the maximum non-expansion work available from a reversible spontaneous process ( < 0) at constant...

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1

The maximum non-expansion work available from a reversible spontaneous process ( < 0) at constant T and p is equal to , that is

ΔGΔG

2

The maximum non-expansion work available from a reversible spontaneous process ( < 0) at constant T and p is equal to , that is

ΔG

exp-nonwΔG

ΔG

3

The maximum non-expansion work available from a reversible spontaneous process ( < 0) at constant T and p is equal to , that is

This is a second important application of .

ΔG

exp-nonwΔG

ΔG

ΔG

4

The maximum non-expansion work available from a reversible spontaneous process ( < 0) at constant T and p is equal to , that is

This is a second important application of . The key constraints are indicated in blue type.

ΔG

exp-nonwΔG

ΔG

ΔG

5

To prove that , start with a summary of

previous results:exp-nonwΔG

6

To prove that , start with a summary of

previous results: G = H – T S (1)

exp-nonwΔG

7

To prove that , start with a summary of

previous results: G = H – T S (1) H = E + pV (2)

exp-nonwΔG

8

To prove that , start with a summary of

previous results: G = H – T S (1) H = E + pV (2) (3)

exp-nonwΔG

wqΔE

9

To prove that , start with a summary of

previous results: G = H – T S (1) H = E + pV (2) (3) (4)

exp-nonwΔG

exp-nonexp www

wqΔE

10

To prove that , start with a summary of

previous results: G = H – T S (1) H = E + pV (2) (3) (4) (5)

exp-nonwΔG

exp-nonexp www

TqΔS rev

wqΔE

11

Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6)

12

Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6) Now take a change in each variable

Δ(TS)Δ(pV)ΔEΔG

13

Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6) Now take a change in each variable

(7)

Δ(TS)Δ(pV)ΔEΔG STΔTSΔpVΔΔVpΔE

14

Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6) Now take a change in each variable

(7) Plug Eq. (4) into Eq. (3) and insert the result into Eq.

(7):

Δ(TS)Δ(pV)ΔEΔG STΔTSΔpVΔΔVpΔE

15

Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6) Now take a change in each variable

(7) Plug Eq. (4) into Eq. (3) and insert the result into Eq.

(7): (8)

Δ(TS)Δ(pV)ΔEΔG

STΔTSΔpVΔΔVpwwqΔG exp-nonexp

STΔTSΔpVΔΔVpΔE

16

Now fix the conditions:

17

Now fix the conditions: (a) constant temperature, so that ,

0ΔT

18

Now fix the conditions: (a) constant temperature, so that , (b) constant pressure, so that ,

0ΔT 0Δp

19

Now fix the conditions: (a) constant temperature, so that , (b) constant pressure, so that , (c) and reversible process,

0ΔT 0Δp

20

Now fix the conditions: (a) constant temperature, so that , (b) constant pressure, so that , (c) and reversible process, then Eq. (8) simplifies to (9)

STΔVpwwqΔG revexp,-nonrevexp,rev

0ΔT 0Δp

21

Now fix the conditions: (a) constant temperature, so that , (b) constant pressure, so that , (c) and reversible process, then Eq. (8) simplifies to (9) which simplifies using Eq. (5) to yield

STΔVpwwqΔG revexp,-nonrevexp,rev

0ΔT 0Δp

22

Now fix the conditions: (a) constant temperature, so that , (b) constant pressure, so that , (c) and reversible process, then Eq. (8) simplifies to (9) which simplifies using Eq. (5) to yield (10)

STΔVpwwqΔG revexp,-nonrevexp,rev

0ΔT 0Δp

ΔVpwwΔG revexp,-nonrevexp,

23

Now fix the conditions: (a) constant temperature, so that , (b) constant pressure, so that , (c) and reversible process, then Eq. (8) simplifies to (9) which simplifies using Eq. (5) to yield (10) For a reversible change , hence

STΔVpwwqΔG revexp,-nonrevexp,rev

0ΔT 0Δp

ΔVpwwΔG revexp,-nonrevexp,

ΔVpw revexp,

24

Now fix the conditions: (a) constant temperature, so that , (b) constant pressure, so that , (c) and reversible process, then Eq. (8) simplifies to (9) which simplifies using Eq. (5) to yield (10) For a reversible change , hence

STΔVpwwqΔG revexp,-nonrevexp,rev

0ΔT 0Δp

ΔVpwwΔG revexp,-nonrevexp,

ΔVpw revexp,

revexp,-nonwΔG

25

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

26

The Gibbs Energy and Equilibrium

27

The Gibbs Energy and Equilibrium When a system goes from an initial to a final state, a

indicates a spontaneous change left to right, a indicates a non-spontaneous process, the reaction is spontaneous right to left.

0ΔG 0ΔG

28

The Gibbs Energy and Equilibrium When a system goes from an initial to a final state, a

indicates a spontaneous change left to right, a indicates a non-spontaneous process, the reaction is spontaneous right to left.

It is possible that , and hence

0ΔG 0ΔG

ΔSTΔH 0ΔG

29

The Gibbs Energy and Equilibrium When a system goes from an initial to a final state, a

indicates a spontaneous change left to right, a indicates a non-spontaneous process, the reaction is spontaneous right to left.

It is possible that , and hence

When , the system is at equilibrium, there is no net change.

0ΔG 0ΔG

ΔSTΔH 0ΔG

0ΔG

30

Example: Consider a mixture of ice and water at 0 oC and 1 bar.

31

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.

32

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:

33

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

34

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 , and therefore the amount of useful work that can be extracted from the system is zero.

0ΔG

35

Predicting the Outcome of Chemical Reactions

36

Predicting the Outcome of Chemical Reactions

Consider the “simple” reaction A B

37

Predicting the Outcome of Chemical Reactions

Consider the “simple” reaction A B How do we tell which is the spontaneous direction:

38

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 ?

39

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 for each reaction gives the answer.ΔG

40

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 for each reaction gives the answer.Suppose A B is spontaneous

ΔG

41

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 for each reaction gives the answer.Suppose A B is spontaneous – will the reaction B A take place to any extent?

ΔG

42

All chemical reactions proceed so as to reach the minimum of the total Gibbs energy of the system.

43

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.

44

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.

45

It is necessary to keep in mind that all reactions for which 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).

ΔG

46

47

48

Standard Gibbs Energy and the Equilibrium Constant

49

Standard Gibbs Energy and the Equilibrium Constant

The Gibbs energy for a species X which is not in its standard state is given by

50

Standard Gibbs Energy and the Equilibrium Constant

The Gibbs energy for a species X which is not in its standard state is given by

X0XX alnRTGG

51

Standard Gibbs Energy and the Equilibrium Constant

The Gibbs energy for a species X which is not in its standard state is given by

where aX is the activity of species X. X

0XX alnRTGG

52

Standard Gibbs Energy and the Equilibrium Constant

The Gibbs energy for a species X which is not in its standard state is given by

where aX is the activity of species X. Recall that

.

X0XX alnRTGG

[X]a XX

53

Standard Gibbs Energy and the Equilibrium Constant

The Gibbs energy for a species X which is not in its standard state is given by

where aX is the activity of species X. Recall that

. In a number of situations the activity coefficient

satisfies , so that ,

X0XX alnRTGG

[X]a XX

1X [X]aX

54

Standard Gibbs Energy and the Equilibrium Constant

The Gibbs energy for a species X which is not in its standard state is given by

where aX is the activity of species X. Recall that

. In a number of situations the activity coefficient

satisfies , so that , so that the above result simplifies to

X0XX alnRTGG

[X]a XX

1X [X]aX

[X]lnRTGG 0XX

55

Standard Gibbs Energy and the Equilibrium Constant

56

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 for a reaction ΔG

57

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 for a reaction

a A + b B c C + d D

ΔG

58

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 for a reaction

a A + b B c C + d D is given by = c GC + d GD – a GA – b GB

ΔG

ΔG

59

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 for a reaction

a A + b B c C + d D is given by = c GC + d GD – a GA – b GB

= + – –

ΔG

ΔG

[C]lnRTcGc 0C [D]lnRTdGd 0

D

[A]lnRTaGa 0A [B]lnRTbGb 0

B

60

0B

0A

0D

0C GbGaGdGcΔG

badc [B]lnRT[A]lnRT[D]lnRT[C]lnRT

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