unit 11 (chp 5,8,19): thermodynamics (∆h, ∆s, ∆g, k)
DESCRIPTION
Chapters 5,8: Energy (E), Heat (q), Work (w), and Enthalpy (∆H) Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapters 5,8: Energy (E), Heat (q), Work (w), and Enthalpy (∆H) John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall, Inc.TRANSCRIPT
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Unit 11 (Chp 5,8,19):Thermodynamics
(∆H, ∆S, ∆G, K)
John D. BookstaverSt. Charles Community College
St. Peters, MO 2006, Prentice Hall, Inc.
Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
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Chapters 5,8:Energy (E), Heat (q),
Work (w), and Enthalpy (∆H)
John D. BookstaverSt. Charles Community College
St. Peters, MO 2006, Prentice Hall, Inc.
Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
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Energy (E)
Enthalpy (H)(kJ)
Entropy (S)(J/K)
Free Energy (G) (kJ)
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Energy (E)• ability to do work OR transfer heat Work (w): transfer of energy by applying a
force over a distance. Heat (q): transfer of energy by DT (high to low)
• unit of energy: joule (J)
• an older unit still in widespread use is… calorie (cal)
What is it?
1 Cal = 1000 cal
2000 Cal ≈ 8,000,000 J ≈ 8 MJ!!!1 cal = 4.18 J
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System and Surroundings
• System:molecules to be studied (reactants & products)
• Surroundings:everything else(container, thermometer)
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• Energy is neither created nor destroyed.
• total energy of an isolated system is constantInternal Energy (E):E = KE + PE (motions)(Thermal Energy)
(calculating E is too complex a problem)
DE = Efinal − Einitial
released or absorbed
(attractions)
1st Law of Thermodynamics
(no transfer matter/energy)(universe) (conserved)
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Changes in Internal Energy• Energy is transferred between
the system and surroundings, as either heat (q) or work (w).
Surroundings
Systemq out (–)
w by (–)
DE = q + w
DE = q + wDE = ?DE = (–) + (+)DE = +
q in (+)
w on (+)
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Changes in Internal Energy
Efinal > Einitial
absorbed energy(endergonic)
Efinal < Einitial
released energy(exergonic)
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Work (w)The only work done by a gas at constant P is change in V by pushing on surroundings.
PDV = −w ΔV
Zn + H+ Zn2+ + H2(g)
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Enthalpy
By substituting at constant P :DH = DE + PDVDH = (q+w) + (−w)DH = q
• (change in) enthalpy IS heat absorbed/released
DH = DE + PDV
DE = q+w PDV = −w
Enthalpy (H) is: H = E + PV internal work done energy
work done by systemheat/work energy
in or out of system
DH = heat
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Enthalpy of Reactionenthalpy is…
…the heat transfer in/out of a system
(at constant P)
DH = DE + PDV
DH = q
endergonicexergonic
endothermicexothermic
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
DH = Hproducts − Hreactants
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Endothermic & Exothermic
• Endothermic: DH > 0 (+)
• Exothermic: DH < 0 (–)
DH(+) = Hfinal − Hinitial
DH(–) is thermodynamically favorable
products reactants
DH(–) = Hfinal − Hinitialproducts reactants
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Enthalpy of Reaction
DHrxn, is the enthalpy of reaction, or “heat” of reaction.
units: kJmolrxn
kJ/molrxn
kJ∙molrxn
Demo
–1
2 H2(g) + O2(g) 2 H2O(g)
DH =–242 kJ/molrxn
–242 kJper 1 mol O2
–242 kJper 2 mol H2
–121 kJper 1 mol H2
(OR)
(OR)
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Enthalpy1. DH depends on amount (moles, coefficients)2. DHreverse rxn = –DHforward rxn
3. DHrxn dependson the state (s, l, g) ofproducts & reactants
DH1 = –802 kJ if 2 H2O(g) because… 2 H2O(l) 2 H2O(g)
DH = +88 kJ/molrxn CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
HW p. 207 #34,35,38,45
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Chp. 5,8: Calculate ∆H (4 Ways)1) Bond Energies
2) Hess’s Law
3) Standard Heats of Formation (Hf )
4) Calorimetry (lab)
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Overlap and Bonding
What attractive forces?
When bonds/attractions form, energy is _________.released
+ +––
What repulsive forces?
Where is energy being released?
Where must energy be added?
++
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(energy absorbed when bonds break)
Potential Energy of BondsHigh PE
Low PE(energy released
when bonds form)
chemical bond
High PE+ +
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Bond Enthalpy (BE)p.330BE: ∆H for the breaking of a bond (all +)
aka…bond dissociation energy
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DHrxn = (BEreactants) (BEproducts)
Enthalpies of Reaction (∆H)
To determine DH for a reaction:• compare the BE of bonds broken (reactants)
to the BE of bonds formed (products).
(bonds broken) (bonds formed)(released)
BE: ∆H for the breaking of a bond (all +)
DH(+) = BEreac − BEprod
DH(–) = BEreac − BEprod
(stronger)
(stronger)
(NOT on equation sheet)
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Enthalpies of Reaction (∆H)
= [4(413) + (242)] [3(413) + (328) + (431)]
= (655) (759)
DHrxn = 104 kJ/molrxn
CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)
DHrxn =
HW p. 339 #66, 68
[4(C—H) + (Cl—Cl)] [3(C—H) + (C—Cl) + (H—Cl)]
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Hess’s Law
DHrxn is independent of
route taken
DHoverall = DH1 + DH2 + DH3 …
DHrxn = sum of DH of all steps
p. 191
DH = Hfinal − Hinitialprod. react.
(NOT on equation sheet)
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C3H8(g) 3 C(gr.) + 4 H2(g) ∆H1= +104 kJ
3 C(gr.) + 3 O2(g) 3 CO2(g) ∆H2= –1182 kJ
4 H2(g) + 2 O2(g) 4 H2O(l) ∆H3= –1144 kJC3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
Calculation of DH by Hess’s LawC3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
3 C(gr.) + 4 H2(g) C3H8(g) ∆H1= –104 kJ
C(gr.) + O2(g) CO2(g) ∆H2= –394 kJ
H2(g) + ½ O2(g) H2O(l) ∆H3= –286 kJ
∆Hcomb =–2222 kJ
∆Hcomb = ?
3( ) 3( )
+
4( ) 4( )
Given:
Used:
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Standard Enthalpy of Formation (DHf )
o
Standard Enthalpy of Formation (DHf ):heat released or absorbed by the formation of a compound from its pure elements in their standard states.
(25oC , 1 atm)
o
3 C(gr.) + 4 H2(g) C3H8(g)∆Hf = –104 kJ
DHf = 0 for all elements in standard stateo
standard
DHf = 0 DHf = 0 DHf = –104 kJ
DH = Hfinal − Hinitialprod. react.
o o o
…therefore --->Recall…
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Calculation of DH by DHf’s
…we can use Hess’s law in this way:
DH = nHf(products) – mHf(reactants)
where n and m are the stoichiometric coefficients.
“sum”
(on equation sheet)
o
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Calculation of DH by DHf’s
DH = [3(DHf CO2) + 4(DHf H2O)] – [1(DHf C3H8) + 5(DHf O2)]
DH = [(–1180.5) + (–1143.2)] – [(–103.85) + (0)]
DH = (–2323.7) – (–103.85)
DH = –2219.9 kJ
Appendix C (p. 1123 )
HW p. 209 #60,63,66,
72,73
∆Hcomb = ?C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
o
DH = nHf(products) – mHf(reactants)
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By reacting in solution in a calorimeter, we indirectly determine DH of system by measuring ∆T & calculating q of the surroundings (calorimeter).
We can’t know the exact enthalpy of reactants and products, so we measure DH by calorimetry, the measurement of heat flow.
q = mcDT (on equation sheet)
Calorimeter nearly
isolated heat (J)
mass (g)[of sol’n]
Tf – Ti (oC)[of surroundings]
Calorimetry
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Specific Heat Capacity (c) • specific heat capacity,(c):
(or specific heat)energy required to raise temp of 1 g by 1C.
(for water)c = 4.18 J/goC + 4.18 J
of heat
Metals have much lower c’s b/c they transfer heat and change temp easily.
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Calorimetry
– q = DHrxn
q = mcDT
HW p. 208 #49, 52, 54
When 4.50 g NaOH(s) is dissolved 200. g of water in a calorimeter, the temp. changes from 22.4oC to 28.3oC. Calculate the molar heat of solution, ∆Hsoln (in kJ/mol NaOH).
(in J) of calorimeter or surroundings
(in kJ/mol) of system
q = (4.50 + 200)(4.18)(28.3–22.4)q = 5040 J
DH = –5.04 kJ4.50 g NaOH x 1 mol = 0.1125 mol 40.00 g
DH = –5.04 kJ 0.1125 mol
= –44.8 kJ mol
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Chp. 5,8: Calculate ∆H (4 Ways)1) Bond Energies
DHrxn = (BEreactants) (BEproducts)
2) Hess’s Law DHoverall = DHrxn1 + DHrxn2 + DHrxn3 …
3) Standard Heats of Formation (Hf )
DH = nHf(products) – mHf(reactants)
4) Calorimetry (lab) q = mc∆T (surroundings or thermometer)–q = ∆H ∆H/mol = kJ/mol (molar enthalpy)
(+ broken) (– formed)(NOT)
(given)
(NOT)
(given)
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Chapter 19:Thermodynamics
(∆H, ∆S, ∆G, K)
Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
John D. BookstaverSt. Charles Community College
St. Peters, MO 2006, Prentice Hall, Inc.
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ΔH = ?ΔH = q(heat)
ΔG = ?
ΔE = q + wPΔV = –w (at constant P)
ΔS = ?
Enthalpy (H)(kJ)
Entropy (S)(J/K)
Free Energy (G) (kJ)
Energy (E)
ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w)
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Big Idea #5: Thermodynamics
Chemical and physical processes are driven by:• a decrease in enthalpy (–∆H), or• an increase in entropy (+∆S), or• both.
Bonds break and formto lower free energy (∆G).
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1st Law of Thermodynamics
• Energy cannot be created nor destroyed (is conserved)
• Therefore, the total energy of the universe is constant.
DHuniv = DHsystem + DHsurroundings = 0if (+) then (–) = 0if (–) then (+) = 0
DHsystem = –DHsurroundings
OR
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Thermodynamically Favorable• Thermodynamically
Favorable (spontaneous) processes are those that can proceed without any outside intervention.
• The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously
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Thermodynamically Favorable
• Processes that are thermodynamically favorable (spontaneous) in one direction are NOT in the reverse direction.
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Thermodynamically Favorable
melting
• Processes that are favorable (spontaneous) at one temperature……may not be at other temperatures.
HW p. 837 #7, 11
freezing
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•(okay but oversimplified) disorder/randomness•(more correct)dispersal of matter & energy among various motions of particles in space at a temperature in J/K.
DS = Sfinal Sinitial
DS = + therm favDS = – not therm fav
(more dispersal)
(less dispersal)(structure/organization)
DS = ∆HT
“The energy of the universe is constant.”
“The entropy of the universe tends
toward a maximum.”
Entropy (S)
(ratio of heat to temp)
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DS = ∆HT
(a part)(the rest)
System A(100 K)
∆S = ____J/K+0.5
same ∆Hdiff. ∆S
Entropy (S)
(quiet library, more disturbed)
(loud restaurant, less disturbed) Cough!
50 J
Surroundings (100 K)
System B(25 K)
∆S = ____J/K+2.0
50 J
Surroundings (25 K)
• change in entropy (DS) depends on heat transferred (∆H) AND temperature (T)
heightAND
weight
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Entropy
6000 J
DHhand = –6000 J
DHice = +6000 J
• Example: melting 1 mol of ice at 0oC.
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Entropy
• The melting of 1 mol of ice at 0oC.DHfusion
T=
(1 mol)(6000 J/mol) 273 K = +22.0 J/K
• Assume the ice melted in your hand at 37oC.DHfusion
T=
(1 mol)(–6000 J/mol) 310 K = –19.4 J/K
DSuniv = DSsystem + DSsurroundings
DSuniv = (22.0 J/K) + (–19.4 J/K) =
(gained by ice)
(lost by hand)
DSice =
DShand =
(ice) (hand) DSuniv+2.6 J/K
+DHice = –DHhand
+DSice > –DShand
DS =∆HT
DHuniv = 0DSuniv = +
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710 J
5290 J+
(in hand) (in ice)+2.6 J/K x 273 K = (∆Suniv) (T)
Initial Energy
Final Energy
Universe (isolated system)
“dispersed” energy(unusable)
(usable E)6000 J
(usable E)(dispersed E)
1st Law: 6000 J = 6000 J
Free energy(useful for work)
2nd Law: +22 J/K > –19 J/KDHuniv = 0DSuniv = +
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2nd Law of Thermodynamics
For thermodynamically favorable (spontaneous) processes…
All favorable processesincrease the entropy of the universe
(DSuniv > 0) HW p. 837 #20, 21
… +∆S gained always greater than –∆S lost,so…
DSuniv = DSsystem + DSsurroundings
DSuniv = DSsystem + DSsurroundings > 0
2nd Law of Thermodynamics (formally stated):
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Entropy (Molecular Scale)
Motion: Translational , Vibrational, Rotational
• Ludwig Boltzmann described entropy with molecular motion.
• He envisioned the molecular motions of a sample of matter at a single instant in time (like a snapshot) called a microstate.
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Entropy (Molecular Scale)
Boltzmann constant1.38 1023
J/K
microstates (max number
possible)
Entropy increases (+∆S) with the number of
microstates in the system.
<<<
S = k lnW
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• The number of microstates and, therefore, the entropy tends to increase with…
↑Temperature (motion as KEavg)
↑Volume (motion in space)
↑Number of particles (motion as KEtotal)
↑Size of particles (motion of bond vibrations)↑Types of particles (mixing)
Entropy (Molecular Scale)S : dispersal of matter & energy at T
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Maxwell-Boltzmann distribution curve:∆S > 0 by adding heat as…
…distribution of KEavg increases
S : dispersal of matter & energy at TEntropy (Molecular Scale)
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S : dispersal of matter & energy at TEntropy increases with the freedom of motion.
(s) + (l) (aq)
solid gas
V
more microstates
H2O(g) H2O(g)
T
S(s) < S(l) < S(g) S(s) < S(l) < S(aq) < S(g)
Entropy (Molecular Scale)
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Standard Entropy (So)• Standard entropies tend to
increase with increasing molecular size. larger
molecules have more microstates
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Entropy Changes (DS)• In general, entropy increases when
gases form from liquids and solidsliquids or solutions form from solidsmoles of gas molecules increasetotal moles increase
Predict the sign of DS in these reactions:1. Pb(s) + 2 HI(aq) PbI2(s) + H2(g)
2. NH3(g) + H2O(l) NH4OH(aq)
DS =
DS =
+
–
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Third Law of ThermodynamicsThe entropy of a pure crystalline substance at absolute zero is 0. (not possible)
increase
Temp.
0 KS = 0
> 0 KS > 0
S = k lnWS = k ln(1)S = 0
only 1 microstate
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Standard Entropy Changes (∆So)
where n and m are the coefficients in the balanced chemical equation.
DSo = nSo(products) – mSo(reactants)
Standard entropies, S.
(on equation sheet)
HW p. 838#29, 31, 40, 42, 48
(Appendix C)
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ΔS = ?ΔS = ΔH T
ΔH = q(heat)
(disorder)(microstates)
(dispersal of matter &
energy at T) ΔE = q + wPΔV = –w (at constant P)
ΔG = ?
Enthalpy (H)(kJ)
Entropy (S)(J/K)
Free Energy (G) (kJ)
Energy (E)
ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w)
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Big Idea #5: Thermodynamics
Bonds break and formto lower free energy (∆G).
Chemical and physical processes are driven by:• a decrease in enthalpy (–∆H), or• an increase in entropy (+∆S), or• both.
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• Thermodynamically Favorable: (defined as)increasing entropy of the universe (∆Suniv > 0)
Thermodynamically Favorable
∆Suniv > 0 (+Entropy Change of the Universe)
DSuniv = DSsystem + DSsurroundings > 0(+) (+)
Chemical and physical processes are driven by:• decrease in enthalpy (–∆Hsys)
• increase in entropy (+∆Ssys)
causes (+∆Ssurr)
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DSuniverse = DSsystem +
(∆Suniv) & (∆Gsys)
DSuniverse = DSsystem + DSsurroundings > 0For all thermodynamically favorable reactions:
multiplying each term by T:
DHsystem
T
–TDSuniverse = –TDSsystem + DHsystem
rearrange terms:–TDSuniverse = DHsystem – TDSsystem
(Boltzmann)(Clausius)
DGsystem = DHsystem – TDSsystem(Gibbs free energy equation)
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–DG is thermodynamically favorable.
• Gibbs defined TDSuniv as the change infree energy of a system (DGsys) or DG.
• Free Energy (DG) is more useful thanDSuniv b/c all terms focus on the system.
• If –DGsys , then +DSuniverse . Therefore…
(∆Suniv) & (∆Gsys)–TDSuniv = DHsys – TDSsys
DGsys = DHsys – TDSsys
(Gibbs free energy equation)
“Bonds break & form to lower free energy (∆G).”
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∆G : free energy transfer of system as workGibbs Free Energy (∆G)
(not react to completion)
–∆G : work done by system (–w) favorably+∆G : work done on system (+w) to cause rxn
–DG+DG
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R P –DG (release), therm. fav.
+DG (absorb), not therm. fav.
DG = 0, system at
equilibrium.(Q = K)
–DG
DG = 0
Q > K
Q < K
–DG
Q & ∆G (not ∆Go)
+DG
Gmin 0
Q =[P][R]
DGo (1 M, 1 atm, 25oC)Q = 1 = K (rare)
(not react to completion)
Q = K
can cause with electricity/light
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Standard Free Energy (∆Go) and Temperature (T)
The temperature dependence of free energy comes from the entropy term (–TDS).
DG = DH – TDS(on
equation sheet)
enthalpy term
(kJ/mol)
entropy term
(J/mol∙K)
free energy (kJ/mol)
energy transferred
as heat
energy dispersed as disorder
max energy used for
work
(consists of 2 terms)
units convert to kJ!!!
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∆Go = (∆Ho) ∆So
( ) –T( )( ) – T ( )
( ) –T( )( ) – T ( )
( ) – T( )( ) – T( )
DG = DH TDS
(high T) –(low T) +
+ –
(high T) +(low T) –
+ –– +
+ +
– –
(unfav. at ALL T)
(fav. at ALL T)
(fav. at high T)(unfav. at low T)
(unfav. at high T)(fav. at low T)
– T( )
=
=
=
=
+ +
– –
Standard Free Energy (∆Go) and Temperature (T)
Thermodynamic Favorability
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Calculating ∆Go (4 ways)
1) Standard free energies of formation, Gf :
2) Gibbs Free Energy equation:
3) From K value (next few slides)4) From voltage, Eo (next Unit)
DG = nG(products) – mG(reactants)f f
(given equation)
DG = DH – TDS(given equation)
(may need to calc. ∆Ho & ∆So first)
HW p. 840 #52, 54, 56, 60
(given equation)(given equation)
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ΔS = ΔH T
ΔH = q(heat)
ΔG = ?
(disorder)(microstates)
(dispersal of matter &
energy at T) ΔE = q + wPΔV = –w (at constant P)
+ =Enthalpy (H)(kJ)
Entropy (S)(J/K)
Free Energy (G) (kJ)
Energy (E)
(energy of ΔH and ΔS at a T)
(max work done by favorable rxn)
ΔG = ΔH – TΔS
(–∆Gsys means +∆Suniv & K>1)
ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w)
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Under any conditions, standard or nonstandard, the free energy change can be found by:
DG = DG + RT lnQ
Free Energy (∆G) & Equilibrium (K)
Q =[P][R]
At equilibrium: Q = K DG = 0
0 = DG + RT lnK
rearrange: DG = –RT lnK
therefore:
RT is “thermal energy”RT = (0.008314 kJ)(298) = 2.5 kJ at 25oC
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Free Energy (∆G) & Equilibrium (K)
DG = –RT ln K
K = e^–∆Go
RT
(on equation sheet)
(NOT on equation sheet)
R = 8.314 J∙mol–1∙K–1
= 0.008314 kJ∙mol–1∙K–1If DG in kJ,then R in kJ………
Solved for K :
–∆Go
RT= ln K
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∆Go = –RT(ln K) K @ Equilibrium–RT ( )
–RT ( )
> 1
< 1
– + product favored
+ – reactant favored
(favorable forward)
(unfavorable forward)
=
=
DG = –RT ln K
Free Energy (∆G) & Equilibrium (K)
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ΔS = ΔH T
ΔH = q(heat)
(disorder)(microstates)
(dispersal of matter &
energy at T) ΔE = q + wPΔV = –w (at constant P)
+ =Enthalpy (H)(kJ)
Entropy (S)(J/K)
Free Energy (G) (kJ)
Energy (E)
(energy of ΔH and ΔS at a T)
(max work done by favorable rxn)
ΔG = ΔH – TΔS
(–∆Gsys means +∆Suniv & K>1)
ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w)
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What is significant at this point?
p. 837 #6
ΔG = 0 (at equilibrium)
What does x quantify?
G of Reactants
G of Products
ΔGo
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What happens at 300 K?
ΔH = TΔSso…ΔG = ΔH – TΔSΔG = 0 (at equilibrium)
In what T range is this favorable? ΔG = –ΔG = ΔH – TΔS
T > 300 K
p. 837 #4
HW p. 840 #63, 72, 74, 76
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Rxn Coupling:Unfav. rxns (+∆Go) combine with Fav. rxns (–∆Go) to make a Fav. overall (–∆Go
overall ).
ZnS(s) Zn(s) + S(s)(zinc ore) (zinc metal)
∆Go = +198 kJ/mol
S(s) + O2(g) SO2(g)
(NOT therm.fav.)
∆Go = –300 kJ/mol
ZnS(s) + O2(g) Zn(s) + SO2(g) ∆Go = –102 kJ/mol(therm.fav.)
∆Go & Rxn Coupling
goes up if
coupled
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∆Go & Biochemical Rxn Coupling
ATP ADP
ATP + H2O ADP + H3PO4 ∆Go = –31 kJ/mol
Alanine + Glycine Alanylglycine ∆Go = +29 kJ/mol(amino acids) (peptide/proteins)
ATP + H2O + Ala + Gly ADP + H3PO4 + Alanylglycine ∆Go = –2 kJ/mol
(weak bond broken, stronger bonds formed)
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∆Go & Biochemical Rxn CouplingRxn 1:Rxn 2:
Overall Rxn:
Glu + Pi Glu-6-P ATP ADP + Pi
Glu + ATP Glu-6-P + ADP
+14 (not fav)–31 (fav)–17 (fav)
Overall Reaction:
∆Govr
∆Govr = ∆G1 + ∆G2
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∆Go & Biochemical Rxn Coupling
Glucose(C6H12O6)
CO2 + H2O
Proteins
Amino Acids
ATP
ADP
Free
Ene
rgy
(G)
–∆G(fav)
+∆G(not fav)
–∆G(fav)
+∆G(not fav)
+ O2
(oxidation)
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Thermodynamic vs Kinetic ControlFr
ee E
nerg
y (G
)
B
C
A
A B ∆Go = +10 Ea = +20
(kinetic product)
A C ∆Go = –50 Ea = +50
(thermodynamic product)–50 kJ
+10 kJ
(initially pure reactant A)
(–∆Go, Temp, Q<<K, time)
(low Ea , Temp , time)
path 1
path 2
Kinetic Control: (path 2: A C )A thermodynamically favored process (–ΔGo) with no measurable product or rate while not at equilibrium, must have a very high Ea .
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Thermodynamic vs Kinetic ControlPa
in
Kinetic Product: ___Thermodynamic Product: ___E DRxn A E will be under ______________ control at low temp and Q > K .
kinetic (high Ea)