Energies: (Free) Helmholtz and Gibbs(Chapter 22.1,22.2)
For an isolated system, dS>0 spontaneous process occurs. But how is
an open beaker isolated? It isnt, so we must consider S of the
surroundings as well. Then, Suniverse= Ssurr+ Ssys>0 will determine
spontaneity.
This is a lot of work to determine if a process is spontaneous.
Is there an easier way?
Is there an easier way?
Consider a real (not ideal!) system under isothermal conditions and at
dU = q + w
constant V:
constant V:
Since: w = P dV = 0 ( constant V )
ext
and
Then:
q TdS
The Helmholtz energy (based solely on the system) can tell us
whether a process will occur spontaneously!
The Helmholtz energy will continue to decrease for any spontaneous
process until it reaches equilibrium, where A will be a minimum.
because
( because dS q / T )
dU = q TdS so dU TdS at constant V
dU TdS 0 at constant V
at constant
Fig 22.1
Lecture 17
S
+ (more disorder)
A
-
Reversible ( equilibrium ) : A = U T S = 0
1
Spontaneous?
Always
+
+
?
At high T, S controls
-
-
?
At low T, U controls
+
-
+
Never
Now for a reversible isothermal process:
for reversible isothermal process:
A = U T S = U qrev = 0
for reversible process
Lecture A
Spontaneous irreversible: 17 = U T S < 0
2
If A is positive, wrev represents the min amount of work required to
cause the process to reversibly occur in the system. Actual work
required for the process to occur is greater!
ogous
sys
An analogous thermodynamic function exists for systems under
constant pressure. Recall:
dH = q p TdS
( = if reversible, < if spontaneous )
dH TdS 0 or d ( H TS ) 0 for isothermal, isobaric processes
G H TS = Gibbs (free) energy
(free) energy
= U + PV - TS
However, if V is not constant, then A = wrev
What does this mean?
dG<0 = Spontaneous process. Will continue to decrease until
system reaches equilibrium when dG=0.
If A is negative, A represents the max work that can be done by
system on surroundings.
G = H T S 0 ( at constant T and P )
A is negative for a spontaneous process, so work done is less,
since
since wrev is the maximum work that can be done.
the maximum work that can be done
Lecture 17
A = U - T S 0
Isothermal:
d (U TS ) 0 at constant T
U
- (energy decreases)
Let's define A U TS = Helmholtz (free) energy
where dA = 0 at equilibrium
dA 0
and dA < 0 for a spontaneous process
3
Lecture 17
4
H
- (enthalpy decreases)
+
S
+ (more disorder)
G
-
+
?
Consider 2O H ( l ) H 2O ( g ) vap G = G ( H 2O ( g ) ) G ( H 2O ( l ) )
Spontaneous?
Always
0
At high T, S controls
-
-
?
At low T, H controls
+
-
+
Never
0
0
0
at 373.15K, vap H =40.65kJ/mol and vap S =108.9J/molK
vap G 0 = 40.65kJ / mol 373.15 ( 0.1089 ) kJ / mol = 40.65 40.64 0
So vapor and liquid are in equilibrium at 373.15K.
vapor and liquid are in equilibrium at 373
Example from text: NH 3 ( g ) + HCl ( g ) NH 4Cl ( s )
r H 0 = 176.2kJ / mol , r S 0 = 0.285 kJ / mol K
r G 0 = 176.2 298.15 ( 0.285 ) = 91.2 kJ / mol = spontaneous!
If r H 0 and r S 0 were temperature independent, at what temperature
would the reaction be at equilibrium?
H 176.2
r G 0 = 0 H = T S T =
=
= 618 K
S 0.285
Within these restrictions, for T>618K, the reverse reaction would
be spontaneous but it requires large change in temperature
be spontaneous, but it requires a large change in temperature.
Lecture 17
vap G 0 = vap H T vap S
0
0
Over a small temperature range vap H , vap S are ~ constant.
So at 363.15K vap G = +1.12kJ / mol ( g l is spontaneous )
0
0
at 385
at 385.15K vap G = 1.08kJ / mol
5
Lecture 17
There is also a concept of maximum work that we can associate with
the Gibbs energy.
Start with G = U TS + PV
with
dG = dU TdS SdT + PdV + VdP
1
Consider Na ( s ) + Cl2 ( g ) NaCl ( s )
2
is spontaneous
( l g is spontaneous )
= (TdS + wrev ) TdS SdT + PdV + VdP
= SdT + VdP + wrev + PdV
wrev = PdV + wrev ,other ( i.e. electrical, gravitational, torsional )
electrical gravitational torsiona
dG = SdT + VdP + ( PdV + wrev ,other ) + PdV
dG = SdT + VdP + wrev ,other
dG = wrev ,other at constant T and P
f G 0 = 394kJ / mol
Electrolysis (which represents non-PV work) of molten NaCl is used to
produce Cl2(g) and Na metal.
If we supply sufficient electrical work, we can drive the reverse reaction:
1
NaCl ( s ) Na ( s ) + Cl2 ( g ) r G 0 = 394kJ / mol = wrev ,elect .
394
2
work/energy for charged systems = Q V = charge voltage
1e- is transferred from Cl- to Na+ in this process:
Q = qe N A = 1.602 1019
G = wrev ,other
As we argued for the Helmholtz energy:
G<0=Spontaneous process. So wrev,other is the maximum amount of
non-pressure-volume work that can be done by system.
Lecture 17
6
7
C
e
C
6.02 1023
= 9.64 104
e
mol
mol
G 394 103 J / mol
=
= 4.08Volts
Q 9.64 104 C / mol
V=4.08 volts is the minimum voltage needed for electrolysis.
Q V = G and V =
Lecture 17
8