chemreviewtest1 - The Best Review Ever For Exam 1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: The Best Review Ever For Exam 1 – Dr. Holcombe CH 302 Spring 2010 By: Paul Alexander Vapor pressure: the pressure exerted by the vapor of a substance that exists in a condensed (liquid or solid) phase. The more surface, the more vapor. Smaller IMF force, greater vapor pressure because less force holding substance together As IMF increases  ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ Dipoles <dipole ­dipole < h ­bonding < ionic Vapor pressure decreases  ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ He > CH4>C3H8>CCl4 > CHCl3>CH3OH>H20>NaCl>CaO Vapor pressure is temperature dependent**** Plot of vapor pressure vs. temp gives us an EXPONENTIAL graph ∆G = RT ln(P) standard free energy of vaporization but: ∆G(vap) = ∆H(vap)  ­ T∆S(vap) Vant Hoff Eq: so: ln P =  ­∆G/RT =  ­∆H/RT + ∆S/R a constant, assuming ideal gas beh., mole of gas has about the same entropy 1 The Best Review Ever For Exam 1 – Dr. Holcombe CH 302 Spring 2010 By: Paul Alexander or if you exponentiate: As ∆H decreases with increasing IMF, boiling point also increases. Think of it as ∆H being the heat of the reaction. If it’s really cool (decrease in heat) then there are probably more IMF (more solid/liquid particles than gas), and therefore the BP goes up because it takes more heat to break the molecules into smaller ones since the IMF is increased. Clausius Clapeyron equation: allows us to use one value of P1 and T1 to predict the vapor pressure anywhere else on the graph Since ∆S/R is a constant, we identify two states of system P1,T1 and P2, T2 to get: Ln (P2/P1) = ∆H/R ( 1/T1  ­ 1/T2 ) 1 torr = 133.3 Pa make sure to convert **** 1 kJ = 1,000 J T = C + 273 Rearranged to solve for boiling pt.: 1/T1 = 1/T2 + R/∆H(Ln (P2/P1)) For phase change diagrams: In general, As T increases, gas forms As T decreases, solid forms As P increases, solid forms As P decreases, gas forms Solid lines identify phase transitions (ex. Solid/liquid). They exist simultaneously and define melting/freezing pt, etc. for compound. Water is weird: as pressure goes up, ice melts. Ex. When you bite ice cube, it liquefies. *** ∆H = ∆H prod  ­ ∆H react ∆S = ∆S prod  ­ ∆S react T = ∆H / ∆S P = K e ­∆H/RT 2 The Best Review Ever For Exam 1 – Dr. Holcombe CH 302 Spring 2010 By: Paul Alexander A solution is the consequence of mixing two or more compounds A solute is the smaller quantity in the solution A solvent is the larger quantity in the solution ∆G is negative means that a solid dissolved in solution or that one liquid is miscible with another liquid ∆H is uncertain with mixing ∆S is always positive when mixing a solution Dissolving Salts in H2O: Solubility depends on the magnitude of ∆H of the solution... ∆H solution = ∆H (solvation/hydration)  ­ ∆H (energy of forming salt) ∆H solution is value that goes in ∆G = ∆H  ­ T∆S ***Most salts increase in solubility with increase in temperature Charge density and ∆H (hydration): *The smaller charge density means weaker IMF, means easier to melt and dissolve *Larger charge density means larger IMF, harder to melt and dissolve Ion Charge Charge Density ∆H (hydr) K+ +1 0.6 350 kJ Na+ +1 0.9 440 kJ Ca++ +2 1.8 1900 kJ Al +++ +3 4.4 4800 kJ So when you look at ions like K2O and want to know if it dissolves in H20, just look at the place on the periodic table of the first element of compound. Since K+ has a smaller charge density, it will more than likely dissolve. In comparison, Al2O3 will not dissolve in H20. Look at the charge density of Al. 3+ is not favorable. Dissolving liquids in liquids: Miscible: two liquids that are mixed together to form a single phase. ∆G =  ­ Immiscible: two liquids do not mix well together, so exist separately: ∆G=+ “like dissolves like” means that the more alike the IMF are the more likely they are to dissolve with one another 3 The Best Review Ever For Exam 1 – Dr. Holcombe CH 302 Spring 2010 By: Paul Alexander Gases Mixing with Liquids: We know gases dissolve in liquids because we can hear them fizz, and we know that fish breathe O2, but some gases are preferred over others. Ex. CO2 is very soluble in H20 because it reacts to make H2CO3. Chemical Reactivity drives solubility. HF is very soluble in H20. Highly soluble because of hydrogen bonding strength. HCl and HI are highly soluble in H20. HCl and HI dissociate completely to make H30+ + H20 + Cl ­ Henry’s Law: as pressure of a gas above a liquid increases, the solubility of the gas in the solvent increases proportionally. At higher temperatures, concentration of a gas (ex. O2) decreases Colligative properties: solution property for which it is the amount of solute dissolved in the solvent that matters, but not the kind of solute Examples: vapor pressure lowering, freezing point depression, osmotic pressure, boiling point elevation Mole fraction X = moles solute/ total moles Molality = same as molarity except we divide by the mass of the solvent in kg rather than liters of the solution = m Molality (m) = moles solute / kg solvent Molarity (M) = moles solute/Liters solution Concept: when adding compounds into a mixture, you must realize that adding 1 mol of NaCl is adding 2 moles of particles in the solvent, 1 mole of Na2S is 3 moles of particles, 1 mole of insoluble CaS is 0 moles of particles in the solvent, 1 mole of sugar is 1 mole of particles in the solution The more moles of particles added to the solution, the more one raises the boiling point Vapor Pressure Depression (Raoult’s Law): the vapor pressure above a solution is proportional to the mole fraction of the solute. ∆P = P*X X = mole fraction of solute added 4 The Best Review Ever For Exam 1 – Dr. Holcombe CH 302 Spring 2010 By: Paul Alexander P = vapor pressure of solvent at T of solvent At ANY liquidʼs normal boiling point, the vapor pressure = 1.0 atm. Volatile - Liquid with high vapor pressure Easily vaporized liquids are called volatile liquids, and they have relatively high vapor pressures. Boiling Point Elevation m = molality of solute added ∆T= temperature increase because of solute Freezing Point Depression: ∆T =  ­Kf*mi ∆T = temperature decrease because of solute m = molality of solute added Osmotic Pressure: osmotic pressure increase because of solute Π = iMRT in atm (760 torr) M = Molarity of solute added R= .08206 T=298K i= moles particles in solution or moles of solute dissolved typically sugars like C6H12O6 = 1, things like NaCl = 2, etc. Plot of mole fraction vs. vapor pressure shows linear correlation Important: For gases that do not react chemically with Water: (exact opposite of what youʼd expect for something that IS soluble in water/has the same bond properties as water) *solubility of the gas in water generally increases with an increase in the pressure of the gas *solubility of the gas in water decreases with increasing temperature. ∆T = Kb*mi K = [prod]/[react] at equilibrium If K > 1 the reaction happened and ∆G is negative If K < 1 the reaction didn’t happen and ∆G is positive Q = [C]c [D]d / [A]a [B]b Brackets refer to concentration of the compounds in molarity units (moles per liter of solution) Coefficients are the exponents in Q Q = K at equilibrium Q < K reaction left  > right 5 The Best Review Ever For Exam 1 – Dr. Holcombe CH 302 Spring 2010 By: Paul Alexander Q > K reaction right  > left The difference between Q and K is that K uses values at equilibrium and Q uses values not at equilibrium for comparison. Ex. Reaction: 3H2 N2 = 2NH3 Initial C C C Change C + [H2] C +[N2] C+[NH3] Equilibrium [H2] [N2] [NH3] *** K = [NH3]^2 / [H2]^3 * [N2] = 3.8 INITIAL is the data used to find Q EQUIL is the data used to find K Make sure you can solve a fourth quadratic equation on a calculator!! It’ll help when you’re solving tough problems given only the concentration of a final product and a K value. Solids and Pure liquids are NOT included in equilibrium expression calculations Free Energy and K are related: R = 8.314 ∆G =  ­RT ln K Rearranged: K = e ­∆G/RT Use the second when given free energy asked to find eq. constant When dealing with gases: Kp =Kc RT∆n ∆n = change in moles of gas (prod – react) Equilibrium constants are DIMENSIONLESS:::Every concentration or pressure that enters into Kc or Kp is really divided by the corresponding concentration or pressure of the substance in its standard state. The amount of each component is in terms of activity, which is the measured amount (concentration, pressure) divided by the amount of that component in its standard state in that unit. The units in the numerator and denominator are identical and cancel out. LeChatelier [leh shat uh leer]’s Principle: tells us how to relieve stress from a system Ex. aA + bB  ­>< ­ cC + dD 6 Kp =Kc(RT)−1 The Best Review Ever For Exam 1 – Dr. Holcombe CH 302 Spring 2010 By: Paul Alexander Concentration stress: If we add A or B to left, reaction shifts right to decrease A and B Remove A or B from left, reaction shifts left, to increase A or B Add C or D from right, reaction shifts left to decrease C or D Remove C or D from right, reaction shifts right to increase C or D Pressure Stress: affects only gases If increase P, system shifts to make fewer gas molecules If decrease P, system shifts to make for more gas molecules p.s. no shift in equilibrium as pressure changes, ∆n = 0 Temperature stress: Endothermic reactions shift left to get warmer, and right to get colder Exothermic shifts right to get warmer, left to get colder Review from HW’s and Test: First law of thermo: ∆Euniv = 0 from CH 301, just a reminder for calorimeter probs: −mmCs,m∆Tm =mwCs,w∆Tw Elements in standard states have standard enthalpies of formation = to zero The second law of thermodynamics states that the entropy of the universe is always increasing. Consequently, only processes which increase the overall entropy of the universe satisfy the second law and happen spontaneously. The Second Law of Thermodynamics also states that in spontaneous reactions, the universe tends toward a state of greater disorder. When the entropy increase in the system is greater than the entropy decrease in the surroundings, the entropy of the universe will increase From ∆S = T since T is in the denomina- tor, ∆S will be larger (more positive) whenever T is smaller. ∆H0 =!n∆Hf0products− n ∆Hf0 reactants ∆S0 =!n∆S0 products −!n∆S0 reactants for enthalpy of vaporization we use: ∆H◦ =T∆S 7 The Best Review Ever For Exam 1 – Dr. Holcombe CH 302 Spring 2010 By: Paul Alexander In dynamic equilibrium the forward and reverse processes of the system continue but because their rates are equal there is no change in the composition of the system. When solving for new equilibrium constants, look at how the equation changed. Ex. 2 HgO(s) → 2 Hg(l) + O2(g) k = 2.8 .5 O2(g) + Hg(l) → HgO(s) since the equation reversed, and is half of the original, we use -.5 as our power, and we put the original K to this power to get our new k. so 2.8^-0.5 = .60 For a pure solid or liquid, the molar free energy always has its standard value. ∆G (of rxn) + 0 Endo/exo? Endothermic Exothermic Neither Spont? Nonspont. Spont. Neither K value? 0 -1 >1 0 Also, this table helped in CH301, but here it is again: ∆S ∆H +  ­ + +  ­  ­  ­ + Spontaneity? All temps. High temps. Low temps (exotherm) No temps. The observation that the solubility of gases in water decreases with increasing temperature means that the enthalpy of solvation (∆Hsolvation) for gases is : exothermic. A high velocity gas molecule must release its excess kinetic energy as heat in order to decelerate and enter the aqueous phase. 8 The Best Review Ever For Exam 1 – Dr. Holcombe CH 302 Spring 2010 By: Paul Alexander Number 6 on HW #4 is good practice in solving for final concentrations given the equilibrium constant Kc = 49, and initial concentrations of 0.5 M and HI of 0.0 M 9 The Best Review Ever For Exam 1 – Dr. Holcombe CH 302 Spring 2010 By: Paul Alexander Which curve best describes the sit- uation where the more volatile component of the mixture CAN be initially collected in the pure form in the distillate if XA in the boiling pot = 0.5? This is because if we look at where the 0.5 of X is located on the graph we can see it falls on the right side of the graph, where the graph is in a decreasing slope. Since this is true, it will continue to decrease until the pure form of 1.0 X of A is reached. You can tell whether a mixture is a colloid or a solution by shining a light through it. Remember the example of the chalk dust in class with a laser. 10 ...
View Full Document

This note was uploaded on 09/26/2010 for the course CH 53215 taught by Professor Mccord during the Spring '09 term at University of Texas at Austin.

Ask a homework question - tutors are online