Chapter 12: Thermodynamics Definitions & Concepts to know
Transcript of Chapter 12: Thermodynamics Definitions & Concepts to know
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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
− − − or + Spontaneous at low T;
Nonspontaneous at high T
+ + − or + Spontaneous at high T;
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
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?
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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)
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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
t1/2 = [A]0
2 𝑘
1 Rate = k[A]1 s– 1
ln[A]t = – kt + ln[A]0
ln [A]t
[A]0 = – kt
t1/2 = 0.693
𝑘
2 Rate = k[A]2 M– 1.s– 1 1
[A]t = kt +
1
[A]0
t1/2 = 1
𝑘[𝐴]0