The three laws of thermodynamics
Lecture 3:
Thermodynamics
Matter and energy are conserved
Entropy always increases
Margaret A. Daugherty
Absolute zero is unattainable
Fall 2003
BIOC 205
System and Surroundings
BIOC 205
1st Law of Thermodynamics
Relationship of heat,work, internal energy & Enthalpy
Total E of a system & its surroundings is a constant
∆E = Efinal - Estart = Q + W
Q: heat absorbed by system; W: work done on system
BIOC 205
BIOC 205
Biological Systems and Enthalpy, ∆H:
Calorimetry & van’t Hoff plots: Determination of enthalpy
Definition: H = E + PV
For a biological system:
Temperature
Pressure
Volume
Keq = [B]/[A]
constant!
∆H = -R
Work is a function of V & P
W = V∆P + P∆V; pressure and volume are constant
W=O
∆H = ∆E = Q
Calorimetry
A <--> B
dlnKeq
d(1/T)
∆H = +533 kJ/mol
Enthalpy is equal to the heat
absorbed in a biological process
BIOC 205
BIOC 205
Second Law:
Entropy and Disorder
2nd Law of Thermodynamics
• The degree of randomness or disorder of a system is
measured by a function of state called the entropy (S);
1M NaCl
0.5M NaCl
Diffusion of
a solute
• The entropy of an isolated system will tend to
increase to a maximum value.
∆S = entropy
1). Systems proceed from an ordered to a
disordered state
2). Reversible processes - entropy of Sys + Sur
is unchanged
Irreversible process = entropy of Sys +
Sur increases
3). All process tend to equilibrium - minimum
potential energy
For an isolated system, the favored direction:
∆S = Sfinal - Sinitial > 0
BIOC 205
BIOC 205
Entropy: the equations
Third Law:
Absolute zero is unattainable
S = k log W
What is absolute zero? A temperature where entropy is zero!
k = Boltzmann’s constant
1.38 x 10-23 J/K
W = # of possible ways to arrange a
system at a given temperature
Absolute zero = 0 K
Conversion to Celsius = degrees K - 273.15C
The entropy of any perfectly crystalline substance
approaches zero as absolute zero is approached
1844-1906
The heat capacity, Cp, allows us to have an absolute entropy
T
scale
or Cp = dH/dT
S = CpdlnT
dSrev = dq/T
0
Relates entropy to heat absorbed
If ∆Cp < 0, molecules become more restricted;
if ∆Cp > 0, molecules aquire new ways to move
BIOC 205
Gibbs Free Energy, ∆G: Is a reaction feasible?
BIOC 205
Three Ways To Have Thermodynamically Favored Reactions
(∆G = ∆H - T∆S; ∆G < 0)
For a reaction A <--> B
∆H negative
∆S positive
∆H very negative
∆S negative
∆H positive
∆S very positive
Keq = [B]/[A]
∆G = -RTlnKeq
Relates
1rst
Gibbs Free Energy
and 2nd Laws of Thermodynamics
J. Willard Gibbs
1839-1903
∆G = ∆H - T∆S
BIOC 205
BIOC 205
Standard State Free Energy, ∆Go
State Functions
Functions of State depend only on the initial and final states - not on
the path taken.
State Functions
∆G
∆H
∆S
Volume
Temperature
Pressure
initial
∆G = Ginit - Gfinal
Standard state:
25C,1 atm, concentration of all solutes = 1 M
∆G = ∆Go + RTln
[C][D]
[A][B]
At equilibrium ∆G = 0 and
final
Not State Functions
Work
Heat
[C][D]
Keq = [A][B]
A + B < --> C + D
our criteria for spontaneity is ∆G
[C][D]
= Keq
[A][B]
∆Go = - RTln Keq
∆Go’ = standard state free energy at pH 7.0
Note that 1 M H+ = pH 0, which is unphysiological
Physiologically, we don’t operate at 1 M solution conditions
[C][D]
[A][B]
ADP + Pi ---> ATP
Phosphocreatine + H20 --> creatine + Pi
∆Go = -42.8 kJ/mol at 37C
∆G = -42.8 + RT ln
Coupled processes: How we actually survive!
Problem: Formation of ATP is energetically unfavorable
A + B < --> C + D
∆G = ∆Go + RTln
BIOC 205
(0.001 x 0.001)
(0.001)
∆G = -60.5 kJ/mol
Muscle uses p-creatine to regenerate ATP from ADP BIOC 205
∆G = + 55 kJ/mol
Solution: Couple this reaction to a favorable reaction
PEP + H20 ----> pyruvate + Pi
∆G = -78 kJ/mol
What we end up with:
∆G = -23 kJ/mol
Note: coupling occurs via enzymes; in this
case, pyruvate kinase (see CH 15)
BIOC 205
What do thermodynamic quantities tell us?
Chemically, why is ATP so good?
1
+ Pi
2
3
Reactants: electrostatic repulsion causes bond strain
(4 neg charges) --- destabilizes molecule
∆G
1). Favorable
2). Unfavorable
3). Unfavorable
∆H
∆S
v. favorable
favorable
unfavorable
unfavorable
unfavorable
unfavorable
Products: stabilized by ionization and resonance
∆Cp
H-bonds less labile
H-bonds more labile
BIOC 205
REVIEW
1). Living organisms are thermodynamically open systems.
2). Living organisms operate under the laws of thermodynamics.
3). First law: ∆H = Q: ∆H < 0 for spontaneous reaction
4). Second law: Systems tend to maximize entropy, ∆S > 0 for spont. rxn.
5). Third law: ∆Cp provides information on molecular order in a reaction.
6). ∆G provide information on spontaneity; relates ∆H & ∆S.
7). Thermodynamic quantities provide chemical information on reactions.
8). Standard state free energy allows us to compare biochemical reactions.
9). Metabolically, we can couple unfavorable reactions to favorable
reactions.
10). High energy phosphate molecules drive metabolic reactions.
11). ATP has an intermediate energy among the high energy phosphate
molecules, which positions it as an energy donor and energy acceptor. Thus
various chemical reactions can be coupled in a controlled manner.
12). Various chemical factors contriubte to the large ∆Go’ for ATP
hydrolysis.
BIOC 205
Entropically favorable: 1 reactant --> 2 products
BIOC 205