a more appropriate definition of k (for the same chemical reaction discussed previously) is with...

1

A more appropriate definition of K (for the same chemical reaction discussed previously) is

With this definition, each concentration term is divided by the units of molarity, M, so K is then dimensionless.

ba

dc

[B]/M[A]/M

[D]/M[C]/MK

2

A more appropriate definition of K (for the same chemical reaction discussed previously) is

With this definition, each concentration term is divided by the units of molarity, M, so K is then dimensionless. Because it is tedious to write the above equation, most people simply write

to save writing time.

ba

dc

[B]/M[A]/M

[D]/M[C]/MK

ba

dc

[B][A]

[D][C]K

3

Relationship between Q and K

N2O4(g) 2 NO2(g)

4

5

Different ways of expressing the equilibrium constant

6

Different ways of expressing the equilibrium constant

Homogeneous equilibrium: A state of equilibrium between reactants and products in the same phase.

7

Different ways of expressing the equilibrium constant

Homogeneous equilibrium: A state of equilibrium between reactants and products in the same phase.

Example: CH3CO2H(aq) CH3CO2-(aq) + H+

(aq)

8

Different ways of expressing the equilibrium constant

Homogeneous equilibrium: A state of equilibrium between reactants and products in the same phase.

Example: CH3CO2H(aq) CH3CO2-(aq) + H+

(aq)

H]CO[CH][H]-CO[CHK

23

23c

9

Example: For the gas phase reaction N2O4(g) 2 NO2(g)

]O[N][NOK42

22c

10

Example: For the gas phase reaction N2O4(g) 2 NO2(g)

When working with gaseous reactions, it is frequently more convenient to use different units.

]O[N][NOK42

22c

11

For the preceding reaction

42

2

ON

2NO

p p

pK

12

For the preceding reaction

The subscript p on K indicates partial pressures are used in the mass action expression.

42

2

ON

2NO

p p

pK

13

For the preceding reaction

The subscript p on K indicates partial pressures are used in the mass action expression.

is the equilibrium partial pressure of N2O4.

is the equilibrium partial pressure of NO2.

42

2

ON

2NO

p p

pK

42ONp

2NOp

14

For the preceding reaction

The subscript p on K indicates partial pressures are used in the mass action expression.

is the equilibrium partial pressure of N2O4.

is the equilibrium partial pressure of NO2.

In general, Kp and Kc (for the same reaction) are not equal.

42

2

ON

2NO

p p

pK

42ONp

2NOp

15

Exercise: For the reaction a A(g) b B(g) find a connection between Kp and Kc. Assume the gases can be treated as ideal gases.

16

Exercise: For the reaction a A(g) b B(g) find a connection between Kp and Kc. Assume the gases can be treated as ideal gases.

aA

bBp p

pK

17

Exercise: For the reaction a A(g) b B(g) find a connection between Kp and Kc. Assume the gases can be treated as ideal gases.

Now from the ideal gas equation PV = n RT,

aA

bBp p

pK

18

Exercise: For the reaction a A(g) b B(g) find a connection between Kp and Kc. Assume the gases can be treated as ideal gases.

Now from the ideal gas equation PV = n RT,

aA

bBp p

pK

[A]RTVRTnpA

AA [B]RT

VRTnpB

BB

19

Exercise: For the reaction a A(g) b B(g) find a connection between Kp and Kc. Assume the gases can be treated as ideal gases.

Now from the ideal gas equation PV = n RT,

Plug these two results into the expression for Kp.

aA

bBp p

pK

[A]RTVRTnpA

AA [B]RT

VRTnpB

BB

20

Hence:

aA

bBp p

pK

21

Hence:

aA

bBp p

pK

a

b

p[A]RT

[B]RTK

22

Hence:

aA

bBp p

pK

a

b

p[A]RT

[B]RTK

a

b

a

b

RT

RT[A][B]

23

Hence:

aA

bBp p

pK

a

b

p[A]RT

[B]RTK

a

b

a

b

RT

RT[A][B]

a - ba

bRT

[A][B]

24

Hence:

where

aA

bBp p

pK

a

b

p[A]RT

[B]RTK

a

b

a

b

RT

RT[A][B]

a - ba

bRT

[A][B]

abΔn na

bRT

[A][B]

25

Hence:

where and therefore

aA

bBp p

pK

a

b

p[A]RT

[B]RTK

a

b

a

b

RT

RT[A][B]

a - ba

bRT

[A][B]

abΔn na

bRT

[A][B]

ncp RTKK

26

Sample Problems

27

Sample Problems Example: 2 NO(g) + O2(g) 2 NO2(g)

At a temperature of 230 oC the concentrations of the various species are [NO] = 0.0542 M, [O2] = 0.127 M, and [NO2] = 15.5 M at equilibrium. Calculate the equilibrium constant Kc at 230 oC.

28

Sample Problems Example: 2 NO(g) + O2(g) 2 NO2(g)

At a temperature of 230 oC the concentrations of the various species are [NO] = 0.0542 M, [O2] = 0.127 M, and [NO2] = 15.5 M at equilibrium. Calculate the equilibrium constant Kc at 230 oC.

][O[NO]][NOK

22

22c

29

Sample Problems Example: 2 NO(g) + O2(g) 2 NO2(g)

At a temperature of 230 oC the concentrations of the various species are [NO] = 0.0542 M, [O2] = 0.127 M, and [NO2] = 15.5 M at equilibrium. Calculate the equilibrium constant Kc at 230 oC.

= 6.44 x 105

][O[NO]][NOK

22

22c

(0.127)(0.0542)(15.5)

2

2

30

Example: The equilibrium constant Kp for the reaction PCl5 (g) PCl3(g) + Cl2(g) is 1.05 at 250 oC.

If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2?

31

Example: The equilibrium constant Kp for the reaction PCl5 (g) PCl3(g) + Cl2(g) is 1.05 at 250 oC.

If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2?

5

23

PCl

ClPClp p

ppK

32

Example: The equilibrium constant Kp for the reaction PCl5 (g) PCl3(g) + Cl2(g) is 1.05 at 250 oC.

If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2?

This is shorthand for

5

23

PCl

ClPClp p

ppK

atmp

atmpatmpK

5

23

PCl

ClPClp

/

//

33

Example: The equilibrium constant Kp for the reaction PCl5 (g) PCl3(g) + Cl2(g) is 1.05 at 250 oC.

If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2?

This is shorthand for

so that

5

23

PCl

ClPClp p

ppK

atmp

atmpatmpK

5

23

PCl

ClPClp

/

//

atmp

Katmpatmp

3

5

2PCl

pPClCl

/

//

34

and so atm1.98p2Cl

atmatm0.463

(1.05)atmatm0.875atmp

2Cl /

//

35

Heterogeneous Equilibria

36

Heterogeneous Equilibria A reaction often involves reactants that are not

present in the same phase – this leads to a heterogeneous equilibrium.

37

Heterogeneous Equilibria A reaction often involves reactants that are not

present in the same phase – this leads to a heterogeneous equilibrium.

Example: Heating CaCO3 in a closed vessel.

38

Heterogeneous Equilibria A reaction often involves reactants that are not

present in the same phase – this leads to a heterogeneous equilibrium.

Example: Heating CaCO3 in a closed vessel.

CaCO3(s) CaO(s) + CO2(g)

39

Heterogeneous Equilibria A reaction often involves reactants that are not

present in the same phase – this leads to a heterogeneous equilibrium.

Example: Heating CaCO3 in a closed vessel.

CaCO3(s) CaO(s) + CO2(g)

][CaCO][CO[CaO]K

3

2c

40

The “concentration” of any pure solid is the ratio of the total number of moles present in the solid, divided by the volume of the solid.

41

The “concentration” of any pure solid is the ratio of the total number of moles present in the solid, divided by the volume of the solid. If part of the solid is removed, the number of moles of solid will decrease – but so will it volume.

42

The “concentration” of any pure solid is the ratio of the total number of moles present in the solid, divided by the volume of the solid. If part of the solid is removed, the number of moles of solid will decrease – but so will it volume. The ratio of moles to volume remains unchanged.

43

Consider the ratio:

molesgramslitersgrams

mass molardensity

44

Consider the ratio:

molesgramslitersgrams

mass molardensity

litersmoles

45

Consider the ratio:

molesgramslitersgrams

mass molardensity

litersmoles

3

33 CaCOmass molar

CaCOdensity][CaCO

46

Consider the ratio:

If the temperature is held fixed, then [CaCO3] is a constant. Similarly for [CaO], which is also a constant.

molesgramslitersgrams

mass molardensity

litersmoles

3

33 CaCOmass molar

CaCOdensity][CaCO

47

So we can rewrite

][CaCO][CO[CaO]K

3

2c

48

So we can rewrite

in the form

][CO[CaO]

][CaCOK 23c

][CaCO][CO[CaO]K

3

2c

49

So we can rewrite

in the form

Now since [CaCO3] and [CaO] are both constant, the left-hand side of the preceding equation is constant.

][CO[CaO]

][CaCOK 23c

][CaCO][CO[CaO]K

3

2c

50

So we can rewrite

in the form

Now since [CaCO3] and [CaO] are both constant, the left-hand side of the preceding equation is constant. Now set

][CO[CaO]

][CaCOK 23c

][CaCO][CO[CaO]K

3

2c

[CaO]

][CaCOKK 3cc

51

Hence Kc = [CO2]

52

Hence Kc = [CO2]

Notice that terms involving pure solids do not appear in the final equilibrium constant expression.

53

Hence Kc = [CO2]

Notice that terms involving pure solids do not appear in the final equilibrium constant expression. This result generalizes to all chemical reactions.

54

Hence Kc = [CO2]

Notice that terms involving pure solids do not appear in the final equilibrium constant expression. This result generalizes to all chemical reactions.

The corresponding expression for Kp for the decomposition of CaCO3 is:

Kp = 2COp

55

The same concentration of CO2 exists in both containers (provided the temperature is the same), even though the amounts of CaO and CaCO3 are different.

56

Summary Comment

Concentration factors for pure solids and pure liquids are ignored in the expression for the equilibrium constant for a chemical reaction.

57

Multiple Equilibria

58

Multiple Equilibria

Consider the two reactions:

59

Multiple Equilibria

Consider the two reactions: A + B C + D C + D E + F

60

Multiple Equilibria

Consider the two reactions: A + B C + D C + D E + F

For the first reaction

[A][B][C][D]Kc