Ch16z5efreeenergy 110115232348-phpapp01

1
Chapter 16 pp
Spontaneity, entropy and free energySpontaneity, entropy and free energy
Note: For online HW, youNote: For online HW, you mightmight need toneed to
use the thermodynamic data sheetuse the thermodynamic data sheet
posted on my downloads website in theposted on my downloads website in the
“helps” section“helps” section ifif the values are notthe values are not
already in the problem.already in the problem.
You need to use ∆G values that haveYou need to use ∆G values that have atat
least oneleast one decimal place and yourdecimal place and your
textbook only goes to the “one’s” place.textbook only goes to the “one’s” place.

2
Z5e 791 16.1 Methane and Oxygen React
The products have lower potential energy than the
reactants, resulting in energy flow (heat) to the
surroundings.

3
16.1 Spontaneous
A reaction that will occurA reaction that will occur withoutwithout
outside intervention.outside intervention.
WeWe can’tcan’t determine how fast.determine how fast.
We needWe need bothboth thermodynamics andthermodynamics and
kinetics to describe a reactionkinetics to describe a reaction
completely.completely.
Thermodynamics comparesThermodynamics compares initial &initial &
finalfinal states (states (i.e.,i.e., potential energy).potential energy).
Kinetics describesKinetics describes pathway betweenpathway between
(i.e.,(i.e., the reaction rate)the reaction rate)..

4
Z5e 792
Figure 16.2
Rate of
Reaction
Rate is a function of theRate is a function of the
pathway (kinetics)pathway (kinetics)
Spontaneity is a function ofSpontaneity is a function of
the potential energiesthe potential energies
(thermodynamics)(thermodynamics)

5
Thermodynamics
1st Law1st Law - the- the energyenergy of the universe isof the universe is
constantconstant..
Keeps track of thermodynamics, butKeeps track of thermodynamics, but
doesn’t correctly predict spontaneity.doesn’t correctly predict spontaneity.
EntropyEntropy (S) is disorder or(S) is disorder or
randomness.randomness.
2nd Law2nd Law - the- the entropyentropy of the universeof the universe
increasesincreases..

6
Entropy
Defined in terms ofDefined in terms of probabilityprobability..
Substances take the arrangementSubstances take the arrangement
that is most likely.that is most likely.
The most likely is the most random.The most likely is the most random.
Calculate the number ofCalculate the number of
arrangements for a system toarrangements for a system to
determine entropy.determine entropy.

7
One particle: 2One particle: 2
possiblepossible
arrangementsarrangements
(microstates)(microstates)
50 % chance of50 % chance of
finding the leftfinding the left
emptyempty

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Two particles: 4Two particles: 4
possiblepossible
microstatesmicrostates
25% chance of25% chance of
emptyempty
them beingthem being
evenlyevenly
disperseddispersed

9
Figure 16.4Figure 16.4
Three PossibleThree Possible
Arrangements (states)Arrangements (states)
of Four Molecules in aof Four Molecules in a
Two-Bulbed FlaskTwo-Bulbed Flask
(Each arrangement(Each arrangement
has severalhas several
microstates as seenmicrostates as seen inin
next slide)next slide)

10
3 possible3 possible
arrangementsarrangements
(11 microstates - top(11 microstates - top
one is duplicative)one is duplicative)
emptyempty
50 % chance of50 % chance of
them beingthem being
evenlyevenly
disperseddispersed

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Gases
Gases completely fill their chamberGases completely fill their chamber
because there are many more waysbecause there are many more ways
to do that than to leave half empty.to do that than to leave half empty.
SSsolidsolid < S< Sliquidliquid << S<< Sgasgas
There are many more ways for theThere are many more ways for the
molecules to be arranged as a liquidmolecules to be arranged as a liquid
than a solid (than a solid (positionalpositional entropy).entropy).
Gases have a huge number ofGases have a huge number of
positions possible (more Entropy S).positions possible (more Entropy S).

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Positional Entropy
For each pair below, choose the one withFor each pair below, choose the one with
the higher positional entropythe higher positional entropy per moleper mole
((i.e.i.e., S is (+)) at a given temperature:, S is (+)) at a given temperature:
Solid COSolid CO22 and gaseous COand gaseous CO22 . . .. . .
COCO2(2(gg)) - more volume so more positions.- more volume so more positions.
NN22 gas at 1 atmgas at 1 atm vs.vs. at 1.0 x 10at 1.0 x 10-2-2
atmatm (hint:(hint:
Boyle’s Law)Boyle’s Law)
1.0 x 101.0 x 10-2-2
atmatm since Boyle’s law sayssince Boyle’s law says
lower P = higher V (more positions).lower P = higher V (more positions).

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16.2 Entropy & 2nd Law of Thermodynamics
SolutionsSolutions formform because there are many morebecause there are many more
possible arrangements of dissolved piecespossible arrangements of dissolved pieces
than if they stay separate from the solvent.than if they stay separate from the solvent.
2nd Law of thermodynamics - the2nd Law of thermodynamics - the entropyentropy ofof
the universe is increasingthe universe is increasing
∆∆SSunivuniv == ∆∆SSsyssys ++ ∆∆SSsurrsurr
IfIf ∆∆SSunivuniv is (+) the process is spontaneous.is (+) the process is spontaneous.
IfIf ∆∆SSunivuniv is (-) the process is spontaneous in theis (-) the process is spontaneous in the
oppositeopposite direction.direction.
IfIf ∆∆SSunivuniv is 0, then atis 0, then at equilibriumequilibrium

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Cell metabolism
In a living cell, complex molecules areIn a living cell, complex molecules are
assembled from simple ones. Is thisassembled from simple ones. Is this
consistent with the 2nd law?consistent with the 2nd law?
Yes. OnlyYes. Only ∆∆SSunivuniv must be (+). We canmust be (+). We can
have (-)have (-) ∆∆SSsyssys as long asas long as ∆∆SSsurrsurr is bothis both
larger and positivelarger and positive

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ForFor exoexothermic processesthermic processes ∆∆SSsurrsurr isis
positivepositive..
ForFor endothermicendothermic processesprocesses ∆∆SSsurrsurr isis
negativenegative..
Consider this processConsider this process
HH22O(l)O(l)→→ HH22O(g)O(g)
∆∆SSsyssys is (+);is (+); systemsystem gains entropy)gains entropy)
∆∆SSsurrsurr is (-);is (-); surroundingssurroundings lose entropylose entropy
(molecular motion decreases)(molecular motion decreases)
∆∆SS depends on temperature.depends on temperature.
pp

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16.3 Temperature and Spontaneity
Entropy changes in the surroundings areEntropy changes in the surroundings are
determined by thedetermined by the heat flowheat flow..
AnAn exoexothermic process is favoredthermic process is favored
because by giving up heat the entropy ofbecause by giving up heat the entropy of
thethe surroundingssurroundings increases.increases.
The size ofThe size of ∆∆SSsurrsurr depends ondepends on
temperature.temperature.
∆∆SSsurrsurr = -= -∆∆H/T (T in Kelvin)H/T (T in Kelvin)

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∆Ssys
∆Ssurr ∆Suniv
Spontaneous?
- - -
+ ++
+ - ?
+- ?
Yes
No, Reverse
At low temp (∆ssurr
magnitude > ∆Ssys
At high temp (∆ssys
magnitude > ∆Ssurr
Z5e 803 Table 16.3Z5e 803 Table 16.3 pppp

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16.4 Gibb's Free Energy
G = H - TS (no subscriptG = H - TS (no subscript →→ “system”)“system”)
Never used this way.Never used this way.
∆∆G =G = ∆∆H - TH - T∆∆SS at constant temperatureat constant temperature
Divide both sides by -TDivide both sides by -T
--∆∆G/T = -G/T = -∆∆H/T +H/T + ∆∆SS
--∆∆G/T =G/T = ∆∆SSsurrsurr ++ ∆∆SS
--∆∆G/T =G/T = ∆∆SSunivuniv
IfIf ∆∆G isG is negativenegative at constant T and P,at constant T and P,
the Process isthe Process is spontaneousspontaneous..

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Let’s Check
For the reaction HFor the reaction H22O(s)O(s) →→ HH22O(l)O(l)
∆∆Sº = 22.1Sº = 22.1 JJ/K mol &/K mol & ∆∆Hº = 6.030Hº = 6.030 kkJJ/mol/mol
CalculateCalculate ∆∆G &G & ∆∆SSunivuniv at -10ºC, 0ºC & 10ºCat -10ºC, 0ºC & 10ºC
We useWe use ∆∆GGoo
== ∆∆HHoo
- T- T∆∆SSoo
((seesee next slide & Table 16.4 p. 760 fornext slide & Table 16.4 p. 760 for
results).results).
We see thatWe see that ∆∆GGoo
= (+) at -10ºC (spontaneous= (+) at -10ºC (spontaneous
in the opposite direction), zero at 0ºC (at =m)in the opposite direction), zero at 0ºC (at =m)
& (-) at 10ºC (spontaneous as written).& (-) at 10ºC (spontaneous as written).

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Z5e 804 Table 16.4
We see that ∆Go
= (+) at -10ºC (spontaneous
in the opposite direction), zero at 0ºC (at =m) &
(-) at 10ºC (spontaneous as written).
So ∆Go
= 0 at fp and mp (since at =m; use for
preptest)

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Predicting Spontaneity
Spontaneity can be predicted from theSpontaneity can be predicted from the signssigns
ofof ∆∆H andH and ∆∆S.S.
SeeSee Table 16.5 p 761 (and next slide).Table 16.5 p 761 (and next slide).
You will have a test question on this table!You will have a test question on this table!
Other hints especially for preptest and test:Other hints especially for preptest and test:
– You can’t add kJ to J - convert
– On MC questions, watch what
temperature units the answer is in. May
need to convert K to Cº

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∆G = ∆H - T∆S pp
∆H∆S Spontaneous?
+ - At all Temp & exotherm
+ +
At high temp & endoth.,
“entropy driven”
- -
At low temperatures,
“enthalpy driven”
+-
Not at any temperature,
Reverse is spontaneous

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16.5 Entropy Changes in Chem. Rxns pp
Predict the Sign ofPredict the Sign of ∆∆Sº for the following:Sº for the following:
Thermal decomposition of solid calciumThermal decomposition of solid calcium
carbonate. . . Hint: write reaction andcarbonate. . . Hint: write reaction and
examineexamine positionalpositional entropy.entropy.
CaCOCaCO3(s)3(s) CaOCaO(s)(s) + CO+ CO2(g)2(g)
Positional entropyPositional entropy increasesincreases since solidsince solid
to gas so S = (+)to gas so S = (+)
OxidationOxidation of SOof SO22 in air. . .in air. . .
2SO2SO2(g)2(g) + O+ O2(g)2(g) 2SO2SO3(g)3(g)
3 molecules of gas become 2 so S is (-).3 molecules of gas become 2 so S is (-).

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Third Law of Thermo
TheThe entropyentropy of a pure crystal at 0 K is 0.of a pure crystal at 0 K is 0.
This gives us a starting point.This gives us a starting point.
All otherAll other substancessubstances must have S > 0.must have S > 0.
Standard Entropies Sº ( at 298 K and 1Standard Entropies Sº ( at 298 K and 1
atm) of substances are listed.atm) of substances are listed.
∑∑Products - ∑reactants to findProducts - ∑reactants to find ∆∆Sº (aSº (a
state function)state function)
More complex molecules - higher Sº.More complex molecules - higher Sº.

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Z5e 810 Figure 16.6
H2O Molecule
HH22O molecule canO molecule can
vibrate and rotate invibrate and rotate in
several waysseveral ways asas
shown here.shown here.
This freedom ofThis freedom of
motion leads tomotion leads to
higher entropy forhigher entropy for
water than for awater than for a
substance likesubstance like
hydrogenhydrogen (a simple(a simple
diatomic moleculediatomic molecule
with fewer possiblewith fewer possible
motions).motions).

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16.6 Free Energy in Reactions
∆∆Gº = standardGº = standard free energyfree energy change.change.
This is the free energy change that willThis is the free energy change that will
occur ifoccur if reactantsreactants in theirin their standardstandard statestate
turn toturn to productsproducts in theirin their standardstandard state.state.
Can’t be measured directly, can beCan’t be measured directly, can be
calculated from other measurements.calculated from other measurements.
∆∆Gº =Gº = ∆∆Hº - THº - T∆∆Sº Watch units of H (inSº Watch units of H (in kkJ)J)
and S (in J). Change temp. to Kelvin.and S (in J). Change temp. to Kelvin.
Use Hess’ Law with known reactions.Use Hess’ Law with known reactions.

7
Calculations Involving Energy
If 2 mol reactantsIf 2 mol reactants →→ 4 mol product + 500 kJ, is the rxn likely spontaneous?4 mol product + 500 kJ, is the rxn likely spontaneous?
YesYes
What is the sign of ∆SWhat is the sign of ∆Srxnrxn when molten wax hardens?when molten wax hardens?
(-)(-)
How would your calculations change if the coefficients were not 1?How would your calculations change if the coefficients were not 1?
∆∆HHff and S values inand S values in ∆∆Gº =Gº = ∆∆Hº - THº - T∆∆Sº multiplied by the coefficients.Sº multiplied by the coefficients.

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Free Energy in Reactions pp
There are tables ofThere are tables of ∆∆GºGºff (pp. A21(pp. A21 ffff))
∆∆GºGºrxnrxn = ∑products - ∑reactants= ∑products - ∑reactants because it isbecause it is
a state function.a state function.
Std. free energy ofStd. free energy of formationformation ((∆∆GºGºff) for any) for any
elementelement in standard statein standard state is 0 (don’t forget!).is 0 (don’t forget!).
Remember: Spontaneity tells us nothingRemember: Spontaneity tells us nothing
about rate.about rate.
Watch for trick questions that give you valuesWatch for trick questions that give you values
ofof combustioncombustion (may need to change the sign(may need to change the sign
- write the reaction to find out)- write the reaction to find out)

29
Free Energy in Reactions pp
Three methods to determine GThree methods to determine Goo
#1#1 ∆∆Gº =Gº = ∆∆Hº - THº - T∆∆SºSº
UseUse ∆∆Hº = ∑nHº = ∑npp ∆∆HºHºf(products)f(products) - ∑n- ∑nrr ∆∆HºHºf(reactants)f(reactants)
UseUse ∆∆Sº = ∑nSº = ∑npp ∆∆SºSºproductsproducts - ∑n- ∑nrr ∆∆SºSºreactantsreactants
find standard values in pp. A21find standard values in pp. A21 ffff
#2#2 Hess’ LawHess’ Law (free energy is state function)(free energy is state function)
#3#3 ∆∆Gº = ∑nGº = ∑npp∆∆GºGºf(products)f(products) - ∑n- ∑nrr ∆∆GºGºf(reactants)f(reactants)
remember:remember: ∆∆Gº of element = 0Gº of element = 0

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Three methods to determine Go
pp
#1#1 ∆∆GºGº == ∆∆HºHº - T- T∆∆SºSº @25ºC & 1 atm@25ºC & 1 atm
2SO2SO2(g)2(g) + O+ O2(g)2(g) 2SO2SO3(g)3(g)
∆Hº∆Hºff (kJ/mol)(kJ/mol) -297-297 00 -396-396
SºSº (J/K•mol)(J/K•mol) 248248 205205 257257
∆∆HºHº = ∑n= ∑npp ∆∆HºHºf(products)f(products) - ∑n- ∑nrr ∆∆HºHºf(reactants)f(reactants)
∆∆HºHº = [2(-396)] - [2(-297) + 0] == [2(-396)] - [2(-297) + 0] = -198 kJ-198 kJ
∆∆SºSº = ∑n= ∑npp ∆∆SºSºproductsproducts - ∑n- ∑nrr ∆∆SºSºreactantsreactants
∆∆Sº =Sº = [2(257)] - [2(248) + 205] =[2(257)] - [2(248) + 205] = -187 J-187 J
(expect (-) since 3 moles gas(expect (-) since 3 moles gas 2 moles)2 moles)
∆∆GºGº = -198= -198 kJkJ -(298-(298KK)(-187)(-187 J/KJ/K)(1)(1 kJkJ/1000/1000 JJ))
∆∆Gº = -142 kJ (spontaneous since (-))Gº = -142 kJ (spontaneous since (-))

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pp
• #2 Hess’ Law (free energy is state function)
• Cdiamond(s) + O2(g) CO2(g) ∆Gº = -397 kJ Cgraphite(s)
+ O2(g) CO2(g) ∆Gº = -394 kJ
• Calculate ∆Gº for Cdiamond(s) Cgraphite(s)
• Steps . . .
• Reverse the 2nd reaction (change the
sign) and add.
• ∆Gº = -397 kJ + 394 kJ = -3 kJ, but slow.
• Diamond is kinetically stable, but
thermodynamically unstable.

32
pp
• #3 ∆Gº = ∑np∆Gºf(products) - ∑nr ∆Gºf(reactants)
• 2CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g) ∆Gºf(kJ/mol)
-163 0 -394 -229
• ∆Gº = [2(-394) + 4(-229)] - [2(-163)]
• ∆Gº = -1378 kJ/mol
• Large magnitude & (-) sign means this is very
favorable thermodynamically.

33
Z5e 16.7: So Far . . . pp
Three methods to determine ∆GThree methods to determine ∆Goo
#1#1 ∆∆Gº =Gº = ∆∆Hº - THº - T∆∆SºSº
remember:remember: ∆∆Gº of element = 0Gº of element = 0
Now, let’s look at free energy & pressure, then freeNow, let’s look at free energy & pressure, then free
energy & equilibrium (energy & equilibrium (2 more methods2 more methods).).

34
16.7 Free energy and16.7 Free energy and PressurePressure pppp
∆∆G =G = ∆∆Gº +Gº + RTRTln(Q) where Q is theln(Q) where Q is the
reaction quotients (Preaction quotients (Pproductsproducts ÷ P÷ Preactantsreactants) and) and
follows the law of mass action.follows the law of mass action.
R = 8.314R = 8.314 JJ/Kmol (but ∆Gº is in/Kmol (but ∆Gº is in kkJ!)J!)

35
16.7 Free energy and Pressure pp
∆∆G =G = ∆∆Gº +Gº +RTRTln(Q)ln(Q)
Given:Given: CO(g) + 2HCO(g) + 2H22(g)(g) →→ CHCH33OH(l)OH(l)
Would the reaction be spontaneous atWould the reaction be spontaneous at
25ºC with P25ºC with PHH22 = 3.0 atm & P= 3.0 atm & PCOCO = 5.0 atm?= 5.0 atm?
∆∆GºGºff CHCH33OH(l) = -166 kJOH(l) = -166 kJ ∆∆GºGºff CO(g) = -137 kJCO(g) = -137 kJ ∆∆GºGºff
HH22(g) = 0 kJ(g) = 0 kJ
Which method (we now know 4) shouldWhich method (we now know 4) should
we use? . . .we use? . . .
Anytime you have a problem involving aAnytime you have a problem involving a
gas andgas and pressurepressure you need an equationyou need an equation
with “with “RR”. So use”. So use ∆∆G =G = ∆∆Gº +Gº +RTRTln(Q)ln(Q)

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Free energy and Pressure pp
CO(g) + 2HCO(g) + 2H22(g)(g) →→ CHCH33OH(l)OH(l)
∆∆GºGºff CHCH33OH(l) = -166 kJOH(l) = -166 kJ ∆∆GºGºff CO(g) = -137 kJCO(g) = -137 kJ
∆∆GºGºff HH22(g) = 0 kJ(g) = 0 kJ
∆∆G =G = ∆∆Gº +Gº +RTRTln(Q)ln(Q)
1st, calc.1st, calc. ∆∆GGºº from standard free energiesfrom standard free energies
ofof formationformation (see pp. A21(see pp. A21ffff or above) . . .or above) . . .
∆∆GGºº = -29 kJ= -29 kJ [-166[-166 kkJ] - [-137J] - [-137 kkJ + 2•0J + 2•0 kkJ]J]
Convert kJ to J (Convert kJ to J (R is in joulesR is in joules) = -2.9 x 10) = -2.9 x 1044
JJ

37
∆∆Gº = -2.9 x 10Gº = -2.9 x 1044
J, PJ, PH2H2 = 3.0 atm & P= 3.0 atm & PCOCO = 5.0 atm= 5.0 atm
∆∆Gº =Gº = -2.9 x 10-2.9 x 1044
J/mol•rxnJ/mol•rxn (from last slide)(from last slide)
2nd, calc.2nd, calc. ∆∆G usingG using ∆∆G =G = ∆∆Gº +Gº + RTRTln(Q),ln(Q),
but first have to calculate Q . . .but first have to calculate Q . . .
Note: methanol not used in calculating Q.Note: methanol not used in calculating Q.
Why? . . .Why? . . .
Pure liquid, so assign value =Pure liquid, so assign value = 11, Q = . . .?, Q = . . .?
Q = 2.2 x 10Q = 2.2 x 10-2-2
11/[(5.0 atm)(3.0 atm)/[(5.0 atm)(3.0 atm)22
]]

38
∆∆Gº = -2.9 x 10Gº = -2.9 x 1044
J, PJ, PH2H2 = 3.0 atm & P= 3.0 atm & PCOCO = 5.0 atm= 5.0 atm @25º C@25º C ∆∆GGºº ==
-2.9 x 10-2.9 x 1044
J/mol•rxnJ/mol•rxn Q = 2.2 x 10Q = 2.2 x 10--
22
1/[(5.0 atm)(3.0 atm)1/[(5.0 atm)(3.0 atm)22
]]
So,So, ∆∆G =G = ∆∆Gº +Gº + RTRTln(Q)ln(Q) == ? . . .? . . .
∆∆G = -38 kJ/molG = -38 kJ/mol Calculation is . . .Calculation is . . .
-2.9 x 10-2.9 x 1044
J/mol•rxn + (8.3145 J/mol)(25 + 273 K)•ln(2.2 x 10J/mol•rxn + (8.3145 J/mol)(25 + 273 K)•ln(2.2 x 10-2-2
))
Is it spontaneous? . . .Is it spontaneous? . . .
Yes.Yes. ∆∆G isG is negativenegative..

39
16.8 Free Energy &16.8 Free Energy & EquilibriumEquilibrium pppp
∆∆G =G = ∆∆Gº +Gº + RTRTln(Q)ln(Q)
∆∆G tells us spontaneity at currentG tells us spontaneity at current
conditions.conditions. But, when will it stop?But, when will it stop?
It will go to the lowest possible freeIt will go to the lowest possible free
energy, which may be an equilibrium.energy, which may be an equilibrium.
At equilibriumAt equilibrium ∆∆G = 0, Q = K, so . . .G = 0, Q = K, so . . .
00 = ∆Gº += ∆Gº + RTRTlnlnKK
So,So, ∆∆Gº =Gº = --RTRTlnKlnK
We now haveWe now have 5 equations for ∆G5 equations for ∆G!!
You know to useYou know to use thisthis equation when youequation when you
have a problem that has G &have a problem that has G & KK

40
5 Methods to Determine Gº pp
#1#1 ∆∆Gº =Gº = ∆∆Hº - THº - T∆∆SºSº
find standard values in pp. A21find standard values in pp. A21ffff ∆∆GºGºelementelement = 0= 0
#4#4 ∆G = ∆Gº +∆G = ∆Gº + RTRTln(Q)ln(Q)
#5#5 ∆Gº = -∆Gº = -RTRTln(K)ln(K)

41
Figure 16.7 Balls Rolling Down into Two Types of HillsFigure 16.7 Balls Rolling Down into Two Types of Hills
a. Goes to pure products since B is lowest free
energy (no equilibrium)
b. Doesn’t go to pure products (i.e., there is an
equilibrium) since intermediate C has lowest G.

42
Z5e 820 16.8 TheZ5e 820 16.8 The
Dependence of FreeDependence of Free
Energy on Partial PressureEnergy on Partial Pressure
AA(g)(g) BB(g)(g)
a.a. InitialInitial free energies as Afree energies as A
goes to Bgoes to B
b.b. Free energy of AFree energy of A lessenslessens &&
free energy of Bfree energy of B increasesincreases
c.c. Finally, partial pressuresFinally, partial pressures
are equal and equilibrium isare equal and equilibrium is
reachedreached

43
Figure 16.9 Free Energy and Equilibrium
a.a. ∆∆Gº to reach =m, beginning with 1.0 molGº to reach =m, beginning with 1.0 mol AA(g(g)) & P& PAA = 2.0 atm= 2.0 atm
b.b. ∆∆Gº to reach =m, beginning with 1.0 molGº to reach =m, beginning with 1.0 mol BB(g)(g) & P& PBB = 2.0 atm= 2.0 atm
c. G profile for Ac. G profile for A(g)(g) BB(g)(g) with 1.0 mole of each at Pwith 1.0 mole of each at Ptotaltotal == 2.02.0
atm.atm. Each pointEach point on the curve corresponds to theon the curve corresponds to the totaltotal freefree
energy of the system for aenergy of the system for a givengiven combinationcombination of A and B.of A and B.

44
∆∆Gº KGº K ∆Gº = -∆Gº = -RTRTln(K)ln(K)
= 0= 0
< 0< 0
> 0> 0
= 1= 1
> 1> 1
< 1< 1
pp

45
Free energy and equilibrium pp
NN2(g)2(g) + 3H+ 3H2(g)2(g) 2NH2NH3(g)3(g) PPaa
1.47 atm .0100 atm1.47 atm .0100 atm 1.00 atm1.00 atm PPbb
1.00 atm 1.00 atm1.00 atm 1.00 atm 1.00 atm1.00 atm
∆Gº = -33.3 kJ/mol of N∆Gº = -33.3 kJ/mol of N22 consumed @ 25º C.consumed @ 25º C.
Predict shift in =m for the above 2 cases.Predict shift in =m for the above 2 cases.
What to use? . . .What to use? . . .
Look for an equation with ∆G &Look for an equation with ∆G & QQ (since(since
wantwant initialinitial conditions), calc. ∆G, if (-) thenconditions), calc. ∆G, if (-) then
shifts to right (spontaneous). . .shifts to right (spontaneous). . .
∆∆G = ∆Gº +G = ∆Gº + RTRTln(Q)ln(Q)

46
NN2(g)2(g) + 3H+ 3H2(g)2(g) 2NH2NH3(g)3(g)
PPaa 1.47 atm .0100 atm1.47 atm .0100 atm 1.00 atm1.00 atm PPbb
1.00 atm 1.00 atm1.00 atm 1.00 atm 1.00 atm1.00 atm
∆Gº = -33.3 kJ/mol of N∆Gº = -33.3 kJ/mol of N22 consumed @ 25º C.consumed @ 25º C.
∆∆G = ∆Gº +G = ∆Gº + RTRTln(Q). For Pln(Q). For Paa: ∆G = . . .?: ∆G = . . .?
∆∆G = 0 (did you convert J & kJ?)G = 0 (did you convert J & kJ?)
(-3.33(-3.33 x 10x 1044
JJ/mol) + (8.3145 J/K•mol)(298 K)ln(6.8 x 10/mol) + (8.3145 J/K•mol)(298 K)ln(6.8 x 1055
))
Be sure to use extra ( ) for lnBe sure to use extra ( ) for ln((6.8 x 106.8 x 1055
)) inin
your calculator! Do sig figsyour calculator! Do sig figs beforebefore adding.adding.
So, PSo, Paa is at =m & no shift occurs.is at =m & no shift occurs.

47
NN2(g)2(g) + 3H+ 3H2(g)2(g) 2NH2NH3(g)3(g)
PPaa 1.47 atm .0100 atm1.47 atm .0100 atm 1.00 atm1.00 atm
PPbb 1.00 atm 1.00 atm1.00 atm 1.00 atm 1.00 atm1.00 atm
∆Gº = -33.3 kJ/mol of N∆Gº = -33.3 kJ/mol of N22 consumedconsumed..
∆∆G = ∆Gº +G = ∆Gº + RTRTln(Q)ln(Q)
For PFor Pbb: everything is at: everything is at standard statestandard state..
∆∆G = ∆Gº +G = ∆Gº + RTRTln(1) = ∆Gº = -33.3kJ/molln(1) = ∆Gº = -33.3kJ/mol
--∆Gº means spontaneous, so products∆Gº means spontaneous, so products
have lower free energy than reactants.have lower free energy than reactants.
So, shifts to right and K > 1.So, shifts to right and K > 1.

48
4Fe4Fe(s)(s) + 3O+ 3O2(g)2(g) 2Fe2Fe22OO ∆Hº∆Hºff (kJ/mol)(kJ/mol)
00 00 -826-826 SºSºff (J/K•mol)(J/K•mol)
27 20527 205 9090 Calc K @Calc K @
25ºC & 1 atm. Which equation?25ºC & 1 atm. Which equation?
Use ∆Gº = -Use ∆Gº = -RTRTln(K) (std state) find K.ln(K) (std state) find K.
First calculate ∆Gº from what??? . . .First calculate ∆Gº from what??? . . .
∆∆Gº = ∆Hº - T∆Sº CalcGº = ∆Hº - T∆Sº Calc ∆Hº∆Hºrxnrxn && SºSºrxnrxn
Your answer for ∆Gº is . . .Your answer for ∆Gº is . . .
∆∆Gº = -1.490 x 10Gº = -1.490 x 1066
JJ (did you convert kJ?)(did you convert kJ?)
Calculations on next slide.Calculations on next slide.

49
00 00 -826-826 SºSº (J/K•mol)(J/K•mol)
27 20527 205 9090
∆∆Gº = ∆Hº - T∆SºGº = ∆Hº - T∆Sº CalcCalc ∆Hº∆Hºrxnrxn && SºSºrxnrxn
∆∆HºHºrxnrxn == {[2(-826 kJ)] - [0]} x 1000 J/kJ ={[2(-826 kJ)] - [0]} x 1000 J/kJ = -1.652 x 10-1.652 x 1066
JJ
SºSºrxnrxn = [(2)(90)] - [(4)(27) + (3)(205)] == [(2)(90)] - [(4)(27) + (3)(205)] = -545 J/K-545 J/K
∆∆GºGº == -1.652 x 10-1.652 x 1066
JJ -- ((25 + 273 K)(25 + 273 K)(-545 J/K-545 J/K) = . . .) = . . .
As we saw, ∆Gº = -1.490 x 10As we saw, ∆Gº = -1.490 x 1066
JJ
(Remember to convert kJ to J and ºC to K)(Remember to convert kJ to J and ºC to K)
Now, find K. . . (next slide)Now, find K. . . (next slide)

50
00 00 -826-826 SºSº (J/K•mol)(J/K•mol)
27 20527 205 9090 (What is K at(What is K at
25ºC & 1 atm?)25ºC & 1 atm?) ∆Gº = -1.490 x 10∆Gº = -1.490 x 1066
JJ
Plug into ∆Gº = -Plug into ∆Gº = -RTRTln(K) & find K . . .?ln(K) & find K . . .?
lnlnK = 601, so K =K = 601, so K = ee601601
= 10= 10261261
Even though a humongous K = 10Even though a humongous K = 10261261
(too large(too large
for my calculator) this means iron rusting isfor my calculator) this means iron rusting is
thermodynamically favored.thermodynamically favored.
This doesThis does notnot tell us the rate of iron rusting.tell us the rate of iron rusting.

51
Temperature dependenceTemperature dependence of Kof K pp
∆∆Gº= -Gº= -RTRTlnK =lnK = ∆∆Hº - THº - T∆∆SºSº
ln(K) = -ln(K) = -∆∆Hº/R(1/T)+Hº/R(1/T)+ ∆∆Sº/RSº/R
i.e.,i.e., y = mx + by = mx + b
Get a straight line of lnKGet a straight line of lnK vs.vs. 1/T with1/T with
slope ofslope of ∆∆Hº/RHº/R
So, if weSo, if we experimentallyexperimentally determine Kdetermine K
at different temperatures, canat different temperatures, can
graphicallygraphically getget ∆∆Hº andHº and ∆∆SºSº

52
16.9 Free energy And Work
Free energy is the energy availableFree energy is the energy available
to do workto do work..
It represents the maximum amountIt represents the maximum amount
of workof work possiblepossible at a givenat a given
temperature and pressure.temperature and pressure.
Never really achieved because someNever really achieved because some
of the free energyof the free energy is changed to heatis changed to heat
during a change, so it can’t be usedduring a change, so it can’t be used
to do work.to do work.

53
Figure 16.10 A Battery
A battery can do work by sending current to a starter
motor. It can be recharged by forcing current through it
in the opposite direction.

54
If the current flow in both processes is infinitesimally
small, w1 = w2 and the process is reversible

55
But, in real world, current flow is finite, so w2 > w1 and
process is irreversible; that is, the universe is different
after the process occurs.

56
All real processes are irreversible.

57
Reversible v. Irreversible
Processes pp
ReversibleReversible: The universe is: The universe is exactly theexactly the
samesame as it was before the cyclic process.as it was before the cyclic process.
IrreversibleIrreversible: The universe is: The universe is differentdifferent
after the cyclic process.after the cyclic process.
All real processes are irreversibleAll real processes are irreversible ----
(some work is changed to heat).(some work is changed to heat).

Ch16z5efreeenergy 110115232348-phpapp01

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