The maximum non-expansion work available from a
reversible spontaneous process (ΔG < 0) at constant
T and p is equal to ΔG, that is
1441
The maximum non-expansion work available from a
reversible spontaneous process (ΔG < 0) at constant
T and p is equal to ΔG, that is
ΔG  wnon - exp
1442
The maximum non-expansion work available from a
reversible spontaneous process (ΔG < 0) at constant
T and p is equal to ΔG, that is
ΔG  wnon - exp
This is a second important application of ΔG.
1443
The maximum non-expansion work available from a
reversible spontaneous process (ΔG < 0) at constant
T and p is equal to ΔG, that is
ΔG  wnon - exp
This is a second important application of ΔG.
The key constraints are indicated in blue type.
1444
To prove that ΔG  wnon - exp , start with a summary of
previous results:
1445
To prove that ΔG  wnon - exp , start with a summary of
previous results:
G=H–TS
(1)
1446
To prove that ΔG  wnon - exp , start with a summary of
previous results:
G=H–TS
(1)
H = E + pV
(2)
1447
To prove that ΔG  wnon - exp , start with a summary of
previous results:
G=H–TS
(1)
H = E + pV
(2)
ΔE  q  w
(3)
1448
To prove that ΔG  wnon - exp , start with a summary of
previous results:
G=H–TS
(1)
H = E + pV
(2)
ΔE  q  w
(3)
w  wexp  wnon -exp
(4)
1449
To prove that ΔG  wnon - exp , start with a summary of
previous results:
G=H–TS
(1)
H = E + pV
(2)
ΔE  q  w
(3)
w  wexp  wnon -exp
(4)
ΔS  qrev
(5)
T
1450
Plug Eq. (2) into Eq. (1) so that
G = E + pV – TS
(6)
1451
Plug Eq. (2) into Eq. (1) so that
G = E + pV – TS
(6)
Now take a change in each variable
ΔG  ΔE  Δ(pV)  Δ(TS)
1452
Plug Eq. (2) into Eq. (1) so that
G = E + pV – TS
(6)
Now take a change in each variable
ΔG  ΔE  Δ(pV)  Δ(TS)
 ΔE  pΔV  VΔp  SΔ T  TΔS
(7)
1453
Plug Eq. (2) into Eq. (1) so that
G = E + pV – TS
(6)
Now take a change in each variable
ΔG  ΔE  Δ(pV)  Δ(TS)
 ΔE  pΔV  VΔp  SΔ T  TΔS
(7)
Plug Eq. (4) into Eq. (3) and insert the result into Eq.
(7):
1454
Plug Eq. (2) into Eq. (1) so that
G = E + pV – TS
(6)
Now take a change in each variable
ΔG  ΔE  Δ(pV)  Δ(TS)
 ΔE  pΔV  VΔp  SΔ T  TΔS
(7)
Plug Eq. (4) into Eq. (3) and insert the result into Eq.
(7):
ΔG q  wexp  wnon -exp  pΔV  VΔp  SΔT  TΔS (8)
1455
Now fix the conditions:
1456
Now fix the conditions:
(a) constant temperature, so that ΔT  0 ,
1457
Now fix the conditions:
(a) constant temperature, so that ΔT  0 ,
(b) constant pressure, so that Δp  0 ,
1458
Now fix the conditions:
(a) constant temperature, so that ΔT  0 ,
(b) constant pressure, so that Δp  0 ,
(c) and reversible process,
1459
Now fix the conditions:
(a) constant temperature, so that ΔT  0 ,
(b) constant pressure, so that Δp  0 ,
(c) and reversible process,
then Eq. (8) simplifies to
ΔG qrev  wexp,rev  wnon -exp,rev  pΔV  TS (9)
1460
Now fix the conditions:
(a) constant temperature, so that ΔT  0 ,
(b) constant pressure, so that Δp  0 ,
(c) and reversible process,
then Eq. (8) simplifies to
ΔG qrev  wexp,rev  wnon -exp,rev  pΔV  TS (9)
which simplifies using Eq. (5) to yield
1461
Now fix the conditions:
(a) constant temperature, so that ΔT  0 ,
(b) constant pressure, so that Δp  0 ,
(c) and reversible process,
then Eq. (8) simplifies to
ΔG qrev  wexp,rev  wnon -exp,rev  pΔV  TS (9)
which simplifies using Eq. (5) to yield
ΔG wexp,rev  wnon-exp,rev  pΔV
(10)
1462
Now fix the conditions:
(a) constant temperature, so that ΔT  0 ,
(b) constant pressure, so that Δp  0 ,
(c) and reversible process,
then Eq. (8) simplifies to
ΔG qrev  wexp,rev  wnon -exp,rev  pΔV  TS (9)
which simplifies using Eq. (5) to yield
ΔG wexp,rev  wnon-exp,rev  pΔV
For a reversible change wexp,rev
(10)
 pΔV , hence
1463
Now fix the conditions:
(a) constant temperature, so that ΔT  0 ,
(b) constant pressure, so that Δp  0 ,
(c) and reversible process,
then Eq. (8) simplifies to
ΔG qrev  wexp,rev  wnon -exp,rev  pΔV  TS (9)
which simplifies using Eq. (5) to yield
ΔG wexp,rev  wnon-exp,rev  pΔV
For a reversible change wexp,rev
(10)
 pΔV , hence
ΔG wnon - exp,rev
1464
A true reversible process takes an infinite amount
of time to complete. Therefore we can never obtain
in any process the amount of useful work predicted
by the value of ΔG.
1465
The Gibbs Energy and Equilibrium
1466
The Gibbs Energy and Equilibrium
When a system goes from an initial to a final state,
a ΔG  0 indicates a spontaneous change left to
right, a ΔG  0 indicates a non-spontaneous
process, the reaction is spontaneous right to left.
1467
The Gibbs Energy and Equilibrium
When a system goes from an initial to a final state,
a ΔG  0 indicates a spontaneous change left to
right, a ΔG  0 indicates a non-spontaneous
process, the reaction is spontaneous right to left.
It is possible that ΔH  T ΔS, and hence
ΔG  0
1468
The Gibbs Energy and Equilibrium
When a system goes from an initial to a final state,
a ΔG  0 indicates a spontaneous change left to
right, a ΔG  0 indicates a non-spontaneous
process, the reaction is spontaneous right to left.
It is possible that ΔH  T ΔS, and hence
ΔG  0
When ΔG  0 , the system is at equilibrium, there is
no net change.
1469
Example: Consider a mixture of ice and water at
0 oC and 1 bar.
1470
Example: Consider a mixture of ice and water at
0 oC and 1 bar. Neither freezing nor melting is
spontaneous, provided no heat is added or
removed from the system.
1471
Example: Consider a mixture of ice and water at
0 oC and 1 bar. Neither freezing nor melting is
spontaneous, provided no heat is added or
removed from the system. There is a dynamic
equilibrium:
1472
Example: Consider a mixture of ice and water at
0 oC and 1 bar. Neither freezing nor melting is
spontaneous, provided no heat is added or
removed from the system. There is a dynamic
equilibrium:
ice
water
1473
Example: Consider a mixture of ice and water at
0 oC and 1 bar. Neither freezing nor melting is
spontaneous, provided no heat is added or
removed from the system. There is a dynamic
equilibrium:
ice
water
The ice lattice is broken down to form liquid water
and water freezes to form ice at every instant.
At equilibrium ΔG  0, and therefore the amount of
useful work that can be extracted from the system
is zero.
1474
Predicting the Outcome of Chemical
Reactions
1475
Predicting the Outcome of Chemical
Reactions
Consider the “simple” reaction
A
B
1476
Predicting the Outcome of Chemical
Reactions
Consider the “simple” reaction
A
B
How do we tell which is the spontaneous direction:
1477
Predicting the Outcome of Chemical
Reactions
Consider the “simple” reaction
A
B
How do we tell which is the spontaneous direction:
A
B or B
A?
1478
Predicting the Outcome of Chemical
Reactions
Consider the “simple” reaction
A
B
How do we tell which is the spontaneous direction:
A
B or B
A?
Examination of ΔG for each reaction gives the answer.
1479
Predicting the Outcome of Chemical
Reactions
Consider the “simple” reaction
A
B
How do we tell which is the spontaneous direction:
A
B or B
A?
Examination of ΔG for each reaction gives the answer.
Suppose A
B is spontaneous
1480
Predicting the Outcome of Chemical
Reactions
Consider the “simple” reaction
A
B
How do we tell which is the spontaneous direction:
A
B or B
A?
Examination of ΔG for each reaction gives the answer.
Suppose A
B is spontaneous – will the reaction
B
A take place to any extent?
1481
All chemical reactions proceed so as to reach the
minimum of the total Gibbs energy of the system.
1482
All chemical reactions proceed so as to reach the
minimum of the total Gibbs energy of the system.
Always between the total Gibbs energy of the
products and the total Gibbs energy of the
reactants, there will be some point where the total
Gibbs energy of a mixture of reactants and products
has a minimum Gibbs energy.
1483
All chemical reactions proceed so as to reach the
minimum of the total Gibbs energy of the system.
Always between the total Gibbs energy of the
products and the total Gibbs energy of the
reactants, there will be some point where the total
Gibbs energy of a mixture of reactants and products
has a minimum Gibbs energy.
The minimum indicates the composition at
equilibrium, i.e. A
B.
1484
It is necessary to keep in mind that all reactions for
which ΔG is positive in the forward direction, take
place to some extent. However the extent of the
reaction may be extremely small (particularly for
many typical inorganic reactions).
1485
1486
1487
Standard Gibbs Energy and the
Equilibrium Constant
1488
Standard Gibbs Energy and the
Equilibrium Constant
The Gibbs energy for a species X which is not in its
standard state is given by
1489
Standard Gibbs Energy and the
Equilibrium Constant
The Gibbs energy for a species X which is not in its
standard state is given by
GX  GX0  RT lnaX
1490
Standard Gibbs Energy and the
Equilibrium Constant
The Gibbs energy for a species X which is not in its
standard state is given by
GX  GX0  RT lnaX
where aX is the activity of species X.
1491
Standard Gibbs Energy and the
Equilibrium Constant
The Gibbs energy for a species X which is not in its
standard state is given by
GX  GX0  RT lnaX
where aX is the activity of species X. Recall that
aX   X[X] .
1492
Standard Gibbs Energy and the
Equilibrium Constant
The Gibbs energy for a species X which is not in its
standard state is given by
GX  GX0  RT lnaX
where aX is the activity of species X. Recall that
aX   X[X] .
In a number of situations the activity coefficient
satisfies  X  1, so that aX  [X] ,
1493
Standard Gibbs Energy and the
Equilibrium Constant
The Gibbs energy for a species X which is not in its
standard state is given by
GX  GX0  RT lnaX
where aX is the activity of species X. Recall that
aX   X[X] .
In a number of situations the activity coefficient
satisfies  X  1, so that aX  [X] , so that the above
result simplifies to
GX  GX0  RT ln[X]
1494
Standard Gibbs Energy and the
Equilibrium Constant
1495
Standard Gibbs Energy and the
Equilibrium Constant
If a reaction is run under conditions such that all of
the reactants and products are not in their standard
states – then ΔG for a reaction
1496
Standard Gibbs Energy and the
Equilibrium Constant
If a reaction is run under conditions such that all of
the reactants and products are not in their standard
states – then ΔG for a reaction
aA + bB
cC + dD
1497
Standard Gibbs Energy and the
Equilibrium Constant
If a reaction is run under conditions such that all of
the reactants and products are not in their standard
states – then ΔG for a reaction
aA + bB
cC + dD
is given by
ΔG = c GC + d GD – a GA – b GB
1498
Standard Gibbs Energy and the
Equilibrium Constant
If a reaction is run under conditions such that all of
the reactants and products are not in their standard
states – then ΔG for a reaction
aA + bB
cC + dD
is given by
ΔG = c GC + d GD – a GA – b GB
= c GC0  cRT ln[C] + dGD0  dRT ln[D]
– aGA0  aRT ln[A] – b GB0  bRT ln[B]
1499
ΔG  c GC0  dGD0  aGA0  b GB0
 RT ln[C]c  RT ln[D]d  RT ln[A]a  RT ln[B]b
1500