Entropy and Free Energy

1111

Entropy and Free Entropy and Free EnergyEnergy

Entropy and Free Entropy and Free EnergyEnergy

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

THERMODYNAMICSTHERMODYNAMICS

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

KINETICSKINETICS

Modified 02/02/12 by Laura Peck to accompanyModified 02/02/12 by Laura Peck to accompanySilberberg: Principles of General Chemistry Chapter 20Silberberg: Principles of General Chemistry Chapter 20

2222ThermodynamicsThermodynamics• If the state of a chemical system is such that a If the state of a chemical system is such that a

rearrangement of its atoms and molecules would rearrangement of its atoms and molecules would

decrease the energy of the system--- decrease the energy of the system---

• ANDAND the K is greater than 1, the K is greater than 1,

• then this is a then this is a product-favoredproduct-favored system.system.

• Most product-favored reactions are Most product-favored reactions are

exothermicexothermic

—but this is not the only criterion—but this is not the only criterion

3333ThermodynamicsThermodynamics• The sign of The sign of ΔΔH cannot predict spontaneous change.H cannot predict spontaneous change.

• CHCH4(g) 4(g) + 2O+ 2O2(g) 2(g) CO CO2(g) 2(g) + 2H+ 2H22OO(g) (g) ΔΔH°H°rxnrxn = - 802kJ K>1 = - 802kJ K>1

• NN22OO5(s) 5(s) 2NO 2NO2(g) 2(g) + 1/2O+ 1/2O2(g) 2(g) ΔΔH°H°rxnrxn = +109.5 kJ K>1 = +109.5 kJ K>1

• *****Both are product favored********Both are product favored***

• Both product- and reactant-favored reactions can proceed to Both product- and reactant-favored reactions can proceed to

equilibrium in a equilibrium in a spontaneousspontaneous process.process.

• AgCl(s) AgCl(s) AgAg++(aq) + Cl(aq) + Cl––(aq) (aq) K<1 Reactant favoredK<1 Reactant favored

Reaction is not product-favored, but it moves Reaction is not product-favored, but it moves spontaneously toward equilibrium.spontaneously toward equilibrium.

• Spontaneous does not imply anything about time for reaction Spontaneous does not imply anything about time for reaction to occur.to occur.

4444Thermodynamics and Thermodynamics and KineticsKinetics

Thermodynamics and Thermodynamics and KineticsKinetics

Diamond is Diamond is thermodynamically thermodynamically favoredfavored to convert to to convert to graphite, but graphite, but not not kinetically favoredkinetically favored..

Paper burns — a Paper burns — a product-favoredproduct-favored reaction. reaction. Also Also kinetically favoredkinetically favored once once reaction is begun.reaction is begun.

5555

Spontaneous ReactionsSpontaneous ReactionsIn general, spontaneous In general, spontaneous

reactions are reactions are exothermicexothermic..

FeFe22OO33(s) + 2 Al(s) ---> (s) + 2 Al(s) --->

2 Fe(s) + Al2 Fe(s) + Al22OO33(s)(s)

∆∆H = - 848 kJH = - 848 kJ

6666

Spontaneous ReactionsSpontaneous ReactionsBut many spontaneous reactions or But many spontaneous reactions or

processes are endothermic or even processes are endothermic or even have ∆H = 0.have ∆H = 0.

NHNH44NONO33(s) + heat ---> (s) + heat --->

NHNH44NONO33(aq)(aq)

Discuss your labs using urea & Discuss your labs using urea & water or vinegar & baking soda!water or vinegar & baking soda!

7777

Entropy, SEntropy, SOne property common to One property common to

spontaneous processes is spontaneous processes is that the final state is more that the final state is more DISORDEREDDISORDERED or or RANDOMRANDOM than the original.than the original.

Spontaneity is related to an Spontaneity is related to an increase in randomness.increase in randomness.

The thermodynamic property The thermodynamic property related to randomness is related to randomness is ENTROPY, SENTROPY, S..

Reaction of K Reaction of K with waterwith water

8888

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

9999

Entropy and the Number of Microstates

• Ludwig Boltzmann defined the entropy (S) of a system in terms of W:

• S= klnW

• K is the universal gas constant R divided by Avogadro’s number. It equals 1.38x10-23J/K

• A system with fewer microstates (smaller W) among which to spread its energy has lower entropy (lower S)

• A system with more microstate (larger W) among which to spread its energy has higher entropy (higher S)

• Smore microstates > Sless microstates

10101010

11111111

How probable is it that reactant How probable is it that reactant molecules will react? molecules will react?

PROBABILITYPROBABILITY suggests that a suggests that a spontaneous reaction will result in spontaneous reaction will result in

the the dispersaldispersal

** ofof energyenergy

** or ofor of mattermatter

** or of or of energy & matterenergy & matter..

Directionality of Directionality of ReactionsReactions

12121212

Probability suggests that a spontaneous Probability suggests that a spontaneous reaction will result in the dispersal of reaction will result in the dispersal of energy or of matter or both.energy or of matter or both.

Matter DispersalMatter Dispersal


13131313

Probability suggests that a spontaneous Probability suggests that a spontaneous reaction will result in the dispersal of reaction will result in the dispersal of energy or of matter or both.energy or of matter or both.

Energy DispersalEnergy Dispersal


14141414


Energy DispersalEnergy Dispersal

Exothermic reactions involve a release of Exothermic reactions involve a release of stored chemical potential energy to the stored chemical potential energy to the surroundings. surroundings.

The stored potential energy starts out in a few The stored potential energy starts out in a few molecules but is finally dispersed over a molecules but is finally dispersed over a great many molecules. great many molecules.

The final state—with energy dispersed—is The final state—with energy dispersed—is more probable and makes a reaction more probable and makes a reaction spontaneous.spontaneous.

15151515

S (gases) > S (liquids) > S (solids)S (gases) > S (liquids) > S (solids)

ΔSΔSsyssys = S = SliquidH2OliquidH2O – S – SgasH2OgasH2O <0 <0

= = 69.95J/K*mol – 188.8J/K*mol = -118.9 J/K*mol69.95J/K*mol – 188.8J/K*mol = -118.9 J/K*mol

ΔsΔssyssys = S = SgasH2OgasH2O – S – SliquidH2OliquidH2O >0 >0

= 188.8J/K*mol – 69.95J/K*mol = +118.9J/K*mol= 188.8J/K*mol – 69.95J/K*mol = +118.9J/K*mol (discuss in light of microstates)(discuss in light of microstates)

SSoo (J/K•mol) (J/K•mol)

HH22O(liq)O(liq) 69.9569.95

HH22O(gas)O(gas) 188.8 188.8


HH22O(liq)O(liq) 69.9569.95

HH22O(gas)O(gas) 188.8 188.8

Entropy, S Entropy, S ((ΔΔsssyssys = S = Sfinalfinal – S – Sinitialinitial))

16161616

Entropy and States of Entropy and States of MatterMatter

S˚(BrS˚(Br22 liq) < S˚(Br liq) < S˚(Br22 gas) gas) S˚(HS˚(H22O sol) < S˚(HO sol) < S˚(H22O liq)O liq)

17171717

Entropy of a substance increases Entropy of a substance increases with temperature.with temperature.

Molecular motions Molecular motions of heptane, Cof heptane, C77HH1616

Molecular motions of Molecular motions of heptane at different temps.heptane at different temps.

Entropy, SEntropy, S

18181818

Increase in molecular Increase in molecular complexity generally complexity generally leads to increase in S.leads to increase in S.


19191919

Entropies of ionic solids depend on Entropies of ionic solids depend on coulombic attractions.coulombic attractions.


MgOMgO 26.926.9

NaFNaF 51.551.5


MgOMgO 26.926.9

NaFNaF 51.551.5


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

20202020

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

Entropy, SEntropy, SEntropy, SEntropy, S

21212121

Practice problems

• Choose the member with the higher entropy in each of the following pairs, and justify your choice (assume constant temperature except E)

• A) 1 mol of SO2(g) or 1 mol of SO3(g)

• B) 1 mol of CO2(s) or 1 mol of CO2(g)

• C) 3 mol of O2(g) or 2 mol of O3(g)

• D) 1 mol of KBr(s) or 1 mol of KBr(aq)

• E) Seawater at 2°C or at 23°C

• F) 1 mol of CF4(g) or 1 mol of CCl4(g)

SOSO3(g) 3(g) >atoms>atoms

COCO2(g) 2(g) s<l<gs<l<g

3 mol O3 mol O2(g) 2(g) >#mols>#mols

KBrKBr(aq) (aq) s<l<gs<l<g

23°C >T23°C >TCClCCl4(g) 4(g) >molar mass>molar mass

22222222

Where does the entropy go?

• The second law of thermodynamics states that entropy of the universe will increase.

• An exothermic reaction will release energy (E) to the environment and thus, entropy within the system will decrease. The system becomes more ordered.

• An endothermic reaction will absorb energy (E) from the environment and thus, entropy within the system will INCREASE. The system becomes more disordered.

• In both cases, the total S will be greater than zero.

• ΔSuniv = ΔSsys + ΔSsurr >0• (discuss combustion & urea/water reactions)

23232323

Standard Molar EntropiesStandard Molar Entropies

24242424Entropy Changes for Phase Entropy Changes for Phase ChangesChanges

For a phase change, For a phase change, ∆S = q/T∆S = q/T

where q = heat transferred in where q = heat transferred in phase changephase change

For HFor H22O (liq) ---> HO (liq) ---> H22O(g)O(g)

∆∆H = q = +40,700 J/molH = q = +40,700 J/mol

S = qT

= 40, 700 J/mol

373.15 K = + 109 J/K • molS =

qT

= 40, 700 J/mol

373.15 K = + 109 J/K • mol

25252525Entropy and TemperatureEntropy and TemperatureEntropy and TemperatureEntropy and Temperature

S increases S increases slightly with Tslightly with T

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

26262626

Consider 2 HConsider 2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)

∆∆SSoo = 2 S = 2 Soo (H (H22O) - [2 SO) - [2 Soo (H (H22) + S) + Soo (O (O22)])]

∆∆SSoo = 2 mol (69.9 J/K•mol) - = 2 mol (69.9 J/K•mol) - [2 mol (130.7 J/K•mol) + [2 mol (130.7 J/K•mol) +

1 mol (205.3 J/K•mol)]1 mol (205.3 J/K•mol)]

∆∆SSoo = -326.9 J/K = -326.9 J/K

Note that there is a Note that there is a decrease in S decrease in S because 3 because 3 mol of gas give 2 mol of liquid.mol of gas give 2 mol of liquid.

Calculating ∆S for a Calculating ∆S for a ReactionReaction

∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)

27272727

Practice Problem• Calculate ΔS°rxn for the combustion of 1 mol

of propane (C3H8) at 25°CWrite a balanced reaction. Since its combustion we know The productsWrite a balanced reaction. Since its combustion we know The products have to be COhave to be CO2(g) 2(g) and Hand H22OO(l)(l), the other reactant Has to be O, the other reactant Has to be O2(g)2(g)

CC33HH8(g) 8(g) + 5O+ 5O2(g)2(g) 3CO 3CO2(g) 2(g) + 4H+ 4H22OO(l)(l)*hint: since there’s 6 mols of gas reactants*hint: since there’s 6 mols of gas reactantsAnd only 3 mols of gas products – what doAnd only 3 mols of gas products – what doYou think entropy should do?You think entropy should do?

ΔΔS°S°rxnrxn = [(3mol CO = [(3mol CO22)(S°CO)(S°CO22) + (4mol H) + (4mol H22O)(S°HO)(S°H22O)] – O)] –

[(1mol C[(1mol C33HH88)(S°C)(S°C33HH88) + (5mol O) + (5mol O22)(S°O)(S°O22)])]

ΔΔS°S°rxnrxn = [(3mol)(213.7J/mol*K) + (4mol)(69.9J/mol*K)] – = [(3mol)(213.7J/mol*K) + (4mol)(69.9J/mol*K)] –

[(1mol)(269.9J/mol*K) + (5mol)(205.0J/mol*K)][(1mol)(269.9J/mol*K) + (5mol)(205.0J/mol*K)]

ΔΔS°S°rxnrxn = -374 J/K = -374 J/K

28282828LE CHATELIER'S PRINCIPLE

Le Chatelier's Principle:If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

• 1) Using Le Chatelier's Principle with a change of concentration

• Suppose you have an equilibrium established between four substances A, B, C and D.

• Increase concentration of A: Increase concentration of B:

29292929

Using Le Chatelier's Principle with a change of pressure

(This only applies to reactions involving gases):

30303030Using Le Chatelier's Principle with a change of

temperature

• For this, you need to know whether heat is given out or absorbed during the reaction. Assume that our forward reaction is exothermic (heat is evolved):

31313131What would happen if you changed the conditions by increasing the

temperature?

• According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the temperature is reduced again.

• Suppose the system is in equilibrium at 300°C, and you increase the temperature to 500°C. How can the reaction counteract the change you have made? How can it cool itself down again?

• To cool down, it needs to absorb the extra heat that you have just put in. In the case we are looking at, the back reaction absorbs heat. The position of equilibrium therefore moves to the left. The new equilibrium mixture contains more A and B, and less C and D.

32323232What would happen if you changed the conditions by decreasing the

temperature?

• The equilibrium will move in such a way that the temperature increases again.

• Suppose the system is in equilibrium at 500°C and you reduce the temperature to 400°C. The reaction will tend to heat itself up again to return to the original temperature. It can do that by favoring the exothermic reaction.

33333333

Le Chateliers principle

343434342nd Law of 2nd Law of ThermodynamicsThermodynamics

A reaction is spontaneous if ∆S for the A reaction is spontaneous if ∆S for the universeuniverse is is positive.positive.

∆∆SSuniverseuniverse = ∆S = ∆Ssystemsystem + ∆S + ∆Ssurroundingssurroundings

∆∆SSuniverseuniverse > 0 for spontaneous process > 0 for spontaneous process

First calc. entropy created by First calc. entropy created by mattermatter dispersal dispersal (∆S(∆Ssystemsystem))

Next, calc. entropy created by Next, calc. entropy created by energyenergy dispersal dispersal (∆S(∆Ssurroundsurround))

35353535

Dissolving NH4NO3 in water—an entropy driven process.


∆∆SSuniverseuniverse = =

∆∆SSsystemsystem + ∆S + ∆Ssurroundingssurroundings

36363636

2 H2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)

∆∆SSoosystemsystem = -326.9 J/K = -326.9 J/K

Sosurroundings =

qsurroundings

T =

-Hsystem

TSo

surroundings = qsurroundings

T =

-Hsystem

T

Sosurroundings =

- (-571.7 kJ)(1000 J/kJ)

298.15 KSo

surroundings = - (-571.7 kJ)(1000 J/kJ)

298.15 K


∆∆SSoosurroundingssurroundings = +1917 J/K = +1917 J/K

Can calc. that ∆HCan calc. that ∆Hoorxnrxn = ∆H = ∆Hoo

systemsystem = -571.7 kJ = -571.7 kJ

37373737Practice problem:Determining Reaction Spontaneity

• At 298K, the formation of ammonia has a negative ΔS°sys. Calculate ΔSuniv, and state whether the reaction occurs spontaneously at this temperature.

• N2(g) + 3H2(g) 2NH3(g) ΔS°sys = -197J/KΔΔHH°°syssys = = ΔΔHH°°rxnrxn

= [(2molNH= [(2molNH33)(-45.9kJ/mol)] – [(3molH)(-45.9kJ/mol)] – [(3molH22)(0 kJ/mol) + (1mol N)(0 kJ/mol) + (1mol N22)(0 kJ/mol)] )(0 kJ/mol)]

= -91.8kJ= -91.8kJΔΔSSsurrsurr = - = - ΔΔHH°°syssys = - = - (-91.8kJ)(1000 J)(-91.8kJ)(1000 J)

TT (298K)(1kJ) = 308 J/K (298K)(1kJ) = 308 J/K ΔΔSSunivuniv = = ΔΔSS°°syssys + + ΔΔSSsurrsurr = -197 J/K + 308 J/K = 111 J/K = -197 J/K + 308 J/K = 111 J/K

ΔΔSSuniv univ > 0, so the rxn is spontaneous at 298K> 0, so the rxn is spontaneous at 298K

38383838

2 H2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)

∆∆SSoosystemsystem = -326.9 J/K = -326.9 J/K

∆∆SSoosurroundingssurroundings = +1917 J/K = +1917 J/K

∆∆SSoouniverse universe = +1590. J/K= +1590. J/K

• The entropy of The entropy of the universe is the universe is increasing, so increasing, so the reaction is the reaction is product-favored.product-favored.


39393939

Spontaneous or Not?Spontaneous or Not?

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.

40404040

Entropy and Gibbs free energy, ΔG = ΔH - TΔS

What happens when one of the potential driving forces What happens when one of the potential driving forces behind a chemical reaction is favorable and the other is behind a chemical reaction is favorable and the other is not? We can answer this question by defining a new not? We can answer this question by defining a new quantity known as the quantity known as the Gibbs free energyGibbs free energy ( (GG) of the ) of the system, which reflects the balance between these forces.system, which reflects the balance between these forces.

Favorable Unfavorable The entropy term is Favorable Unfavorable The entropy term is ΔH° < 0ΔH° < 0 ΔH° > 0 ΔH° > 0 therefore subtractedtherefore subtractedΔS° > 0ΔS° > 0 ΔS° < 0 from the enthalpy term ΔS° < 0 from the enthalpy term when calculating ΔG°when calculating ΔG° ΔG°< 0 ΔG° > 0 for a reaction ΔG°< 0 ΔG° > 0 for a reaction

41414141

When to use Gibbs

Documents

Entropy and Free Energy