lect w11 152 - entropy and free energy_alg
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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)