### Review of Thermodynamics - University of Alabama at Birmingham

```Thermodynamics of Biological Systems
Champion Deivanayagam
Center for Biophysical Sciences and Engineering
University of Alabama at Birmingham.
Outline:
 Laws of thermodynamics
 Enthalpy
 Entropy
 Gibbs free energy
 some examples
What you need to know for your exam:
1.
2.
3.
4.
5.
Definition of a thermodynamic system
Three laws of Thermodynamics
Definitions of Enthalpy, Entropy
Gibbs Free energy
ATP’s ionization states and its potential
Energy:
Energy is the capacity to do work
1. Kinetic Energy
2. Potential Energy
• Kinetic energy is the form of energy expended by objects in motion
• A resting object still possess energy in the form of potential energy
Energy can be converted from one form into another
Thermodynamics:
A study of energy changes in systems:
System:
1. Isolated
2. Closed
3. Open
First law of thermodynamics:
Energy is neither created nor destroyed; the energy of the universe is a constant
The total internal energy of an isolated system in conserved.
E = E2 – E1 = q + w
q – heat absorbed by the system from surroundings
w – work done on the system by the surroundings
Mechanical work is defined as
movement through some distance
caused by the application of force
Internal energy is independent of path
and represents the present state of the
system and is referred to as a State function
Mechanical Work:
At constant pressure work can be defined as
w = -PV where V = V2 – V1
Work may occur in multiple forms:
1. Mechanical
2 . Electrical
3. magnetic
4. Chemical
The calorie (cal, kcal) are traditional units
Joule’s is the recommended SI unit.
Table of important thermodynamic units and constants
Enthalpy:
H = E + PV
In a constant pressure system (as in most biological systems)
one can then define it as
H = E + PV
When you expand on this equation:
H = q (simply put the heat energy of the system)
Enthalpy changes can be measured using a calorimeter:
For a system at equilibrium for any process where AB
the standard enthalpy can be determined from the temperature dependence
using:
R- is the gas constant
R= 8.314 J/mol . K
Notice the ° sign: These are used to denote standard state:
For solutes in a solution, the standard state is normally unit activity (simplified to 1M concentration)
Protein denaturation:
Study of temperature induced reversible denaturation of chymotrypsinogen
At pH 3.0
T(K)
Keq
324.4
326.6
0.041
0.12
327.5
0.27
329.0
0.68
330.7
1.9
332.0
333.8
5.0
21.0
van’t Hoff Plot
Native state (N)  Denatured state (D)
Keq = [D] / [N]
H° at any given temperature is the negative
of the slope of the plot:
H° = -[14.42]/[-0.027] x 10-3 = +533 kJ/mol
Positive values for H° would be expected to break bonds and expose hydrophobic groups
During the unfolding process and raise the energy of the protein in solution.
Second law of thermodynamics:
Every energy transfer increases the entropy (disorder) of the universe
System tends to proceed from ordered states to disordered states
Some definitions:
Reversible: a process where transfer of energy happens in both directions
Irreversible: a process where transfer of energy flows in one direction
Equilibrium: A  B
(all naturally occurring process tend to equilibrium)
Entropy:
S = k ln W
S = k ln Wfinal – k ln Winitial
k- is the Boltzmann’s constant
W – number of microstates
Relationship between entropy and temperature
dSreversible = dq/T
Entropy changes measure the dispersal of energy in a process.
Third law of thermodynamics:
Entropy of any crystalline substance must approach zero as temperature approaches 0° K
The absolute entropy can be calculated from this equation:
Cp is the heat capacity, defined as the amount of heat 1 mole of it can store as the
temperature of that substance is raised by 1 degree.
For biological systems entropy changes are more useful than absolute entropies
Gibb’s free energy ‘G’
Determines the direction of any reaction from the equation:
G = H – TS
For a constant pressure and temperature system (as most biological systems) then the
Equation becomes easier to handle
G = H - TS
The enthalpy and entropy are now defined in one equation.
G is negative for exergonic reactions (release energy in the form of work)
is positive for endergonic reactions (absorbing energy in the form of work)
Consider a reaction: A + B  C + D
 [ C ][ D ] 
 G   G  RT ln 

 [ A ][ B ] 
At Equilibrium: G° = RT ln Keq
and Keq = 10 - G°/2.3RT
Example of chymotrypsinogen denaturation
From the van’t Hoff plot we calculated H° = +533 kJ/mol
At pH 3.0
T(K) 324.4 326.6 327.5 329.0 330.7 332.0 333.8
Keq
0.041 0.13 0.27
0.68 1.9
5.0 21.0
The equilibrium constant at 54.5 °C (327.5K) is 0.27
Then G° = (-8.314 J/mol·K) (327.5K) ln (0.27) = - 3.56 kJ/mol
Similarly calculating S° = - (G - H°) / T = 1620 J/mol·K
For a process to occur spontaneously
the system must either give up energy (decrease H)
or give up order (increase in S)
or both
In general for the process to be spontaneous:
G must be negative
The more negative the G value, the greater the amount of work the process can perform
Exergonic reactions:
G is negative and the reaction is spontaneous
Example: Cellular respiration
C6H12O6 + 6O2
6C02 + 6 H20
G = -686 kcal/mol (-2870 kJ/mol)
(For each molecule of glucose broken 686 kcal energy is made available for work)
Endergonic reactions:
G is positive and requires large input of energy:
Example: Photosynthesis where the energy is derived from the sun.
Table: Variation of Reaction Spontaneity (Sign of G) with the signs of H and S.
A cell does three kinds of work:
1. Mechanical work: beating of cilia, muscle contraction etc.
2. Transport work: Moving substances across membranes
3. Chemical work: Enabling non-spontaneous reactions to occur spontaneously
e.g. Protein synthesis.
The molecule that powers most kinds of work in the cell is ATP
Energy is released when one or more phosphate
groups are hydrolyzed
ATP + H2O → ADP + Pi (G° = -35.7 kJ/mol)
G° = RT ln Keq
The activation energies for
phosphoryl group-transfer
reactions (200 to 400 kJ/mol) are
substantially larger than the free
energy of hydrolysis of ATP (30.5 kJ/mol).
ΔGo` = -30.5 kJ/mole
= -7.3 kcal/mole
ΔG` = -52 kJ/mole
= -12.4 kcal/mole
Cellular conditions:
Ionization States of ATP
• ATP has five dissociable protons
• pKa values range from 0-1 to 6.95
• Free energy of hydrolysis of ATP is relatively
constant from pH 1 to 6, but rises steeply at
high pH
• Since most biological reactions occur near pH
7, this variation is usually of little
consequence
The pH dependence of the free
energy of hydrolysis of ATP.
Because pH varies only slightly in
biological environments, the effect
on G is usually small.
The free energy of
hydrolysis of ATP as a
function of total Mg2+ ion
concentration at 38°C and
pH 7.0.
and Veech, R. L., 1973. The
equilibrium constants of the
triphosphate-citrate lyase
reactions. Journal of
Biological Chemistry
248:6966–6972.)
The free energy of hydrolysis of
ATP as a function of
concentration at 38°C, pH 7.0.
The plot follows the relationship
described in Equation (3.36),
with the concentrations [C] of
ATP, ADP, and Pi assumed to be
equal.
What is the Daily Human Requirement for ATP?
• The average adult human consumes approximately 11,700 kJ of food energy per
day
• Assuming thermodynamic efficiency of 50%, about 5860 kJ of this energy ends
up in form of ATP
• Assuming 50 kJ of energy required to synthesize one mole of ATP, the body must
cycle through 5860/50 or 117 moles of ATP per day
• This is equivalent to 65 kg of ATP per day
• Thus each ATP molecule must be recycled nearly 1300 times per day
Isothermal titration calorimetry
Reference and experimental cell
Heat energy required to maintain both of them
At the same level is measured and integrated.
This allows for the measurement of Kd, G and S
ITC studies on A123 + VP3 regions
1 39
S
201
448 578
A1
A2
A3
A1
A2
A3
V-region
828 840
P1
P2
960
P3
V-region
P1
P2
P3
1486 1561
WMC
Time (min)
0
Large release of Heat energy indicates
Ordering of structures
Hydrogen bonding
The Kd and the energy released indicate
that the interaction between these two
regions is strong.
20
30
40
50
60
70
µcal/sec
-8
-10
Titration of 12 M VP3
by 10 L x 25 injections
of 96 M A1-3
-12
o
at 25 C
kcal/mole of injectant
1:1 Stoichiometric ratio
(N=0.946 ± 0.006) and
Kd ~ 40 nm
10
-14
0
-50
-100
-150
N
Ka
0.946 + 0.006
7
-1
(2.6 + 0.4) x 10 M
G
H
T S
-10.1 kcal/mol
(-212 + 2) kcal/mol
-202 kcal/mol
-200
0.0
0.5
1.0
Molar Ratio
1.5
What happens in a cell if G = 0 ?
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