a little thermodynamics
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CH339K. (which is probably more than anybody wants). A little thermodynamics. Thermodynamics (Briefly). Systems est divisa in partes tres Open Exchange energy and matter Closed Exchange energy only Isolated Exchange nothing. More Thermodynamics. - PowerPoint PPT PresentationTRANSCRIPT
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A little thermodynamics
(which is probably more than anybody wants)
CH339K
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Thermodynamics (Briefly)
• Systems est divisa in partes tres– Open
• Exchange energy and matter
– Closed• Exchange energy only
– Isolated• Exchange nothing
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More Thermodynamics• Energy can be exchanged as heat (q) or work
(w)• By convention:
– q > 0:heat has been gained by the system from the surroundings
– q < 0:heat has been lost by the system to the surroundings
– w > 0:work has been done by the system on the surroundings
– w < 0: work has been done on the system by the surroundings
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First Law of Thermo• ESYSTEM = q – w or, alternatively, q = E + w
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First law of Thermo (cont.)
C16H32O2 + 23O2 (g) 16CO2 (g) + 16H2O (l)
• Under Constant Volume:
q = -9941.4 kJ/mol.• Under Constant Pressure:
q = -9958.7 kJ/mol
Example: Oxidation of a Fatty Acid (Palmitic):
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First Law of Thermo (cont.)
• Why the difference?
• Under Constant Volume, q = E + w = -9941.4 kJ/mol + 0 = -9941.4 kJ/mol
• Under Constant Pressure, W is not 0! Used 23 moles O2, only produced 16 moles CO2
W = PΔVΔV = ΔnRT/PW = ΔnRT = (-7 mol)(8.314 J/Kmol)(298 K) = -17.3 kJ q = -9941.4 kJ/mol + (-17.3 kJ/mol) = -9958.7 kJ/mol
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Enthalpy• Technically speaking, most cells operate under
constant pressure conditions• Practically, there’s not much difference most of the
time• Enthalpy (H) is defined as:
H = E + PV or
H = E + PV
• If H > 0, heat is flowing from the surroundings to the system and the process is endothermic
• if H < 0, heat is being given off, and the process is exothermic.
• Many spontaneous processes are exothermic, but not all
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Endothermic but spontaneous• Ammonium Nitrate spontaneously dissolves
in water to the tune of about 2 kg/liter• Ammonium nitrate has a Hsolution of +25.7
kJ/mol• Remember positive enthalpy = endothermic• This is the basis of instant cold packs
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Second law of Thermo
• Any spontaneous process must be accompanied by a net increase in entropy (S).
• What the heck is entropy?• Entropy is a measure of the “disorderliness”
of a system (and/or the surroundings).• What the heck does that mean?• Better, it is a measure of the number of states
that a system can occupy.• Huh?...let me explain
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Entropy
S = k x ln(W) where• W is the number of possible states• k is Boltzmann’s constant, = R/N
Two states of 5 “atoms” in 50 possible “slots.”
State 1… State 2… etc…
X
X
X
X X
X
X
X
X
X
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What happens if the volume increases?
K
K
K K
K
Adding volume increases the number of “slots,” therefore increasing W, the number of states, thereby increasing entropy.
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• We can quantify that:– Number of atoms dissolved = Na– Number of original slots = no
– Number of original states = Wo– Number of final slots = nf
– Number of final states = Wf
o o o o oW = n (n 1)(n 2)...(n Na)
f f f f fW = n (n 1)(n 2)...(n Na) • Since Na << Wo and Na << Wf (dilute solution), then:
o on Na n f fn Na n and
• So we can simplify the top equations to:Na
o oW = n Naf fW = n
and
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• Okay, so what (quantitatively) is the change in entropy from increasing the volume?
f oS = S - S• Substituting and solving:
f oS = k ln(W ) k ln(W )
f
o
WS = k ln
W
NafNao
nS = k ln
n
Na
f
o
nS = k ln
n
f
o
nS = Na k ln
n
So S is logarithmically related to the change in the number of “slots.”
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• Let’s make the assumption that we are dealing with 1 mole (i.e. N atoms) of solute dissolved in a large volume of water.
• Since Boltzmann’s constant (k) = R/N, our equation resolves to:
f
o
nS = R ln
n
• Since the number of “slots” is directly related to the volume:
f
o
VS = R ln
V
• And since the concentration is inversely related to the volume:
o
f
CS = R ln
C
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Entropy (cont.)• Entropy change tells us whether a reaction is
spontaneous, but…• Entropy can increase in the System, the
Surroundings, or both, as long as the total is positive.
• Can’t directly measure the entropy of the surroundings.
• HOWEVER, the change in enthalpy of the system is an indirect measure of the change in entropy of the surroundings – an exothermic reaction contributes heat (disorder) to the universe.
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Gibbs Free Energy• We can coin a term called the Free Energy (G) of the
system which tells us the directionality of a reaction.
G = H – TSΔG = ΔH - T ΔS
If ΔG < 0, free energy is lost exergonic – forward rxn favored.
If ΔG > 0, free energy is gained endergonic – reverse rxn favored.
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Different ΔG’s
• ΔG is the change in free energy for a reaction under some set of real conditions.
• ΔGo is the change in free energy for a reaction under standard conditions (all reactants 1M)
• ΔGo’ is the change of free energy for a reaction with all reactants at 1M and pH 7.
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Partial Molar free Energies• The free energy of a mixture of stuff is equal to the
total free energies of all its components• The free energy contribution of each component is
the partial molar free energy:
]ln[0 GRTGG xx • Where:
solutionor mixture in thecomponent theof the][
component theofenergy free standard the0
activityG
G
x
x
• In dilute (i.e. biochemical) solutions, • the activity of a solute is its concentration• The activity of the solvent is 1
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Free Energy and Chemical EquilibriumTake a simple reaction:
A + B C + D⇌Then we can figure the Free Energy Change:
reactantsproducts GG G BRTln - G - ARTln - G - DRTln G CRTln G G o
BoA
oD
oC
Rearranging BRTln - ARTln - DRTln CRTln G G G G G o
BoA
oD
oC
Combining
Factoring Bln - Aln - Dln ClnRT G G o
B A
D CRTln G G o
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Freee Energy and Equilibrium (cont.)
Hang on a second!
[A][B] is the product of the reactant concentrations
[C][D] is the product of the product concentrations
Remembering Freshman Chem, we have a word for that ratio.
B A
D CRTln G G o
o ProductsG G RTln
Reactants
B A
D C K eq
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Free Energy and Equilibrium (cont.)
SO: ΔGo for a reaction is related to the equilibrium constant for that reaction.
ΔGo = -RTlnKeq
Or
Keq = e-ΔGo/RT
If you know one, you can determine the other.
Note: things profs highlight with colored arrows are probably worth remembering
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Real Free Energy of a Reaction
As derived 2 slides previously:
G is related to Go’, adjusted for the concentration of the reactants:
][Reactants
[Products]RTln'ΔGΔG o
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Example:Glucose-6-Phosphate ⇄ Glucose + Pi ∆Go’ = -13.8 kJ/mol
At 100 μM Glucose-6-Phosphate 5 mM Phosphate 10 mM Glucose
15587J/mol1787J/mol)(13800J/molΔG
(.0001M)
5M)(.01M)(.00ln310Kl8.315J/Kmo13800J/molΔG
Phosphate]6[Glucose
Pi] [Glucose][RTln'ΔGΔG o
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Measuring H, S, and G
We know
ΔG = ΔH - T ΔS
And
ΔGo = -RTlnKeq
So
ΔH - T ΔS = -RTlnKeq
Or
R
S
T
1
R
H- Kln
oo
eq
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Measuring H, S, and G
• This is the van’t Hoff Equation• You can control T• You can measure Keq• If you plot ln(Keq) versus 1/T, you get a line
– Slope = -ΔHo/R– Y-intercept = ΔSo/R
R
S
T
1
R
H- Kln
oo
eq
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Van’t Hoff Plot
ΔHo = -902.1* 8.315 = -7500 J/molΔSo = +3.61 * 8.315 = 30 J/Kmol
y = -902.09x + 3.6084
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037
1/T (K-1)
ln(K
eq)
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Why the big Go’ for Hydrolyzing Phosphoanhydrides?
• Electrostatic repulsion betwixt negative charges
• Resonance stabilization of products
• pH effects
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pH Effects – Go vs. Go’
mol
kJ 5.41ln'
10lnln'
M10 7,pH At
lnln
ln
ln
7
7-
2
2
ReactantsProducts
ATP
PiADPRTGG
RTATP
PiADPRTGG
H
OH
HRT
ATP
PiADPRTGG
OHATP
HPiADPRTGG
RTGG
oo
oo
o
o
o
G in kcal/mol)
WOW!
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Cellular Gs are not Go’ sGo’ for hydrolysis of ATP is about -31 kJ/mol
Cellular conditions are not standard, however:
In a human erythrocyte,
[ATP]≈2.25 mM, [ADP] ≈0.25 mM, [PO4] ≈1.65 mM
mol
kJ
mol
kJ
mol
kJG
M
MMK
molK
J
mol
kJG
ATP
PiADPRTGG
Hyd
Hyd
oHyd
52)21(31
)00225(.
)00165)(.00025(.ln298315.831
][
]][[ln'
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Unfavorable Reactions can be Subsidized with Favorable Ones
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Hydrolysis of Thioesters can also provide a lot of free energy
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Acetyl Coenzyme A
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Sample Go’Hydrolysis