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General Chemistry IIGeneral Chemistry IICHEM 152 Unit 3CHEM 152 Unit 3

Week 11

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Week 11 Reading Assignment

Chapter 17 – Sections 17.2 (spontaneous), 17.3 (entropy), 17.4 (S), 17.5 (Gibbs)

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You Predict –

IF left on the stove will sugar burn to

make CO2 and Water?

Yes, the ReactionHas a clearDirection.

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Product-Favored Product-Favored ReactionsReactions

FeFe22OO33(s) + 2 Al(s) (s) + 2 Al(s) 2 Fe(s) + Al 2 Fe(s) + Al22OO33(s)(s)

What do these reactions have in common?

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g)

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

They release heat -- exothermic

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Product-Favored Product-Favored ReactionsReactions

Many product-favored reactions or Many product-favored reactions or processes are exothermic processes are exothermic

((but not allbut not all).).

HH22O(s) + heatO(s) + heat HH22O(liq)O(liq)

All product-favored processes All product-favored processes increase the:increase the:

•Dispersal of energyDispersal of energyenergy is dispersed over a larger

number of particles•Dispersal of matterDispersal of matter

Atoms and molecules become more Atoms and molecules become more disordereddisordered

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Directionality of Directionality of ReactionsReactions

Energy DispersalEnergy Dispersal

Exothermic reactions involve a Exothermic reactions involve a release of stored chemical potential release of stored chemical potential

energy to the surroundings. energy to the surroundings.

The stored potential energy starts out The stored potential energy starts out in a few molecules but is finally in a few molecules but is finally

dispersed over a great many dispersed over a great many molecules. molecules.

The final state—with energy dispersedThe final state—with energy dispersed—is more probable and makes a —is more probable and makes a

reaction product-favored.reaction product-favored.

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You PREDICT

In our classroom what is the chance that all the oxygen in the air will move only to the teacher’s desk and all thenitrogen will move to the students resulting in choking?

.

O2 O2

O2

O2

O2

O2

N2

N2

N2N2

N2 N2

N2

N2

.

DoesNot

Happen

Non-spontaneous

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Product-Favored ReactionsProduct-Favored ReactionsOne property common to One property common to

thermodynamically favored thermodynamically favored processes is that the final state is processes is that the final state is

more more DISORDEREDDISORDERED or or RANDOMRANDOM

than the originalthan the original Why does a gas tend to expand

Into an empty chamber and not

The opposite?

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Will spilled chemicals by themselves jump backinto the tank?

Systems tend towardGreater disorderThan to organization.

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Entropy, SEntropy, SThe degree of dispersal of matter and energy (ENTROPY) in a system can be

quantified experimentally

What is the entropy of ice at 0 K?

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Which has more entropy?

1. Liquid water at 0 ºC2. Ice at 0 ºC3. They are the same

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Which has more entropy?

1. Water at 5 ºC2. Water at 50 ºC3. They are the same

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

Entropy, SEntropy, S

Entropy of a substance increases Entropy of a substance increases with temperaturewith temperature

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

Entropy, SEntropy, S

Which of these

systems has a greater entropy?

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Entropies (S) of ionic solids are greater the weaker the attractions are among the ions

Entropy, SEntropy, S

SSoo (J/K•mol) (J/K•mol)

MgOMgO 26.926.9

NaFNaF 51.551.5

NaCl 72.13NaCl 72.13

SSoo (J/K•mol) (J/K•mol)

MgOMgO 26.926.9

NaFNaF 51.551.5

NaCl 72.13NaCl 72.13

MgMg2+2+ & O & O2-2-

NaNa++ & F & F--

NaNa++ & Cl & Cl--

Why?

Order the following ionic compounds in order of increasing

entropy: NaCl, MgO, NaF

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

Entropy, SEntropy, S

Water + propyl alcohol mixture

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Increases in Entropy

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What happens to

The ENTROPY (S)

When a gas dissolves

In a liquid?

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has been detected in gas clouds between stars. The

predicted C-N-H bond angle is about

O C N H

1. 902. 1093. 1204. 180

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Predicting Entropy Changes, SFor each process, predict whether

the entropy of the system increases (Ssys>0) or decreases (Ssys<0)

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Which of the following would you predict to have a POSITIVE S?You can answer more than 1…1. 2CO(g) + O2(g) 2CO2(g)

2. NaCl(s) NaCl(aq)3. MgCO3(s) MgO(s) + CO2(g)

4. Ag+(aq) + I-(aq) AgI(s)5. 2H2(g) + O2(g) 2H2O(l)

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Calculating Entropy Calculating Entropy Changes, Changes, SS

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Consider Consider 2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(l)O(l)

∆∆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

there is a there is a decrease in S decrease in S because 3 mol of gas because 3 mol of gas give 2 mol of liquid. Also Sgive 2 mol of liquid. Also S00 for an element for an element is NOT zero.is NOT zero.

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)

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Entropy Changes for Phase Entropy Changes for Phase ChangesChanges

For a For a phase changephase change, ,

∆∆S = q/TS = q/Twhere q = heat transferred where q = heat transferred

in phase changein phase change

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

For For HH22O (liq) O (liq) HH22O(g)O(g)

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

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Entropy 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

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Thermodynamics:Directionality of

Chemical Reactions

THERMODYNAMICSTHERMODYNAMICS predicts if a predicts if a

reaction can occur, reaction can occur, given enough timegiven enough time

KINETICSKINETICS predicts if a predicts if a reaction can occur reaction can occur

at a reasonable at a reasonable raterate

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Thermodynamics and Thermodynamics and KineticsKinetics

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

C(diamond) C(graphite)

This reaction is This reaction is thermodynamically thermodynamically favored (favored (product-product-favoredfavored reaction). reaction).

Also kinetically Also kinetically favoredfavored once once

reaction is begun.reaction is begun.2K(s) + 2H2O(l) 2KOH(aq) +

H2(g)

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Favoring factors

•Energy dispersal – exothermic reactions tend to be product favored.

•Material dispersal - positive ∆S drives reactions.

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2nd Law of 2nd Law of ThermodynamicsThermodynamics

A reaction is product-favored A reaction is product-favored (spontaneous) if the total entropy of (spontaneous) if the total entropy of

the the universeuniverse increases increases∆∆SSuniverseuniverse = ∆S = ∆Ssystemsystem + ∆S + ∆Ssurroundingssurroundings

∆S∆Suniverseuniverse > 0 > 0

for product-favored processfor product-favored process

Studying every known product-favored

(spontaneous) reaction we observe

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2nd Law of 2nd Law of ThermodynamicsThermodynamics

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

∆S∆Suniverseuniverse > 0 > 0

0 < ∆S0 < ∆Ssystemsystem + ∆S + ∆Ssurroundingssurroundings

but but ∆S∆Ssurroundings =surroundings =

∆∆HHsurroudingssurroudings/T = -/T = -∆H∆Hsystemsystem/T/T

Because heat into the system is equal Because heat into the system is equal but opposite to heat out of the but opposite to heat out of the

surroundings.surroundings.

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2nd Law of 2nd Law of ThermodynamicsThermodynamics

0 < ∆S0 < ∆Ssystemsystem - ∆H - ∆Hsystemsystem/T/T

oror

0 > ∆H0 > ∆Hsystemsystem- T∆S- T∆Ssystemsystem

These are all These are all variables of the systemvariables of the system!!

They tell us if a process is product They tell us if a process is product favored without measuring the effect favored without measuring the effect

on the universe directly!on the universe directly!

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2nd Law of 2nd Law of ThermodynamicsThermodynamics

∆ ∆HHsystemsystem- T∆S- T∆Ssystemsystem < 0

Let’s defineLet’s define

∆∆G = G = ∆H - T∆S∆H - T∆S

If ∆G is negative (∆G is negative (decreasesdecreases) -) - the process is spontaneous the process is spontaneous

(product-favored).(product-favored).

This tells us nothing about the This tells us nothing about the speedspeed of the of the reaction, but that it is product-favored reaction, but that it is product-favored ((spontaneousspontaneous). ).

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Gibbs Free Energy, GGibbs Free Energy, GThe change of Gibbs free energy is

defined as:

Gsys=-TSuniv= Hsys- TSsys

The free energy of the Universe decreases in every spontaneous

(product-favored) process.

G represents the maximum useful work that can be done by a product-favored

system on its surroundings.

G also represents the minimum work that must be done to force a reactant-

favored process to occur.

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Trends with ∆G = Trends with ∆G = ∆H-T ∆S∆H-T ∆S

Predict the effect (product-favored or not) on ∆G:

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Exothermic reaction with increasing (+) S

1. Always product-favored2. Always reactant-favored3. Product-favored at high

temp4. Product-favored at low

temp

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Endothermic reaction with decreasing (-) S

1. Always product-favored2. Always reactant-favored3. Product-favored at high

temp4. Product-favored at low

temp

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Free EnergyFree Energy

2Fe2Fe22OO33(s) + 3 C(gr) (s) + 3 C(gr) 4 Fe(s) + 3 4 Fe(s) + 3 COCO22(g)(g)

∆∆HHoorxnrxn = +468.3 kJ = +468.3 kJ ∆S∆Soo

rxnrxn = +560.3 J/K = +560.3 J/K

∆∆GGoorxnrxn = +301.2 kJ at 298K = +301.2 kJ at 298K

Reaction is Reaction is reactant-favoredreactant-favored at 298 K at 298 KFor any reaction

Sosys= ΣSo(products) -ΣSo(reactants)

Hosys= ΣHf

o(products) -ΣHfo(reactants)

Gosys= Ho

sys - TSosys

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What does ∆G tell us

1) the direction of the reaction2) the maximum work that can be

recovered from the reaction (net work)

3) for ∆G(+) the minimum work needed to force the reaction to occur.

4) ∆G = -50 and ∆G=-100 both tell us the reactions are spontaneous, but do not predict the speed of the reaction.

Next time we’ll start to calculate values for the GIBBS FREE ENERGY

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Summary Activity

Predict the sign of S for each of the following. Does each process have a tendency to be spontaneous based on entropy change?

2SO2(g) + O2(g) 2SO3(g)

Ba(OH)2(s) BaO(s) + H2O(g)

CO(g) + 2H2(g) CH3OH(l)

FeCl2(s) + H2(g) Fe(s) + 2HCl(g)