Chapter 12: Thermodynamics Definitions & Concepts to know

Spring 2017_CH1020_Dr. Kreider-Mueller

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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


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


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

Chapter 12: Thermodynamics Definitions & Concepts to know

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