Spring 2017_CH1020_Dr. Kreider-Mueller
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 = 1.38×10−23 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) 1st Law of Thermodynamics: in any process, spontaneous or nonspontaneous, the total energy of a system & its surroundings is constant
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2nd 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 3rd 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
Spontaneous at low T; − − − or + Nonspontaneous at high T Spontaneous at high T; + + − or + Nonspontaneous at low T At equilibrium: G = 0 = H − TS
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o We can calculate the temperature at which two phases are in equilibrium (Crossover ∆푯 Temperature): T = ∆푺 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?
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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 0.1-2 nm o Colloids: homogenous mixture containing particles with diameters of 2-500 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
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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
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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)
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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-, 1st-, and 2nd-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
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o If the reaction is 1st-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 2nd-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-1s-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 1st order with respect to compound A, 2nd order with respect to compound B, and 3rd 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-, 1st-, and 2nd- order rxns . Half-life: the time required for the reactant concentration to drop to one- half of its initial value . For a 1st-order rxn, each successive half-life is an equal period of time . For a 2nd-order rxn, each successive half-life is 2X as long as the preceding one Be able to explain how rates of zero, 1st, & 2nd 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 1st order reaction, a plot of ln[A] vs. time will yield a straight line (slope = −k; intercept = ln[A]0) For a 2nd order reaction, a plot of 1/[A] vs. time will yield a straight line (slope = k; intercept = 1/[A]0)
8 of 8 Spring 2017_CH1020_Dr. Kreider-Mueller 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 – 1 [A]0 0 Rate = k[A] M.s [A]t = – kt + [A]0 t1/2 = 2 푘
ln[A]t = – kt + ln[A]0 1 – 1 0.693 1 Rate = k[A] s t1/2 = [A] 푘 ln t = – kt [A]0
2 – 1 – 1 1 1 1 2 Rate = k[A] M .s = kt + t1/2 = [A]t [A]0 푘[퐴]0
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