Chapter 17

CHAPTER 17 Chemical Equilibrium

1

Chapter Goals     

Basic Concepts The Equilibrium Constant Variation of Kc with the Form of the Balanced Equation The Reaction Quotient Uses of the Equilibrium Constant, Kc

Disturbing a System at Equilibrium: Predictions  The Haber Process: A Practical 

2

Chapter Goals Disturbing a System at Equilibrium: Calculations  Partial Pressures and the Equilibrium Constant  Relationship between Kp and Kc 

Heterogeneous Equilibria  Relationship between ∆Gorxn and the Equilibrium Constant  Evaluation of Equilibrium Constants at Different Temperatures 

3

Basic Concepts 

Reversible reactions do not go to completion.  

They can occur in either direction Symbolically, this is represented as:

→cC +dD a A ( g ) + b B( g ) ← ( g) ( g)

4

Basic Concepts  Chemical equilibrium exists when two

opposing reactions occur simultaneously at the same rate. 

A chemical equilibrium is a reversible reaction that the forward reaction rate is equal to the reverse reaction rate.

 Chemical equilibria are dynamic equilibria. 

Molecules are continually reacting, even though the overall composition of the reaction mixture does not change. 5

Basic Concepts  One example of a

dynamic equilibrium can be shown using radioactive 131I as a tracer in a saturated PbI2 solution.

{1} Place solid PbI*2 in a saturated PbI 2 solution. 2O H  → 2+ PbI 2(s) ← Pb (aq) + 2 I -(aq)

{ 2} Stir for a few minutes, then filter the solution. Some of the radioactive iodine will go into solution. 6

Basic Concepts  This movie depicts a dynamic equilibrium.

7

Basic Concepts  Graphically, this is a representation of the

rates for the forward and reverse reactions for this general reaction.

→c C + d D a A ( g ) + b B( g ) ← (g) (g)

8

Basic Concepts  One of the fundamental ideas of chemical

equilibrium is that equilibrium can be established from either the forward or reverse direction.

9

Basic Concepts

10

Basic Concepts

11

The Equilibrium Constant 

For a simple one-step mechanism reversible reaction such as:

A (g) 

→ + B(g) ← C (g) + D (g)

The rates of the forward and reverse reactions can be represented as:

Rate f = k f [A ][B] which represents the forward rate. Rate r = k r [C][D ] which represents the reverse rate.

12

The Equilibrium Constant  When system is at equilibrium:

Ratef = Rater Substitute for the rate relationship to give :

k f [ A ][ B] = k r [ C][ D]

which rearranges to k f [ C][ D] = k r [ A ][ B]

13

The Equilibrium Constant  Because the ratio of two constants is a

constant we can define a new constant as follows :

kf = Kc and kr

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

14

The Equilibrium Constant 

Similarly, for the general reaction:

a A (g)

→ + b B(g) ← c C (g) + d D (g)

we can define a constant

[ C] [ D ] Kc = a b [ A ] [ B] c

d

← products ← reactants

This expression is valid for all reactions. 15

The Equilibrium Constant  Kc is the equilibrium

constant .  Kc is defined for a reversible reaction at a given temperature as the product of the equilibrium concentrations (in M) of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation, divided by the product of the equilibrium concentrations (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the 16 balanced equation.

The Equilibrium Constant  Example 17-1: Write equilibrium constant

expressions for the following reactions at 500oC. All reactants and products are gases at 500oC.

→ PCl 5 ← PCl 3 + Cl 2

[ PCl 3 ][ Cl 2 ] Kc = [ PCl5 ]

17

The Equilibrium Constant

→ H 2 + I 2 ← 2 HI You do it! 18

The Equilibrium Constant

→ H 2 + I 2 ← 2 HI

[ HI] Kc = [ H 2 ][ I 2 ] 2

19

The Equilibrium Constant

→ 4 NH 3 + 5 O 2 ← 4 NO + 6 H 2 O You do it!

20

The Equilibrium Constant

→ 4 NH 3 + 5 O 2 ← 4 NO + 6 H 2 O

[ NO] [ H 2 O] Kc = 4 5 [ NH 3 ] [ O 2 ] 4

6

21

The Equilibrium Constant  Equilibrium

constants are dimensionless because they actually involve a thermodynamic quantity called activity. 

Activities are directly related to molarity

22

The Equilibrium Constant 

Example 17-2: One liter of equilibrium mixture from the following system at a high temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate Kc for the reaction.

→ PCl 5 ← PCl 3 + Cl 2



Equil []’s 0.028 M

0.172 M You do it!

0.086 M

23

The Equilibrium Constant

[ PCl 3 ][ Cl 2 ] Kc = [ PCl5 ] ( 0.172 )( 0.086 ) Kc = ( 0.028) K c = 0.53 24

The Equilibrium Constant  Example 17-3: The decomposition of PCl5

was studied at another temperature. One mole of PCl5 was introduced into an evacuated 1.00 liter container. The system was allowed to reach equilibrium at the new temperature. At equilibrium 0.60 mole of PCl3 was present in the container. Calculate the equilibrium constant at this temperature. 25

The Equilibrium Constant

PCl 5( g ) Initial

1.00 M

→ ← PCl 3( g ) + Cl 2( g ) 0

0

26

The Equilibrium Constant

PCl 5( g ) Initial Change

1.00 M

→ ← PCl 3( g ) + Cl 2( g ) 0

0

- 0.60 M + 0.60 M + 0.60 M

27

The Equilibrium Constant

PCl 5( g ) Initial Change

1.00 M

→ ← PCl 3( g ) + Cl 2( g ) 0

0

- 0.60 M + 0.60 M + 0.60 M

Equilibrium 0.40 M

0.60 M

0.60 M 28

The Equilibrium Constant PCl 5( g ) Initial Change

1.00 M 0 0 - 0.60 M + 0.60 M + 0.60 M

Equilibrium 0.40 M K

' c

→ ← PCl 3( g ) + Cl 2( g )

0.60 M

0.60 M

( 0.60 )( 0.60 ) = = 0.90 at another T ( 0.40)

29

The Equilibrium Constant  Example 17-4: At a given temperature

0.80 mole of N2 and 0.90 mole of H2 were placed in an evacuated 1.00-liter container. At equilibrium 0.20 mole of NH3 was present. Calculate Kc for the reaction. You do it!

30

The Equilibrium Constant N 2(g) Initial Change

→ + 3 H 2(g) ← 2 NH 3(g)

0.80 M 0.90 M - 0.10 M - 0.30 M

Equilibrium 0.70 M

0.60 M

0 + 0.20 M 0.20 M

[ ( NH 3 ] 0.20 ) Kc = = = 0.26 3 3 [ N 2 ][ H 2 ] ( 0.70)( 0.60) 2

2

31

Variation of Kc with the Form of the Balanced Equation  

The value of Kc depends upon how the balanced equation is written. From example 17-2 we have this reaction:

→ PCl 5 ← PCl 3 + Cl 2 

This reaction has a Kc=[PCl3][Cl2]/[PCl5]=0.53

32

Variation of Kc with the Form of the Balanced Equation 

Example 17-5: Calculate the equilibrium constant for the reverse reaction by two methods, i.e, the equilibrium constant for this reaction.

→ PCl 3 + Cl 2 ← PCl 5

Equil. []’s 0.172 M 0.086 M 0.028 M The concentrations are from Example 17-2.

33

Variation of Kc with the Form of the Balanced Equation ' Kc

PCl5 ] [ = [ PCl3 ][ Cl2 ]

34

Variation of Kc with the Form of the Balanced Equation ' Kc

PCl5 ] 0.028) ( [ = = = 19 . . )( 0.086) [ PCl3 ][ Cl2 ] ( 0172

35

Variation of Kc with the Form of the Balanced Equation PCl5 ] 0.028) ( [ ' Kc = = = 1.9 [ PCl3 ][ Cl2 ] ( 0.172)( 0.086) ' 1 1 1 Kc = or K = = = 1 . 9 ' c Kc 0.53 Kc  

Large equilibrium constants indicate that most of the reactants are converted to products. Small equilibrium constants indicate that only small amounts of products are formed. 36

The Reaction Quotient

37

The Reaction Quotient

38

The Reaction Quotient



The mass action expression or reaction quotient has the symbol Q. 



Q has the same form as Kc

The major difference between Q and Kc is that the concentrations used in Q are not necessarily equilibrium values.

For this general reaction : → aA + bB← cC + dD

[ C] [ D ] Q= a b [ A ] [ B] c

d

39

The Reaction Quotient  Why do we need another “equilibrium

constant” that does not use equilibrium concentrations?  Q will help us predict how the equilibrium will respond to an applied stress.  To make this prediction we compare Q with Kc.

40

The Reaction Quotient

Q = Kc

When : The system is at equilibrium.

Q > K c The reaction occurs to the left to a greater extent. Q < K c The reaction occurs to the right to a greater extent. To help understand this think of Q and K c as fractions.

41

The Reaction Quotient 

Example 17-6: The equilibrium constant The concentrat ions given in the problemfor the following reaction is 49 at 450oC. If 0.22 mole of not necessaril equilibriu s. were put I2, 0.22are mole of H2, andy 0.66 molemof[]'HI We can calculate Q. into an evacuated 1.00-liter container, would the system be H at equilibrium? → 2 IfHInot, what must + I ← 2(g) 2(g) (g) occur to establish equilibrium?

0.22 M 0.22 M 0.66 M

2 2 ( [ HI] 0.66) Q= = = 9.0 [ H 2 ][ I 2 ] ( 0.22)( 0.22)

Q = 9.0 but K c = 49 Q < Kc

42

Uses of the Equilibrium Constant, Kc 

Example 17-7: The equilibrium constant, Kc, is 3.00 for the following reaction at a given temperature. If 1.00 mole of SO2 and 1.00 mole of NO2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium?

SO 2(g)

+

NO 2(g)

→ ←

SO 3(g) + NO (g)

43

Uses of the Equilibrium Constant, Kc SO 2(g) Initial

0.500 M

+

NO 2(g) → ←

0.500 M

SO 3(g) + NO (g) 0

0

44

Uses of the Equilibrium Constant, Kc SO 2(g) Initial Change

0.500 M - xM

+

NO 2(g) → ←

0.500 M - xM

SO 3(g) + NO (g) 0 +xM

0 +xM

45

Uses of the Equilibrium Constant, Kc SO 2(g) Initial Change

0.500 M - xM

+

NO 2(g) → ←

0.500 M - xM

Equilibrium ( 0.500 − x ) M ( 0.500 − x ) M

SO 3(g) + NO (g) 0

0

+xM

+xM

xM

xM

46

Uses of the Equilibrium Constant, Kc SO 2(g) Initial Change

0.500 M - xM

+

NO 2(g) → ← 0.500 M - xM

Equilibrium ( 0.500 − x ) M ( 0.500 − x ) M

SO 3(g) + NO (g) 0 +xM

0 +xM

xM

xM

[ ( SO 3 ][ NO] x )( x ) Kc = = 3.00 = ( 0.500 − x )( 0.500 − x ) [SO 2 ][ NO 2 ] This equation is a perfect square. We can take the

of both sidesof the equation.

47

Uses of the Equilibrium Constant, Kc SO 2(g) Initial

+

0.500 M

Change

- xM

NO 2(g) → ←

SO 3(g) +

0.500 M - xM

0 +xM

Equilibrium ( 0.500 − x ) M ( 0.500 − x ) M xM [SO 3 ][ NO] = 3.00 = ( x )( x ) Kc = [SO 2 ][ NO 2 ] ( 0.500 − x )( 0.500 − x )

NO (g) 0 +xM xM

This equation is a perfect square. We can take the 1.73 =

of both sides.

x 0.500 − x

0.865 - 1.73 x = x;

0.865 = 2.73 x; x = 0.865

x = 0.316M = [SO 3 ] = [ NO]

( 0.500 − x ) M

= 0.184M = [SO 2 ] = [ NO 2 ]

2.73 48

Uses of the Equilibrium Constant, Kc  Example 17-8: The equilibrium

constant is 49 for the following reaction at 450oC. If 1.00 mole of HI is put into an evacuated 1.00-liter container and allowed to reach equilibrium, what will be the equilibrium concentration of each substance?

→ H 2(g) + I 2(g) ← 2 HI (g) You do it!

49

Uses of the Equilibrium Constant, Kc → 2 HI H 2(g) + I 2(g) ← (g)

Initial

0

Change

0

+ x M + xM

Equilibrium xM

1.00M - 2x M

xM (1.00 - 2 x ) M

2 2 ( [ HI] 1.00 - 2 x ) Kc = = 49 = ( x )( x ) [ H 2 ][ I 2 ] ( 1.00 - 2 x ) K = 7.0 = c

x 7.0 x = 1.00 − 2 x; 9 x = 1.00; x = 0.11 M

[ H 2 ] = [ I 2 ] = xM = 0.11 M [ HI] = (1.00 − 2 x ) M = 0.78 M

50

Disturbing a System at Equlibrium: Predictions LeChatelier’s Principle - If a change of conditions (stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress in reaching a new state of equilibrium.



We first encountered LeChatelier’s Principle in Chapter 14.



Some possible stresses to a system at equilibrium are:



1. 2. 3.

Changes in concentration of reactants or products. Changes in pressure or volume (for gaseous reactions) 51 Changes in temperature.

Disturbing a System at Equlibrium: Predictions 



For convenience we express the amount of PVmay = nRT a gas in terms of its partial pressure rather than n  its concentration. P =  RT  V  we must solve the To derive this relationship, ideal gas equation. n

Because  has the units mol/L, V P = []RT

Thus at constant T, the partial pressure of a gas is directly proportional to its concentration. 52

Disturbing a System at Equlibrium: Predictions Changes in Concentration of Reactants and/or Products

1

• 

Also true for changes in pressure for reactions involving gases. Look at the following system at equilibrium at 450oC.

→ 2 HI H 2 + I2 ← ( g)

[ HI] Kc = = 49 [ H 2 ][ I 2 ] 2

53

Disturbing a System at Equlibrium: Predictions Changes in Concentration of Reactants and/or Products

1

• 

Also true for changes in pressure for reactions involving gases. Look at the following system at equilibrium at 450oC.

→ 2 HI H 2 + I2 ← ( g)

[ HI] Kc = = 49 [ H 2 ][ I 2 ] 2

If some H 2 is added, Q < K c . This favors the forward reaction. Equilbrium will shift to the right or product side.54

Disturbing a System at Equlibrium: Predictions Changes in Concentration of Reactants and/or Products

1

• 

Also true for changes in pressure for reactions involving gases. Look at the following system at equilibrium at 450oC.

→ 2 HI H 2 + I2 ← (g)

[ HI] Kc = = 49 [ H 2 ][ I 2 ] 2

If we remove some H 2 , Q > K c This favors the reverse reaction. Equilbrium will shift to the left, or reactant side. 55

Disturbing a System at Equlibrium: Predictions 2

Changes in Volume • (and pressure for reactions involving gases)  Predict what will happen if the volume of this system at equilibrium is changed by changing the pressure at constant temperature:

→ 2 NO 2( g ) ← N 2 O 4( g )

[ N 2O 4 ] Kc = 2 [ NO 2 ]

56

Disturbing a System at Equlibrium: Predictions →N O 2 NO 2 ( g ) ← 2 4( g ) Kc =

[ N 2O 4 ] [ NO 2 ] 2

If the volume is decreased, which increases the pressure, Q < K c . This favors product formation or the forward reaction. The forward reaction produces fewer moles of gas.

57

Disturbing a System at Equlibrium: Predictions →N O 2 NO 2( g ) ← 2 4( g )

[ N 2O 4 ] Kc = 2 [ NO 2 ]

If the volume is increased, which decreases the pressure, Q > K c . This favors the reactants or the reverse reaction. More moles of gas are produced.

58

Disturbing a System at Equlibrium: Predictions

59

Disturbing a System at Equlibrium: Predictions 3

Changing the Temperature

60

Disturbing a System at Equlibrium: Predictions 3

Changing the Reaction Temperature 

Consider the following reaction at equilibrium:

o → → → 2 SO + O ← 2 SO kJ kJ/mol 2 SO + O ← 2 SO ∆ H =198 −198 2 SO O 2 ( 2g()g← SO 3(3g()grxn + kJ )2(g) ) 23(g) ) +198 2(g) 2 ( g2 () g+

Heat isaaproduct product thisreaction reaction. Heat ofofthis Is heat aisreactant or product in this reaction? Increasing Decreasingthe thereaction reactiontemperature temperaturestresses stressesthe theproducts. reactants. This oror reactant reaction. Thisfavors favorsthe the reverse reactants forward reaction.

61

Disturbing a System at Equlibrium: Predictions 

Introduction of a Catalyst 



Catalysts decrease the activation energy of both the forward and reverse reaction equally.

Catalysts do not affect the position of equilibrium. 



The concentrations of the products and reactants will be the same whether a catalyst is introduced or not. Equilibrium will be established faster with a catalyst.

62

Disturbing a System at Equlibrium: Predictions 

Example 17-9: Given the reaction below at equilibrium in a closed container at 500oC. How would the equilibrium be influenced by the following?

→ 2 NH ∆Hooooo = −92 kJ/mol N 2(g) + 3 H 2(g) ← → 22 NH → 3(g) ∆H rxn = −92 kJ/mol N 2(g) + 3 H ← ← NH3(g) ∆ 92 kJ/mol 3 H ∆HH Hrxn =−−−92 92kJ/mol kJ/mol 3(g) rxn 2(g) 2(g) 3(g) rxn 2(g) 2(g) 3(g)∆ rxn== Factor Effect on reaction procedure Effect Factor Effect on reaction procedure Effect on reaction procedure Effecton onreaction reactionprocedure procedure a. Increasing the reaction temperature ← left a. Increasing ← Increasing the the reaction reaction ttemperature ← left temperature emperature ←left left b. Decreasing the reaction temperature → right b. Decreasing the reaction → reaction ttemperature emperature →right right c. Increasing the pressure by decreasing the volume → right c. Increasing the pressure by decreasing the volume → right d. Increase the concentration of H 2 → right d. Increase the concentration of H 2 → right e. Decrease the concentration of NH 3 → right e. Decrease the concentration of NH 3 → right 63 f. Introducing a platinum catalyst no effect

Disturbing a System at Equlibrium: Predictions 

Example 17-10: How will an increase in pressure (caused by decreasing the volume) affect the equilibrium in each of the following reactions? Reaction Effect on Equilibriu m Effect on Equilibriu mm Effect on Equilibriu mm Reaction Effect on Equilibriu Reaction Effect on Equilibriu → → →222HI a. H nono effect ← HI no effect H222(((ggg))) +++III222((g(gg) ) )← ← HI no effect effect ( g((gg)( g)) ) → →444NO → b. ← 6 H 2H OO ←← left b. 44 NH NH333( (g(gg) ))+++555OO O2(g) ← NO ←left left NO ( g()g( + )( g()g ) 2(g) ← g) )++66H 2(g) 2 2( gO → → c. PCl + Cl →→ right c. PCl3( g ) + Cl2( g ) ← ←PCl PCl5( g ) right 3( g )

2( g )

5( g )

→2H O d. 2 H 2( g ) + O 2( g ) ← 2 ( g)

→ right

64

Disturbing a System at Equlibrium: Predictions 

Example 17-11: How will an increase in temperature affect each of the following reactions? Reaction Effect onon Equilibriu m mm Reaction Effect Equilibriu Effect on Equilibriu o oo → → → a. 2 NO ← N O ∆ H 0<00 ←← leftleft a. 2 NO2(g) ← ←N N2 O O4(g) ∆∆H H ← left rxn < < 2(g) 2(g)

22

4(g) 4(g)

rxn rxn

→ → 22 HCl b. H 2( g ) + Cl 22((gg)) ← ← ← HCl( (gg) ) ++92 92kJ kJ ←left left c. H 2( g ) + I 2( g ) → 2 HI ( g ) ∆H = +25 kJ → right

65

The Haber Process: A Practical Application of Equilibrium  The Haber process is used for the

commercial production of ammonia. 



This is an enormous industrial process in the US and many other countries. Ammonia is the starting material for fertilizer production.

 Look at Example 17-9.

What conditions did we predict would be most favorable for the production of ammonia? 66

The Haber Process: A Practical Application of Equilibrium Fe & metal oxides N 2 ( g ) + 3 H 2( g ) ←  → 2 NH 3( g ) ∆H o = −92.22 kJ

N 2( g ) is obtained from liquid air.

H 2( g ) obtained from coal gas.

This reaction is run at a T = 450 o C and P of N 2 = 200 to 1000 atm. ∆G < 0 which is favorable.

∆H < 0 also favorable.

∆S < 0 which is unfavorable.

However the reaction kinetics are very slow at low temperatures. Haber' s solution to this dilemma.

67

The Haber Process: A Practical Application of Equilibrium Haber' s solution to this dilemma. (1) Increase T to increase rate, but yield is decreased. ( 2) Increase reaction pressure to ⇒ right. ( 3) Use excess N 2 to ⇒ right.

( 4) Remove NH 3 periodically to ⇒ right.

The reaction system never reaches equilibrium because NH 3 is removed. This increases the reaction yield and helps with the kinetics!

68

The Haber Process: A Practical Application of Equilibrium  This diagram

illustrates the commercial system devised for the Haber process.

69

Disturbing a System at Equilibrium: Calculations 

To help with the calculations, we must determine the direction that the equilibrium will shift by comparing Q with Kc.



Example 17-12: An equilibrium mixture from the following reaction was found to contain 0.20 mol/L of A, 0.30 mol/L of B, and 0.30 mol/L of C. What is the value of Kc for this reaction?

A( g)

→ ← B( g ) + C ( g )

70

Disturbing a System at Equilibrium: Calculations A( g) → ← B( g ) + C ( g ) Equil. []' s 0.20 M 0.30 M 0.30 M [ B][ C] ( 0.30 )( 0.30 ) Kc = = = 0.45 ( 0.20) [ A]

71

Disturbing a System at Equilibrium: Calculations 

1

If the volume of the reaction vessel were suddenly doubled while the temperature remained constant, what would be the new equilibrium concentrations? Calculate Q, after the volume has been doubled

A( g)

→ ← B( g ) + C ( g )

Equil. []' s 0.10 M 0.15 M 0.15 M

[ B][ C] ( 0.15)( 0.15) Q= = = 0.22 ( 0.10) [ A]

72

Disturbing a System at Equilibrium: Calculations  2

Since Q
A( g) New initial []' s 0.10 M

→ ←

B( g ) + C ( g ) 0.15 M

0.15 M

73

Disturbing a System at Equilibrium: Calculations  2

Since Q
A( g)

→ ←

B( g ) + C ( g )

New initial []' s 0.10 M

0.15 M

0.15 M

Change

+xM

+xM

-x M

74

Disturbing a System at Equilibrium: Calculations  2

Since Q
( g)

( g)

New initial []' s 0.10 M

0.15M

0.15M

Change

+ xM

+ xM

- xM

New Equil. []' s ( 0.10 - x ) M

( 0.15 + x ) M ( 0.15 + x ) M 75

Disturbing a System at Equilibrium: Calculations  2

Since Q
New initial []' s 0.10 M Change - xM

0.15M + xM

New Equil. []' s ( 0.10 - x ) M ( 0.15 + x ) M ( [ B][ C] 0.15 + x )( 0.15 + x ) Kc = = 0.45 = ( 0.10 − x ) [ A]

0.15M + xM

( 0.15 + x ) M 76

Disturbing a System at Equilibrium: Calculations

Solve this quadratic equation 0.045 - 0.45 x = 0.0225 + 0.30 x + x

2

x + 0.75 x − 0.0225 = 0 2

77

Disturbing a System at Equilibrium: Calculations - b ± b − 4ac x= 2a 2

x=

− 0.75 ±

( 0.75)

− 4(1)( − 0.0225) 2(1) 2

− 0.75 ± 0.81 x= = −0.78 and 0.03 M 2 78

Disturbing a System at Equilibrium: Calculations Since 0 < x < 0.10, we can discard - 0.78 as an answer. The only posible value is x = 0.03 M.

[ A] = (0.10 − x) M = 0.07 M [ B] = [ C] = ( 0.15 + x ) M = 0.18 M

These are the new concentrations after the equilibrium has been disturbed. 79

Disturbing a System at Equilibrium: Calculations  Example 17-13: Refer to example 17-12.

If the initial volume of the reaction vessel were halved, while the temperature remains constant, what will the new equilibrium concentrations be? Recall that the original concentrations were: [A]=0.20 M, [B]=0.30 M, and [C]=0.30 M. You do it! 80

Disturbing a System at Equilibrium: Calculations A( g) → ← B( g ) + C ( g ) Instantaneous []' s 0.40 M

0.60 M 0.60 M

(1) Calculate Q, after the volume is halved [ B][ C] ( 0.60 )( 0.60 ) Q= = = 0.90 ( 0.40) [ A] Q > K c thus the equilibrium shifts to the left or reactant side. (2) Set up the algebraic expressions to determine the equilibrium concentrations. 81

Disturbing a System at Equilibrium: Calculations → A( g) ← New initial []' s 0.40 M Change +xM

B( g ) 0.60M -xM

+

C( g ) 0.60M -xM

New Equil.

(0.40 + x) M (0.60 - x) M (0.60 - x) M [ B][ C] (0.60 - x)(0.60 - x) Kc = = 0.45 = [ A] (0.40 + x) After the algebra is completed, this equation reduces to x 2 − 1.65 x + 0.18 = 0 82

Disturbing a System at Equilibrium: Calculations Solve the quadratic equation for this expression. 1.65 ± (−1.65) 2 − 4(1)(0.18) x= 2(1) 1.65 ± 1.42 x= = 1.5 and 0.12 2 Because the limits are 0 < x < 0.60 we can discard 1.5 as an answer. Thus 0.12 is the only possible answer.

[ A] = (0.40 + x) M = 0.52 M [ B] = [ C ] = (0.60 − x) M = 0.48 M

83

Disturbing a System at Equilibrium: Calculations 

Example 17-14: A 2.00 liter vessel in which the following system is in equilibrium contains 1.20 moles of COCl2, 0.60 moles of CO and 0.20 mole of Cl2. Calculate the equilibrium constant.

→ COCl CO ( g ) + Cl 2( g ) ← 2( g ) You do it!

84

Disturbing a System at Equilibrium: Calculations → COCl CO ( g ) + Cl 2( g ) ← 2( g ) Equil. []' s 0.30 M 0.10 M 0.60 M ( [ COCl 2 ] 0.60 ) Kc = = = 20 [ CO][ Cl 2 ] ( 0.30)( 0.10)

85

Disturbing a System at Equilibrium: Calculations  An additional 0.80 mole of Cl2

is added to the vessel at the same temperature. Calculate the molar concentrations of CO, Cl2, and COCl2 when the new equilibrium is established. You do it!

86

Disturbing a System at Equilibrium: Calculations → CO ( g ) + Cl 2 ( g ) ← Orig. Equil. 0.30 M (Stress) Add

0.60 M

+ 0.40 M

New Initial 0.30 M Change

0.10 M

COCl 2 ( g )

-xM

0.60 M Q < K c ∴ shift right

0.50M -xM

New Equil. (0.30 - x) M (0.50 - x) M

+xM (0.60 + x) M

( [ COCl 2 ] 0.60 + x ) Kc = = 20 = ( 0.30 − x )( 0.50 − x ) [ CO][ Cl 2 ]

87

Disturbing a System at Equilibrium: Calculations 2

equation reduces to 20 x − 17 x + 2.4 = 0

17 ± (−17) − 4(20)(2.4) X= = 0.67 & 0.18 2(20) limits are 0 < x < 0.30 thus we can discard 0.67 2

[ CO] = (0.30 − x) M = 0.12M [ Cl 2 ] = (0.50 − x) M = 0.32M [ COCl 2 ] = (0.60 + x) M = 0.78M

88

Partial Pressures and the Equilibrium Constant 

 

For gas phase reactions the equilibrium constants can be expressed in partial pressures rather than concentrations. For gases, the pressure is proportional to the concentration. We can see this by looking at the ideal gas law.    

PV = nRT P = nRT/V n/V = M P= MRT and M = P/RT 89

Partial Pressures and the Equilibrium Constant  Consider this system

at equilibrium at

5000C.

→ 4 HCl + O 2 Cl 2( g ) + 2 H 2 O ( g ) ← ( g) 2( g )

( PHCl ) ( PO ) [ HCl] [ O 2 ] Kc = and K p = 2 2 2 2 [ Cl 2 ] [ H 2O] ( PCl ) ( PH O ) 4

4

2

2

2

90

Partial Pressures and the Equilibrium Constant

Kc

( )( )

( P ) ( ) = = × ( ) ( ) (P ) (P ) ( ) ( PHCl )

PHCl 4 PO2 RT RT 2 P 2 P Cl2

RT

4

O2

2

H 2O

Cl 2

RT

1 5 RT 1 4 RT

2

H 2O

1 ) so for this reaction K c = K p ( RT -1

1

K c = K p (RT) or K p = K c (RT) L atm Must use R = 0.0821 mol K

91

Relationship Between Kp and Kc  From the previous slide we can see that

the relationship between Kp and Kc is: K p = K c ( RT )

∆n

or K c = K p ( RT )

− ∆n

∆n = (# of moles of gaseous products) - (# of moles of gaseous reactants)

92

Relationship Between Kp and Kc  Example 17-15: Nitrosyl bromide, NOBr, is

34% dissociated by the following reaction at 25oC, in a vessel in which the total pressure is 0.25 atmosphere. What is the value of Kp?

2 NOBr( g )

→ ← 2 NO ( g ) + Br2( g ) 93

Relationship Between Kp and Kc 2 NOBr( g ) → ← 2 NO ( g ) + Br2( g ) Initial [] Change

x atm 0 0 - 0.34 x atm + 0.34 x atm + 0.17 x atm

Equilibrium ( x - 0.34 x ) atm 0.34 xatm

0.17 x atm

94

Relationship Between Kp and Kc PTot = PNOBr + PNO + PBr2

0.25 atm = ( x - 0.34 x ) atm + 0.34 x atm + 0.17 x atm 0.25 atm = 1.17 x atm, thus x = 0.21 atm. Because NOBr is 34% dissociated, it is 66% undissociated.

95

Relationship Between Kp and Kc PNOBr = ( x - 0.34 x ) = 0.66 x

PNOBr = ( 0.66 )( 0.21 atm ) = 0.14 atm

PNO = 0.34 x = ( 0.34 )( 0.21 atm ) = 0.071 atm

PBr2 = 0.17 x = ( 0.17 )( 0.21 atm ) = 0.036 atm

( PNO ) ( PBr Kp = 2 ( PNOBr ) 2

2

) = ( 0.071) ( 0.036) = 9.3 ×10 2

( 0.14)

−3

2

96

Relationship Between Kp and Kc  The numerical value of Kc

for this reaction can be determined from the relationship of Kp and Kc.

K p = Kc ( RT) Kc = 9.3 × 10

∆n

−3

or Kc = K p ( RT)

[ ( 0.0821)( 298) ]

−1

− ∆n

∆n = 1

= 38 . × 10

−4

97

Relationship Between Kp and Kc 

Example 17-16: Kc is 49 for the following reaction at 450oC. If 1.0 mole of H2 and 1.0 mole of I2 are allowed to reach equilibrium in a 3.0-liter vessel,

H 2( g )

→ + I 2( g ) ← 2 HI ( g )

(a) How many moles of I2 remain unreacted at equilibrium? You do it! 98

Relationship Between Kp and Kc H 2( g ) Initial Change

0.33M -xM

+

→ 2 HI I 2( g ) ← (g)

0.33M -xM

Equilibrium ( 0.33 - x ) M ( 0.33 - x ) M

0 + 2x M 2x M

[ ( HI] 2x) 2x Kc = = 49 = ∴ 7. 0 = 2 [ H 2 ][ I 2 ] 0.33 - x ( 0.33 - x ) 2

2

9 x = 2.3; x = 0.256 M [H 2 ] = [I 2 ] = (0.33 − x) M = 0.074 M

[ HI] = 2 x M = 0.51M

? mol I 2 = 3.0 L × 0.074 mol L = 0.21 mol

99

Relationship Between Kp and Kc (b) What are the equilibrium partial pressures of H2, I2 and HI? You do it!

100

Relationship Between Kp and Kc L atm ) PH 2 = PI 2 = MRT = ( 0.074 mol L ) ( 0.0821 mol K ( 723 K ) = 4.4 atm L atm ) PHI = MRT = ( 0.51 mol L ) ( 0.0821 mol K ( 723 K ) = 30 atm

101

Relationship Between Kp and Kc (c) What is the total pressure in the reaction vessel? You do it!

102

Relationship Between Kp and Kc PTot = PH 2 + PI2 + PHI = ( 4.4 + 4.4 + 30) atm = 39 atm

103

Heterogeneous Equlibria 

Heterogeneous equilibria have more than one phase present. 

For example, a gas and a solid or a liquid and a gas.

CaCO 3( s ) 

→ ← CaO ( s ) + CO 2( g )

o

at 500 C

How does the equilibrium constant differ for heterogeneous equilibria?  



Pure solids and liquids have activities of unity. Solvents in very dilute solutions have activities that are essentially unity. The Kc and Kp for the reaction shown above are:

K c = [CO 2 ]

K p = PCO 2 104

Heterogeneous Equlibria SO 2 ( aq )

For this reaction : → + H 2 O ( ) ← H 2SO 3( aq )

o

(at 25 C)

H 2 O ( ) is the solvent. What are the forms of K c and K p ? You do it!

[ H 2SO 3 ] Kc = [SO 2 ]

K p is undefined 105

Heterogeneous Equlibria  What are Kc

CaF2( s )

and Kp for this reaction?

2+ 1→ ← Ca ( aq ) + 2 F( aq )

o

(at 25 C)

You do it!

[

Kc = Ca

2+

][ F ]

− 2

K p is undefined 106

Heterogeneous Equlibria  What are Kc

and Kp for this reaction? → Fe O + 4 H o 3 Fe( s ) + 4 H 2 O ( g ) ← ( at 500 C) 3 4( s ) 2( g )

Kc =

[ H2 ]

4

[ H 2 O]

P ) ( = (P )

4

4

Kp

H2

4

H 2O

107

Relationship Between ∆Gorxn and the Equilibrium Constant 

∆Gorxn is the standard free energy change. 



∆G is the free energy change at nonstandard conditions •



∆Gorxn is defined for the complete conversion of all reactants to all products. For example, concentrations other than 1 M or pressures other than 1 atm.

∆G is related to ∆Go byo the following relationship.

∆G = ∆G + RT lnQ or

∆G = ∆G o + 2.303 RT log Q R = universal gas constant T = absolute temperature Q = reaction quotient

108

Relationship Between ∆Gorxn and the Equilibrium Constant  At equilibrium, ∆G=0 and Q=Kc.  Then we can derive this relationship:

0 = ∆G + RT ln K or 0

0 = ∆G + 2.303 RT log K which rearranges to : 0

∆G = - RT ln K or 0

∆G = - 2.303 RT log K 0

109

Relationship Between ∆Gorxn and the Equilibrium Constant  For the following generalized reaction, the

thermodynamic equilibrium constant is defined as follows:

→ aA + bB ← cC + dD

( aC ) ( a D ) K= a b ( aA ) ( aB ) c

d

where

aA is the activity of A aB is the activity of B aC is the activity of C aD is the activity of D 110

Relationship Between ∆Gorxn and the Equilibrium Constant 

The relationships among ∆Gorxn, K, and the spontaneity of a reaction are: ∆Gorxn <0

K >1

Spontaneity at unit concentration Forward reaction spontaneous

=0

=1

System at equilibrium

>0

<1

Reverse reaction spontaneous 111

Relationship Between ∆Gorxn and the Equilibrium Constant

112

Relationship Between ∆Gorxn and the Equilibrium Constant 

Example 17-17: Calculate the equilibrium constant, Kp, for the following reaction at 25oC from thermodynamic data in Appendix K.

N 2O 4( g ) 

→ ← 2 NO 2( g )

Note: this is a gas phase reaction.

113

Relationship Between ∆Gorxn and the Equilibrium Constant N 2O 4( g )

→ ← 2 NO 2( g )

1. Calculate ∆G ∆G ∆G

o rxn

o rxn

= 2∆G

o rxn

= 2( 51.30 kJ ) − ( 97.82 kJ ) ∆G

∆G

o rxn

o rxn

o f NO 2 ( g )

− ∆G

o f N 2O 4 ( g )

= 4.78 kJ mol rxn

= 4.78 ×10

3 J

mol rxn

This reaction is nonspontaneous.

114

Relationship Between ∆Gorxn and the Equilibrium Constant 2. Calculate K from ∆G orxn = − RT ln K p ∆G 4.78 × 10 mol ln K p = = = −1.93 − RT - ( 8.314 J mol K )( 298 K ) o rxn

3 J

( P ) = 0.145 = (P ) 2

Kp = e

−1.93

NO 2

N 2O 4

A very common mistake is to not convert the ∆G

o rxn

from kJ to J! 115

Relationship Between ∆Gorxn and the Equilibrium Constant  Kp

for the reverse reaction at 25oC can be calculated easily, it is the reciprocal of the above reaction. → 2 NO 2(g) ← N 2 O 4(g) ∆G orxn = −4.78 kJ/mol

( PN 2O4 ) 1 1 K = = = 6.90 = 2 K p 0.145 ( PNO2 ) ' p

116

Relationship Between ∆Gorxn and the Equilibrium Constant 

Example 17-18: At 25oC and 1.00 atmosphere, Kp = 4.3 x 10-13 for the decomposition of NO2. Calculate ∆Gorxn at 25oC.

2 NO 2( g )

→ ← 2 NO ( g ) + O 2( g )

You do it.

117

Relationship Between ∆Gorxn and the Equilibrium Constant ∆G

o rxn

= − RT ln K p

∆G

o rxn

= −(8.314

∆G

o rxn

= −(2480 J mol)(−28.47)

∆G

o rxn

= 7.06 ×10

J

mol K

4 J

)(298 K ) ln 4.3 ×10

-13

kJ = 70 . 6 mol rxn mol rxn

118

Relationship Between ∆Gorxn and the Equilibrium Constant  The relationship for K at conditions other

than thermodynamic standard state conditions is derived from this equation. ∆G = ∆G o + RT lnQ or ∆G = ∆G + 2.303 RT log Q o

119

Evaluation of Equilibrium Constants at Different Temperatures  From the value of ∆Ho

and K at one temperature, T1, we can use the van’t Hoff equation to estimate the value of K at another temperature, T2. K T2 ∆H o (T2 − T1 ) ln = K T1 R T2 T1

or ln

K T2 K T1

∆H o = R

1 1  −   T1 T2 

120

Evaluation of Equilibrium Constants at Different Temperatures  Example 17-19: For the reaction in

example 17-18, ∆Ho = 114 kJ/mol and Kp = 4.3 x 10-13 at 25oC. Estimate Kp at 250oC. 2 NO2(g) ↔ 2 NO(g) + O2(g)

121

Evaluation of Equilibrium Constants at Different Temperatures

LetLet T1 T= 298 andTT 523 298K K and == 523 K K 1 = 2 2 apply the Hoffequation equation apply thevan' van'tt Hoff 5 J K T2 (1.14 ×10 mol)( 523 − 298) ln = K T1 ( 8.314 J mol K )( 523 K )( 298 K ) ln

K T2 K T1

= 19.795 122

Evaluation of Equilibrium Constants at Different Temperatures Take the antilog of both sides of equation. K T2 = e19.80 = 4.0 ×108 K T1

Solve for K T2 & substitute the known value of K T1

(

)(

K T2 = 4.0 ×10 K T1 = 4.0 ×10 4.3 ×10 8

8

-13

)

K T2 = 1.7 ×10 −4 @ 250o C vs K T1 = 4.3 ×10 -13 @ 25o C The reaction is more product favored at the higher T. 123

Synthesis Question  Mars is a reddish colored planet because it has

numerous iron oxides in its soil. Mars also has a very thin atmosphere, although it is believed that quite some time ago its atmosphere was considerably thicker. The thin atmosphere does not retain heat well, thus at night on Mars the surface temperatures are 145 K and in the daytime the temperature rises to 300 K. Does Mars get redder in the daytime or at night? 124

Synthesis Question  The formation of iron

oxides from iron and oxygen is an exothermic process. Thus the equilibrium that is established on Mars shifts to the iron oxide (red) side when the planet is cooler - at night. Mars gets redder at night by a small amount.

125

Group Question  If you are having trouble getting a fire

started in the barbecue grill, a common response is to blow on the coals until the fire begins to burn better. However, this has the side effect of dizziness. This is because you have disturbed an equilibrium in your body. What equilibrium have you affected? 126

End of Chapter 17  This chapter is the key to the

understanding of Chapters 18, 19, & 20.  Make sure you understand this chapter’s concepts!

127