1501. 1502 where Q is the reaction quotient. 1503

1

0B

0A

0D

0C GbGaGdGcΔG

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

0ΔG

ba

dc

[B][A]

[D][C]lnRT

2

0B

0A

0D

0C GbGaGdGcΔG

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

0ΔG

ba

dc

[B][A]

[D][C]lnRT

QlnRTΔGΔG 0

3

where Q is the reaction quotient.

0B

0A

0D

0C GbGaGdGcΔG

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

0ΔG

ba

dc

[B][A]

[D][C]lnRT

QlnRTΔGΔG 0

4

where Q is the reaction quotient. For a reaction system at equilibrium and Q = Kc

0B

0A

0D

0C GbGaGdGcΔG

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

0ΔG

ba

dc

[B][A]

[D][C]lnRT

0ΔG

QlnRTΔGΔG 0

5

where Q is the reaction quotient. For a reaction system at equilibrium and Q = Kc

and hence (a very key result)

0B

0A

0D

0C GbGaGdGcΔG

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

0ΔG

ba

dc

[B][A]

[D][C]lnRT

0ΔG

QlnRTΔGΔG 0

c0 KlnRTΔG

6

For a reaction in which all the species are in the gas phase,

p0 KlnRTΔG

7

It follows directly from the result

c0 KlnRTΔG

8

It follows directly from the result

for Kc > 1

c0 KlnRTΔG

0ΔG0

9

It follows directly from the result

for Kc > 1

for Kc < 1

c0 KlnRTΔG

0ΔG0

0ΔG0

10

It follows directly from the result

for Kc > 1

for Kc < 1

for Kc = 1 In this case the position of the

equilibrium does not favor either products or reactants forming.

c0 KlnRTΔG

0ΔG0

0ΔG0 0ΔG0

11

Example: The standard Gibbs energy for the reaction ½ N2(g) + 3/2 H2(g) NH3(g)

is -16.45 kJmol-1. Calculate the equilibrium constant for the reaction at 25.00 oC. In a certain experiment the initial concentrations are [H2] = 0.00345 M, [N2] = 0.00956 M, and [NH3] = 0.00678 M. Predict the direction of the reaction.

12

Example: The standard Gibbs energy for the reaction ½ N2(g) + 3/2 H2(g) NH3(g)

is -16.45 kJmol-1. Calculate the equilibrium constant for the reaction at 25.00 oC. In a certain experiment the initial concentrations are [H2] = 0.00345 M, [N2] = 0.00956 M, and [NH3] = 0.00678 M. Predict the direction of the reaction.

From we have

c0 KlnRTΔG

13

Example: The standard Gibbs energy for the reaction ½ N2(g) + 3/2 H2(g) NH3(g)

is -16.45 kJmol-1. Calculate the equilibrium constant for the reaction at 25.00 oC. In a certain experiment the initial concentrations are [H2] = 0.00345 M, [N2] = 0.00956 M, and [NH3] = 0.00678 M. Predict the direction of the reaction.

From we have

c0 KlnRTΔG

RTΔGKln

0c

14

Example: The standard Gibbs energy for the reaction ½ N2(g) + 3/2 H2(g) NH3(g)

is -16.45 kJmol-1. Calculate the equilibrium constant for the reaction at 25.00 oC. In a certain experiment the initial concentrations are [H2] = 0.00345 M, [N2] = 0.00956 M, and [NH3] = 0.00678 M. Predict the direction of the reaction.

From we have

c0 KlnRTΔG

RTΔGKln

0c

K) )(298.15molJK(8.314)molJ10x16.45(Kln

1-1-

1-3c

15

Hence ln Kc = 6.636

6.636cK e

16

Hence ln Kc = 6.636

therefore Kc = 762

6.636cK e

17

Hence ln Kc = 6.636

therefore Kc = 762

(Note almost one significant digit is lost in evaluating the exponential.)

6.636cK e

18

Hence ln Kc = 6.636

therefore Kc = 762

(Note almost one significant digit is lost in evaluating the exponential.)

To predict the direction of the reaction, use

6.636cK e

QlnRTΔGΔG 0

19

Now

3/22

1/22

3][H][N

][NHQ

20

Now

3/22

1/22

3][H][N

][NHQ

3/21/2(0.00345)(0.00956)0.00678

21

Now

= 342

3/22

1/22

3][H][N

][NHQ

3/21/2(0.00345)(0.00956)0.00678

.000203)(0.0978)(00.00678

22

Now

= 342 Since Q < Kc the reaction will move in the direction

to produce more NH3.

3/22

1/22

3][H][N

][NHQ

3/21/2(0.00345)(0.00956)0.00678

.000203)(0.0978)(00.00678

23

EmployingQlnRTΔGΔG 0

24

EmployingQlnRTΔGΔG 0

(342)lnK) )(298.15molKJ(8.314molkJ16.45ΔG 1-1-1-

25

Employing

= – 16.45 kJ mol-1 + 14.46 kJ mol-1 = – 1.99 kJ mol-1

QlnRTΔGΔG 0

(342)lnK) )(298.15molKJ(8.314molkJ16.45ΔG 1-1-1-

26

Employing

= – 16.45 kJ mol-1 + 14.46 kJ mol-1 = – 1.99 kJ mol-1

Since < 0, the reaction takes place in the forward direction.

QlnRTΔGΔG 0

(342)lnK) )(298.15molKJ(8.314molkJ16.45ΔG 1-1-1-

ΔG

27

Are diamonds forever?

28

Are diamonds forever?

In chemical jargon, is the transition Cdiamond Cgraphite

spontaneous at room temperature?

29

Are diamonds forever?

In chemical jargon, is the transition Cdiamond Cgraphite

spontaneous at room temperature? Tabulated data: S0(diamond) = 2.44 J K-1 mol-1

S0(graphite) = 5.69 J K-1 mol-1

30

Are diamonds forever?

In chemical jargon, is the transition Cdiamond Cgraphite

spontaneous at room temperature? Tabulated data: S0(diamond) = 2.44 J K-1 mol-1

S0(graphite) = 5.69 J K-1 mol-1

for the reaction Cgraphite Cdiamond = 1.90 kJ

0ΔH

31

For the reaction Cdiamond Cgraphite

32

For the reaction Cdiamond Cgraphite

= 1 mol x 5.69 J K-1 mol-1 – 1 mol x 2.44 J K-1 mol-1

= 3.25 J K-1

0ΔS

33

For the reaction Cdiamond Cgraphite

= 1 mol x 5.69 J K-1 mol-1 – 1 mol x 2.44 J K-1 mol-1

= 3.25 J K-1 Using = - 1.90 x 103J - 298 K (3.25 J K-1)

0ΔS

000 ΔST-ΔHΔG

34

For the reaction Cdiamond Cgraphite

= 1 mol x 5.69 J K-1 mol-1 – 1 mol x 2.44 J K-1 mol-1

= 3.25 J K-1 Using = - 1.90 x 103J - 298 K (3.25 J K-1) = - 1.90 x 103J - 969 J = -2.87 kJ

0ΔS

000 ΔST-ΔHΔG

35

For the reaction Cdiamond Cgraphite

= 1 mol x 5.69 J K-1 mol-1 – 1 mol x 2.44 J K-1 mol-1

= 3.25 J K-1 Using = - 1.90 x 103J - 298 K (3.25 J K-1) = - 1.90 x 103J - 969 J = -2.87 kJ Conclusion: The reaction Cdiamond Cgraphite

is

spontaneous!

0ΔS

000 ΔST-ΔHΔG

36

Electrochemistry

37

Electrochemistry

Two Key Ideas in this section:

38

Electrochemistry

Two Key Ideas in this section: (1) Use spontaneous redox chemistry to generate

electricity.

39

Electrochemistry

Two Key Ideas in this section: (1) Use spontaneous redox chemistry to generate

electricity.(2) Use electric current to drive non-spontaneous

redox reactions.

40

Oxidation-reduction reactions

41

Oxidation-reduction reactions

Oxidation number: A number equal to the charge an atom would have if its shared electrons were held completely by the atom that attracts them more strongly.

42

Oxidation-reduction reactions

Oxidation number: A number equal to the charge an atom would have if its shared electrons were held completely by the atom that attracts them more strongly.

Need to be able to distinguish reactions where there is a change in oxidation number (these are redox reactions) from reactions where there is no change in oxidation number.

43

44

Example:

Na2SO4(aq) + BaCl2(aq) 2 NaCl(aq) + BaSO4(s)

45

Example:

Na2SO4(aq) + BaCl2(aq) 2 NaCl(aq) + BaSO4(s) + 2- 2+ -+ 2-2+-

46

Example:

Na2SO4(aq) + BaCl2(aq) 2 NaCl(aq) + BaSO4(s)

There is no change in oxidation number, there is no electron transfer taking place in this reaction.

Example: 2 Na(s) + Cl2(g) 2 NaCl(s)

+ 2- 2+ -+ 2-2+-

47

Example:

Na2SO4(aq) + BaCl2(aq) 2 NaCl(aq) + BaSO4(s)

There is no change in oxidation number, there is no electron transfer taking place in this reaction.

Example: 2 Na(s) + Cl2(g) 2 NaCl(s)

+ 2- 2+ -+ 2-2+-

0 0 -+

48

Example:

Na2SO4(aq) + BaCl2(aq) 2 NaCl(aq) + BaSO4(s)

There is no change in oxidation number, there is no electron transfer taking place in this reaction.

Example: 2 Na(s) + Cl2(g) 2 NaCl(s)

In this example, there is a change in oxidation number, so electron transfer is taking place.

+ 2- 2+ -+ 2-2+-

0 0 -+

49

Half-Equations:

50

Half-Equations:

Na Na+ + e- (1)

51

Half-Equations:

Na Na+ + e- (1)

2e- + Cl2 2 Cl- (2)

52

Half-Equations:

Na Na+ + e- (1)

2e- + Cl2 2 Cl- (2) -

+0

0

53

Half-Equations:

Na Na+ + e- (1)

2e- + Cl2 2 Cl- (2)

These are called the half-equations.

-

+0

0

54

Half-Equations:

Na Na+ + e- (1)

2e- + Cl2 2 Cl- (2)

These are called the half-equations. Multiply equation (1) by 2 and add equation (2) leads to the overall balanced equation.

2 Na + Cl2 2 NaCl

-

+0

0

55

Oxidation: A process in which electrons are lost. Na Na+ + e-

56

Oxidation: A process in which electrons are lost. Na Na+ + e-

Reduction: A process in which electrons are gained. 2e- + Cl2 2 Cl-

57

Oxidation: A process in which electrons are lost. Na Na+ + e-

Reduction: A process in which electrons are gained. 2e- + Cl2 2 Cl-

Oxidizing Agent: A substance that gains electrons (in the reduction step). It undergoes a decrease in oxidation number. Example: Cl2 in the above eq.

58

Oxidation: A process in which electrons are lost. Na Na+ + e-

Reduction: A process in which electrons are gained. 2e- + Cl2 2 Cl-

Oxidizing Agent: A substance that gains electrons (in the reduction step). It undergoes a decrease in oxidation number. Example: Cl2 in the above eq.

Reducing Agent: A substance that donates electrons in a redox reaction (specifically in the oxidation step). It undergoes an increase in oxidation number. Example: Na in the above oxidation.

59

Redox reaction: A reaction that can be split apart into an oxidation step and a reduction step.

60

Redox reaction: A reaction that can be split apart into an oxidation step and a reduction step.

Example: 2 Na + Cl2 2 NaCl