Unit 11 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K)

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)




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)

• Surroundings:everything else(container, thermometer)

• 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 (+)

Changes in Internal Energy

Efinal > Einitial

absorbed energy(endergonic)

Efinal < Einitial

released energy(exergonic)

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)

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

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

Endothermic & Exothermic

• Endothermic: DH > 0 (+)

• Exothermic: DH < 0 (–)

DH(+) = Hfinal − Hinitial

DH(–) is thermodynamically favorable

products reactants

DH(–) = Hfinal − Hinitialproducts reactants

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)

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

Chp. 5,8: Calculate ∆H (4 Ways)1) Bond Energies

2) Hess’s Law

3) Standard Heats of Formation (Hf )

4) Calorimetry (lab)

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?

++

(energy absorbed when bonds break)

Potential Energy of BondsHigh PE

Low PE(energy released

when bonds form)

chemical bond

High PE+ +

Bond Enthalpy (BE)p.330BE: ∆H for the breaking of a bond (all +)

aka…bond dissociation energy

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)

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)]

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.


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:

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…

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

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


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

DH = –5.04 kJ 0.1125 mol

= –44.8 kJ mol

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 )


4) Calorimetry (lab) q = mc∆T (surroundings or thermometer)–q = ∆H ∆H/mol = kJ/mol (molar enthalpy)

(+ broken) (– formed)(NOT)

(given)

(NOT)

(given)

Chapter 19:Thermodynamics

(∆H, ∆S, ∆G, K)




ΔH = ?ΔH = q(heat)

ΔG = ?

ΔE = q + wPΔV = –w (at constant P)

ΔS = ?

Enthalpy (H)(kJ)

Entropy (S)(J/K)


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.


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.


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)


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)


Energy (E)


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)


∆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)

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)

–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).”

∆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

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

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!!!

∆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

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)

ΔS = ΔH T

ΔH = q(heat)

ΔG = ?




+ =Enthalpy (H)(kJ)

Entropy (S)(J/K)


Energy (E)

(energy of ΔH and ΔS at a T)

(max work done by favorable rxn)

ΔG = ΔH – TΔS

(–∆Gsys means +∆Suniv & K>1)


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


DG = –RT ln K

K = e^–∆Go

RT

(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

∆Go = –RT(ln K) K @ Equilibrium–RT ( )

–RT ( )

> 1

< 1

– + product favored

+ – reactant favored

(favorable forward)

(unfavorable forward)

=

=

DG = –RT ln K


ΔS = ΔH T

ΔH = q(heat)




+ =Enthalpy (H)(kJ)

Entropy (S)(J/K)


Energy (E)

(energy of ΔH and ΔS at a T)

(max work done by favorable rxn)

ΔG = ΔH – TΔS

(–∆Gsys means +∆Suniv & K>1)


What is significant at this point?

p. 837 #6

ΔG = 0 (at equilibrium)

What does x quantify?

G of Reactants

G of Products

ΔGo

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

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)

∆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 .

kinetic (high Ea)

Unit 11 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K)

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Transcript of Unit 11 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K)