Download (Chp 5,8,19).

Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 11 (Chp 5,8,19):
Thermodynamics
(∆H, ∆S, ∆G, K)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.
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.
Energy (E)
Enthalpy (H)
(kJ)
Entropy (S)
(J/K)
Free Energy (G)
(kJ)
Energy (E)
What is it?
• 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)
1 Cal = 1000 cal 1 cal = 4.18 J
2000 Cal ≈ 8,000,000 J
≈
8 MJ!!!
System and Surroundings
• System:
molecules to be studied
(reactants & products)
• Surroundings:
everything else
(container, thermometer,…)
1st Law of Thermodynamics
• Energy is neither created nor destroyed.
• total energy of an isolated system is constant
(universe)
(no transfer matter/energy) (conserved)
Internal Energy (E):
E = KE
+
PE
(motions) (attractions)
(Thermal Energy)
(calculating E is too
complex a problem)
DE = Efinal − Einitial
released or absorbed
Changes in Internal Energy
• Energy is transferred between
the system and surroundings,
as either heat (q) or work (w). DE = q + w
DE = ?
DE = (–) + (+)
Surroundings
System
DE = q + w
q in (+)
q out (–)
w on (+)
w by (–)
DE = +
Changes in Internal Energy
Efinal > Einitial
Efinal < Einitial
absorbed energy
released energy
(endergonic)
(exergonic)
Work (w)
The only work done by a gas at constant P
is change in V by pushing on surroundings.
PDV = −w
ΔV
“–” b/c work
done BY system
ON surroundings
Zn + H+
Zn2+ + H2(g)
Enthalpy
Enthalpy (H) is: H = E + PV
internal work done
energy
work done
heat/work energy DH = DE + PDV
by system
in or out
DE = q+w
PDV = −w
of system
(change in) H or
DH (at constant P) :
DH = DE + PDV
DH = (q+w) + (−w)
DH = heat
DH = q
• (change in) enthalpy IS heat absorbed/released
Enthalpy of Reaction
enthalpy is…
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
DH = DE + PDV
DH = q
…the heat transfer
in/out of a system
(at constant P)
DH = Hproducts − Hreactants
exergonic
endergonic
exothermic
endothermic
Endothermic &
Exothermic
• Endothermic: DH > 0 (+)
DH(+) =
Hfinal
− Hinitial
products
reactants
• Exothermic: DH < 0 (–)
DH(–) = Hfinal −
products
Hinitial
reactants
DH(–) is thermodynamically favorable
Enthalpy of Reaction
2 H2(g) + O2(g)  2 H2O(g)
DHrxn, is the
enthalpy of reaction,
or “heat” of reaction. Demo
–242 kJ
units:
per 1 mol O2
kJ
(OR)
molrxn
–242 kJ
kJ/molrxn
–1
kJ∙molrxn
per 2 mol H2
(OR)
–121 kJ
per 1 mol H2
DH =
–242 kJ/molrxn
Enthalpy
HW p. 207
#34,35,38,45
1. DH depends on amount (moles, coefficients)
2. DHreverse rxn = –DHforward rxn
3. DHrxn depends
on the state (s, l, g) of
products & 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)
Chp. 5,8: Calculate ∆H (4 Ways)
1) Bond Energies
2) Hess’s Law
3) Standard Heats of Formation (Hf )
4) Calorimetry (lab)
Overlap and Bonding
When bonds/attractions form, energy is _________.
released
–
+
++
What repulsive forces?
–+
What attractive forces?
Where is energy
being released?
Where must
energy be added?
Potential Energy of Bonds
High PE
Low PE
(energy released
when bonds form)
+
High PE
(energy absorbed
when bonds break)
+
p.330
Bond Enthalpy (BE)
BE: ∆H for the breaking of a bond (all +)
aka…
bond
dissociation
energy
Enthalpy of Reaction (∆H)
BE: ∆H for the breaking of a bond (all +)
To determine DH for a reaction:
• compare the BE of bonds broken (reactants)
to the BE of bonds formed (products).
DHrxn = (BEreactants)  (BEproducts)
(bonds broken)
(stronger)
DH(+) = BEreac − BEprod
DH(–) =
BEreac −
BEprod
(stronger)
(bonds formed)
(released)
(NOT on
equation sheet)
Enthalpy of Reaction (∆H)
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
DHrxn =
[4(C—H) + (Cl—Cl)]  [3(C—H) + (C—Cl) + (H—Cl)]
= (4x413 + 242)
 (3x413 + 328 + 431)
= (655)  (759)
DHrxn = 104 kJ/molrxn
HW p. 339
#66, 68
Chp. 5,8: Calculate ∆H (4 Ways)
1) Bond Energies
(NOT DH
rxn = (BEreactants)  (BEproducts)
given)
(+ broken)
(– formed)
2) Hess’s Law
3) Standard Heats of Formation (Hf )
4) Calorimetry (lab)
Hess’s Law
DH = Hfinal − Hinitial
prod.
react.
DHrxn is
independent of
route taken
DHrxn = sum of
DH of all steps
DHoverall = DH1 + DH2 + DH3 …
(NOT on
equation sheet)
Calculation of DH by Hess’s Law
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) ∆Hcomb = ?

Given:
+
3 C(gr.) + 4 H2(g)  C3H8(g) ∆H1= –104 kJ
3(C(gr.) + O2(g)  CO2(g) ) 3( ∆H2= –394 kJ )
4( H2(g) + ½ O2(g)  H2O(l) ) 4( ∆H3= –286 kJ )
Used:
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 kJ
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) ∆Hcomb =
–2222 kJ
Standard Enthalpy of Formation
o
standard
(DHf )
Standard Enthalpy of Formation (DHof ):
heat released or absorbed by the formation
of a compound from its pure elements in
their natural states.
(25oC , 1 atm)
DHf = 0 for all elements in natural state
o
3 C(gr.) + 4 H2(g)  C3H8(g)
o
o
DHf = 0 DHf = 0
o
DHf = –104 kJ
∆Hf = –104 kJ
Recall… DH = Hfinal − Hinitial …therefore --->
prod.
react.
Calculation of DH by DHf’s
o
…we can use Hess’s law in this way:
“sum”

DH = nHf(products) – nHf(reactants)
(on equation sheet)
n (mol) is the stoichiometric coefficient.
Calculation of DH by DHf’s
o
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) ∆Hcomb = ?


DH = nHf(products)
– nHf(reactants)
Appendix C (p. 1123 )
DH = [3(DHf CO2) + 4(DHf H2O)] – [1(DHf C3H8) + 5(DHf O2)]
DH = (3 ∙ –393.5 + 4 ∙ –285.83) – (–103.85 + 5 ∙ 0)
DH = (–2323.7) – (–103.85)
DH = –2219.9 kJ
HW p. 209
#60,63,66,
72,73
Chp. 5,8: Calculate ∆H (4 Ways)
1) Bond Energies
(NOT DH
rxn = (BEreactants)  (BEproducts)
given)
(+ broken)
(– formed)
2) Hess’s Law
(NOT DH
overall = DHrxn1 + DHrxn2 + DHrxn3 …
given)
3) Standard Heats of Formation (Hf )
(given)

DH  = nHf(products) – nHf(reactants)
4) Calorimetry (lab)
Calorimetry
We can’t know the exact enthalpy
of reactants and products, so we
calculate DH by calorimetry,
the measurement of heat flow.
Calorimeter
nearly
isolated
By reacting (in solution) in a calorimeter,
we indirectly determine DH of system
by measuring ∆T & calculating q of the
surroundings (calorimeter).
heat (J)
q = mcDT
mass (g)
[of sol’n]
(on equation
sheet)
Tf – Ti (oC)
[of surroundings]
Specific Heat Capacity (c)
• specific heat capacity,(c):
(or specific heat)
energy required to
raise temp of 1 g by 1C.
(for water)
c = 4.18 J/goC
Metals have much lower
c’s b/c they transfer heat
and change temp easily.
+ 4.18 J
of heat
HW p. 208
#49, 52, 54
Calorimetry
(in
J)
of
calorimeter
q = mcDT
or surroundings
– q = DHrxn (in kJ/mol) of system
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).
q = (4.50 + 200)(4.18)(28.3–22.4)
qsurr = 5040 J
DH = –5.04 kJ
DHsys = –5.04 kJ
0.1125 mol
4.50 g NaOH x 1 mol = 0.1125 mol NaOH = –44.8 kJ
mol
40.00 g
Chp. 5,8: Calculate ∆H (4 Ways)
1) Bond Energies
(NOT DH
rxn = (BEreactants)  (BEproducts)
given)
(+ broken)
(– formed)
2) Hess’s Law
(NOT DH
overall = DHrxn1 + DHrxn2 + DHrxn3 …
given)
3) Standard Heats of Formation (Hf )
(given)

DH  = nHf(products) – nHf(reactants)
4) Calorimetry (lab)
(given)
q = mc∆T (surroundings or thermometer)
–q = ∆H ∆H/mol = kJ/mol (molar enthalpy)
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 19:
Thermodynamics
(∆H, ∆S, ∆G, K)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.
Energy (E)
Enthalpy (H)
(kJ)
Entropy (S)
(J/K)
Free Energy (G)
(kJ)
ΔH = ΔE + PΔV
internal work by
energy system
(KE + PE) (–w)
ΔE = q + w
PΔV = –w
(at constant P)
ΔH = ?
q
(heat)
ΔS = ?
ΔG = ?
Big Idea #5: Thermodynamics
Bonds break and form
to lower free energy (∆G).
Chemical and physical processes
are driven by:
• a decrease in enthalpy (–∆H), or
• an increase in entropy (+∆S), or
• both.
1st Law of Thermodynamics
• Energy cannot be created nor destroyed
(is conserved)
• or…total energy of the universe is constant.
DHsystem = –DHsurroundings
OR DHuniv = DHsystem + DHsurroundings = 0
if (+)
then (–) = 0
if (–)
then (+)
= 0
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
• Processes that are favorable (spontaneous)
at one temperature…
…may not be at other temperatures.
HW p. 837 #7, 11
melting
freezing
Entropy (S)
•(okay but oversimplified)
disorder/randomness
“The energy of the
•(more correct)
dispersal of matter & energy universe is constant.”
“The entropy of the
among various motions of
universe tends
particles in space at a
toward a maximum.”
(ratio of heat to temp)
temperature in J/K.
∆H
DS = Sfinal  Sinitial
DS =
T
(more
dispersal)
DS = + therm fav
DS = – not therm fav (less dispersal)
(structure/organization)
Entropy (S)
∆H
DS =
T
(a part)
(the rest)
height
AND
weight
• change in entropy (DS) depends on…
…heat (∆H)…AND…temp. (T)
System A 50 J
(100 K)
Surroundings (100 K)
same ∆H
diff. ∆S
System B 50 J
(25 K)
Surroundings (25 K)
∆S = ____J/K
+0.5
∆S = ____J/K
+2.0
(loud restaurant,
less disturbed)
(quiet library,
more disturbed)
Cough!
Entropy
• Example: melting 1 mol of ice at 0oC.
DShand = ?
DHhand = –6000 J
6000 J
DHice = +6000 J
DSice = ?
∆H
+DH
=
–DH
DH
=
0
ice
hand
univ
DS =
Entropy
T
+DSice > –DShand DSuniv = +
• The melting of 1 mol of ice at 0oC.
DHfusion
DSice =
T
(1 mol)(6000 J/mol)
=
= +22.0 J/K
273 K
(gained by ice)
• Assume the ice melted in your hand at 37oC.
DHfusion (1 mol)(–6000 J/mol) –19.4 J/K
DShand =
=
=
310
K
T
(lost by hand)
DSuniv = DSsystem + DSsurroundings
(hand)
(ice)
DSuniv
DSuniv = (22.0 J/K) + (–19.4 J/K) = +2.6 J/K
DHuniv = 0
2nd Law: +22 J/K > –19 J/K DSuniv = +
1st Law: 6000 J = 6000 J
(usable E)
6000 J
(usable E)
(dispersed E)
(in hand)
(in ice)
5290 J
+
710 J
x
Universe (isolated system)
(∆Suniv)
=
(T)
“dispersed” energy
(unusable)
Free energy
(useful for work)
2nd Law of Thermodynamics
DSuniv = DSsystem + DSsurroundings
For thermodynamically favorable
(spontaneous) processes…
… +∆S gained always greater than –∆S lost,
so… DSuniv = DSsystem + DSsurroundings > 0
2nd Law of Thermodynamics (formally stated):
All favorable processes
increase the entropy of the universe
(DSuniv > 0)
HW p. 837 #20, 21
Entropy (Molecular Scale)
• Ludwig Boltzmann described entropy
with molecular motion.
Motion: Translational , Vibrational, Rotational
• 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)
S = k lnW
Boltzmann
constant
1.38  1023
J/K
microstates
(max number
possible)
Entropy increases (+∆S)
with the number of
microstates in the system.
<
<
<
Entropy (Molecular Scale)
S : dispersal of matter & energy at T
• 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
Entropy (Molecular Scale)
S : dispersal of matter & energy at T
Entropy increases with the freedom of motion.
S(s) < S(l) < S(g)
solid
H2O(g)
S(s) < S(l) < S(aq) < S(g)
gas
T
V
more
microstates (s) + (l)  (aq)
H2O(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 solids
liquids or solutions form from solids
moles of gas molecules increase
total moles increase
Predict the sign of DS in these reactions:
1. Pb(s) + 2 HI(aq)  PbI2(s) + H2(g)
DS = +
2. NH3(g) + H2O(l)  NH4OH(aq)
DS = –
3rd Law of Thermodynamics
The entropy of a pure crystalline substance
at absolute zero is 0. (not possible)
S = k lnW
S = k ln(1)
S=0
increase
only 1
microstate
0K
S=0
Temp.
>0K
S>0
Standard Entropy Changes (∆So)
HW p. 838
#29, 31, 40, 42, 48
Standard entropies, S.
(Appendix C)
DSo = nSo(products) – nSo(reactants)
(on equation sheet)
n (mol) is the stoichiometric
coefficient.
Energy (E)
Enthalpy (H)
(kJ)
ΔH = ΔE + PΔV
internal work by
energy system
(KE + PE) (–w)
ΔE = q + w
PΔV = –w
Entropy (S)
(J/K)
(disorder)
microstates
dispersal of
matter &
energy at T
(at constant P)
∆Suniv = +
ΔH = q
ΔS
ΔS==ΔH
?
T
(heat)
Free Energy (G)
(kJ)
ΔG = ?
Big Idea #5: Thermodynamics
Bonds break and form
to 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
Chemical and physical processes are driven by:
•
decrease in enthalpy (–∆Hsys) causes (+∆Ssurr)
•
increase in entropy (+∆Ssys)
(+) + DS (+)
DSuniv = DSsystem
surroundings > 0
•
Thermodynamically Favorable: (defined as)
increasing entropy of the universe (∆Suniv > 0)
∆Suniv > 0
(+Entropy Change of the Universe)
(∆Suniv) ↔ (∆Gsys)
For all thermodynamically favorable reactions:
DSuniverse = DSsystem + DSsurroundings > 0
DSuniverse = DSsystem +
(Boltzmann)
DHsystem (Clausius)
T
multiplying each term by T:
–TDSuniverse = –TDSsystem + DHsystem
rearrange terms:
–TDSuniverse = DHsystem – TDSsystem
DGsystem = DHsystem – TDSsystem
(Gibbs free energy equation)
(∆Suniv) & (∆Gsys)
–TDSuniv = DHsys – TDSsys
DGsys = DHsys – TDSsys
(Gibbs free energy equation)
• Gibbs defined TDSuniv as the change in
free energy of a system (DGsys) or DG.
• Free Energy (DG) is more useful than
DSuniv b/c all terms focus on the system.
• If –DGsys , then +DSuniverse . Therefore…
–DG is thermodynamically favorable.
“Bonds break & form to lower free energy (∆G).”
Gibbs Free Energy (∆G)
∆G : free energy transfer of system as work
–∆G : work done by system (–w) favorably
+∆G : work done on system (+w) to cause rxn
(not react to
completion)
–DG
+DG
Q & ∆G (not ∆Go)
Q<K
[P]
R  P Q=
[R]
(not react to
completion)
can cause with
electricity/light
Q>K
+DG
–DG
Gmin  0
–DG
Q=K
–DG (release),
therm. fav.
+DG (absorb),
not therm. fav.
DG = 0,
system at
equilibrium.
(Q = K)
DG = 0
DGo (1 M, 1 atm, 25oC)
Q = 1 = K (rare)
o
(∆G )
Standard Free Energy
and Temperature (T)
(on
equation
sheet)
(consists DG = DH – TDS
of 2 terms)
free
enthalpy entropy units
energy
term
term
convert
(kJ/mol) (kJ/mol) (J/mol∙K) to kJ!!!
max energy
energy
energy
used for
transferred dispersed
work
as heat
as disorder
The temperature dependence of free energy
comes from the entropy term (–TDS).
o
(∆G )
Standard Free Energy
and Temperature (T)
DG
Thermodynamic
o
∆G
Favorability
(fav. at high T) (high T) –
(unfav. at low T) (low T) +
(unfav. at ALL T)
+
(fav. at ALL T)
–
(unfav. at high T) (high T) +
(fav. at low T) (low T) –
= DH  TDS
= (∆Ho) – T(∆So)
( + ) –T( + )
= ( + ) – T( + )
= ( + ) – T( – )
= ( – ) – T( + )
– ) –T( – )
(
= ( – ) – T( – )
Energy (E)
Enthalpy (H)
(kJ)
+
ΔH = ΔE + PΔV
internal work by
energy system
(KE + PE) (–w)
ΔE = q + w
PΔV = –w
Entropy (S)
(J/K)
(disorder)
microstates
–T∆Suniv as: ΔHsys
& ΔSsys at T
dispersal of
matter &
energy at T
max work done
by favorable rxn
(at constant P)
∆Suniv = +
ΔH = q
ΔS = ΔH
T
(heat)
=
Free Energy (G)
(kJ)
ΔGΔG
= ΔH
= ?– TΔS
Calculating ∆Go (4 ways)
1) Standard free energies of formation, Gf :
DG = nG
f (products) – mG
f (reactants)
(given equation)
2) Gibbs Free Energy equation:
HW p. 840
#52, 54, 56, 60
DG = DH – TDS
(given equation)
(may need to calc. ∆Ho & ∆So first)
3) From K value (next few slides) (given equation)
4) From voltage, Eo (next Unit) (given equation)
Free Energy (∆G) & Equilibrium (K)
Under any conditions, standard or nonstandard,
the free energy change can be found by:
[P]
DG = DG + RT lnQ
Q=
[R]
At equilibrium:
Q=K
DG = 0
therefore:
rearrange:
RT is “thermal energy”
RT = (0.008314 kJ)(298)
= 2.5 kJ at 25oC
0 = DG + RT lnK
DG = –RT lnK
Free Energy (∆G) & Equilibrium (K)
DG = –RT ln K
(on equation sheet)
–1∙K–1
R
=
8.314
J∙mol
If DG in kJ,
then R in kJ……… = 0.008314 kJ∙mol–1∙K–1
–∆Go = ln K
RT
Solved
for K :
–∆Go
RT
K = e^
(NOT on
equation sheet)
Free Energy (∆G) & Equilibrium (K)
DG = –RT ln K
∆Go = –RT(ln K)
– = –RT ( + )
K
@ Equilibrium
> 1 product favored
(favorable forward)
+ = –RT ( – ) < 1
reactant favored
(unfavorable forward)
Energy (E)
Enthalpy (H)
(kJ)
+
ΔH = ΔE + PΔV
internal work by
energy system
(KE + PE) (–w)
ΔE = q + w
PΔV = –w
Entropy (S)
(J/K)
(disorder)
microstates
–T∆Suniv as: ΔHsys
& ΔSsys at a T
dispersal of
matter &
energy at T
max work done
by favorable rxn
(at constant P)
ΔH = q
(heat)
=
Free Energy (G)
(kJ)
ΔS = ΔH
T
K > 1 means
–∆Gsys & +∆Suniv
ΔG = ΔH – TΔS
p. 837 #6
G of Reactants
What does x
quantify?
What is
significant at
this point?
ΔGo
G of Products
ΔG = 0 (at equilibrium)
p. 837 #4
What
happens
at 300 K?
In what T range
is this favorable?
ΔG = –
ΔG = ΔH – TΔS
T > 300 K
ΔH = TΔS
so…
ΔG = ΔH – TΔS
ΔG = 0 (at equilibrium)
HW p. 841
#62, 63, 76
∆Go & Rxn Coupling
Rxn Coupling:
Unfav. rxns (+∆Go) combine with
Fav. rxns (–∆Go) to make a
Fav. overall (–∆Gooverall ).
(zinc ore)  (zinc metal)
ZnS(s)  Zn(s) + S(s)
S(s) + O2(g)  SO2(g)
goes up
if
coupled
(NOT therm.fav.)
∆Go = +198 kJ/mol
∆Go = –300 kJ/mol
ZnS(s) + O2(g)  Zn(s) + SO2(g) ∆Go = –102 kJ/mol
(therm.fav.)
∆Go & Biochemical Rxn Coupling
(weak bond broken,
stronger bonds formed)
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
∆Go & Biochemical Rxn Coupling
Rxn 1: Glu + Pi  Glu-6-P
ATP  ADP + Pi
Rxn 2:
Overall Rxn: Glu + ATP  Glu-6-P + ADP
+14 (not fav)
–31 (fav)
–17 (fav)
Overall Reaction:
∆Govr
∆Govr = ∆G1 + ∆G2
∆Go & Biochemical Rxn Coupling
Free Energy (G)
Glucose
(C6H12O6)
Proteins
ATP
+ O2
(oxidation)
–∆G
(fav)
CO2 + H2O
+∆G
(not fav)
ADP
–∆G
(fav)
+∆G
(not fav)
Amino Acids
Thermodynamic vs Kinetic Control
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 .
∆Go = +10
Ea = +20
(kinetic product)
Free Energy (G) 
A  B
(initially pure reactant A)
path 1
A
B
+10 kJ
–50 kJ
(low Ea , Temp , time)
∆Go = –50
path 2
Ea = +50
C (thermodynamic product)
A  C
(–∆Go, Temp, Q<<K, time)
Thermodynamic vs Kinetic Control
Thermodynamic Product: ___
E
Kinetic Product: ___
D
Pain 
Rxn A  E will be under ______________
kinetic (high Ea)
control at low temp and Q > K .