Ch16 z5e free energy

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Chapter 16 pp

Spontaneity, entropy and free energySpontaneity, entropy and free energy

Note: For online HW, you Note: For online HW, you mightmight need to need to use the thermodynamic data sheet use the thermodynamic data sheet posted on my downloads website in the posted on my downloads website in the “helps” section “helps” section ifif the values are not the values are not already in the problem. already in the problem.

You need to use ∆G values that have You need to use ∆G values that have at at least oneleast one decimal place and your decimal place and your textbook only goes to the “one’s” place.textbook only goes to the “one’s” place.

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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.

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16.1 Spontaneous A reaction that will occur A reaction that will occur withoutwithout

outside intervention.outside intervention. We We can’tcan’t determine how fast. determine how fast. We need We need bothboth thermodynamics and thermodynamics and

kinetics to describe a reaction kinetics to describe a reaction completely.completely.

Thermodynamics compares Thermodynamics compares initial & initial & finalfinal states ( states (i.e.,i.e., potential energy). potential energy).

Kinetics describes Kinetics describes pathway betweenpathway between (i.e., (i.e., the reaction rate)the reaction rate)..

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Z5e 792 Figure 16.2Rate of

Reaction Rate is a function of the Rate is a function of the pathway (kinetics)pathway (kinetics)

Spontaneity is a function of Spontaneity is a function of the potential energies the potential energies (thermodynamics)(thermodynamics)

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Thermodynamics 1st Law1st Law - the - the energyenergy of the universe is of the universe is

constantconstant.. Keeps track of thermodynamics, but Keeps track of thermodynamics, but

doesn’t correctly predict spontaneity.doesn’t correctly predict spontaneity. EntropyEntropy (S) is disorder or randomness. (S) is disorder or randomness. 2nd Law2nd Law - the - the entropyentropy of the universe of the universe

increasesincreases..

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Entropy

Defined in terms of Defined in terms of probabilityprobability.. Substances take the arrangement Substances 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 of Calculate the number of

arrangements for a system to arrangements for a system to determine entropy.determine entropy.

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One particle: 2 One particle: 2 possible possible arrangements arrangements (microstates)(microstates)

50 % chance of 50 % chance of finding the left finding the left emptyempty

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Two particles: 4 Two particles: 4 possible possible microstatesmicrostates

25% chance of 25% chance of finding the left finding the left emptyempty

50% chance of 50% chance of them being them being evenly evenly disperseddispersed

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

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3 possible 3 possible arrangements arrangements (11 microstates - top (11 microstates - top one is duplicative)one is duplicative)

8% chance of 8% chance of finding the left finding the left emptyempty

50 % chance of 50 % chance of them being them being evenly evenly disperseddispersed

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Gases Gases completely fill their chamber Gases completely fill their chamber

because there are many more ways because 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 the There are many more ways for the molecules to be arranged as a liquid molecules to be arranged as a liquid than a solid (than a solid (positionalpositional entropy). entropy).

Gases have a huge number of Gases 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 with For each pair below, choose the one with

the higher positional entropy the 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 CO and gaseous CO22 . . . . . . COCO2(2(gg)) - more volume so more positions. - more volume so more positions. NN22 gas at 1 atm gas at 1 atm vs.vs. at 1.0 x 10 at 1.0 x 10-2-2 atm atm (hint: (hint:

Boyle’s Law) Boyle’s Law) 1.0 x 101.0 x 10-2-2 atm atm since Boyle’s law says since 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 form form because there are many more because there are many more

possible arrangements of dissolved pieces than possible arrangements of dissolved pieces than if they stay separate from the solvent.if they stay separate from the solvent.

2nd Law of thermodynamics - the 2nd Law of thermodynamics - the entropyentropy of the of the universe is increasinguniverse is increasing

SSunivuniv = = SSsyssys + + SSsurrsurr

IfIfSSunivuniv is (+) the process is spontaneous.is (+) the process is spontaneous.

IfIfSSuniv univ is (-) the process is spontaneous in the is (-) the process is spontaneous in the oppositeopposite direction. direction.

If If SSunivuniv is 0, then at is 0, then at equilibriumequilibrium

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Cell metabolism In a living cell, complex molecules are In a living cell, complex molecules are

assembled from simple ones. Is this assembled from simple ones. Is this consistent with the 2nd law?consistent with the 2nd law?

Yes. Only Yes. Only SSunivuniv must be (+). We can must be (+). We can have (-) have (-) SSsyssys as long as as long as SSsurrsurr is both is both larger and positivelarger and positive

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For For exoexothermic processes thermic processes SSsurrsurr is is positivepositive. .

For For endothermicendothermic processes processes SSsurrsurr is is negativenegative..

Consider this processConsider this processHH22O(l)O(l)HH22O(g)O(g)

SSsyssys is (+); is (+); systemsystem gains entropy) gains entropy) SSsurrsurr is (-); is (-); surroundingssurroundings lose entropy lose entropy

(molecular motion decreases)(molecular motion decreases) SSunivuniv depends on temperature.depends on temperature.

pp

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16.3 Temperature and Spontaneity

Entropy changes in the surroundings are Entropy changes in the surroundings are determined by the determined by the heat flowheat flow..

An An exoexothermic process is favored thermic process is favored because by giving up heat the entropy of because by giving up heat the entropy of the the surroundingssurroundings increases. increases.

The size of The size of SSsurr surr depends on depends on temperature.temperature.

SSsurrsurr = -= -H/T (T in Kelvin)H/T (T in Kelvin)

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SsysSsurr 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 subscript G = H - TS (no subscript “system”) “system”) Never used this way.Never used this way. G = G = H - TH - TSS at constant temperature at 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 + + S S --G/T = G/T = SSunivuniv If If G is G is negativenegative at constant T and P, at constant T and P,

the Process is the Process is spontaneousspontaneous..

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Let’s Check

For the reaction HFor the reaction H22O(s) O(s) H H22O(l)O(l) Sº = 22.1 Sº = 22.1 JJ/K mol & /K mol & Hº = 6.030 Hº = 6.030 kkJJ/mol/mol

Calculate Calculate G & G & SSunivuniv at -10ºC, 0ºC & 10ºC at -10ºC, 0ºC & 10ºC We use We use GGoo = = HHoo - T - TSSoo ((seesee next slide & Table 16.4 p. 760 for next slide & Table 16.4 p. 760 for

results).results). We see that We 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 the Spontaneity can be predicted from the signssigns

of of H and H 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 - TS pp

HS 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 of Predict the Sign of Sº for the following:Sº for the following: Thermal decomposition of solid calcium Thermal decomposition of solid calcium

carbonate. . . Hint: write reaction and carbonate. . . Hint: write reaction and examine examine positionalpositional entropy. entropy.

CaCOCaCO3(s)3(s) CaOCaO(s)(s) + CO + CO2(g)2(g) Positional entropy Positional entropy increasesincreases since solid since solid

to gas so S = (+)to gas so S = (+) OxidationOxidation of SO of 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 The The 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 other All other substancessubstances must have S > 0. must have S > 0. Standard Entropies Sº ( at 298 K and 1 Standard Entropies Sº ( at 298 K and 1

atm) of substances are listed.atm) of substances are listed. ∑∑Products - ∑reactants to find Products - ∑reactants to find Sº (a Sº (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 can O molecule can vibrate and rotate in vibrate and rotate in several waysseveral ways as as shown here. shown here.

This freedom of This freedom of motion leads to motion leads to higher entropy for higher entropy for water than for a water than for a substance like substance like hydrogenhydrogen (a simple (a simple diatomic molecule diatomic molecule with fewer possible with fewer possible motions).motions).

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16.6 Free Energy in Reactions Gº = standard Gº = standard free energyfree energy change. change. This is the free energy change that will This is the free energy change that will

occur if occur if reactantsreactants in their in their standardstandard state state turn to turn to productsproducts in their in their standardstandard state. state.

Can’t be measured directly, can be Can’t be measured directly, can be calculated from other measurements.calculated from other measurements.

Gº = Gº = Hº - THº - TSº Watch units of H (in Sº 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.

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Calculations Involving Energy

If 2 mol reactants If 2 mol reactants 4 mol product + 500 kJ, is the rxn likely spontaneous? 4 mol product + 500 kJ, is the rxn likely spontaneous?

Yes Yes

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 in and S values in Gº = Gº = Hº - THº - TSº multiplied by the coefficients.Sº multiplied by the coefficients.

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Free Energy in Reactions pp

There are tables of There are tables of GºGºff (pp. A21 (pp. A21 ffff))

GºGºrxnrxn = ∑products - ∑reactants = ∑products - ∑reactants because it is because it is a state function.a state function.

Std. free energy of Std. free energy of formationformation ( (GºGºff) for any ) for any element element in standard statein standard state is 0 (don’t forget!). is 0 (don’t forget!).

Remember: Spontaneity tells us nothing Remember: Spontaneity tells us nothing about rate.about rate.

Watch for trick questions that give you values Watch for trick questions that give you values of of 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)

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Free Energy in Reactions pp

Three methods to determine GThree methods to determine Goo

#1 #1 Gº = Gº = Hº - THº - TSºSº Use Use Hº = ∑nHº = ∑np p HºHºf(products)f(products) - ∑n - ∑nr r HºHºf(reactants)f(reactants) Use Use Sº = ∑nSº = ∑np p SºSºproductsproducts - ∑n - ∑nr r SºSºreactantsreactants

find standard values in pp. A21 find 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 = ∑nppGºGºf(products)f(products) - ∑n - ∑nr r GºGºf(reactants)f(reactants)find standard values in pp. A21 find standard values in pp. A21 ffff

remember:remember: Gº of element = 0Gº of element = 0

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Three methods to determine Go pp

#1 #1 GºGº = = HºHº - T - TSºSº @25ºC & 1 atm @25ºC & 1 atm 2SO2SO2(g)2(g) + O + O2(g)2(g) 2SO 2SO3(g)3(g)

∆Hº∆Hºff (kJ/mol)(kJ/mol) -297 -297 00 -396 -396 SºSº (J/K•mol)(J/K•mol) 248 248 205 205 257 257

HºHº = ∑n = ∑np p HºHºf(products)f(products) - ∑n - ∑nr r HºHºf(reactants)f(reactants)

HºHº = [2(-396)] - [2(-297) + 0] = = [2(-396)] - [2(-297) + 0] = -198 kJ-198 kJ Sº Sº = ∑n= ∑np p SºSºproductsproducts - ∑n - ∑nr r 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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Three methods to determine Go 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.

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Three methods to determine Go pp

• #3 Gº = ∑npGºf(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.

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Z5e 16.7: So Far . . . pp

Three methods to determine ∆GThree methods to determine ∆Goo

#1 #1 Gº = Gº = Hº - THº - TSºSº Use Use Hº = ∑nHº = ∑np p HºHºf(products)f(products) - ∑n - ∑nr r HºHºf(reactants)f(reactants) Use Use Sº = ∑nSº = ∑np p SºSºproductsproducts - ∑n - ∑nr r SºSºreactantsreactants

find standard values in pp. A21 find 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 = ∑nppGºGºf(products)f(products) - ∑n - ∑nr r GºGºf(reactants)f(reactants)find standard values in pp. A21 find standard values in pp. A21 ffff

remember:remember: Gº of element = 0Gº of element = 0 Now, let’s look at free energy & pressure, then free Now, let’s look at free energy & pressure, then free

energy & equilibrium (energy & equilibrium (2 more methods2 more methods).).

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16.7 Free energy and 16.7 Free energy and PressurePressure pppp

G = G = Gº + Gº + RTRTln(Q) where Q is the ln(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 = 8.314 R = 8.314 JJ/Kmol (but ∆Gº is in /Kmol (but ∆Gº is in kkJ!)J!)

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16.7 Free energy and Pressure pp

G = G = Gº +Gº +RTRTln(Q)ln(Q) Given: Given: CO(g) + 2HCO(g) + 2H22(g) (g) CH CH33OH(l)OH(l)

Would the reaction be spontaneous at Would 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 CH CH33OH(l) = -166 kJ OH(l) = -166 kJ GºGºff CO(g) = -137 kJ CO(g) = -137 kJ GºGºff HH22(g) = 0 kJ(g) = 0 kJ

Which method (we now know 4) should we Which method (we now know 4) should we use? . . .use? . . .

Anytime you have a problem involving a Anytime you have a problem involving a gas and gas and pressurepressure you need an equation you 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) CH CH33OH(l)OH(l) GºGºff CH CH33OH(l) = -166 kJ OH(l) = -166 kJ GºGºff CO(g) = -137 kJ CO(g) = -137 kJ

GºGºff H H22(g) = 0 kJ(g) = 0 kJ G = G = Gº +Gº +RTRTln(Q)ln(Q)

1st, calc. 1st, calc. GGºº from standard free energies from standard free energies of of formationformation (see pp. A21 (see pp. A21ffff or above) . . . or above) . . .

GGºº = -29 kJ = -29 kJ [-166 [-166 kkJ] - [-137 J] - [-137 kkJ + 2•0 J + 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 J J

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Free energy and Pressure pp

CO(g) + 2HCO(g) + 2H22(g) (g) CH CH33OH(l)OH(l) Gº = -2.9 x 10Gº = -2.9 x 1044 J, P J, 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•rxn J/mol•rxn (from last slide) (from last slide)

2nd, calc. 2nd, calc. G using G 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]]

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Free energy and Pressure pp

CO(g) + 2HCO(g) + 2H22(g) (g) CH CH33OH(l)OH(l)

Gº = -2.9 x 10Gº = -2.9 x 1044 J, P J, 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•rxn J/mol•rxn Q = Q = 2.2 x 102.2 x 10-2-2 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 10 J/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 is G is negativenegative..

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16.8 Free Energy & 16.8 Free Energy & EquilibriumEquilibrium pppp

G = G = Gº + Gº + RTRTln(Q)ln(Q) G tells us spontaneity at current G tells us spontaneity at current

conditions. conditions. But, when will it stop?But, when will it stop? It will go to the lowest possible free It will go to the lowest possible free

energy, which may be an equilibrium.energy, which may be an equilibrium. At equilibrium At equilibrium G = 0, Q = K, so . . .G = 0, Q = K, so . . . 00 = ∆Gº + = ∆Gº + RTRTlnlnKK So, So, Gº = Gº = --RTRTlnKlnK We now have We now have 5 equations for ∆G5 equations for ∆G!! You know to use You know to use thisthis equation when you equation when you

have a problem that has G & have a problem that has G & KK

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5 Methods to Determine Gº pp

#1 #1 Gº = Gº = Hº - THº - TSºSº Use Use Hº = ∑nHº = ∑np p HºHºf(products)f(products) - ∑n - ∑nr r

HºHºf(reactants)f(reactants) Use Use Sº = ∑nSº = ∑np p SºSºproductsproducts - ∑n - ∑nr r

SºSºreactantsreactants find standard values in pp. find standard values in pp. A21 A21 ffff

#2 #2 Hess’ LawHess’ Law (free energy is state function)(free energy is state function)

#3 #3 Gº = ∑nGº = ∑nppGºGºf(products)f(products) - ∑n - ∑nr r GºGºf(reactants)f(reactants)

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)

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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.

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Z5e 820 16.8 The Z5e 820 16.8 The Dependence of Free Dependence of Free Energy on Partial Pressure Energy on Partial Pressure

AA(g)(g) BB(g)(g) a.a. InitialInitial free energies as A free energies as A

goes to Bgoes to B

b.b. Free energy of A Free energy of A lessenslessens & & free energy of B free energy of B increasesincreases

c.c. Finally, partial pressures Finally, partial pressures are equal and equilibrium is are equal and equilibrium is reachedreached

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Figure 16.9 Free Energy and Equilibrium

a. a. Gº to reach =m, beginning with 1.0 mol Gº 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 mol Gº 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) B B(g)(g) with 1.0 mole of each at P with 1.0 mole of each at Ptotaltotal = = 2.02.0 atm. atm. Each pointEach point on the curve corresponds to the on the curve corresponds to the totaltotal free free energy of the system for a energy of the system for a givengiven combination combination of A and B. of A and B.

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Gº K Gº K ∆Gº = -∆Gº = -RTRTln(K)ln(K)

= 0= 0

< 0< 0

> 0> 0

= 1= 1

> 1> 1

< 1< 1

pp

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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 atm 1.00 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

want want initialinitial conditions), calc. ∆G, if (-) then conditions), calc. ∆G, if (-) then shifts to right (spontaneous). . .shifts to right (spontaneous). . .

∆∆G = ∆Gº + G = ∆Gº + RTRTln(Q)ln(Q)

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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 atm 1.00 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 J J/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)) in in your calculator! Do sig figs your calculator! Do sig figs beforebefore adding. adding.

So, PSo, Paa is at =m & no shift occurs. is at =m & no shift occurs.

47

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 atm 1.00 atm 1.00 atm 1.00 atm1.00 atm

∆Gº = -33.3 kJ/mol of N∆Gº = -33.3 kJ/mol of N22 consumed consumed..

∆∆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

Free energy and equilibrium pp

4Fe4Fe(s)(s) + 3O + 3O2(g)2(g) 2Fe 2Fe22O O ∆Hº∆Hºff (kJ/mol)(kJ/mol) 0 0 00 -826 -826 SºSºff (J/K•mol)(J/K•mol) 27 20527 205 90 90 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 Calc Gº = ∆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

Free energy and equilibrium pp

4Fe4Fe(s)(s) + 3O + 3O2(g)2(g) 2Fe 2Fe22O O ∆Hº∆Hºff (kJ/mol)(kJ/mol) 0 0 00 -826 -826 SºSº (J/K•mol)(J/K•mol) 27 20527 205 90 90

∆∆Gº = ∆Hº - T∆SºGº = ∆Hº - T∆Sº Calc Calc ∆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 J J

SºSºrxnrxn = [(2)(90)] - [(4)(27) + (3)(205)] = = [(2)(90)] - [(4)(27) + (3)(205)] = -545 J/K-545 J/K ∆∆Gº Gº = = -1.652 x 10-1.652 x 1066 J J - - ((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 J J (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

Free energy and equilibrium pp

4Fe4Fe(s)(s) + 3O + 3O2(g)2(g) 2Fe 2Fe22O O ∆Hº∆Hºff (kJ/mol)(kJ/mol) 0 0 00 -826 -826 SºSº (J/K•mol)(J/K•mol) 27 20527 205 90 90 (What is (What is K at 25ºC & 1 atm?)K at 25ºC & 1 atm?) ∆Gº = -1.490 x ∆Gº = -1.490 x 101066 J J

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 is for my calculator) this means iron rusting is thermodynamically favored.thermodynamically favored.

This does This does notnot tell us the rate of iron rusting. tell us the rate of iron rusting.

51

Temperature dependenceTemperature dependence of K of K pp

Gº= -Gº= -RTRTlnK = lnK = Hº - THº - TSºSº 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 lnK Get a straight line of lnK vs.vs. 1/T with 1/T with

slope of slope of Hº/RHº/R So, if we So, if we experimentallyexperimentally determine K determine K

at different temperatures, can at different temperatures, can graphicallygraphically get get Hº and Hº and SºSº

52

16.9 Free energy And Work Free energy is the energy available Free energy is the energy available

to do workto do work.. It represents the maximum amount It represents the maximum amount

of work of work possiblepossible at a given at a given temperature and pressure.temperature and pressure.

Never really achieved because some Never really achieved because some of the free energy of the free energy is changed to heatis changed to heat during a change, so it can’t be used during 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

Figure 16.10 A Battery

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

55

Figure 16.10 A Battery

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

Figure 16.10 A Battery

All real processes are irreversible.

57

Reversible v. Irreversible Processes pp

Reversible v. Irreversible Processes pp

ReversibleReversible: The universe is : The universe is exactly the exactly the same same 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 irreversible All real processes are irreversible -- -- (some work is changed to heat).(some work is changed to heat).