Chapter 19 – Principles Chapter 19 – Principles of Reactivity: Entropy of Reactivity: Entropy

and Free Energyand Free Energy

Objectives:1) Describe terms: entropy and

spontaneity.2) Predict whether a process will be

spontaneous.3) Describe: free energy.4) Describe the relationship

between G, K, and product favorability.

ThermodynamicsThermodynamics

• Thermodynamics is _______________ ____________________.

• First Law of Thermodynamics– The law of conservation of energy: ______

________________________________.E = q + w

– The change in internal energy of a system is the sum of the heat transferred to or from the system and the work done on or by the system.

Spontaneous ChangeSpontaneous Change

• Chemical changes, physical changes

• Spontaneous change: occurs _____________________. It leads to ____________.

• Example: heat transfers spontaneously from a hotter object to a cooler object.

• ____________ is reached in product-favored and in reactant-favored processes.

Spontaneous Chemical Spontaneous Chemical ReactionsReactions

2 H2 + O2 2 H2O

CH4 + 2 O2 CO2 + 2 H2O

2 Na + Cl2 2 NaCl

HCl + NaOH NaCl + H2O• Common feature: _____________ • But many processes are ____________

and spontaneous.• H2 + I2 2 HI (g) _______________ can

be approached from either direction.

Spontaneous ProcessesSpontaneous Processes

Spontaneous ProcessesSpontaneous Processes

• Dissolving NH4NO3 in water: H = +25.7 KJ/mol

• Expansion of a gas into a vacuum: energy neutral, heat is neither evolved nor required.

• Phase changes: melting of ice requires ~ 6 kJ/mol; but only occurs if T > 0oC.– ______________ determines whether a process is

spontaneous.

• Heat transfer: The T of a cold substance in a warm environment will rise until the substance reaches the ambient T.– The required heat comes from the _____________.

EntropyEntropy

• To predict whether a process will be spontaneous.

• Entropy, S is a thermodynamic function– State function: a quantity whose value is

determined only by the initial and final states of a system.

• Second Law of Thermodynamics– _________________________________________________

___________________________.– _________________________________________________

___________________________.

Dispersal of EnergyDispersal of Energy

• By statistical analysis:• Energy is distributed of a number of

particles• Most often case is when energy is

distributed over all particles and to a large number of states.

• As the number of particles and the number of energy levels grows, one arrangement turns out to be vastly more probable than all others.

Dispersal of EnergyDispersal of Energy

• Dispersal of __________ often contributes to energy dispersal.

Boltzmann EquationBoltzmann Equation

• Ludwig Boltzmann (1844-1906)• Look at the distribution of energy over

different energy states as a way to calculate ____________.S = k log W

• K – Boltzmann constant• W – represent the number of different

ways that the energy can be distributed over the available energy levels.

• A maximum entropy will be achieved at _________________ , a state in which W has the maximum value.

Matter and Energy DispersalMatter and Energy Dispersal

Matter and Energy DispersalMatter and Energy Dispersal

Summary: Matter and Energy Summary: Matter and Energy DispersalDispersal

• A final state of a system can be more probable than the initial state if:– The atoms and molecules can be more

____________ and/or– ___________ can be dispersed over a greater

number of atoms and molecules.• If energy and matter are both dispersed in a

process, it is _______________.• If only matter is dispersed, then

quantitative information is needed to decide whether the process is spontaneous.

• If energy is not dispersed after a process occurs, then that process will ____________ _____________________.

EntropyEntropy

• Entropy is used to __________________ ___________ resulting from dispersal of energy and matter. The greater the _______ in a system, the greater the value of S.

• Third Law of Thermodynamics• There is no disorder in a perfect

crystal at 0K, S=0.• The entropy of a substance at any T can be

obtained by measuring the heat required to raise the T from 0K, where the conversion must be carried by a reversible process (very slow addition of heat in small amounts).

EntropyEntropy• The entropy of a substance at any T can be

obtained by measuring the ________ required to raise the T from 0K, where the conversion must be carried by a reversible process (very slow addition of heat in small amounts).

• The entropy added by each incremental change is: S =

• Adding the entropy changes gives the total entropy.

• All substances have ___________ entropy values at temperatures above 0K.

Standard Molar Entropy ValuesStandard Molar Entropy Values

Standard Molar Entropy ValuesStandard Molar Entropy Values

ThermodynamicsThermodynamics

• First Law: The total energy of the universe is a constant.

• Second Law: The total entropy of the universe is always increasing.

• Third Law: The entropy of a pure, perfectly formed crystalline substance at 0K is zero.- A local decrease in entropy (the assembly of large molecules) is offset by an increase in entropy in the rest of the universe -.

Standard EntropyStandard Entropy

• So, is the entropy gained by converting it from a perfect crystal at 0K to standard state conditions (1 bar, 1 molal solution).

• Units: J/Kmol• Entropies of gases are ____________than

those for liquids, entropies of liquids are ____________ than those for solids.

• Larger molecules have a _________ entropy than smaller molecules, molecules with more complex structures have ________entropies than simpler molecules.

EntropyEntropy

• The entropy of liquid water is ___________ than the entropy of solid water (ice) at 0˚ C.

S˚(HS˚(H22O sol) < S˚(HO sol) < S˚(H22O liq)O liq)

EntropyEntropy

Entropies of ionic solids depend on Entropies of ionic solids depend on ___________________________.___________________________.

SSoo (J/K•mol) (J/K•mol)

MgOMgO 26.926.9

NaFNaF 51.551.5

SSoo (J/K•mol) (J/K•mol)

MgOMgO 26.926.9

NaFNaF 51.551.5

MgMg2+2+ & O & O2-2- NaNa++ & F & F--

The larger coulombic attraction on MgO than NaF leads to a lower entropy.

Which substance has the Which substance has the higher entropy, why?higher entropy, why?

• O2 (g) or 03 (g)

• SnCl4 (l) or SnCl4 (g)

Arrange the substances in order of increasing Arrange the substances in order of increasing entropy. Assume 1 mole of each at standard entropy. Assume 1 mole of each at standard

conditions.conditions.

HCOOH(l)CO2(g)

Al(s)CH3COOH(l)

Predict whether S for each reaction would be Predict whether S for each reaction would be greater than zero, less than zero, or too close to greater than zero, less than zero, or too close to

zero to decide.zero to decide.

CO(g) + 3 H2(g) CH4(g) + H2O(g)

2 H2O(l) 2 H2(g) + O2(g)

I2(g) + Cl2(g) 2 ICl(g)

Entropy ChangeEntropy Change

S increases S increases slightly with Tslightly with T

S increases a S increases a large amount large amount with phase with phase changeschanges

Entropy ChangeEntropy Change

• Entropy usually increases when a Entropy usually increases when a pure liquid or solid ______________ in pure liquid or solid ______________ in a solvent.a solvent.

• Entropy of a substance ____________ Entropy of a substance ____________ with temperature.with temperature.

Entropy ChangeEntropy Change

• The entropy change is the sum of the entropies of the products minus the sum of the entropies of reactants:

S0system = S0 (products) – S0 (reactants)

You will find So values in the Appendix L of your book.

Calculate the standard entropy changes for the Calculate the standard entropy changes for the evaporation of 1.0 mol of liquid ethanol to ethanol evaporation of 1.0 mol of liquid ethanol to ethanol

vapor.vapor.

C2H5OH(l) C2H5OH(g)

Calculate the standard entropy change for Calculate the standard entropy change for forming 2.0 mol of NHforming 2.0 mol of NH33(g) from N(g) from N22(g) and (g) and

HH22(g)(g)

N2(g) + 3 H2(g) 2 NH3 (g)

Using standard absolute entropies at 298K, Using standard absolute entropies at 298K, calculate the entropy change for the system when calculate the entropy change for the system when 2.35 moles of NO(g) react at standard conditions.2.35 moles of NO(g) react at standard conditions.

2 NO(g) + O2(g) 2 NO2(g)

Calculate the standard entropy change for Calculate the standard entropy change for the oxidation of ethanol vapor (CHthe oxidation of ethanol vapor (CH22HH55OH OH

(g)).(g)).

Entropy in the UniverseEntropy in the Universe

S0univ = S0

sys + S0surr

• 2nd Law of Thermodynamics: Suniv is positive for a spontaneous process.

• For a nonspontaneous process: S0

univ < 0 (negative)• If Suniv = 0 the system is at equilibrium.• Calculate first the S0

sys, then S0surr.

S0surr = qsurr/T = -H0

sys/T

H0sys = H0 (products) – H0 (reactants)

Show thatShow that SS00univuniv is positive (>0) for is positive (>0) for

dissolving NaCl in waterdissolving NaCl in waterS0

univ = S0sys + S0

surr

1) Determine S0sys

2) Determine S0surr

NaCl(s) NaCl (aq)

Show thatShow that SS00univuniv is positive (>0) for is positive (>0) for

dissolving NaCl in waterdissolving NaCl in water

Predicting whether a Process Predicting whether a Process will be Spontaneous – Table will be Spontaneous – Table

19.219.2Based on the values of H0sys and S0sys there are 4 types:1) H0sys < 0 Exothermic & S0sys > 0 Less order S0univ > 0 ________________ under all conditions.

2) H0sys < 0 Exothermic & S0sys < 0 More order Depends on values, more favorable at __________ temperatures.

3) H0sys > 0 Endothermic & S0sys > 0 Less order Depends on values, more favorable at __________ temperatures.

4) H0sys > 0 Endothermic & S0sys < 0 More orderS0univ < 0 ___________________ under any conditions.

Remember that –∆H˚Remember that –∆H˚syssys is proportional to ∆S˚ is proportional to ∆S˚surrsurr

An exothermic process has ∆S˚An exothermic process has ∆S˚surrsurr > 0. > 0.

Remember that –∆H˚Remember that –∆H˚syssys is proportional to ∆S˚ is proportional to ∆S˚surrsurr

An exothermic process has ∆S˚An exothermic process has ∆S˚surrsurr > 0. > 0.

Classify the following as one of Classify the following as one of the four types of Table 19.2the four types of Table 19.2

H0 (kJ) S0 (J/K)

CH4 (g) + 2 O2 (g) 2 H2O (l) + CO2 (g) -890 -242.8

2 FeO3(s) + 3 C (graphite) 4 Fe(s) + 3 CO2 (g) +467 +560.7

Calculate the entropy change of the UNIVERSE when 1.890 moles of CO2(g) react under standard

conditions at 298.15 K.

Consider the reaction

6 CO2(g) + 6 H2O(l) C6H12O6 + 6O2(g)for which Ho = 2801 kJ and So = -259.0 J/K at 298.15 K.

• Is this reaction reactant or product favored under standard conditions?

Gibbs Free EnergyGibbs Free Energy

Suniv = Ssurr + Ssys

Ssurr -Hsys/T

Suniv = -Hsys/T + Ssys

Multiply equation by –T-T Suniv = Hsys –TSsys

J. Willard Gibbs (1839-1903)Gsys = -T Suniv

Gsys = Hsys –TSsys

Gsys < 0, a reaction is ____________

Gsys = 0, a reaction is _____________ Gsys > 0, the reaction is ____________

Gibbs Free Energy and Gibbs Free Energy and SpontaneitySpontaneity

• J. Willard Gibbs (1839-1903)• Gibbs free energy, G, “free energy”, a

thermodynamic function associated with the ________________.G = H –TS

H- EnthalpyT- Kelvin temperatureS- Entropy• Changes during a process: G• Use to determine whether a reaction is __________. G is ___________related to the value of the

_____________________________ , and hence to product favorability.

““Free” EnergyFree” EnergyG = w max

• The free energy represents the maximum energy ____________________________.

Example: C(graphite) + 2 H2 (g) CH4 (g)

H0rx = -74.9 kJ; S0rx = -80.7 J/KG0rx = H0 – TS0

= -74.9 kJ – (298)(-80.7)/1000 kJ= -74.9 kJ + 24.05 kJG0rx = - 50.85 kJ• Some of the energy liberated by the reaction is

needed to “order” the system. The energy left is energy available energy to do_________, “free” energy.

G < 0, the reaction is _______________.

Calculate Go for the reaction below at 25.0 C.

P4(s) + 6 H2O(l) → 4 H3PO4(l)

G0rx = H0 – TS0

Species fH

(kJ/mol) fS (J/K·mol)

P4(s) 0 22.80 H2O(l) -285.8 69.95 H3PO4(l) -1279.0 110.5

Standard Molar Free Energy of Standard Molar Free Energy of FormationFormation

• The standard free energy of formation of a compound, G0

f, is the free energy change when forming __________of the compound from the __________________, with products and reactants in their __________________.• Then, G0

f of an element in its standard states is _________.

Gibbs Free EnergyGibbs Free Energy

G0rxn is the increase or decrease in free

energy as the reactants in their standard states are converted completely to the products in their standard states.

* Complete reaction is not always ________________.* Reactions reach an _____________.

G0system = G0 (products) – G0 (reactants)

Calculating Calculating GG00rxn rxn from from GG00

ff

G0system = G0 (products) – G0 (reactants)

Calculate the standard free energy change for the oxidation of 1.0 mol of SO2 (g) to form SO3 (g).

G0system =

Gf0 (kJ/)

SO2(g) -300.13SO3(g) -371.04

Free Energy and Free Energy and TemperatureTemperature

• G = H – TS• G is a function of T, G will change as T

changes.• Entropy-favored and enthalpy-

disfavored• Entropy-disfavored and enthalpy-

favored

Changes in Changes in GG00 with T with T

Consider the reaction below. What is Consider the reaction below. What is GG00 at 341.4 K and at 341.4 K and will this reaction be product-favored spontaneously at will this reaction be product-favored spontaneously at

this T?this T?

CaCO3(s) CaO(s) + CO2(g)

Thermodynamic values:

Hf0 (kJ/mol) S0 (kJ/Kmol)

CaCO3(s) -1206.9 +0.0929

CaO(s) -635.1 + 0.0398

CO2(g) -393.5 + 0.2136

Estimate the temperature required to Estimate the temperature required to decompose CaSOdecompose CaSO44(s) into CaO(s) and SO(s) into CaO(s) and SO33(g).(g).

CaSO4(s) CaO(s) + SO3(g)

H0sys = H0 (products) – H0 (reactants)

H0sys = S0 (products) – S0 (reactants)

Thermodynamic values: Hf

0 (kJ/mol) S0 (J/Kmol)CaSO4(s) -1434.52 +106.50CaO(s) -635.09 + 38.20SO3(g) -395.77 + 256.77

For the reaction: 2H2O(l) 2H2(g) + O2(g)Go = 460.8 kJ and Ho = 571.6 kJ at 339 K and 1

atm.

• This reaction is (reactant,product) _____________ favored under standard conditions at 339 K.

• The entropy change for the reaction of 2.44 moles of H2O(l) at this temperature would be _________J/K.

Gorxn = Ho

rxn - T Sorxn

So = (Ho - Go)/T

GG00, K, and Product , K, and Product FavorabilityFavorability

• Large K – ____________ favored• Small K – ____________favored

• At any point along the reaction, the reactants are not under standard conditions.

• To calculate G at these points:

G = G0 + RT ln QR – Universal gas constantT - Temperature (kelvins)Q - Reaction quotient

GG00, K, and Product , K, and Product FavorabilityFavorability

G = G0 + RT ln QFor a A + b B c C + d DQ = [C]c [D]d [A]a [B]b

G of a mixture of reactants and products is determined by G0 and Q.When G is _____________ (“descending”) the reaction is spontaneous . At ______________ (no more change in concentrations), G = 0.

0 = G0 + RT ln K (at equilibrium) G0 = - RT ln K For G0 to be negative, K must be larger

than 1 and the reation is product favored.

Summary Summary GG00 and K and K • The free energy at equilibrium is ________ than the free

energy of the pure reactants and of the pure products.

G0 rxn can be calculated from:

G0rxn = G0 (products) – G0 (reactants)

Gorxn = Ho

rxn - T Sorxn

Gorxn = - RT ln K

• Grxn describes the direction in which a reaction proceeds to reach ___________, it can be calculated from:

Grxn = G0rxn + RT ln Q

– When Grxn < 0, Q < K, reaction proceeds spontaneously to convert ______________________ until equilibrium is reached.

– When Grxn > 0, Q > K, reaction proceeds spontaneously to convert ______________________ until equilibrium is reached.

– When Grxn = 0 , Q = K, reaction is ___________________.

The formation constant for [Ag(NHThe formation constant for [Ag(NH33))22]]++ is is 1.6 x101.6 x1077. Calculate . Calculate GG00 for the reaction for the reaction

below.below.

Ag+ (aq) + 2 NH3 (aq) [Ag(NH3)2]+ (aq)

G0 = -RTlnK

The reaction below has a The reaction below has a GG00 = -16.37 = -16.37 kJ/mol. Calculate the equilibrium constant.kJ/mol. Calculate the equilibrium constant.

1/2 N2 (g) + 3/2 H2 (g) NH3 (g)G0

rxn = G0f NH3 (g)

G0 = -RTlnK

The value of Ksp for AgCl (s) at 25oC is 1.8 x 10-10. Determine Go for the process:

Ag+ (aq) + Cl- (aq) AgCl (s) at 25oC.

The standard free energy change for a chemical reaction is -18.3 kJ/mole. What is the equilibrium

constant for the reaction at 87 C? (R = 8.314 J/K·mol)

End of ChapterEnd of Chapter

• Go over all the contents of your textbook.

• Practice with examples and with problems at the end of the chapter.

• Practice with OWL tutor.• Practice with the quiz on CD of

Chemistry Now.• Work on your OWL assignment for

Chapter 19.