Transcript No Slide Title

Spontaneous Processes
The Second Law:
DS  0
The entropy of a closed system can only increase.
If a process will decrease entropy in a closed system, then it does not occur
spontaneously. Its opposite will occur spontaneously.
But we very rarely work with closed systems...
Spontaneous Processes
…except for the universe as a whole!
DSsys + DSsurr  0
The entropy of a system can spontaneously decrease, as long as the entropy
of the surroundings increases by at least as much.
Do we have to keep calculating DSsurr ?
Not necessarily!
Spontaneous Processes
Let’s stay at constant T and P:
DSsurr  
qP
DH

T
T
DStot  DSsys 
DH
T
Now everything is in terms of the system. The criterion for
spontaneity given by the Second Law becomes:
DStot  0  DSsys 
DH
0
T
Gibbs Free Energy
The Gibbs Free Energy is a new state
function, defined as:
G  H  TS
At constant temperature and pressure, DG is
DG  DH  TDS
From the previous slide, we end up with
DStot  0  DG  0
Josiah Willard Gibbs
(1839-1903)
Gibbs Free Energy
G is extremely useful for chemistry and biochemistry, since so much
takes place at constant temperature and pressure.
The condition of constant T and P is very important when using G.
Otherwise, the entropy change of the surroundings might be different
leading to a different result.
• G is still defined and can be calculated for any change of state, including
changing P and T.
• We can also define another state variable, Helmholtz Free Energy (A = E - TS),
which has similar characteristics as G, but relates more directly to constant T and V
processes.
At constant T and P, consideration of DG will answer the question
“Will a given reaction be spontaneous?”
Gibbs Free Energy Summary
The Gibbs Free Energy is a direct measure of spontaneity:
G = H - TS
It sums up, in a way, the competition between energy considerations and
“configurational” barriers.
We have also learned that a process is spontaneous if
DG < 0
Thus, if
DH < 0 the process is exothermic (downhill)
DS > 0 the process is increases disorder
So
H dominates spontaneity at low temperatures
S dominates spontaneity at high temperatures
Calculation of DG
In many cases, we can build on calculations we have already done
in order to get DG.
DG  DH  TDS
Example: ice melting at 100° C
DH = 6.75 kJ mol–1
DS = 45.5 J K–1 mol–1
so
DG = 6750 – (373)(45.5)
= –10.2 kJ mol–1.
In other cases, like chemical reactions, standard values have
been found. We need only to add them up properly.
Calculation of DG
H2O(g) 
H2(g)
+
1/2 O2(g)
Is DSo298 greater than, less than, or equal to zero?
DSo298= -(188.82)
+ 130.684 +
= 44.4 J / (K mol)
Spontaneous?
1/2 (205.14) J/(K mol)
DHo298=-(-241.82)
1/2 (0) kJ/mol = 241.82
+ 0
+
DGo298= DHo298-T DSo298 = 241.82 kJ/mol - (298 K)*0.0444 kJ/(K mol)
= 228.56 kJ/mol
The process is non-spontaneous!
Calculation of DG
Example:
DG°f
CH4(g) + 2O2(g)
–50.7
0
CO2(g) + 2H2O(g)
–394.36
DG°r = –800 kJ mol–1
–228.6
Calculation of DG
We can perform calculus on G directly:
dG = dH – TdS – SdT
= dE + pdV + Vdp – TdS – SdT
but dE = –pdV +TdS
(dw = -pdV and dqrev = TdS)
In other words,
G( p2 )  G( p1 ) 
p2
 V( p)dp
constant T
p1
and
T2
G(T2 )  G(T1 )   S(T)dT
T1
constant P
Calculation of DG
A puzzle:
at constant T and p,
DG  DE  pDV  TDS
Assume everything is reversible.
DE  w q
DS q T
so DG  w pDV
but
w   pDV
Hence, DG  0
???
According to this, we can’t ever have DG < 0 if everything is
reversible at constant T and p. But what about all those chemical
reactions? Surely they can be run reversibly! But DG  0
Where is the mistake?
Calculation of DG
A puzzle:
at constant T and p,
DG  DE  pDV  TDS
Assume everything is reversible.
DE  w q
DS q T
so DG  w pDV
but
w   pDV
Hence, DG  0
???
Not all work is PV work! For example, electrochemical, mechanical, etc.
“Free” means free to do non-PV work!
Temperature Dependence
Simplest approximation:
DG = DH - T DS
assume both DH and DS do not change for moderate changes of T
DG(T) = DG(298 K) - (T - 298 K)DS
If we are assuming DH and DS independent of T, why not just ignore
DG dependence?
Explicit dependence on T in DG, whereas only implicit for DH and DS
Temperature Dependence
General expression (reversible path, only PV work):
dG = dH - TdS - SdT
= dE +PdV + VdP - TdS - SdT
but
dE = –PdV +TdS
since
dw = -PdV and dqrev = TdS
and
dE = dw + dq (we are on a specified path so this is ok)
so
…
dG = (-PdV + PdV) + (TdS - TdS) + VdP - SdT
= VdP - SdT
At constant pressure:
dG = -SdT
T2
T2
T1
T1
 dG    S(T )dT
T2
G(T2 )  G(T1 )    S(T)dT
T1
DG T1   DG(298) 
T
 DSTdT
298
Temperature Dependence
Gibbs-Helmholtz equation
2
DG T2  DGT1 
DH T 


2 dT
T
T2
T1
T
T
1
If DH changes little with temperature
 1 1 
DG T2  DGT1 

 DH   
T2 T1 
T2
T1
See text p. 93 for mathematical derivation.
Pressure Dependence
For the pressure dependence we hold T constant:
dG = VdP-SdT=VdP
Thus,
P2
P2
P1
P1
 dG  GP1   GP2    VdP
For solids and liquids, V does not vary with temperature so,
and for an ideal gas:
GP2   GP1   V P2  P1 
GP2   GP1  
P2

P1
P 
nRT
dP  nRT ln  2 
P
P1 
Pressure Dependence Example
Can we force graphite to diamond by increasing the pressure? We will use:
GP2   GP1   V P2  P1 
and the fact that molar volumes of graphite and diamond are known:
Vgraphite  5.33cm 3 / mol
Vdiamond  3.42cm 3 / mol
DG(P)  DG(1atm)  DV * (P  1)
DG(P)  2.84  1.935 10 4 * (P 1)
Where we have used a conversion
factor to convert from cm3 atm to kJ.
Now, we want the pressure that makes DG = 0:
Why?
0 = 2.84 - 1.935 10-4 (P-1) kJ/mol
P = 15,000 atm
Experimentally, the required pressure is more! Why?
Example: Reversible Process
H2O (l)
100 °C
H2O (g)
What is the free energy change?
Well, this is at constant T and P. Its reversible. So,
DG°vap= 0 = DH°vap-T DS°vap
Note that this implies:

DH vap
T

 DS vap
DG of Mixing
Consider the isobaric, isothermal mixing of two gases:
Gas A
at
1 Atm
Gas B
at
1 Atm
Gas A+B
at
1 Atm (total)
DGmixing  DGexpansion of A  DGexpansion of B
DGmixing
P2, A 
P2, B 






 n A RT ln
 n B RT ln
1atm
1atm
Is this reaction spontaneous?
Well, both P2,A and P2,A are less than 1 atm…so yes, this process is spontaneous.
Protein Unfolding
Proteins have a native state. (Really, they tend to have a tight cluster of native
states.)
Denaturation occurs when heat or denaturants such as guanidine, urea or detergent
are added to solution. Also, the pH can affect folding.
When performing a denaturation process non-covalent interactions are broken.
Ionic, van der-Waals, dipolar, hydrogen bonding, etc.
Solvent is reorganized.
Protein Unfolding
heat
Let’s consider denaturation with heat. We can determine a great deal about the
nature of the protein from such a consideration.
The experimental technique we use for measuring thermodynamic changes here is
the differential scanning calorimeter.
Basic experiment: Add heat to sample, measure its temperature change.
Protein Unfolding
T1
T2
Heat
Protein
+
Solvent
Solvent
T1-T2
In differential scanning calorimetry you have two samples:
Your material of interest
Control
You put in an amount of heat to raise the temperature of the control at a constant
rate, then measure the rate of change in temperature of the other sample as a
function of the input heat.
This is a measure of the heat capacity!
Protein Unfolding
pH8.0
Bacillus stearothermophilus
Rabbit
E. coli
pH6.0
Data for glyceraldehyde-3-phosphate dehydrogenase.
Is the protein more stable at pH 8 or 6? Why is B. stear. more stable?
Protein Unfolding
We are given the following data for the denaturation of lysozyme:
DG° kJ/mol
DH° kJ/mol
DS° J/ K mol
TDS° kJ/mol
10
25
60
100
67.4
137
297
69.9
60.7
236
586
175
27.8
469
1318
439
-41.4
732
2067
771
°C
Where is the denaturation temperature?
What then is special about the temperature at which the denaturation is spontaneous?