chapter 19 chemical thermodynamics · chapter 19 chemical thermodynamics kinetics how fast a rxn....
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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