S = k ln W
A vignette….
Let’s consider a simpler case first
Thought experiment: Consider a beaker with a partition right in the middle that
starts out holding molecules in only one half. What happens when you remove
the partition?
How about
the reverse
process?
• What is DH for this process?
• Zero (at least if molecules don’t interact, “ideal solution”)
So why does it happen?
[Relevance
to drug
distribution
]
How about the reverse process?
Thought experiment: Molecules spontaneously moving to only one half of
beaker?
• DH is still zero
• This seems unlikely, except maybe if there are very few particles
• Let’s calculate the probabilities quantitatively
Probability of being on one side
1 mol
N=1
p = 1/2
N=2
p = 1/4
N
p = (1/2)N
N ~ 6 x 1023
p = (1/2)N =
very very small
This fits with our intuition (and the second law) which says that things tend toward
disorder, i.e., higher entropy [unless there is an input of energy to counteract …]
Statistical thermodynamics tells us how to
compute entropy
• There are multiple ways to define entropy, including the one
on Boltzmanns tombstone
In modern notation, this is
typically written S = kB ln W
W is the number of
configurations available to a
system, and is related to the
probabilities we just
calculated.
Note that S is a state function, and positive values of DS are favorable,
not negative like DH. Units on S are J/K (more on this later).
Let’s compute DS for this process
This is not just a thought experiment. This is equivalent to entropy change upon
dilution, or mixing two solutions.
State 1
State 2
S1 = kB ln W1
S2 = kB ln W2
We can define the number of “configurations” of the
system in different ways, but regardless, W clearly
doubles for a single molecule when we double the
volume.
N=1
DS = kB ln 2
= 1.38x10-23 J/K * 0.693 ~ 10-23 J/K
To get units of energy, multiply by temperature (300 K),
which gives 3x10-21 J. TINY amount of energy.
We assume that the 2 molecules are
uncorrelated, i.e., the configuration of
one is unrelated to the other.
(This is again the definition of an “ideal
solution”.)
N=1
DS = kB ln 2
N=2
DS = kB ln 4
This then implies that the total number
of configurations is the product (not
sum) of the configurations for each
molecule.
N=1
DS = kB ln 2
N=2
DS = kB ln 4
N molecules
DS = kB ln 2N =NkBln 2
N=1
DS = kB ln 2
N=2
DS = kB ln 4
N = NA = 1 mole
DS = NAkBln 2 = R ln 2
N = # of molecules
NA = Avogadro’s number
kB = Boltzmann’s constant
R = NAkB = gas constant
N=1
DS = kB ln 2
N=2
DS = kB ln 4
Generalizing to any volume change,
Molar entropy:
æ V2 ö
DS = R ln ç ÷
è V1 ø
N = NA = 1 mole
DS = NAkBln 2 = R ln 2
N = # of molecules
NA = Avogadro’s number
kB = Boltzmann’s constant
R = NAkB = gas constant
We can also think of this in terms of concentrations
State 1
æn
ö
æ V2 ö
æ M1 ö
M2 ÷
ç
DS = R ln ç ÷ = R ln ç
= R ln ç
÷
÷
n
ç
÷
è V1 ø
è M2 ø
M
è
1ø
State 2
Convert to concentrations,
in molarity units: M = n/V,
so V = n/M
V = volume
N = # of molecules
n = # of moles
R = NAkB = gas constant
There are other flavors of entropy
• We focused here on “translational” entropy,
which is equivalent to entropy of dilution, or
entropy of mixing
• There is also “rotational” entropy and
“vibrational entropy”, very important in drug
binding as well
Free Energy
• At constant pressure, the correct criterion is
DG < 0
where G = H – TS
• If we additionally assume constant T, you get the famous
equation
DG = DH – TDS
• Key point: note the sign on the entropy term
• When DG = 0, this is called “equilibrium”
• Molar free energy for a substance also referred to as
“chemical potential”:
An intuitive feel for free energy?
• Enthalpy = energy
• Entropy = disorder
• Free energy = ??
• Best understood as a statement of the 2nd law
of thermodynamics
Free energy of dilution/mixing
State 1
Recall that …
State 2
DH = 0
æ V2 ö
æ M1 ö
DS = R ln ç ÷ = R ln ç
÷
è V1 ø
è M2 ø
æ V2 ö
æ M1 ö
æ M2 ö
DG = -RT ln ç ÷ = -RT ln ç
÷ = RT ln ç
÷
è V1 ø
è M2 ø
è M1 ø
Derivation of DG=-RTln K for a
simple AB reaction
At equilibrium (constant T, P):
A  B
Assume ideal solution,
MA
mA (aq) = m (aq) + RT ln
M°
0
A
Mº = 1 molar (to
make argument of the
logarithm unitless)
MB
mB (aq) = m (aq) + RT ln
M°
0
B
Equating these two:
MA
MB
0
m (aq) + RT ln
= mB (aq) + RT ln
M°
M°
0
A
MA
MB
0
m (aq) + RT ln
= mB (aq) + RT ln
M°
M°
MA
MB
0
0
mB (aq) - mA (aq) = RT ln
- RT ln
M°
M°
0
A
Constant that depends
only on nature of
reactants and products
DEFINE KM= (MB/MA)eq
Extension to more general equilibria:
aA + bB cC + dD
At equilibrium (constant T, P; closed system):
c C  d D  a A  b B
æM M ö
KM = ç
÷
è M M øeq
c
C
a
A
d
D
b
B
Alternate Notation
KC
 [C ]c [ D ]d

 [ A ] a [ B ]b




 eq
Other concentration units
Derivations are extremely similar.
For molality (mol/kg solvent), the chemical potential can be
written as
mi
mi (aq) = m (aq) + RT ln
m°
o
i
where the standard state is now defined by mº = 1 molal. The
equilibrium constant winds up looking like
D G   -RT ln K m
0
Km 
mB
mA
Expression for gasses is also very similar, with partial pressures
replacing concentrations.