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Thermodymanics
• Lecture 3 8/31/2004
Units
Energy
Ultimate source of energy is the sun
E = hn 57 Kcal/mol of photons green light
or
238 KJ/mol
1 cal = 4.184 joules
0.239 cal = 1 J
ATP energy carrier, for hydrolysis to ADP + Pi
= 7.3 kcal/mole or 30.5 KJ/mol
While vibrational energy infrared) = 0.6 kcal/mol or 2.5 KJ/mol
C - C bond = 83 Kcal/mol or 346 KJ/mol
the framework of a carbon skeleton is thermally stable
but
non-covalent bonds are only a few kcal/mol or
10-20 KJ/mole
Thermal
Noncovalent
1
ATP Green
bond
light
10
100
C-C glucose
bond
1000
Kcal/mol
KJ/mol
1
10
100
1000
Biomolecule shapes and interactions are mediated by 4 types
of non-covalent bonds.
These bonds are responsible for the overall shape and interaction among
biomolecules and can be modified by thermal energy.
Thermodynamics:
Allows a prediction as to the spontaneous nature of
a chemical reaction:
Will this reaction proceed in a forward direction as the
reaction is written:
A
+
B
C
Will A react with B to form C or not?
Is this reaction going from higher to lower energy?
ENERGY not necessarily just heat!!!
Definitions:
System:
a defined part of the universe
a chemical reaction
a bacteria
a reaction vessel
a metabolic pathway
Surroundings: the rest of the universe
Open system: allows exchange of energy and matter
Closed system: no exchange of matter or energy.
i.e. A perfect insulated box.
Direction of heat flow by definition is most important.
q = heat absorbed by the system from surroundings
If q is positive reaction is endothermic
system absorbs heat from surroundings
If q is negative exothermic
system gives off heat.
A negative W is work done by the system on the
surroundings i.e expansion of a gas
First Law of Thermodynamics: Energy is Conserved
Energy is neither created or destroyed.
In a chemical reaction, all the energy must be accounted for.
Equivalence of work w and energy (heat) q
Work (w) is defined as w = F x D (organized motion)
Heat (q) is a reflection of random molecular motions (heat)
U = Ufinal - Uinitial = q - w
Exothermic system releases heat = -q
Endothermic system gains heat = +q
U = a state function dependent on the current properties only.
U is path independent while q and w are not state functions
because they can be converted from one form of energy to the
other. (excluding other forms of energy, e.g. electrical, light
and nuclear energy, from this discussion.)
Enthalpy (H)
At constant pressure
w = PV + w1
w1 = work from all means other than pressure-volume work.
PV is also a state function.
By removing* this type of energy from U, we get enthalpy or
‘to warm in’
*remember the signs and direction
H = U + PV
H = U + PV = qp -w + PV = qp - w1
Enthalpy (H)
When considering only pressure/volume work
H = qp - PV + PV = qp
H = qp when other work is 0
qp is heat transferred at constant pressure.
In biological systems the differences between
U and H are negligible (e.g. volume changes)
The change in enthalpy in any hypothetical reaction
pathway can be determined from the enthalpy
change in any other reaction pathway between the
same products and reactants.
This is a calorie (joule) “bean counting”
Hot
Cold
Which way does heat travel? This directionality, is
not mentioned in First Law
Entropy (S) is the arrow of time
•The entropy of the universe is always increasing.
•A change in Entropy does not change the total energy.
•What does this mean in a closed system?
•What does this mean in an open system?
Gas on its own will expand to the available volume.
• Disorder increases
• N identical molecules in a
bulb, open the stop cock you
get 2N equally probable
ways that N molecules can
be distributed in the bulb.
As the gas molecules are indistinguishable only N+1
different states exist in the bulb or 0, 1, 2, 3, 4,…., (N-1)
molecules in left bulb
WL = number of probable ways of placing L of the N
molecules in the left bulb is
N!
WL 
L! (N - 1)!
The probability of occurrence of such a state
(Is its fraction of the total number of states)
WL
 N
2
Direction occurs because the aggregate probability of all
other states is only the most probable state will occur.
WN  10
NLn2
2
for N = 1023
W5x10 22  10
7 x10 22
Possible arrangements
Entropy (S)
measure the degree of randomness
S  k b LnW
Each molecule has an inherent amount of energy which
drives it to the most probable state or maximum disorder.
kb or Boltzman’s constant equates the arrangement
probability to calories (joules) per mole.
Entropy is a state function and as such its value depends
only on parameters that describe a state of matter.
The process of diffusion of a gas from the left bulb initially W2 = 1 and
S = 0 to Right (N/2) + Left (N/2) at equilibrium gives a :
S that is (+) with a constant energy process such that
U = 0 while S>0
This means if no energy flows into the bulbs from the outside
expansion will cool the gas! Conservation of Energy says that
the increase in Entropy is the same as the decrease in thermal
(kinetic) energy of the molecules!!
It is difficult or (impossible) to count the number of
arrangements or the most probable state!
So how do we measure entropy?
final
dq
S  
T
inital
It takes 80 kcal/mol of heat to change ice at zero °C to water at zero °C
80,000 = 293 ev’s or entropy units
273
A Reversible process means at equilibrium during the change. This is
impossible but makes the calculations easier but for irreversible process
q
S 
T
At constant pressure we have changes in qp (Enthalpy) and
changes in order - disorder (Entropy)
A spontaneous process gives up energy and becomes more
disordered.
G = H - TS
Describes the total usable energy of a system, thus a change from one
state to another produces:
G = H - TS = qp - TS
If G is negative, the process is spontaneous
S
H
+
-
All favorable
at all temperature
spontaneous
-
-
Enthalpy favored.
Spontaneous at
temperature below
T = H
S
+
+
Entropy driven,
enthalpy opposed.
Spontaneous at
Temperatures above
T = H
S
Non-spontaneous
-
+
STP
Standard Temperature and Pressure and at 1M concentration.
We calculate G’s under these conditions.
aA + bB
cC + dD
We can calculate a G for each component
__
(1)
(2)
__
__
__
G  c G c  d G d - a G a - b G b
o
a
G a - G  RT ln[A]
combining (1) and (2)
(3)
c
d
[C]
[D]
o
G  G  RTln
[A]a [B] b
Now if we are at equilibrium or G = 0
Then
c
d
[C] [D]
G  0  G  RTln
a
b
[A] [B]
o
c
d
[C]
[D]
o
G   RTln
[A]a [B] b
OR
G  RTlnKeq
o
So what does Go really mean?
c
d
[C]
[D]
G  G o  RTln
[A]a [B] b
G
Go
Keq
If Keq = 1 then G = 0
Go equates to how far Keq varies from 1!!
Keq can vary from 106 to 10-6 or more!!!
Go is a method to calculate two reactions whose Keq’s are different
However
The initial products and reactants maybe far from their equilibrium
concentrations
c
so
d
[C] [D]
a
b
[A] [B]
Must be used
Keq 
c
eq
a
eq
[C] [D]
[A] [B]
d
eq
b
eq
e
 G o
RT
Le Chatelier’s principle
Any deviation from equilibrium stimulates a process which restores
equilibrium. All closed systems must therefore reach equilibrium
What does this mean?
Keq varies with 1/T
If Keq varied with temperature things would be very unstable.
Exothermic reactions would heat up causing an increase in Keq
generating more heat etc….
Fortunately, this does not happen in nature.
- H  1  S

 
R T R
o
lnK eq
o
R = gas constant for a 1M solution
Plot lnKeq vs. 1/T ( remember T is in absolute degrees Kelvin)
Van’t Hoff plot
- H o
= Slope
R
lnKeq
S o
R
1
T
= Intercept
The Variation of Keq with Go at 25 oC
Keq
G
o
(kJ·mole-1
106
104
102
101
100
10-1
10-2
10-4
10-6
-34.3
-22.8
-11.4
-5.7
0.0
5.7
11.4
22.8
34.3
G   G (products) -  G (reactants )
o
o
f
o
f
The f = for formation.
By convention, the free energy of the elements is taken
as zero at 25 oC
The free energies of any compound can be measured as
a sum of components from the free energies of
formation.
Standard State for Biochemistry
Unit Activity
25 oC
pH = 7.0 (not 0, as used in chemistry)
[H2O] is taken as 1, however, if water is in the Keq equation
then [H2O] = 55.5
The prime indicates Biochemical standard state
K eq
G
o
G
G  G
o
Most times
o
However, species with either H2O or H+ requires consideration.
For A + B
n
[C][D][H
O]
2
G o  - RTln
[A][B]
C + D + nH2O
[C][D]
G  - RTln Keq  - RTln
[A][B]
o
This is because water is at unity. Water is 55.5 M
and for 1 mol of H2O formed:
G  G  nRT ln[H2O]
o
o
o
o
-1

G  G  9.96kJ  mol
For:
A+B
C + HD
K
H+ + D Where a proton is in the equation
Go  G o

[H ][D ]
K
[HD]

1 - [H ] 
- RT ln
 RT ln[ H

o
K

This is only valid for
-

]o
[H ]o  10 M
-7
Coupled Reactions
A+B
D+E
If
C + D G1
F +G G2
(1)
(2)
G1  0 reaction 1 will not occur as written.
However, if
G 2 is sufficiently exergonic so G1  G 2  0
Then the combined reactions will be favorable through
the common intermediate D
A+B+E
C+F+G
G3
G 3  G1  G 2  0
As long as the overall pathway is exergonic,
it will operate in a forward manner.
Thus, the free energy of ATP hydrolysis, a
highly exergonic reaction, is harnessed to
drive many otherwise endergonic biological
processes to completion!!