Unit 11 (Chp 5,8,19) : Thermodynamic s ( H, S, G, K) ∆ ∆ ∆ John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall, Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
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Unit 11 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K)
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.
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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
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
Energy (E)
Enthalpy (H)(kJ)
Entropy (S)(J/K)
Free Energy (G) (kJ)
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
System and Surroundings
• System:molecules to be studied (reactants & products)
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
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.
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
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.
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)
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).
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
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
Thermodynamically Favorable
• Processes that are thermodynamically favorable (spontaneous) in one direction are NOT in the reverse direction.
Thermodynamically Favorable
melting
• Processes that are favorable (spontaneous) at one temperature……may not be at other temperatures.
HW p. 837 #7, 11
freezing
•(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)
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
Entropy
6000 J
DHhand = –6000 J
DHice = +6000 J
• Example: melting 1 mol of ice at 0oC.
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 = +
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 = +
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):
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.
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
• 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
Maxwell-Boltzmann distribution curve:∆S > 0 by adding heat as…
…distribution of KEavg increases
S : dispersal of matter & energy at TEntropy (Molecular Scale)
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)
Standard Entropy (So)• Standard entropies tend to
increase with increasing molecular size. larger
molecules have more microstates
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 =
+
–
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
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)
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)
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.
• 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)
ATP + H2O + Ala + Gly ADP + H3PO4 + Alanylglycine ∆Go = –2 kJ/mol
(weak bond broken, stronger bonds formed)
∆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
∆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)
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 .
Thermodynamic vs Kinetic ControlPa
in
Kinetic Product: ___Thermodynamic Product: ___E DRxn A E will be under ______________ control at low temp and Q > K .