Ch. 16: Spontaneity, Entropy, and Free Energy 16.1 Spontaneous Processes and Entropy

Ch. 16: Spontaneity, Ch. 16: Spontaneity, Entropy, and Free Entropy, and Free

Energy Energy

16.1 Spontaneous Processes 16.1 Spontaneous Processes and Entropyand Entropy

SpontaneousSpontaneous

Occurs without outside interventionOccurs without outside intervention

Thermodynamics tell us the Thermodynamics tell us the _____________ not the speed _____________ not the speed

Thermodynamics only consider initial Thermodynamics only consider initial and final states, not pathwayand final states, not pathway

Must use kinetics and thermodynamics Must use kinetics and thermodynamics to understand reaction completelyto understand reaction completely

What makes a reaction spontaneous?What makes a reaction spontaneous?

EntropyEntropy

SS - measure of disorder or - measure of disorder or ________________________

the driving force for a spontaneous the driving force for a spontaneous reaction is an increase in entropyreaction is an increase in entropy

Natural tendency: low to high SNatural tendency: low to high S

Entropy also describes the number of Entropy also describes the number of possible positions of a moleculepossible positions of a molecule

ExampleExampleWhich has higher positional S?Which has higher positional S? Solid COSolid CO22 or gaseous CO or gaseous CO22?? NN22 gas at 1 atm or N gas at 1 atm or N22 gas at 0.01 gas at 0.01

atm?atm?

Predict the sign of Predict the sign of SS Solid sugar is added to waterSolid sugar is added to water

Iodine vapor condenses on cold Iodine vapor condenses on cold surface to form crystals surface to form crystals


Energy Energy

16.2 Entropy and 216.2 Entropy and 2ndnd law of law of thermodynamicsthermodynamics

2nd Law of 2nd Law of ThermodynamicsThermodynamics

In any spontaneous reactions, there is In any spontaneous reactions, there is always an increase in entropy of the always an increase in entropy of the universeuniverse

∆∆SSunivuniv = = ∆∆SSsyssys + + ∆∆SSsurrsurr

If If ∆∆SSsyssys is negative, it can still be spont. is negative, it can still be spont. as long as the as long as the ∆∆SSsurrsurr is larger and is larger and positivepositive

∆∆SSunivuniv > 0 : > 0 :

∆∆SSunivuniv = 0 : = 0 :

∆∆SSunivuniv < 0 : < 0 :


Energy Energy

16.3 The Effect of 16.3 The Effect of Temperature on SpontaneityTemperature on Spontaneity

HH22O(l) O(l) H H22O(g)O(g)

∆∆SSsyssys has ____ sign b/c of the increase in has ____ sign b/c of the increase in # of positions# of positions

∆∆SSsurrsurr is determined mostly by heat flow is determined mostly by heat flow

Vaporization is _______________ so it Vaporization is _______________ so it removes heat from the surroundingsremoves heat from the surroundings

So ____________ random motion of So ____________ random motion of surroundingssurroundings

____________ ____________ ∆∆SSsurrsurr

Temperature’s EffectsTemperature’s Effects

If the If the ∆∆SSsyssys and and ∆∆SSsurrsurr have different have different signs, the ______________ determines signs, the ______________ determines the the ∆∆SSunivuniv

For vaporization of waterFor vaporization of water Above 100Above 100°°C, C, ∆∆SSunivuniv is ____________ is ____________ Below 100Below 100°°C, C, ∆∆SSunivuniv is ____________ is ____________

Impact of the transfer of heat will be Impact of the transfer of heat will be greater at lower temperaturesgreater at lower temperatures

Determining Determining ∆∆SSsurrsurr

Sign of Sign of ∆∆SSsurrsurr depends on direction of heat flowdepends on direction of heat flow ∆∆SSsurrsurr ____ for exothermic reactions ____ for exothermic reactions ∆∆SSsurrsurr ____ for endothermic reactions ____ for endothermic reactions

Magnitude of Magnitude of ∆∆SSsurrsurr Depends on temperatureDepends on temperature Heat flow = Heat flow = ∆∆H at constant PH at constant P Very small at high T, increases as T Very small at high T, increases as T

decreasesdecreases

Summary Summary

ExampleExample

A process has a A process has a H of +22 kJ and a H of +22 kJ and a S of -S of -13 J/K. At which temperatures is the 13 J/K. At which temperatures is the process spontaneous?process spontaneous? if there is no subscript, if there is no subscript, S = S = SSsyssys

SSunivuniv ___ 0 to be spontaneous ___ 0 to be spontaneous

SSsyssys + + SSsurrsurr > 0 > 0

ExampleExampleFor methanol, the enthalpy of vaporization is For methanol, the enthalpy of vaporization is 71.8 kJ/mol and the entropy of vaporization is 71.8 kJ/mol and the entropy of vaporization is 213 J/K. What is the normal boiling point of 213 J/K. What is the normal boiling point of methanol?methanol? SSsyssys= 213 J/K and = 213 J/K and H =H = 71.8 kJ/mol K71.8 kJ/mol K at the boiling point, the vaporization begins to be at the boiling point, the vaporization begins to be

spontaneousspontaneous SSunivuniv = 0 to be at bpt = 0 to be at bpt SSsyssys + + SSsurrsurr = 0 = 0


EnergyEnergy16.4: Free Energy16.4: Free Energy

Free EnergyFree EnergyG: helps you determine the temperature G: helps you determine the temperature dependence of spontaneitydependence of spontaneity

G G ≡ H – TS≡ H – TS definition of Gdefinition of G

for constant Tfor constant T

Free Energy and EntropyFree Energy and Entropy

So what sign would the ∆G of So what sign would the ∆G of reaction with a + ∆Sreaction with a + ∆Sunivuniv have? have?

So for a reaction at constant T and P:

ExampleExample

HH22O(s) O(s) H H22O(l)O(l) If If ∆∆HH°°=6.03x10=6.03x1033 J/mol and J/mol and ∆∆SS°°=22.1 J/mol=22.1 J/mol..K K

and the temperature is at -10and the temperature is at -10°°C, is it C, is it spontaneous? spontaneous?

Example 1Example 1

What could be another way to check for What could be another way to check for spontaneity of a reaction?spontaneity of a reaction? Check to see if the Check to see if the ∆∆SSunivuniv is positive is positive How can we solve for How can we solve for ∆∆SSsurrsurr?? use use ∆∆SSsurr surr = - = - ∆∆HH°°//TT

At -10At -10°°C, is it spontaneous?C, is it spontaneous?

Example 1Example 1Is it spontaneous at 0Is it spontaneous at 0°°C?C?

Example 1Example 1

Is it spontaneous at +10Is it spontaneous at +10°°C?C?

Example 1Example 1

When ∆Suniv° is 0, will ∆G° always be 0 too?

How do the signs of ∆Suniv° and ∆G° relate?

Enthalpy and EntropyEnthalpy and EntropyWhen H and S are in opposition, the When H and S are in opposition, the spontaneity depends on Tspontaneity depends on T In opposition: +In opposition: +∆∆S and +S and +∆∆H H

OR – OR –∆∆S and -S and -∆∆HH Exothermic direction is spontaneous at low TExothermic direction is spontaneous at low T

Example 2Example 2At what T is this reaction spontaneous, at At what T is this reaction spontaneous, at 1 atm of pressure?1 atm of pressure? BrBr22(l) (l) Br Br22(g)(g) (aka Boiling) (aka Boiling) ∆∆H=31.0 kJ/mol, ∆S=93.0 J/molKH=31.0 kJ/mol, ∆S=93.0 J/molK Looking for boiling point: ∆G=0= Looking for boiling point: ∆G=0=

equilibriumequilibrium


EnergyEnergy16.5: Entropy Changes in 16.5: Entropy Changes in

Chemical ReactionsChemical Reactions

What about Chemical Changes?What about Chemical Changes?We have been working with only We have been working with only physical changes so far…physical changes so far…

Compare using # of independent Compare using # of independent statesstates NN22(g) + 3H(g) + 3H22(g) (g) 2NH 2NH33(g)(g) Entropy increases/decreasesEntropy increases/decreases

Compare using # molecules in higher Compare using # molecules in higher entropy statesentropy states 4NH4NH33(g) + 5O(g) + 5O22(g) (g) 4NO(g) + 6H 4NO(g) + 6H22O(g)O(g) Entropy increases/decreasesEntropy increases/decreases

ExampleExample

Predict the sign of Predict the sign of ∆S for the ∆S for the following reactions:following reactions:

CaCOCaCO33(s) (s) CaO(s) + CO CaO(s) + CO22(g)(g)

2SO2SO22(g) + O(g) + O22(g) (g) 2SO 2SO33(g)(g)

Assigning S valuesAssigning S values

Normally deal with changes in S but can Normally deal with changes in S but can assign actual S values to substancesassign actual S values to substances

33rdrd law of thermodynamics: law of thermodynamics: _______ for a perfect crystal at 0 K_______ for a perfect crystal at 0 K A standard value to help us set S A standard value to help us set S

valuesvalues

Appendix 4 contains SAppendix 4 contains S° for common ° for common substances at 298 K and 1 atmsubstances at 298 K and 1 atm

Since S is state function, can find ∆S by Since S is state function, can find ∆S by subtracting final - initialsubtracting final - initial

Example 3Example 3

Find the Find the ∆S° at 25°C for:∆S° at 25°C for: 2NiS(s) + 3O2NiS(s) + 3O22(g) (g) 2SO 2SO22(g) + (g) +

2NiO(s)2NiO(s)

5353 205 205 248 248 38 38


EnergyEnergy16.6: Free Energy and 16.6: Free Energy and Chemical ReactionsChemical Reactions

Standard Free Energy Standard Free Energy ChangeChange

∆∆G°: change when reactants and products G°: change when reactants and products are both in standard statesare both in standard states

Cannot measure directly but can calculate Cannot measure directly but can calculate it from other measured valuesit from other measured values

Standard states: p. 260 chartStandard states: p. 260 chart

The more negative ∆G° is, the further a The more negative ∆G° is, the further a reaction’s ____________ position lies to the reaction’s ____________ position lies to the __________________

3 Ways to Calculate 3 Ways to Calculate ∆G°∆G°

__________________________________________________________________ Can calculate ∆ H ° and ∆ S ° using appendix Can calculate ∆ H ° and ∆ S ° using appendix

and products – reactantsand products – reactants Then put in the equation to find ∆Then put in the equation to find ∆GG°°

________________________________________ Rearrange equations to get the goal Rearrange equations to get the goal

equationequation Add the Add the ∆∆GG°° values for the equations values for the equations

___________ : standard free energy of ___________ : standard free energy of formationformation


EnergyEnergy16.7: Free Energy and 16.7: Free Energy and

PressurePressure

Dependence on PressureDependence on Pressure∆∆H is not dependent on the pressure of the H is not dependent on the pressure of the system but ∆S is- Why?system but ∆S is- Why?∆ ∆ G is dependent on both ∆ H and ∆ S so is G is dependent on both ∆ H and ∆ S so is also dependent on pressurealso dependent on pressurecan calculate the ∆G of a can calculate the ∆G of a substancesubstance at a at a pressure other than 1 atm using:pressure other than 1 atm using:

wherewhere ∆∆G° is the change in free energy at 1 atm G° is the change in free energy at 1 atm

and ∆G is the change in free energy at the and ∆G is the change in free energy at the PP

where R (_______________) and T is Kelvinwhere R (_______________) and T is Kelvin

Dependence on PressureDependence on Pressure

Can also determine the ∆G for a Can also determine the ∆G for a reactionreaction using the reaction quotient (Q)using the reaction quotient (Q)

where R is gas law constant (8.314 where R is gas law constant (8.314 J/molK) and Q is the reaction quotient in J/molK) and Q is the reaction quotient in pressures (gases only) or concentrations pressures (gases only) or concentrations aka [Products/Reactants]aka [Products/Reactants]

ExampleExample

Find the ∆G for the following reaction Find the ∆G for the following reaction at 25°C and with CO at 5.0 atm and at 25°C and with CO at 5.0 atm and HH22 at 3.0 atm. at 3.0 atm.

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

must find ∆G° firstmust find ∆G° first using product and reactants values from using product and reactants values from

appendixappendix

calculate Q from pressurecalculate Q from pressure

Example 1Example 1CO(g) + 2HCO(g) + 2H22(g) (g) CH CH33OH(l)OH(l)

∆∆G°=G°=

∆∆G= ∆G° + RT ln QG= ∆G° + RT ln Q

2

2 ))((

1

HCO ppQ

What does What does ∆G mean?∆G mean?

a negative a negative ∆G means∆G means1.1. __

2.2. __


EnergyEnergy16.8: Free Energy and 16.8: Free Energy and

EquilibriumEquilibrium


no matter how much of the reactants or no matter how much of the reactants or products are initially mixed, at a given set products are initially mixed, at a given set of conditions, the equilibrium position (K) of conditions, the equilibrium position (K) will be the samewill be the same

at equilibrium, ∆G= 0 and Q= Kat equilibrium, ∆G= 0 and Q= K

QRTGG ln


∆G°=0: K=1 G°react= G°prod

∆G°<0: K>1 G°react> G°prod

∆G°>0: K<1 G°react< G°prod

Example 2Example 2NN22(g) + 3H(g) + 3H22(g) (g) 2NH 2NH33(g)(g)

For the reaction above, ∆G° =-33.3 kJ/mol. For the reaction above, ∆G° =-33.3 kJ/mol. For each of the mixtures below at 25°C, For each of the mixtures below at 25°C, predict the direction in which the system predict the direction in which the system will shift to reach equilibriumwill shift to reach equilibrium

ppNH3NH3 = 1.00 atm, P = 1.00 atm, PN2N2 = 1.47, P = 1.47, PH2H2 = 0.0100 = 0.0100

∆∆G = ∆G° + RT ln KG = ∆G° + RT ln K∆∆G = G =

Example 2Example 2ppNH3NH3 = 1.00 atm, P = 1.00 atm, PN2N2 = 1.00, P = 1.00, PH2H2 = 1.00 = 1.00

∆∆G = ∆G° + RT ln KG = ∆G° + RT ln K

∆∆G=G=

Temperature Dependence of KTemperature Dependence of K

A plot of ln K vs. 1/T is linear where

y = ln K and x = 1/T (in Kelvin)

m = -∆H°/R

b = ∆ S°/R


EnergyEnergy16.9: Free Energy and Work16.9: Free Energy and Work

WorkWork

Can calculate the maximum amount of Can calculate the maximum amount of work that can be done by a processwork that can be done by a process

wwmaxmax = ∆G = ∆G

if the ∆G is +, then w is the amount of if the ∆G is +, then w is the amount of work expended to make a process work expended to make a process occuroccur

impossible to achieve maximum work impossible to achieve maximum work because of wasted energybecause of wasted energy

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Ch. 16: Spontaneity, Entropy, and Free Energy 16.1 Spontaneous Processes and Entropy