Chapter 19 Entropy and Free Energy Objectives: 1)Define entropy and spontaneity. 2)Predict whether a process will be spontaneous. 3)Describe free energy.

Chapter 19Entropy and Free Energy

Objectives:1) Define entropy and spontaneity.2) Predict whether a process will be

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

between DG, K, and product favorability.

How to predict if a reaction can How to predict if a reaction can occur, given enough time?occur, given enough time?

THERMODYNAMICSTHERMODYNAMICS

How far will it proceed?How far will it proceed? (K) (K)

How to predict if a How to predict if a reaction can occur at a reaction can occur at a reasonable rate?reasonable rate?

KINETICS (Ea)

Introduction

Thermodynamics• Energy relationships• 1st law of thermodynamics

– Energy is conserved: energy is not created nor destroyed

– E = q + w– qp = H Enthalpy: _______ transferred in a

process at constant pressure

• 2nd law of thermodynamics– S Entropy: increases in ________

processes– Direction of the process

spontaneous

non

spon

tane

ous

• Lighter reaction:

C4H8 + 6 O2 4 CO2 + 4 H2O

• Burning candle• Drop a pen• Gas expansion• Heat transfer

SpontaneousProcesses

• Proceed on its own without outside assistance.

• Occurs in a define _______• Leads to ___________• Maybe determined by T, P• The reverse process is

____________________

Identify spontaneous processesPredict whether the following are a) spontaneous as described, b) spontaneous

in the reverse direction, c) in equilibrium1) When a piece of metal heated to 150oC is added to water at 40oC, the

water gets hotter.

2) Benzene vapor, C6H6(g), at P= 1atm, condenses to liquid benzene at the normal boiling point of benzene, 80.1oC

3) Water at room temperature decomposes into H2(g) and O2(g)

4) AgCl(s) Ag+(aq) + Cl- (aq) ; K = 1.8 x 10-10

Reversible vs Irreversible Processes

• Sadi Carnot (French engineer -1824)– Efficiency of heat to work (steam engines)– Significant amount of heat is lost to surroundings

• Rudolph Clausius (German physicist ~ 1924)– Ratio of heat in an ideal engine and temperature at

which is delivered (q/T)– Entropy– Amount of work extracted from spontaneous

processes depends on the manner in which the process is carried (pathway)

Reversible vs Irreversible Processes• Reversible: a process which reverse direction

whenever an infinitesimal change is made in some property of the system. Often at equilibrium.– Here entropy can be obtained at any T by measuring

the heat required rise the temperature from 0K, with the slow addition of heat in very small amounts.

– S = qrev/T

• Any spontaneous process is Irreversible, they often involve non equilibrium conditions; in order to reverse the surroundings must do some work on the system (so the surroundings change).

Entropy

• Associated with the ___________ in a system

• Associated with the extent to which energy is ___________among the various motions of molecules of the system.

• Related to heat transfer and __________.

Entropy• S, is a ________ function (like internal energy, E, and

enthalpy, H).• The value of S is characteristic of the state of a

system (and a property of the bulk matter).• S, change in entropy, depends only on the initial and

final states of the system, and not in the path taken from one state to the other: S = Sfinal – Sinitial

• For isothermal processes: S = qrev/T– q: heat absorbed/released

– T: temperature in K

Adding entropy changes = total entropy

S for phase changes

• Melting of a substance at its m.p. and vaporization of a substance at its b.p. are isothermal processes.

• Change can be achieved by adding heat to/from the system to/from surroundings.

• qrev = H fusion (melting)

• T = 273 K (normal m.p. 1 atm and 273K).

• S fusion = qrev/T = H fusion /T

Calculate S• Calculate Ssystem and the Ssurroundings when 1 mol of ice

(~size of an ice cube) melts in your hand.– Process is not reversible (different T’s).– S can be calculated whether rev or irrev.– Hfusion H2O = 6.01 kJ/mol (melting is endothermic processes,

H is positive).Ssystem = qrev/T = =Ssurroundings = qrev/T (heat gained = - heat lost and T~ 37oC) = =Stotal = Ssystem + Ssurroundings

=Spontaneous or non spontaneous?If T’s were about the same, process would be reversible –

overall S = 0

2nd Law of Thermodynamics

• Any irreversible process results in an overall (increase/decrease) ___________ in entropy, whereas a reversible process results in no overall change in entropy.

• The sum of entropy of a system + entropy of the surroundings = total entropy change = S universe.

• Irreversible processes occur of their own accord are _________ (spontaneous/non spontaneous).

• The total entropy of the universe increases (Suniverse is __________ (positive/negative)) in any spontaneous process.

• Ssystem decreases as Fe + O2 form rust

• Ssurroundings?

• Stotal = Ssystem + Ssurroundings

Entropy- Molecular Interpretation• Ludwig Boltzmann (1844-1906)• Molecules store energy

– KMT– Higher T, higher KE and broader distribution of

molecular speeds– 3 kinds of motion: translational, vibrational , rotational

= motional energy of the molecule

Entropy- Molecular Interpretation

– Molecules moving – snapshot – microstate

W= number of microstates of a system

S = k ln W

k= Boltzmann’s constant 1.38 x 10-23 J/K

• S = k ln Wfinal – k lnWinitial = k ln Wfin/Winit

• Change leading to increase in number of microstates leads to a ________ (positive/negative) value of S.

• Related to probability.

Entropy and probability

Most often case is when energy is distributed over all particles and to a large number of states.

Entropy and probability

Probability and the locations of gas molecules. The two molecules are colored red and blue.a) Before the stopcock is opened, both molecules are in the left-hand flask. b) After the stopcock is opened, there are 4 possible arrangements of the two molecules. The greater number of possible arrangements corresponds to greater disorder in the system, in general, the probability that the molecules will stay in the original flask is (1/2)n, where n is the number of molecules.

Dispersal of Energy

• Dispersal of matter often contributes to energy dispersal.

W increases when:

• Change involves an:– Increase in _______– Increase in __________– Increase in _______________

• Entropy change, S sign will be ________

• The maximum entropy will be achieved at ____________ – a state in which W has the maximum value.

Explain why melting of ice is spontaneous process

Explain formation of solutions of some ionic solids

• If energy and matter are both dispersed in a process, the process is (spontaneous/non spontaneous) _______________.• If only matter is dispersed, __________________________.• If energy is NOT dispersed, the process will (be/never be) ______________ spontaneous.

Predict the sign of S – each process occurs at constant T

a) Ag+(aq) + Cl- (aq) AgCl(s)

b) H2O(l) H2O(g)

c) N2(g) + O2(g) 2 NO (g)

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

3rd Law of Thermodynamics• Lower the temperature until there is only a single

microstate:• The entropy of a pure crystalline substance at absolute

zero is ________: S (0K) = ____• The units of the lattice have no thermal motion.• S= k ln W =

• As T increases, S ___________:

S solid is (larger/smaller) than S liquid is (larger/smaller) than S gas

• Then, all substances have ________ (positive/negative) entropy values at T > 0K.

Entropy 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

Standard Molar Entropy Values

Standard Molar Entropy Values

Diamond is Diamond is thermodynamically thermodynamically (favored/not favored) (favored/not favored) ___________ to convert ___________ to convert to graphite, and _______ to graphite, and _______ (favored/not favored) (favored/not favored) kinetically.kinetically.

Thermodynamics vs Kinetics

• 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.

Standard Entropy

Entropy

• 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)

Entropy of Solids

• Entropy values of solids depend on:– Columbic attractions

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 sample has the higher entropy

a) 1 mol NaCl(s) or 1 ml HCl(g) at 25oC

b) 2 mol HCl(g) or 1 mol HCl(g) at 25oC

c) 1 mol HCl(g) or 1 mol Ar(g) at 298 K

d) O2 (g) or O3 (g)

Entropy 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 evaporation of 1.0 mol of liquid ethanol to ethanol vapor.

C2H5OH(l) C2H5OH(g)

Calculate the standard entropy change for forming 2.0 mol of NH3(g) from N2(g) and H2(g)

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

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

react at standard conditions.

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

Calculate the standard entropy change for the oxidation of ethanol vapor (CH2H5OH (g)).

Show that S0univ is positive (>0) for

dissolving NaCl in water

S0univ = S0

sys + S0surr

1) Determine S0sys

2) Determine S0surr

NaCl(s) NaCl (aq)

Classify the following as one of the 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 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 spontaneous

Gsys = 0, a reaction is at equilibrium

Gsys > 0, the reaction is not spontaneous

Gibbs Free Energy and Spontaneity

• 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

spontaneous.G is ___________related to the value of the

equilibrium constant K, and hence to product favorability.

“Free” 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 work, “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 Formation

• The standard free energy of formation of a compound, G0f, is the

free energy change when forming one mole of the compound from the component elements, with products and reactants in their standard states.• Then, G0

f of an element in its standard states is _________.

Gibbs 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 observed.

* Reactions reach an ______________.

G0system = G0 (products) – G0 (reactants)

Calculating G0rxn from G0

f

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 Temperature

• 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 G0 with T

Consider the reaction below. What is G0 at 341.4 K and will this reaction be product-favored spontaneously at 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 decompose CaSO4(s) into CaO(s) and SO3(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

G0, K, and Product Favorability

• Large K – product favored• Small K – reactant favored

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

• To calculate DG at these points:

G = G0 + RT ln QR – Universal gas constant

T - Temperature (kelvins)

Q - Reaction quotient

G0, K, and Product Favorability

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 negative (“descending”) the reaction is _____________ . At equilibrium (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 G0 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(NH3)2]+ is 1.6 x107. Calculate G0 for the reaction below.

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

G0 = -RTlnK

The reaction below has a G0 = -16.37 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)

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 -.

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.• Work on your OWL assignment for

Chapter 19.