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. (moving an object) 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 constant

Internal Energy (E):E = KE + PE (motions)(Thermal Energy)

(calculating E is too complex a problem)

DE = Efinal − Einitialreleased or absorbed

(attractions)

1st Law of Thermodynamics

(no transfer matter/energy)(universe) (conserved)

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)

“–” b/c work done BY system ON surroundings

Enthalpy

(change in) H or DH (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

–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?

+ +––

What repulsive forces?

Where is energy being released?

++

releasedWhen bonds/attractions form, energy is _________.

internuclear distance

E

(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)

Enthalpy 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)

Enthalpy of Reaction (∆H)

= (4x413 + 242) (3x413 + 328 + 431)

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)]

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

DHrxn = (BEreactants) (BEproducts)

2) Hess’s Law

3) Standard Heats of Formation (Hf )

4) Calorimetry (lab)

(+ broken) (– formed)(NOT given)

Hess’s Law

DHrxn is independent of

path taken

DHoverall = DH1 + DH2 + DH3 …

DHrxn = sum of DH of all steps

DH = Hfinal − Hinitialprod. react.

(NOT on equation sheet)

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 natural states.

(25oC , 1 atm)

o

3 C(gr.) + 4 H2(g) C3H8(g)∆Hf = –104 kJ

DHf = 0 for all elements in natural 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) – nHf(reactants)

n (mol) is the stoichiometric coefficient.

“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 = (3 ∙ –393.5 + 4 ∙ –285.83) – (–103.85 + 5 ∙ 0)

DH = (–2323.7) – (–103.85)

DH = –2220.0 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) – nHf(reactants)

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) – nHf(reactants)

4) Calorimetry (lab)

(+ broken) (– formed)(NOT given)

(given)

(NOT given)

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 calculate 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 surroundings

(in kJ/mol) of system

q = (4.50 + 200)(4.18)(28.3–22.4)qsurr = 5040 J

DHsys = –5.04 kJ4.50 g NaOH x 1 mol = 0.1125 mol NaOH 40.00 g

DH = –5.04 kJ 0.1125 mol

= –44.8 kJ mol

(thermometer)

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) – nHf(reactants)

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

(+ broken) (– formed)(NOT given)

(given)

(NOT given)

(given)

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.

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)

• or…total energy of the universe is constant.

DHuniv = DHsystem + DHsurroundings = 0if (+) then (–) = 0if (–) then (+) = 0

DHsystem = –DHsurroundings

OR

Thermodynamically Favorable• Thermodynamically

Favorable (spontaneous) processes occur with no outside intervention.

• If Favorable in one direction, then UNfavorable in reverse.

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 = – therm UNfav

(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 (∆H)…AND…temp. (T)

heightAND

weight

Entropy

6000 J

DHhand = –6000 J

DSice = ?

• Example: melting 1 mol of ice at 0oC.

DShand = ?

DHice = +6000 J

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)

↑Particle number (motion as KEtotal)

↑Particle Size (motion of bond vibrations)↑Particle Type (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)

(T)

Entropy 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)S : dispersal of matter & energy at T

Standard Entropy (So)• Standard entropies tend to

increase with increasing molecular size. larger

molecules have more microstates

Entropy Changes (DS)

• In general, entropy increases whenliquids or solutions form from solidsmoles of gas increasestotal 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 =

+

–

3rd 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)

n (mol) is the stoichiometric coefficient.

DSo = nSo(products) – nSo(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)

Enthalpy (H)(kJ)

Entropy (S)(J/K)

Free Energy (G) (kJ)

Energy (E)

ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w)

∆Suniv = +

ΔG = ?

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)

• 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)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

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)

–T∆Suniv as: ΔHsys & ΔSsys at T

max work done by favorable rxn

ΔG = ΔH – TΔS

ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w)

∆Suniv = +

–T∆Sunivsys sys

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) – nG(reactants)f f

(given equation)

DG = DH – TDS(given equation)

(may need to calc. ∆Ho & ∆So first)

HW p. 840 #52, 54, 60

(given equation)(given equation)

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

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

∆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)

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)

–T∆Suniv as: ΔHsys & ΔSsys at a Tmax work done by favorable rxn

ΔG = ΔH – TΔS

K > 1 means –∆Gsys & +∆Suniv

ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w)

–T∆Sunivsys sys

∆Suniv = +

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. 841 #62, 63, 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

∆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)