Spontaneous Processes
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Transcript of Spontaneous Processes
Spontaneous Processes
The Second Law:
S 0The 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!
The entropy of a system can spontaneously decrease, as long as the entropy of the surroundings increases by at least as much.
Ssys + Ssurr 0
Do we have to keep calculating Ssurr ?
Not necessarily!
Spontaneous Processes
Let’s stay at constant T and P:
ΔSsurr=−qP
T=−
ΔHT
ΔStot=ΔSsys−ΔHT
ΔStot≥0⇔ ΔSsys−ΔHT
≥0
Now everything is in terms of the system. The criterion for spontaneity given by the Second Law becomes:
Gibbs Free Energy
The Gibbs Free Energy is a new state function, defined as:
G ≡H −TS
ΔG =ΔH −TΔS
ΔStot≥0⇔ ΔG ≤0
Josiah Willard Gibbs(1839-1903)
At constant temperature and pressure, G is
From the previous slide, we end up with
Gibbs Free Energy
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 extremely useful for chemistry and biochemistry, since so much takes place at constant temperature and pressure.
At constant T and P, consideration of G will answer the question
“Will a given reaction be spontaneous?”
• 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.
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
G < 0
Thus, ifH < 0 the process is exothermic (downhill) S > 0 the process is increases disorder
So H dominates spontaneity at low temperaturesS dominates spontaneity at high temperatures
Calculation of G
In many cases, we can build on calculations we have already done in order to get G.
ΔG =ΔH −TΔS
In other cases, like chemical reactions, standard values have been found. We need only to add them up properly.
Example: ice melting at 100° C
H = 6.75 kJ mol–1
S = 45.5 J K–1 mol–1
soG = 6750 – (373)(45.5) = –10.2 kJ mol–1.
Calculation of G
H2O(g) H2(g) + 1/2 O2(g)
Is So298 greater than, less than, or equal to zero?
So298= -(188.82) + 130.684 + 1/2 (205.14) J/(K mol)
= 44.4 J / (K mol)Spontaneous?
Ho298=-(-241.82) + 0 + 1/2 (0) kJ/mol = 241.82
Go298= Ho
298-T So298 = 241.82 kJ/mol - (298 K)*0.0444 kJ/(K mol)
= 228.56 kJ/mol
The process is non-spontaneous!
Calculation of G
Example: CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
G°f –50.7 0 –394.36 –228.6
G°r = –800 kJ mol–1
Calculation of G
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, constant TG(p2)−G(p1) = V(p)dpp1
p2
∫
G(T2)−G(T1) = S(T)dTT1
T2
∫and constant P
Calculation of GA puzzle:
ΔG =ΔE +pΔV −TΔS
ΔS=q T ΔE =w+q
ΔG =w+pΔV
w=−pΔV
Hence, ΔG =0
so
but
Assume everything is reversible.
at constant T and p,
???
According to this, we can’t ever have G < 0 if everything is reversible at constant T and p. But what about all those chemical reactions? Surely they can be run reversibly! But ΔG ≠0
Where is the mistake?
Calculation of G
A puzzle:
ΔG =ΔE +pΔV −TΔS
ΔS=q T ΔE =w+q
ΔG =w+pΔV
w=−pΔV
Hence, ΔG =0
so
but
Assume everything is reversible.
at constant T and p,
???
Not all work is PV work! For example, electrochemical, mechanical, etc.
“Free” means free to do non-PV work!
Temperature Dependence
Simplest approximation:
G = H - T S
assume both H and S do not change for moderate changes of T
G(T) = G(298 K) - (T - 298 K)S
If we are assuming H and S independent of T, why not just ignoreG dependence?
Explicit dependence on T in G, whereas only implicit for H and S
Temperature DependenceGeneral expression (reversible path, only PV work):
dG = dH - TdS - SdT = dE +PdV + VdP - TdS - SdT
but dE = –PdV +TdS since dw = -PdV and dqrev = TdSand 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
dGT1
T2
∫ =− S(T)dTT1
T2
∫
G(T2)−G(T1) =− S(T)dTT1
T2
∫
ΔG T1( )=ΔG(298)− ΔS T( )dT298
T
∫
Temperature Dependence
Gibbs-Helmholtz equation
ΔG T2( )
T2
−ΔG T1( )
T1
=−ΔH T( )
T2 dTT1
T2
∫
If H changes little with temperature
ΔG T2( )
T2
−ΔG T1( )
T1
=ΔH1T2
−1T1
⎛
⎝ ⎜
⎞
⎠ ⎟
See text p. 93 for mathematical derivation.
Pressure Dependence
For the pressure dependence we hold T constant:
dG = VdP-SdT=VdPThus,
For solids and liquids, V does not vary with temperature so,
and for an ideal gas:
dG=G P1( )−G P2( ) =P1
P2
∫ VdPP1
P2
∫
G P2( )−G P1( )=V P2 −P1( )
G P2( )−G P1( )=nRTPP1
P2
∫ dP=nRTlnP2
P1
⎛
⎝ ⎜
⎞
⎠ ⎟
Pressure Dependence ExampleCan we force graphite to diamond by increasing the pressure? We will use:
and the fact that molar volumes of graphite and diamond are known:
V graphite=5.33cm3 /mol
V diamond=3.42cm3 /mol
ΔG(P) =ΔG(1atm)+ΔV*(P −1)
ΔG(P) =2.84−1.935∗10−4*(P −1)
Where we have used a conversion factor to convert from cm3 atm to kJ.
G P2( )−G P1( )=V P2 −P1( )
Now, we want the pressure that makes G = 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) H2O (g)100 °C
What is the free energy change?
Well, this is at constant T and P. Its reversible. So,
G°vap= 0 = H°vap-T S°vap
Note that this implies:
oo
vapvap S
T
H=
G of Mixing
Consider the isobaric, isothermal mixing of two gases:
Gas Aat
1 Atm
Gas Bat
1 Atm
Gas A+Bat
1 Atm (total)
Is this reaction spontaneous?
ΔGmixing=ΔGexpansion of A+ΔGexpansion of B
ΔGmixing=nARTlnP2,A
1atm⎛ ⎝ ⎜ ⎞
⎠ ⎟ +nBRTln
P2,B
1atm⎛ ⎝ ⎜ ⎞
⎠ ⎟
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
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.
heat
Protein Unfolding
In differential scanning calorimetry you have two samples: Your material of interestControl
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+
SolventSolvent
HeatT1
T2
T1-T2
Protein Unfolding
Data for glyceraldehyde-3-phosphate dehydrogenase.
pH8.0
pH6.0
Bacillus stearothermophilus
E. coliRabbit
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:
10 25 60 100 °C
G° kJ/mol 67.4 60.7 27.8 -41.4H° kJ/mol 137 236 469 732S° J/ K mol 297 586 1318 2067TS° kJ/mol 69.9 175 439 771
Where is the denaturation temperature?
What then is special about the temperature at which the denaturation is spontaneous?