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Chapter 19Chemical Thermodynamics

Kinetics

How fast a rxn. proceeds

Equilibrium

How far a rxn proceedstowards completion

Thermodynamics

Study of energy relationships &changes which occur duringchemical & physical processes

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A) First Law of Thermodynamics

Law of Conservation of Energy :

Energy can be neither created nor destroyed but may be converted fromone form to another.

Energy lost = Energy gained

by system by surroundings

1) Internal Energy, E

E = total energy of the system

Actual value of E cannot be determined

Only )E

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2) Relating )E to Heat & Work

2 types of energy exchanges occurbetween system & surroundings

Heat & Work

+ q : heat absorbed, endothermic

! q : heat evolved, exothermic

+ w : work done on the system

! w : work done by the system

a) First Law

)E = q + w

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3) System and Surroundings

System = portion we single out for study

- focus attention on changes which occur w/in definite boundaries

Surroundings = everything else

System : Contents of rx. flask

Surround. : Flask & everything outside it

Agueous soln. rx :

System : dissolved ions & moleculesSurround : H2O that forms the soln.

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B) Thermodynamic State & State Functions

Thermodynamic State of a System

defined by completely specifyingALL properites of the system

- P, V, T, composition, physical st.

1) State Function

prop. of a system determinedby specifying its state.

depends only on its presentconditions & NOT how it got there

)E = Efinal ! Einitial

independent of path takento carry out the change

- Also is an extensive prop.

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C) Enthalpy

In ordinary chem. rx., work generallyarises as a result of pressure-volumechanges

Inc. vol. & system does work againstpressure of the atmosphere

Constant Pressure

w = ! P )V

Negative because work done by system

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)E = q ! P )V

1) )E at Constant Volume

)E = qv

2) )E at Constant Pressure :

)E = qp ! P )V

qp = )E + P)V

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3) Enthalpy, H

H = E + PV

Change in enthalpy at constant P is:

)H = )E + P )V

&

)H = qp

Can think of as heat content

state fnc. & is extensive

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D) Enthalpies of Reaction

)Hrxn = Hproducts ! Hreactants

2 H2(g) + O2(g) 6 2 H2O(R) )H = !572 kJ

Thermochemical eqn.

Physical states are given and energyassociated w. rx. written to right

- MUST give physical states

If product is H2O(g), )H = !484kJ

)H corresponds to molar quantities given in eqn. as written

H2(g) + ½ O2(g) 6 H2O(R) )H = !286 kJ

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1) Guidelines

a) Enthalpy is extensive

Multiply a rxn by some factor the)H is multiplied by that factor

b) )Hreverse = ! )Hforward

c) Enthalpy is a state function

)H depends on the states ofreactants and products.

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E) Hess’s Law

)H is a state fnc.

Same whether the process occurs asa single step or as a series of steps

The )Hrxn is the sum of the )H’sfor the individual steps.

)Hrx = G )Hsteps Steps

* Add chem. eqn’s for stepsto get overall rxn.

* Add )Hsteps | )Hrxn

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1) Note:

In using Hess’s Law:

a) If an eqn. is multiplied by afactor, )H is multiplied bythe same factor.

b) If an eqn. is reversed,sign of )H changes

c) All substances NOT appearing indesired eqn. MUST cancel

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F) Enthalpy of Formation

Enthalpy change for the formationof a compound from its elements

)Hf

1) Standard enthalpy change

Enthalpy change when all reactants and and products are in their standard states

)H°

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2) Standard State

Most stable state of a substance in itspure form under standard pressure(1 atm) & some specified temp. ofinterest (usually 25 °C)

3) Thermochemical Standard States

A) solid or liquid

Pure substance at 1 atm

b) gas

pressure of 1 atm

c) species in solution

Conc. of 1 M

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4) Standard Enthalpy of Formation

)H for the rxn in which 1 mole of acmpd. is formed from its elementswith ALL substances in theirstandard states (in kJ/mol)

)Hf°

Note: )Hf° = 0 for an element inits standard state

)Hf°

1/2 N2(g) + 3/2 H2(g) 6 NH3(g) ! 46.2

Na(s) + 1/2 Cl2(g) 6 NaCl(s) ! 411.0

C(graphite) 6 C(diamond) + 1.897

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C) Determine )Hrxn° from )Hf°

)Hrxn° = 3 n )Hf° ! 3 m )Hf°prod. react.

n = coef. in bal. eqn. for each product

m = coef. in bal. eqn. for each reactant

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I) Spontaneous Processes

Spontaneous Process

Process which proceeds on ownwithout outside assistance

- occurs in one direction only

NONspontaneous Process

Can NOT proceed w/o outside assistance

- reverse of spont. process

Processes which are spont. inone direction are nonspont. inthe reverse direction.

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Mix 2 ideal gases

Open stopcock

Spontaneously mix

Reverse process is NOT spont.

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Will all spont. rxns. beexothermic or vice versa?

Will all nonspont. rxns. beendothermic or vice versa?

NO!

Spontaneity can dependon temp. & pressure

T > 0 °C : melting spont.

T < 0 °C : freezing spont.

T = 0 °C : equilibrium and neitherconversion occurs spont.

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A) Reversible & IRreversible Processes

1) Reversible Process

Change made in a way that system &surroundings can be restored to theiroriginal states by exactly reversing chg

NO net chg. to system or its surr.

2) IRreversible Process

Can NOT simply be reversed torestore system & surroundingsto their original states

A reversible change produces the max.amt. of work which can be done by asystem on its surroundings.

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3) Isothermal Expansion of a Gas

Expansion of ideal gas atconstant temperature.

Remove partition - gas spont. expandsto fill the whole cylinder. No work isdone by the system.

Compressing the gas back to theoriginal state requires surroundingsto do work on the system.

Process is IRreversible

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Can be done reversibly by doing theexpansion and compression infinitelyslowly so the external and internalpressures are always in equilibrium.

- can not actually bedone in real processes

- ALL real processes areIRreversible

4) Conclusions

a) A spont. process is anIRreversible process

b) Driving force is tendency to goto a less ordered state or stateof higher probability.

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II) Entropy, S & 2nd Law of Thermodynamics

Measure of randomness ordisorder in a system

or

Extent to which energy is distributedamong the various motions of themolecules in the system.

State Fnc: )S = Sfinal ! Sinitial

)S > 0 Sf > Si inc. in disorder

)S < 0 Sf < Si dec. in disorder

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A) Isothermal Process

Constant temp. process

qrev)S = ------

T

qrev = heat which would betransferred if processwere reversible

S is a state function

- Thus this eqn. can be used to calc. )S for any isothermalprocess (whether reversibleor irreversible).

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1) )S and Phase Changes

Phase changes occur at constant T

qrev = )Hphase change

)Hphase change)Sphase change = ---------------

T

a) Ex: boiling water at 100 °C

)Hvap = 44.01 kJ/mol

)Hvap 44.01 kJ/mol )Svap = --------- = -----------------

T 373.15 K

= 0.1178 kJ/molCK

= + 117.8 J/molCK

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B) 2nd Law of Thermodynamics and S

Irreversible Process

Results in inc. in Stotal ()S > 0)

Reversible Process

Results in no change in Stotal ()S = 0)

2nd Law of Thermodynamics

Total entropy of a system &its surroundings always inc.for a spont. process

Irreversible (Spont.) Process

)Suniv = )Ssys + )Ssurr > 0

Reversible Process

)Suniv = )Ssys + )Ssurr = 0

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1) Ex: 1 mole of Liquid water is leftoutside where it’s -10.0 °C and itfreezes. Calc. )Ssys & )Ssurr.Will it spontaneously freeze?

)Hfus = + 6.01 kJ/mol

)Hfreezing = ! )Hfus = ! 6.01 kJ/mol

)Hfrz (1 mol)(- 6.01 kJ/mol) )Ssys = -------- = -----------------------------

T 273 K

= ! 22.0 J/mol

)Hsurr (1 mol)(+ 6.01 kJ/mol) )Ssurr = --------- = -----------------------------

T 263 K

= + 22.9 J/mol

)Suniv = )Ssys + )Ssurr

= (! 22.0 J/mol) + (+ 22.9 J/mol)

= + 0.9 J/mol Spont.

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This shows that even though the entropyof the system DEC. the process was stillspont. because the entropy of thesurroundings increased more.

)Suniv > 0

Spont. processes occur w.an overall INC. in )Suniv

Practically speaking we can’talways easily determine )Suniv.

Want to focus on the system.Will see how to determinespontaneity based on )Ssys.

Simplify notation and referto )Ssys simply as )S

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III) Molecular Inerpretation of Entropy

A) Expansion of Gas at Molecular Level

Have the following system:

Open stopcock:

We know gas expands spont.

Why?

Look at possible arrangementsof the particles

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Probability that blue molecule is in leftflask is ½. Same for the red particle.

Probability that both moleculesremain in the left flask is:

(1/2)2 = 1/4

For 3 particles it would be

(1/2)3 = 1/8

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Consider a mole of gas:

Prob. all the molecules are in theleft flask at the same time is:

(1/2)N where N = 6.02 x 1023

Infinitesimally small!!!

Essentially ZERO prob.of all molecules in leftflask at same time.

- Gas SPONT. expandsto fill both flasks

- Does NOT spont. goback to left flask

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Most probable arrangements arethose with essentially equal numbersof particles in both flasks.

Gas spreads out and the arrangementof gas particles is more disorderedthan when confined to one flask

Greater entropy

Just seen disorder onthe molecular level

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B) Boltzmann’s Eqn. & Microstates

Statistical Thermodynamics

Uses statistics and probability to linkmicroscopic & macroscopic worlds.

Microstate

Single possible arrangement ofthe positions and kinetic energiesof the molecules - snapshot

Exceptionally LARGE # microstates

Can use probability and statisticsto determine total # microstatesfor a thermodynamic state

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W = # microstates

Very Large # for a mole of particles

Related to Entropy

Boltzmann Eqn.

S = k C ln W

k = Boltzmann’s constant

= 1.38 x 10!23 J/K

S is a measure of # microstatesassociated w. a particularmacroscopic state

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)S = k C ln Wfinal ! k C ln Winitial

Wf = k C ln ------

Wi

When Wf > Wi )S > 0

Entropy inc. w. # microstates

Ex: Inc volume of a gas

greater vol - greater # of positionsavailable to the particles andgreater # microstates

ˆ Entropy inc. as vol. inc.

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C) Molecular Motions and Energy

Ideal gas particles are idealizedpoints w. no vol. and no bonds

- translational motion only

Real molecules

- translational motion

- rotational motions

spin about an axis

Linear: 2 axes of spin

Nonlinear: 3 axes of spin

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- vibrational motions

Atoms periodically move toward& away from each other

Linear: 3N - 5Nonlinear: 3N - 6

N = # atoms in molecule (N > 2)

# microstates inc. as complexityof molecule inc.

- there are many more vibrational motions

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D) Predicting Sign of )S

1) Phase changes

Solid ----> Liquid ----> Gas

)S > 0 )S > 0

2) Number of Molecules Inc.

F2(g) ----> 2 F(g)

)S > 0

3) Inc. # Atoms in a Molecule

Inc. degrees of freedom

)S > 0

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4) Mixing of Substances

Generally, )Ssoln > 0

5) Temp. Changes

Inc. Temp., KE inc.

- molecules move faster

- broadens distribution of speeds

)S > 0

6) Vol. Inc.

Vol. inc. - greater # positionsavailable to atoms

)S > 0

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E) Ex: The )Hf° of liquid acetone is- 247.6 kJ/mol at 25 °C. The )Hf° forthe vapor is -216.6 kJ/mol at 25 °C.What is the entropy change when1.00 mol of liquid acetone vaporizesat 25 °C?

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F) Ex: A sample of 2.00 mol of an idealgas expands from a vol. of 1.0 L to10.0 L at constant temperature. Whatis the entropy change, )S? Is the signof )S consistent w. your expectations?

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IV) Third Law & Standard Entropy

A) 3rd Law

A perfectly crystalline substanceat 0 K has entropy of zero

Can measure Absolute entropy,also called standard entropy, So

- entropy value for standard state of species

Standard State

Pure substance: 1 atm pressure

Species in Soln: 1 M

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Can calculate from heat capacities

T Cp(T) dTST° = m0 -------------

T

1) Values for compounds doNOT correspond to formationfrom the elements

2) Absolute entropy of an elementin its solid state … 0

3) Values on order of 10's of joules(not kJ like enthalpy)

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B) Entropy Change for a Rxn.

)Srxn° = 3 n S° ! 3 m S°prod. react.

n = coef. in bal. eqn. for each product

m = coef. in bal. eqn. for each reactant

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1) Ex: Calculate the entropy changefor the formation of H2O from itselements at 25 °C.

So(H2) = 130.58 J/molCK

So(O2) = 205.0 J/molCK

So(H2O, liq) = 69.91 J/molCK

2 H2 (g) + O2 (g) W 2 H2O (R)

)Srxn° = 2 So(H2O, liq) ! [2 So

(H2) + So(O2)]

= (2 mol) (69.91 J/molCK) ! [(2 mol) (130.58 J/molCK) + (1 mol)(205.0 J/molCK)]

= ! 326.3 J/molCK

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V) Gibbs Free Energy & Spontaneity

G = H ! TS

State Fnc: )G = Gfinal ! Ginitial

At constant T & P

)G = )H ! T )S

Under standard state conditions,

)Go = )Ho ! T )So

How does this relate to spontaneity?

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! qsys ! )Hsys)Ssurr = --------- = -----------

T T

)Suniv = )Ssys + )Ssurr

! )Hsys

= )Ssys + ------------ T

Rearrange:

! T )Suniv = )Hsys ! T )Ssys

)G = ! T )Suniv

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Now have an eqn. whichrelates spont. to the system.

At constant T & P

)G = )H ! T )S

)G < 0 spont.

)G > 0 NONspont.

)G = 0 equilibrium