CH1020 Exam #1 Study Guide

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1 CH1020 Exam #1 Study Guide For reference see Chemistry: An Atoms-focused Approach by Gilbert, Kirss, and Foster Chapter 12: Thermodynamics Definitions & Concepts to know: Thermodynamics: the study of the interconversion of heat & other forms of energy Enthalpy (H): the sum of the internal energy & the pressure-volume product of a system (H = E + PV) o Exothermic Process (H = ): one in which energy (usually in the form of heat) flows from the system into the surroundings o Endothermic Process (H = +): one in which energy (usually in the form of heat) flows from the surroundings into the system o Be able to predict the sign of H for a given chemical equation or physical change o Recall: ΔHrxn = ΔHf (Products) ΔHf (Reactants) Entropy (S): a measure of how dispersed the energy in a system is at a specific temperature o Entropy is a state function: S = Sfinal Sinitial o S = +, randomness of the system increases o S =, randomness of the system decreases o Systems move toward an increase in randomness because a random arrangement of particles is more probable than an ordered arrangement The Boltzmann Equation explains this concept: S = klnw Where S = entropy of a state; W = # of ways the state can be achieved; k = R/NA = j/k (You do not have to perform calculations using the Boltzmann equation) o As temperature increases: random molecular motion increases, kinetic energy of molecules increase, entropy increases Standard Molar Entropy (S ): the absolute entropy of 1 mole of a substance in its standard state (P = 1 atm & usually T = 25 C) o Allows us to directly compare the entropies of different substances under the same set of temperature & pressure conditions o As molecular weight increases, S increases o S (gas) > S (liquid) > S (solid) o As molecular complexity increases, S increases o Be able to calculate S from tabulated values using the equation: S = S (Products) S (Reactants) 1 st Law of Thermodynamics: in any process, spontaneous or nonspontaneous, the total energy of a system & its surroundings is constant 1 of 8

2 2 nd Law of Thermodynamics: the principle that the total entropy of the universe increases in any spontaneous process Suniverse = Ssys + Ssurr where Ssurr = ( Hsys/T) o Suniverse > 0 spontaneous rxn o Suniverse < 0 nonspontaneous rxn o Suniverse = 0 rxn at equilibrium o A nonspontaneous rxn in forward direction is spontaneous in the reverse direction Spontaneous Process: a process that proceeds without outside intervention Whether a reaction is spontaneous or not has nothing to do with how fast a reaction occurs Nonspontaneous Process: a process that takes place only in the presence of some continuous external influence 3 rd Law of Thermodynamics: the entropy of a perfect crystal is zero at absolute zero Gibbs Free Energy (G): The maximum amount of energy released by a process occurring at constant temperature & pressure that is available to do useful work G = H TS o G < 0 spontaneous reaction o G > 0 nonspontaneous reaction o G = 0 equilibrium o G Tells us about the position & direction of a reaction o In any spontaneous reaction at constant temperature & pressure, the free energy of the system always decreases o Free energy is dependent upon i) temperature; ii) pressure; iii) physical state; iv) concentration (for solutions) H S G = H TS Reaction Spontaneity + Spontaneous at all a T + + Nonspontaneous at all T or or + At equilibrium: G = 0 = H TS Spontaneous at low T; Nonspontaneous at high T Spontaneous at high T; Nonspontaneous at low T 2 of 8

3 o We can calculate the temperature at which two phases are in equilibrium (Crossover Temperature): T = H S Standard Free-Energy Change (G ): the change in free energy that occurs when reactants in their standard states are converted to products in their standard states Standard Free-Energy of Formation (G f): the free-energy change for the formation of 1 mol of the substance in its standard state from the most stable form of its constituent elements in their standard states G = G f(products) G f(reactants) Be able to calculate H, S and G from tabulated values o If G f = ( ) the substance is stable & it does not readily decompose to its elements o If G f = (+) the substance is unstable & it can potentially decompose to its elements Be able to predict the sign of S for given process The entropy of a system generally increases when a reaction results in an increase in the # of gaseous particles. For example: N2O4 (g) 2 NO2 (g) Dissolution of Ionic Compounds: o Typically Sdissolution = (+) when ions have small charges o Typically Sdissolution = ( ) when ions have high charges Phase Changes: physical form, but not the chemical identity of a substance changes o Fusion (melting): sl H = + S = + o Freezing: ls H = S = o Vaporization: lg H = + S = + o Condensation: gl H = S = o Sublimation: sg H = + S = + o Deposition: gs H = S = Entropy increases as follows: solid < liquid < gas Be familiar with the entropy vs. temperature graph Why are there large discontinuous jumps in the graph? 3 of 8

4 Chapter 11: Properties of Solutions Definitions & Concepts to know: Mixture: any combination of 2 or more pure substances blended together in some proportion without chemically changing the individual substances o Heterogeneous: mixing of components is visually non-uniform o Homogeneous: mixing of components is visually uniform o Solution: homogenous mixture containing particles with diameters of nm o Colloids: homogenous mixture containing particles with diameters of nm o Suspensions: mixture containing particles with diameters of > 500 nm o Solute: dissolved substance in a solution (minor component of the soln) o Solvent: major component of the soln o 3 types of interactions among particles must be taken into account for the formation of a solution: Solvent-solvent (usually Hsolvent-solvent = + (endothermic) because energy must be absorbed to break up intermolecular forces between solvent molecules) Solute-solute (usually Hsolute-solute = + (endothermic) because energy must be absorbed to break up intermolecular forces between solute molecules) Solvent-solute (usually Hsolvent-solute = (exothermic) because solvent molecules cluster around solute particles, forming intermolecular forces) Hsoln = Hsolute-solute + Hsolvent-solvent+ Hsolvent-solute Hsoln = if solvent-solute interactions are dominant (strong IMFs form) Hsoln = + if solvent-solute interactions are not dominant (weak IMFs form) o Like dissolves like! (nonpolar molecules dissolve nonpolar molecules; polar molecules dissolve polar molecules) o You must know the equations to calculate: Molarity, molality, % mass, mole fractions (X) Molarity = (moles solute)/(liters of Solution) Molality = (moles solute)/(kg of solvent) % Mass = (Mass of component)/(total mass of solution) 100% X = (moles of component)/(total moles making up the solution) You must be able to convert between each of the above concentrations o Saturated Solution: a solution containing the maximum possible amount of dissolved solute at equilibrium (temperature dependent) o Miscible: mutually soluble in all proportions o Solubility: the amount of a substance that dissolves in a given volume of solvent at a given temperature 4 of 8

5 o Be able to read/interpret a solubility vs. temperature graph As temperature increases, the solubility of a solid or liquid usually increases Gases become less soluble in a liquid solvent as temperature increases Pressure has a profound effect on the solubility of a gas Henry s Law: solubility of a gas in a liquid (at a given temperature) is directly proportional to the partial pressure of the gas over the solution Solubility = k P where k = constant characteristic of a specific gas (mol/l atm); P = partial pressure of gas over the solution (atm) As pressure increases, the gas becomes more soluble Vapor Pressure: The partial pressure of a gas in equilibrium with liquid at a constant temperature o For a pure solvent, weaker IMF s between solvent molecules lead to a higher vapor pressure (easier to get the molecules into the vapor phase) o As temperature increases, vapor pressure increases Be able to read/interpret a distribution curve for the kinetic energy of molecules o At low temperature the curve is sharp, and only a few molecules have a high KE o At a higher temperature the curve is broad & more molecules have a higher value of KE General Chemistry: Atom s First, McMurry & Fay Normal boiling point: the temperature at which a liquid boils at P = 1 atm Be able to read/interpret a phase diagram 5 of 8

6 o Be able to locate/label: Triple point, critical point, 3 phases, normal B.P/F.P. o What does the slope of the solid/liquid line tell us? o For example, below is the phase diagram for H2O: Colligative Properties and Factors that affect them: o Colligative properties: depend only on the amount of dissolved solute rather than on the chemical identity of the solute Vapor-Pressure Lowering A soln of a nonvolatile solute has a lower vapor pressure, Pvap, than the pure solvent A soln always evaporates more slowly than a pure solvent because its vapor pressure is lower & its molecules escape less readily Raoult s Law: vapor pressure of a soln, Psoln, containing a nonvolatile solute is equal to the vapor pressure of the pure solvent, Psolv, times the mole fraction of the solvent, Xsolv. Psoln = Psolv Xsolv Where Xsolv = (moles of solvent/moles of solvent + i moles of solute) Boiling-point elevation Temperature at which Pvap reaches atmospheric pressure is higher for the soln than for the solvent Boiling point of the soln is higher by an amount Tb Tb = Kb m Where Tb = Tsoln Tsolv Where Kb = molal boiling-point-elevation constant characteristic of a given solvent; m = molal concentration of solute (You may need to consider the Van t hoff factor for this type of problem, so Tb = i Kb m) 6 of 8

7 Van t Hoff Factor: measure of the extent of dissociation of a substance i = (moles of particles in soln)/(moles of solute dissolved) For electrolytes: ideal Van t Hoff factor is the total number of ions dissociating. For nonelectrolytes: Ideal Van t Hoff factor is equal to 1. Freezing-point depression Solid/liquid phase transition line is lower for a soln Freezing point of the soln is lower by an amount Tf Tf = Kf m Where Tf = Tsolv Tsoln Where Kf = molal freezing-point-depression constant characteristic of a given solvent; m = molal concentration of solute (You may need to consider the Van t Hoff Factor for this type of problem, so Tf = i Kf m) Osmotic Pressure Osmosis: passage of solvent through a membrane from the less concentrated side to the more concentrated side Osmotic pressure (): amount of pressure necessary to cause osmosis to stop (achieve equilibrium) = i MRT where M = molar concentration of solute; R = gas constant; T = temperature (K), i=van t Hoff Factor Chapter 13: Kinetics Definitions & Concepts to know: Chemical Kinetics: study of the rate of change of concentrations of substances involved in chemical reactions A reaction occurs when reactants collide in the correct orientation, with enough energy o The rate of a given chemical reaction depends on concentration of reactants & temperature Average Rate of Reaction = [concentration]/ time o concentration of a product increases over time rate of formation = positive # o concentration of a reactant decreases over time rate of decomposition = negative # Instantaneous rate: rxn rate at time t Initial rate: rxn rate at time t = 0 Be familiar with 0-, 1 st -, and 2 nd -order rxns Given the reaction: A B o If the reaction is 0-order: the rate is independent of the concentration of [A] Rate = k k has units of M/s integrated rate law: [A]t = -kt + [A]0 7 of 8

8 o If the reaction is 1 st -order: as [A] is doubled, the rate doubles Rate = k[a] k has units of s 1 integrated rate law: ln[a]t = kt + ln[a]0 o If the reaction is 2 nd -order: as [A] is doubled, [A] 2 quadruples, & the rate increases by a factor of 4 Rate = k[a] 2 k has units of M -1 s -1 integrated rate law: 1/[A]t = kt + (1/[A]0) o Be able to determine order with respect to each reactant & the overall reaction order when given a rate law For example, given the reaction: A + B C + D where Rate = k[a][b] 2 The reaction is 1 st order with respect to compound A, 2 nd order with respect to compound B, and 3 rd order overall. The values of exponents in a rate law must be determined by experiment, they cannot be deduced from the stoichiometry of the rxn o Be able to answer conceptual questions regarding these rxns, as well as calculation based questions o Be able to use the integrated rate laws to calculate the following: a rate constant, k the time it took to go from [A]0 to [A]t initial concentration, [A]0 concentration at time t, [A]t o Be able to use initial rate of rxn to determine rxn order, the rate law, & perform related calculations Be able to use the ½-life equations (t1/2) for 0-, 1 st -, and 2 nd - order rxns Half-life: the time required for the reactant concentration to drop to onehalf of its initial value For a 1 st -order rxn, each successive half-life is an equal period of time For a 2 nd -order rxn, each successive half-life is 2X as long as the preceding one Be able to explain how rates of zero, 1 st, & 2 nd order reaction can be determined graphically For a 0 order reaction, a plot of [A] vs. time will yield a straight line (slope = k; intercept = [A]0) For a 1 st order reaction, a plot of ln[a] vs. time will yield a straight line (slope = k; intercept = ln[a]0) For a 2 nd order reaction, a plot of 1/[A] vs. time will yield a straight line (slope = k; intercept = 1/[A]0) 8 of 8

9 Summary of Zero- First- and Second-Order Reactions, Dr. Houjeiry Order Rate Law Units of k Integrated Rate Law Straight line Plot Half-Life Expression 0 Rate = k[a] 0 M.s 1 [A] t = kt + [A] 0 t 1/2 = [A] 0 2 k 1 Rate = k[a] 1 s 1 ln[a] t = kt + ln[a] 0 ln [A] t [A] 0 = kt t 1/2 = k 2 Rate = k[a] 2 M 1.s 1 1 = kt + 1 t 1/2 = [A] t [A] 0 1 k[a] 0 9 of 8