq) + SO - 4 (aq) Classify each of the following as a heterogeneous mixture or homogeneous mixture (a solution). Air. Mixture (homogeneous) Tomato juice Mixture (heterogeneous) Iodine crystals.. Pure substance Mud... Mixture (heterogeneous) 8 Carat White Gold Mixture (homogeneous)

3 Predicting Solubility Patterns Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride (CCl 4 ) or in water: C 7 H a hydrocarbon, so it is molecular and nonpolar. 6 therefore more soluble in the nonpolar CCl 4 Na SO a compound containing a metal and nonmetals, is 4 ionic therefore more soluble in polar water a polar diatomic molecule therefore more soluble HCl in polar water I a nonpolar diatomic molecule therefore more soluble in the nonpolar CCl 4 3 Expressing Concentration (based on weight) weight of solute weight% x0 weight of solution weight of solute ppm x0 weight of solution weight of solute ppb x0 weight of solution 6 9 Mole Fraction (x), Molarity (M) and Molality (m) A Molarity Molality Total Moles of A number of moles Moles of solute Liters of SOLUTION Moles of solute = Kilograms of SOLVENT A solution is made by dissolving 4.35 g glucose (C 6 H O 6 ) in 5.0 ml of water at 5 C. Calculate the molality of glucose in the solution. (5.0 ml of water)(.00 g/ml) = 5.0 g = kg solvent A solution with a density of g/ml contains 5.0 g of toluene (C 7 H 8 ) and 5 g of benzene. Calculate the molarity of the solution Calculate the concentration of CO in a soft drink that is bottled with a partial pressure of CO of 4.0 atm over the liquid at 5 C. The Henry s law constant for CO in water at this temperature is mol/l-atm Henry s law: S g = k P g The volume of the solution is obtained from the mass of the solution= 5.0 g + 5 g = 30 g and its density S CO = kp CO = (3.4 0 mol/l-atm)(4.0 atm) S CO = 0.4 mol/l S CO = 0.4 M 3

4 Colligative Properties van t Hoff i Factor i represents the number of ions in solution Vapor Pressure LOWERING P = i x P Boiling Point ELEVATION T b = i K b m Freezint Point LOWERING T f = i K f m Osmotic Pressure π = i MRT i = i = i = 3 i = 4 sugar, ethylene glycol, etc NaCl ; KNO 3 ; CsC H 3 O ; etc MgCl ; (NH 4 ) ClO 4 ; etc Can you think of examples? Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of.6 g/ml. Calculate the vapor pressure of a solution made by adding 50.0 ml of glycerin to ml of water at 5 C. The vapor pressure of pure water at 5 C is 3.8 torr Raoult s law: P solution = solvent P o solvent Automotive antifreeze consists of ethylene glycol, CH (OH)CH (OH), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 50g of ethylene glycol in 750 g water T b = k b m and T f = k f m T b = k b m = (0.5)(5.37) =.7 therefore T b = 0.7 T f = k f m = (.86)(5.37) = 0.0 so T f = 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 5 C was found to be.54 torr. Treating the protein as a nonelectrolyte, calculate its molar mass. Osmotic pressure = MRT 4

5 Chapter 4 Factors that affect reaction rates Reaction Rates Rate Law Expression Rate constant (units) Exponents in Rate Law (order of reaction) 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 = k x (concentration of reactants) Reaction Order The EPONENTS in the rate law rate = k x (concentration of reactants) 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 S O 8 (aq) + 3 I (aq) SO 4 (aq) + I 3 (aq) Determine rate law, order and rate constant Rate = k [S O 8 ] [I ] Y How many unknowns in rate equation? for Determine the rate law the order of reaction and the rate constant NH 4+ (aq) + NO (aq) N (g) + H O (liq) The rate law is Rate = k [NH 4+ ] x [NO ] y Find the three (3) unknowns k,, and Y Using the method of Initial Rates Run the reaction NH 4+ (aq) + NO (aq) N (g) + H O (liq) at least three (3) times NH 4+ (aq) + NO (aq) N (g) + H O (liq) Rate = k [NH 4+ ] x [NO ] y Comparing Experiments and, when [NH 4+ ] doubles the initial rate doubles Therefore = in Rate = k [NH 4+ ] x [NO ] y Comparing Experiments 5 and 6 when [NO - ] doubles the initial rate doubles Therefore Y = in Rate = k [NH 4+ ] [NO ] y 5

6 NH 4+ (aq) + NO (aq) N (g) + H O (liq) Rate = k [NH 4+ ] x [NO ] y Rate = k [NH + ] [NO ] The equation is called the rate law, the reaction is nd order and k is the rate constant Which is found from any of the experiments Reaction of peroxydisulfate ion (S O 8 ) with iodide ion (I ) is S O 8 (aq) + 3 I (aq) SO 4 (aq) + I 3 (aq) Rate = k [S O 8 ] [I ] Y Use the following data to Determine the rate law the order and rate constant Exp [S O 8 ] [I - ] Rate(M/s) x x x x 0-5 Rate = k [S O 8 ] [I ] Y Rate =.6 x 0-6 = k [0.08] x [0.036] y Rate = 3.9 x 0-6 = k [0.07] x [0.036] y Rate 3 = 7.8 x 0-6 = k [0.036] x [0.054] y Rate 4 =.4 x 0-5 = k [0.050] x [0.07] y Rate =.6 x 0-6 = k [0.08] x [0.036] y Rate = 3.9 x 0-6 = k [0.07] x [0.036] y Rate 3 = 7.8 x 0-6 = k [0.036] x [0.054] y Rate 4 =.4 x 0-5 = k [0.050] x [0.07] y Divide Rate by Rate to find -6 R 3.9 x 0 k R.6 x 0 k Therefore = Y Y (.5) Rate =.6 x 0-6 = k [0.08] x [0.036] y Rate = 3.9 x 0-6 = k [0.07] x [0.036] y Rate 3 = 7.8 x 0-6 = k [0.036] x [0.054] y Rate 4 =.4 x 0-5 = k [0.050] x [0.07] y How do you find y? Remember that is no longer unknown! Rate = k [S O 8 ] [I - ] y Divide Rate 4 by Rate to find Y -6 Y R4 4 x 0 k (.78)() -6 Y R.6 x 0 k y For the Reaction S O 8 (aq) + 3 I (aq) SO 4 (aq) + I 3 (aq) Rate law : Rate = k [S O 8 ] [I - ] Order : + Y = + = Rate constant :. x 0-6 = k (0.08)(0.036) k =.5 x 0-3 Units??? 6

7 For First Order Reactions: A B Extra. Integrated form of the rate equation. Half life final initial d A] [ t 0 k dt [A] 0 = initial concentration at time t = 0 [A] = concentration at time t 0 ln k t For st Order Reactions 0 ln k t When ln P is plotted as a function of time, for reaction CH 3 NC CH 3 CN 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. Therefore CH 3 NC CH 3 CN is a st order process and k is the negative of the slope = s. II. Reaction Rates & Half-Life The half-life (t / ) of a reaction is the TIME required for the concentration of a reactant to decrease to one half its initial value Half-life {t / } is defined as the time required for one-half of a reactant to react For A B at t /, the concentration of A is one-half the initial concentration of A 7

8 Half-life is defined as the time required for one-half of a reactant to react [A] at t / is one-half of the original [A], at 3,000 sec [CH3NC] = 0.5 [A] 0 All radioactive decay is st order [ A ] 0 ln k t [ A ] At t / concentration is ½ initial ln 0 k t / A 0 ln = k t / Plants take up atmospheric CO by photosynthesis, and are ingested by animals, so every living thing is constantly exchanging 4 C with its environment as long as it lives Once it dies this exchange stops, and the amount of 4 C gradually decreases through radioactive beta decay What is the rate constant for the radioactive C 4 decay? Carbon-4 has a half-life of 5,730 ± 40 year All radioactive decay is st order ln = k t / k = ln / t / = / 5730 k =. x 0 4 yr - 8