Chapter 14Chemical Equilibrium

.

The Equilibrium State

Chemical equilibrium is the state reached when the concentrations of the products and reactants remain constant over time. The mixture of reactants and products in the equilibrium state is the equilibrium mixture.

2

Until now, we have shown chemical reactions go to completion with a single-headed arrow ().

For many reactions, this assumption is not true. Consider

CO (g) + 3 H2 (g) CH4 (g) + H2O (g)

CO (g) + 3 H2 (g) CH4 (g) + H2O (g)

This reaction “does not go to completion”… OR“ The reaction occurs in both directions at the

same time.”

3

.

The terms reactants and products are arbitrary. We must always refer to a balanced equation, as it is written, to be completely understood.

N2O4 (g) 2 NO2 (g)reactant product

2 NO2 (g) N2O4 (g)reactant product

4

FigureAs time progresses, the

concentrations of the “reactants” decreases while the concentrations of the “products” increases. This is what we saw in the Chapter on Reaction Rates

•

5

Figure At some point in time

the concentrations of both the reactant and the product stop changing.

We say this system is in equilibrium.

•

6

CO (g) + 3 H2 (g) CH4 (g) + H2O (g)

The reaction occurs in both directions, in the same system, at the same time.

It is reversible.

When we say CO (g) + 3 H2 (g) CH4 (g) + H2O (g)we really mean,CO (g) + 3 H2 (g) CH4 (g) + H2O (g)and CO (g) + 3 H2 (g) CH4 (g) + H2O (g)

7

8

Each reaction occurs at its own rate as defined by its own rate law.

One of the two reactions probably starts out with a faster initial rate. It is dominant.

We expect to lose the reactants and gain the products that we see in the in the dominant reaction. (Kinetics)

9

The opposite reaction in the system will be dominated.

Since the reactants of the dominated reaction are the same as the products of the dominant reaction, their concentration increases.

Since the products of the dominated reaction are the same as the reactants of the dominant reaction, their concentration decreases.

We’ve seen that the rate of a reaction depends on a rate law that often includes concentrations of reactants.

As the concentration of the reactants decreases, we expect the rate to decrease (the reaction slows down).

This happens to the dominant reaction!

10

We saw in Kinetics that the rate of a reaction depends on a rate law that often includes concentrations of reactants.

As the concentration of the reactants increases, we expect the rate to increase (the reaction speeds up).

This happens to the dominated reaction!

11

CO (g) + 3 H2 (g) CH4 (g) + H2O (g)

In this experiment the forward reaction is dominant and slows down while the reverse reaction is dominated and speeds up.

•

12

13

If the dominant reaction always slows down and the dominated reaction always speeds up, then at some point in time, they must have the same rate.

What happens then?

CO (g) + 3 H2 (g) CH4 (g) + H2O (g)

This is equilibrium. The concentrations of all chemicals remain constant because they are made at the same rate they are consumed.

•

14

At equilibrium the rates of the forward and reverse reactions are the same, BUT are not equal to zero.

•

15

.

The system is in a dynamic state, so while no visible change is

occurring,any individual molecule may still be undergoing

change (Reactions are still

happening.).

16

a A + b B c C + d DThe concentrations of an equilibrium mixture are

related to each other through the equilibrium equation.

This equilibrium constant Kc is for the specified reaction with a given balanced equation, and at a given temperature.

17

The Equilibrium Constant Kc

18

The equilibrium concentrations will not always be the same because we start with different numbers C, O, and H atoms in each of the three experiments.

However the value of Kc is always the same!

{All experiments occur at the

same temperature!

N2O4 (g) 2 NO2 (g)

If we change the temperature that a reaction occurs at, the value of the equilibrium constant will change. For example,

Kc = 1.53 at 127 C.

Notice we generally choose not to show the units for equilibrium constants.

19

C25at 10 x 4.64

ON

NOK 3

42

22

c

Reversing an equation

• a A + b B c C + d D

REVERSE the balanced equation

c C + d D a A + b B

Kc’ DOES NOT EQUAL Kc 20

ba

dc

cBA

DCK

c

ba

dcdc

ba'c K

1

BA

DC1

DC

BAK

Multiplying an equation

• a A + b B c C + d D

MULTIPLY the balanced equation by 2

2a A + 2b B 2c C + 2d D

Kc” DOES NOT EQUAL Kc21

ba

dc

cBA

DCK

2c

2

ba

dc

2b2a

2d2c"c K

BA

DC

BA

DCK

When we add up two or more reactions, the equilibrium constant for the resultant reaction

EQUALS the product of the equilibrium constants of the

original reactions, and in general

Knew = K1 x K2 x K3 x …

22

Consider

a A + b B c C + d D PLUSa A + b B c C + d DGIVES2a A + 2b B 2c C + 2d D

ba

dc

cBA

DCK

2b2a

2d2c"c

BA

DCK

23

ba

dc

cBA

DCK

2ccc

"c KKKK

Kinetics and equilibrium

Consider the reaction N2O4 (g) 2 NO2 (g)

42

22

c ON

NOK

24

Kinetics and equilibrium

For the forward elementary reaction

N2O4 (g) 2 NO2 (g)

42forward ONkrate 25

Kinetics and equilibrium

For the reverse elementary reaction

N2O4 (g) 2 NO2 (g)

22reverse NOkrate

26

Kinetics and equilibrium

At equilibrium, these elementary reaction rates are EQUAL

22reverse42forward NOkONk

!c

reverse

forward

42

22 K

k

k

ON

NO

27

ProblemThe following equilibrium concentrations were

measured at 800 K for the reaction:

2 SO2 (g) + O2 (g) 2 SO3 (g)

[SO2] = 3.0 x 10-3 molL-1

[O2] = 3.5 x 10-3 molL-1

[SO3] = 5.0 x 10-2 molL-1

Calculate the equilibrium constant at 800 K.

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The Equilibrium Constant Kp

Gas phase equilibrium constants are often expressed in terms of partial pressures (easy to measure).

Recall, for a gas APAV = nART

PA = (nART) / VPA = (nA/V) RT

What is n / V? It is moles over volume, which is concentration. So nA/V is [A] and

PA = [A] RT

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N2O4 (g) 2 NO2 (g)

We can express an equilibrium constant in terms of partial pressures because they are related to concentrations!

Again, the equilibrium constant is unitless. 30

42

2

ON

2NO

p P

PK

Kc and Kp are related

a A + b B c C + d D (all are GASES!)

31

ba

dc

cba

dc

ba

dc

ba

dc

bB

aA

dD

cC

p

RTRT

RTRTK

RTRT

RTRT

BA

DC

[B]RT[A]RT

[D]RT[C]RT

PP

PPK

b

Ba

A

dD

cC

pba

dc

cPP

PPKor

BA

DCK

Kc and Kp are related

b)(ad)(cΔn

here w(RT) K

(RT) KK

gas

Δnc

b)(ad)(ccp

gas

32

ba

dc

cpRTRT

RTRTKK

We should expect that some of the RT terms might cancel out.

Kc and Kp are related

b)(ad)(cΔn

here w(RT) K

(RT) KK

gas

Δn

c

b)(ad)(ccp

gas

33

Use R = 0.08206 (Latm)(Kmol)-1 because it relates molarity to pressure at a given

temperature.

N2O4 (g) 2 NO2 (g)

ngas = (2-1) = 1 so

At 25 C Kc = 4.64 x 10-3 34

(RT)K(RT)KK c1)(2

cp

N2O4 (g) 2 NO2 (g)

At 25 C Kc = 4.64 x 10-3

Kp = Kc(RT)

Kp = (4.64 x 10-3)(0.08206)(298)note the lack of units

and T is expressed in Kelvin!

Kp = 0.113 at 25 C

NOTE - If ngas = zero then Kc = Kp35

ProblemThe chemical equation for the so-called water-gas shift reaction is

CO (g) + H2O (g) CO2 (g) + H2 (g)

What is the value of Kp at 700 K if the partial pressures in an equilibrium

mixture at 700 K are

1.31 atm of CO

10.0 atm of water

6.12 atm of carbon dioxide, and

20.3 atm of hydrogen gas?

36

ProblemNitric oxide reacts with oxygen to give nitrogen

dioxide:

2 NO (g) + O2 (g) 2 NO2 (g)

If Kc = 6.9 x 105 at 227 C, what is the value of Kp at this temperature?

If Kp = 1.3 x 10-2 at 1000 K, what is the value of Kc at this temperature?

37

Heterogeneous Equilibria

Homogeneous equilibria occur in systems where all compounds in the equilibrium mixture are in the same state.

Heterogeneous equilibria occur in systems where some of the chemicals of the equilibrium mixture are in different states.

38

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

Since one of the products is a gas, while the other two compounds are solids, this is a heterogeneous

equilibrium.Now, if we were to express the equilibrium constant for

this reaction, we would probably say

What is the concentration of a solid?

39

3

2c CaCO

COCaO"K"

What is the concentration of a solid or liquid?

Concentration is moles per unit volume. Also, density is mass divided by volume, and molar mass is mass per number of moles. So, for a pure substance

(mass / volume) / (mass / moles) = moles / volume

density / molar mass = (concentration)

40

What is the concentration of a solid or liquid?

density / molar mass = (concentration)

The density and molar mass of a solid or liquid doesn’t change. Therefore the concentration doesn’t change either!

41

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

We choose not to include the concentrations of

solids and liquids in the calculation of Kc

because they don’t change even when the

system IS NOT at equilibrium!•

or Kp = (PCO2) 42

2c COK

ProblemFor each of the following reactions, write the

equilibrium constant expression for Kc.

a) 2 Fe (s) + 3 H2O (g) Fe2O3 (s) + 3 H2(g)

b) 2 H2O (l) 2 H2 (g) + O2 (g)

c) SiCl4 (g) + 2 H2 (g) Si (s) + 4 HCl (g)

d) Hg22+ (aq) + 2 Cl- (aq) Hg2Cl (s)

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Using the Equilibrium Constant

Judging the extent of a reaction: The magnitude of the constant Kc (or Kp) gives an idea of the extent to which reactants are converted to products.

We can make general statements about the “completeness” of a given equilibrium reaction based on the value of the constant.

44

K much larger than 103 means the written reaction proceeds nearly to completion (all

products, no reactants).

2 H2 (g) + O2 (g) 2 H2O (g) •

45

K 500at 10 x 2.4

OH

OHK 47

22

2

22

c

K much smaller than 10-3 means the written reaction barely proceeds at all (all reactants

and no products).

2 H2O (g) 2 H2 (g) + O2 (g)•

K 500at 10 x 4.1OH

OH'K 48

22

22

2c

46

K is between 10-3 and 103, means the written reaction achieves an equilibrium mixture with relatively significant amounts of the reactants

and products.

H2 (g) + I2 (g) 2 HI (g)•

47

K 700at 57.0

IH

HIK

22

2

c

Predicting the direction of a reaction

If we know a system IS NOT at equilibrium, what happens if we put the concentrations of the system into the equilibrium constant

expression?

You would get a value that does not equal the value of the equilibrium constant.

48

We define the reaction quotient Qc (or Qp) in exactly the same way we define the equilibrium constant Kc (or Kp).

When the system is not at equilibrium, then

Qc Kc49

ba

dc

cBA

DCQ

If Qc > Kc the reaction needs to create more reactants (and use up products) to get to equilibrium, so the reaction will be going

from right to left.

If Qc < Kc the reaction needs to create more products (and use up reactants) to get to equilibrium, so the reaction will be going

from left to right.

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Figure

51

H2 (g) + I2 (g) 2 HI (g)

K 700at 57.0

IH

HIK

22

2

c

52

If [H2]t = 0.80 mol/L,

[I2]t = 0.25 mol/L, and

[HI]t = 10.0 mol/L

H2 (g) + I2 (g) 2 HI (g)

53

Qc Kc, so the system is not at equilibrium.

Qc > Kc, the reaction will proceed from

right to left.

500

5)(0.80)(0.2

(10.0)

IH

HIQ

2

22

2

c

ProblemThe equilibrium constant Kc for the reaction

2 NO (g) + O2 (g) 2 NO2 (g)

is 6.9 x 105 at 500 K. A reaction vessel at this temperature was filled with 0.012 M of NO, 0.20 M of O2, and 0.16 M of NO2.

a) Is the reaction mixture at equilibrium? If not, which direction does the reaction proceed?

b) What is the direction of the reaction if the initial amounts are 1.0 x 10-3 M of NO, 0.04 M of O2, and 0.80 M of NO2?

54

Calculating equilibrium concentrations

If we know an equilibrium constant (which must be referred to a balanced equation!) and all but one of the equilibrium concentrations of the species involved, we can find the unknown concentration.

55

2 NO (g) + O2 (g) 2 NO2 (g) Kc = 6.9 x 105 at 500 K

The system is at equilibrium and we know [O2] = 1.0 mol/L and [NO2] = 0.80 mol/L

so we can calculate [NO].

56

L

mol9.6x10

)(1.0)(6.9x10

(0.64)[NO]

)(1.0)(6.9x10

(0.80)

OK

NO[NO] so

ONO

NOK

45

5

2

2c

22

22

22

c

2 NO (g) + O2 (g) 2 NO2 (g) Kc = 6.9 x 105 at 500 K

Since concentrations are always positive, we can throw out the negative answer.

The [NO] in the equilibrium mixture is 9.6 x 10-4 mol/L. •

Let’s check this answer

57

5724

2

22

22

c 10 x 6.9)(1.0)(9.3x10

(0.64)

(1.0))(9.6x10

(0.80)

ONO

NOK

ProblemThe reactionCO (g) + H2O (g) CO2 (g) + H2 (g)

has an equilibrium constant Kc = 4.24 at 800 K.

Calculate the equilibrium concentrations of all species at 800 K if only CO and H2O are present initially at concentrations of 0.150 mol/L.

58

ICE tables (also applies for Kp and pressures)

Using a balanced equation, we create a table of the Initial concentrations,

and the Change in concentrations, (an unknown amount of change related by the

stoichiometry of the balanced equation), and Equilibrium concentrations (sum of initial

concentration and change in concentration).We can substitute our Equilibrium concentrations into

our equilibrium constant expression.

59

Problem

Taking the square root of both sidesx)x)(0.150(0.150

(x)(x)4.24 soK 800at

O][CO][H

]][H[CO4.24K

2

22c

60

1-1-

9999

9999

99

Lmol 0.292 or x Lmol 0.101 x

x.051- 0.308-or x 3.050.308

x x 2.050.308-or x x 2.050.308

x x0.1502.05 so x0.150

x2.05

x0.150

x

x)x)(0.150(0.150

(x)(x)4.24

Problem

61

If we put both values of x back into all our

Equilibrium concentration expressions, we’ll see

one value of x will give at least one negative

equilibrium concentration.

This isn’t physically possible!

Throw that value of x out and use the other.

ProblemWe can check our results by inserting these equilibrium

concentrations into the equilibrium equation.

Our result is (within rounding error) the equilibrium constant we were given, so our answers for the equilibrium concentrations are correct.

62

4.250.002401

0.01020

049)(0.049)(0.

101)(0.101)(0.

O][CO][H

]][H[COK

2

22c

Problem

The equilibrium constant Kp is 2.44 at 1000 K for the reaction

C(s) + H2O (g) CO (g) + H2 (g)

• What are the equilibrium partial pressures of

H2O, CO, and H2 if the initial partial pressures are PH2O = 1.20 atm, PCO = 1.00 atm, and PH2

= 1.40 atm?

63

Problem

Since this question starts with both reactants and products in the initial mixture, it makes sense to first check the reaction quotient to see in which direction the reaction will proceed.

64

1.17 1.20

1.401.00

P

PP Q

OH

HCOp

2

2

Reaction occurs left to right!

65

(all in atm) C(s) + H2O (g) CO (g) + H2 (g) Initial press. N/A 1.20 1.00 1.40 Press. change N/A -x +x +x Equil. press. N/A 1.20 – x 1.00 + x 1.40 + x

x)(1.20

x)x)(1.40(1.002.44so

1000KatP

PP2.44K

OH

HCOp

2

2

Problem

Rearranging, we get

This is a quadratic equation of the form ax2 + bx + c = 0

which has solutions given by 66

0x4.84x1.528

0]x2.40x[1.402.44x][2.928

0x)x)(1.40(1.00x)2.44(1.20

2

2

Problem

atm 0.30xor atm 5.14xso2.00

.595xor

2.00

10.27x

2.00

29.5384.84xor

2.00

29.5384.84x

67

1.00)2(

8)1.00)(1.524(4.84)(4.84)(xor

1.00)2(

8)1.00)(1.524(4.84)(4.84)(x

1.00)2(

8)1.00)(1.524(4.84)(4.84)(xso

2a

4acbbx

22

22

Problem

atm 0.30xor atm 5.14x

68

Our equilibrium partial pressures must all be positive.

For x = -5.14 atm at equilibrium

PH2O = (1.20 atm – [-5.14 atm]) = 6.34 atm,

PCO = (1.00 atm + [-5.14 atm]) = -4.14 atm, PH2

= (1.40 atm + [-5.14 atm]) = -3.74 atm. Two of these equilibrium pressures are not physically possible, since

they are negative!

Problem

For x = 0.30 atm at equilibrium

PH2O = (1.20 atm – 0.30 atm) = 0.90 atm, PCO = (1.00 atm + 0.30 atm) = 1.30 atm,PH2

= (1.40 atm + 0.30 atm) = 1.70 atm.

atm 0.30xor atm 5.14x

69

Our equilibrium partial pressures must all be positive.

All these equilibrium pressures are physically real. This is the correct equilibrium mixture!

Problem

We should check our answer:

which is the equilibrium constant we were given, within rounding errors.

70

2.460.90

1.701.30

P

PPK

OH

HCOp

2

2

Factors That Alter the Composition of an Equilibrium Mixture

We like to maximize a product yield for a reaction with a minimum of energy (and money) input.

If a reaction doesn’t go to near completion

we must adjust experimental conditions so the reaction proceeds as favourably as possible!

71

Le Chatelier’s PrincipleThree factors can be changed to affect an

equilibrium: the concentrations of the chemicals involved, the pressure and/or volume of the system, or the temperature.

We can describe the effect of a change on the system as it relates to the equilibrium through

Le Chatelier’s Principle.

72

Le Chatelier’s Principle

73

Le Chatelier’s Principle states that if a stress (a change) is applied to a reaction mixture at equilibrium, reaction will occur in the direction that minimizes the change.

Le Chatelier’s Principle

When a system at equilibrium is changed the system will respond by

re-establishing the equilibrium.

However this new equilibrium mixture will likely be different from the original equilibrium mixture.

74

Altering an Equilibrium Mixture: Changes in Concentration

N2 (g) + 3 H2 (g) 2 NH3 (g)Kc = 0.291 at 700 K.

75

Since we have added a reactant, the reaction should proceed towards products to minimize the amount of “extra”

reactant in the system.

76

77

In general, if we increase the concentration of a reactant, the reaction will proceed from reactants to products.

If we increase the concentration of a product, the reaction will go from products to reactants.

78

In general, if we decrease the concentration of a reactant, the reaction will proceed from products to reactants.

If we decrease the concentration of a product, the reaction will go from reactants to products.

Fe3+ (aq) + SCN- (aq) FeNCS2+ (aq)

a) The original equilibrium mixture for this reaction is orange, which is the added colours of pale yellow Fe3+ and the red FeNCS2+. SCN- is colorless.

79

Fe3+ (aq) + SCN- (aq) FeNCS2+ (aq)

b) If we add FeCl3 to the solution, we see the mixture gets more red, meaning more FeNCS2+. Why?

The increase in the reactant Fe3+ concentration prompts the reaction to shift towards the products to reach a new

equilibrium with more red FeNCS2+.

80

Fe3+ (aq) + SCN- (aq) FeNCS2+ (aq)

If we add KSCN to the solution, we see the mixture gets more red, meaning more FeNCS2+. Why?

The increase in the reactant SCN- concentration prompts the reaction to shift towards the products to reach a new equilibrium with more red FeNCS2+.

81

Fe3+ (aq) + SCN- (aq) FeNCS2+ (aq)d) If we add H2C2O4 to the solution, we see the mixture gets

more yellow, meaning less FeNCS2+. Why?

H2C2O4 removes Fe3+ by creating a new compound.

The decrease in the reactant Fe3+ concentration prompts the reaction to shift towards the reactants, meaning less red FeNCS2+ at the new equilibrium.

82

Fe3+ (aq) + SCN- (aq) FeNCS2+ (aq)e) If we add HgCl2 to the solution, we see the mixture gets more

yellow, meaning less FeNCS2+. Why?

Hg2+ removes SCN- by creating a new compound.

The decrease in the reactant SCN- concentration prompts the system to shift towards the reactants, meaning less red FeNCS2+ at the new equilibrium.

83

Problem

Consider the equilibrium :

CO (g) + H2O (g) CO2 (g) + H2 (g)

Use Le Chatelier’s Principle to predict how the concentration of H2 will change when the equilibrium is disturbed by:

a) Adding COb) Adding CO2

c) Removing H2Od) Removing CO2

84

Altering an Equilibrium Mixture: Changes in Pressure and Volume

What happens when pressure is changed as a result of a change in volume?

N2 (g) + 3 H2 (g) 2 NH3 (g) Kc = 0.291 at 700 K•

Since PV = nRT then P = (nRT) / Van increase in the volume decreases the total

pressure of a system, or a decrease in volume increases the total

pressure of the system. 85

Pressure change through a change in volume

The stress on the equilibrium is the change in the total pressure. The system will respond by minimizing the change in total pressure of the system until a

new equilibrium mixture is achieved.

86

Pressure change through a change in volume

The total pressure is a direct result of the number of moles of gas in the system.

To minimize the stress of an increase in total pressure, the system shifts to the side of the reaction with less moles of gas.

To minimize the stress of an decrease in total pressure, the system shifts to the side of the reaction with more moles of gas.

87

Figure CO (g) + 3 H2 (g) CH4 (g) + H2O (g)

88

Other facts to noteIf the number of moles of gas on the reactants side

of a balanced equation equals that on the products side, then a total pressure change due to volume change will not affect the equilibrium.

If we increase the total pressure by adding an inert gas to the system, there is no effect on the equilibrium of the system.

There is NO real stress on the equilibrium system in this case!

89

ProblemDoes the number of moles of products

increase, decrease, or remain the same when each of the following equilibria is subject to an increase in total pressure by decreasing the volume?

a) CO (g) + H2O (g) CO2 (g) + H2 (g) b) 2 CO (g) C (s) + O2 (g)c) N2O4 (g) 2 NO2 (g)

90

Altering an Equilibrium Mixture: Changes in Temperature

. Our reaction for the formation of ammonia

N2 (g) + 3 H2 (g) 2 NH3 (g) H = -92.2 kJ.

91

We see as the temperature increases, the value of Kc decreases, so the reactionshifts towards the reactants with increasing T

Is there some relationship between H and Kc?

Yes!N2 (g) + 3 H2 (g) 2 NH3 (g) + 92.2 kJWe can think of heat as a “product” in the reaction.• As we increase the temperature of the system, we increase

the “concentration” of this “product”. The system will minimize this increase in “product” by shifting from right to left (towards the reactants).

The new equilibrium will have less products and more reactants, giving a smaller value of Kc.

92

Is there some relationship between H and Kc?

In an endothermic reaction, heat will be a “reactant” so increasing the temperature will shift the reaction from the left to the right, increasing the value of Kc. Overall,

Exothermic reaction:

T then KC Endothermic reaction:

T then KC 93

Problem

When air is heated at very high temperatures in an engine, the air pollutant nitric oxide is produced by the reaction

N2 (g) + O2 (g) 2 NO (g) H = 180.5 kJ

How does the equilibrium amount of NO vary with an increase in temperature?

94

The Effect of Catalysis on Equilibrium

A catalyst increases the rate of the reaction, because a pathway through a lower energy transition state is available.

Lowering the activation energy increases the rate of both the forward and reverse reactions by the same amount.

Since equilibrium occurs when the rates are already the same, adding a catalyst

DOES NOT affect equilibrium.95

.

96

Catalysis and equilibrium

Another way to think of it is that a catalyst does not appear in the overall

balanced equation for a reactionand therefore it won’t appear in the equilibrium constant

equationmeaning no change in the equilibrium constant

will be seen when you add a catalyst.

97

Catalysis and equilibriumIn our ammonia reaction we saw Kc decreases as T

increases. We get less ammonia at high temperatures.

The rate of formation of ammonia increases with temperature. (Arrhenius Equation).

To make the reaction occur quickly enough to be useful requires us to run it at a temperature where we get very little of the desired product.

98

Catalysis and equilibrium

A catalyst allows us to run the reaction at a lower temperature with an acceptable rate, where we can get more of the desired product

from an exothermic reaction.

99

ProblemA platinum catalyst is used in automobile catalytic

converters to hasten the oxidation of carbon monoxide:

Will the amount of CO increase, decrease, or remain the same when:

a) A platinum catalyst is added? b) The temperature is increased? c) The pressure is increased by decreasing the volume? d) The pressure is increased by adding argon gas? e) The pressure is increased by adding O2 gas?

100

kJ 566ΔH (g)CO 2(g)OCO(g) 2 22

Pt