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Chapter 12 Gaseous Chemical Equilibrium

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Chapter 12 Gaseous Chemical Equilibrium. The Concept of Equilibrium. Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. The Concept of Equilibrium. As a system approaches equilibrium, both the forward and reverse reactions are occurring. - PowerPoint PPT Presentation
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Chapter 12 Gaseous Chemical Equilibrium
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Page 1: Chapter 12 Gaseous Chemical Equilibrium

Chapter 12GaseousChemical

Equilibrium

Page 2: Chapter 12 Gaseous Chemical Equilibrium

The Concept of Equilibrium

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

Page 3: Chapter 12 Gaseous Chemical Equilibrium

The Concept of Equilibrium• As a system

approaches equilibrium, both the forward and reverse reactions are occurring.

• At equilibrium, the forward and reverse reactions are proceeding at the same rate.

Page 4: Chapter 12 Gaseous Chemical Equilibrium

A System at Equilibrium

Once equilibrium is achieved, the amount of each reactant and product remains constant.

Page 5: Chapter 12 Gaseous Chemical Equilibrium

Depicting Equilibrium

In a system at equilibrium, both the forward and reverse reactions are being carried out (it’s a dynamic equilibrium); as a result, we write its equation with a double arrow

N2O4 (g) 2 NO2 (g)

Page 6: Chapter 12 Gaseous Chemical Equilibrium

The Equilibrium Constant

• Consider the reaction

• The equilibrium expression for this reaction would be

Kc = [C]c[D]d

[A]a[B]b

aA + bB cC + dD

Page 7: Chapter 12 Gaseous Chemical Equilibrium

The Equilibrium Constant

Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

Kp =(PC)c (PD)d

(PA)a (PB)b

Page 8: Chapter 12 Gaseous Chemical Equilibrium

Relationship between Kc and Kp

• From the ideal gas law we know that

• Rearranging it, we get

PV = nRT

P = RTnV

Page 9: Chapter 12 Gaseous Chemical Equilibrium

Relationship between Kc and Kp

Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes

Where

Kp = Kc (RT)n

n = (moles of gaseous product) − (moles of gaseous reactant)

Page 10: Chapter 12 Gaseous Chemical Equilibrium

Equilibrium Can Be Reached from Either Direction

As you can see, the ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

Page 11: Chapter 12 Gaseous Chemical Equilibrium

Equilibrium Can Be Reached from Either Direction

This is the data from the last two trials from the table on the previous slide.

Page 12: Chapter 12 Gaseous Chemical Equilibrium

Equilibrium Can Be Reached from Either Direction

It does not matter whether we start with N2 and H2 or whether we start with NH3. We will have the same proportions of all three substances at equilibrium.

Page 13: Chapter 12 Gaseous Chemical Equilibrium

What Does the Value of K Mean?

• If K >> 1, the reaction is product-favored; product predominates at equilibrium.

Page 14: Chapter 12 Gaseous Chemical Equilibrium

What Does the Value of K Mean?

• If K >> 1, the reaction is product-favored; product predominates at equilibrium.

• If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

Page 15: Chapter 12 Gaseous Chemical Equilibrium

Manipulating Equilibrium Constants

The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

10.212

=

Kc = = 0.212 at 100C[NO2]2

[N2O4]N2O4 (g) 2 NO2 (g)

Kc = = 4.72 at 100C

[N2O4][NO2]2

N2O4 (g)2 NO2 (g)

Page 16: Chapter 12 Gaseous Chemical Equilibrium

Manipulating Equilibrium ConstantsThe equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.

Kc = = 0.212 at 100C[NO2]2

[N2O4]N2O4 (g) 2 NO2 (g)

Kc = = (0.212)2 at 100C[NO2]4

[N2O4]22 N2O4 (g) 4 NO2 (g)

Page 17: Chapter 12 Gaseous Chemical Equilibrium

Manipulating Equilibrium Constants

The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

A + B C K = 1.9 x 10-4

C + D E + A K = 8.5 x 105

B + D E K =

Page 18: Chapter 12 Gaseous Chemical Equilibrium

Heterogeneous Equilibrium

Page 19: Chapter 12 Gaseous Chemical Equilibrium

The Concentrations of Solids and Liquids Are Essentially Constant

Both can be obtained by dividing the density of the substance by its molar mass—and both of these are constants at constant temperature.

Page 20: Chapter 12 Gaseous Chemical Equilibrium

The Concentrations of Solids and Liquids Are Essentially Constant

Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression

Kc = [Pb2+] [Cl−]2

PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq)

Page 21: Chapter 12 Gaseous Chemical Equilibrium

As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.

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

Page 22: Chapter 12 Gaseous Chemical Equilibrium

What Are the Equilibrium Expressions for These Equilibria?

Page 23: Chapter 12 Gaseous Chemical Equilibrium

Equilibrium Calculations

Page 24: Chapter 12 Gaseous Chemical Equilibrium

Equilibrium Calculations

A closed system initially containing

1.000 x 10−3 M H2 and 2.000 x 10−3 M I2

At 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at 448C for the reaction taking place, which is

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

Page 25: Chapter 12 Gaseous Chemical Equilibrium

What Do We Know?

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change

At equilibrium

1.87 x 10-3

Page 26: Chapter 12 Gaseous Chemical Equilibrium

[HI] Increases by 1.87 x 10-3 M

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At equilibrium

1.87 x 10-3

Page 27: Chapter 12 Gaseous Chemical Equilibrium

Stoichiometry tells us [H2] and [I2]decrease by half as much

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium

1.87 x 10-3

Page 28: Chapter 12 Gaseous Chemical Equilibrium

We can now calculate the equilibrium concentrations of all three compounds…

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium

6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

Page 29: Chapter 12 Gaseous Chemical Equilibrium

…and, therefore, the equilibrium constant

Kc =[HI]2

[H2] [I2]

= 51

=(1.87 x 10-3)2

(6.5 x 10-5)(1.065 x 10-3)

Page 30: Chapter 12 Gaseous Chemical Equilibrium

The Reaction Quotient (Q)

• To calculate Q, one substitutes the initial concentrations of reactants and products into the equilibrium expression.

• Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

Page 31: Chapter 12 Gaseous Chemical Equilibrium

Only given Initial Concentrations and the value of K

Page 32: Chapter 12 Gaseous Chemical Equilibrium

If Q = K,

the system is at equilibrium.

Page 33: Chapter 12 Gaseous Chemical Equilibrium

If Q > K,there is too much product and the

equilibrium shifts to the left.

Page 34: Chapter 12 Gaseous Chemical Equilibrium

If Q < K,there is too much reactant, and the

equilibrium shifts to the right.

Page 35: Chapter 12 Gaseous Chemical Equilibrium

Le Châtelier’s Principle

Page 36: Chapter 12 Gaseous Chemical Equilibrium

Le Châtelier’s Principle

“If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”

Page 37: Chapter 12 Gaseous Chemical Equilibrium

If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

Le Châtelier’s Principle

•Changes in Concentration

N2 (g) + 3H2 (g) 2NH3 (g)

AddNH3

Equilibrium shifts left to offset stress

Page 38: Chapter 12 Gaseous Chemical Equilibrium

Le Châtelier’s Principle

•Changes in Concentration continued

Change Shifts the Equilibrium

Increase concentration of product(s) left

Decrease concentration of product(s) right

Decrease concentration of reactant(s)

Increase concentration of reactant(s) right

left

aA + bB cC + dD

AddAddRemove Remove

Page 39: Chapter 12 Gaseous Chemical Equilibrium

N2O4(g) 2NO2(g)(colorless) (brown)

Page 40: Chapter 12 Gaseous Chemical Equilibrium

Le Châtelier’s Principle

•Changes in Volume and Pressure

A (g) + B (g) C (g)

Change Shifts the Equilibrium

Increase pressure Side with fewest moles of gas

Decrease pressure Side with most moles of gas

Decrease volume

Increase volume Side with most moles of gas

Side with fewest moles of gas

Page 41: Chapter 12 Gaseous Chemical Equilibrium

N2O4(g) 2NO2(g)(colorless) (brown)

Is this rxn exo- or endothermic?

Page 42: Chapter 12 Gaseous Chemical Equilibrium

Le Châtelier’s Principle

•Changes in Temperature

Change Exothermic Rx

Increase temperature K decreases

Decrease temperature K increases

Endothermic Rx

K increases

K decreases

colder hotter

Page 43: Chapter 12 Gaseous Chemical Equilibrium

van’t Hoff equation

• ΔHrxn= enthalpy change for a reaction.

• ΔHrxn = ΣΔHfo

(products) - ΣΔHfo

(reactants) see appendix

ln K1

K2

= ΔHrxn

R [ ]1T2

1T1

-

Page 44: Chapter 12 Gaseous Chemical Equilibrium

uncatalyzed catalyzed

Catalyst lowers Ea for both forward and reverse reactions.

Catalyst does not change equilibrium constant or shift equilibrium.

•Adding a Catalyst•does not change K•does not shift the position of an equilibrium system•system will reach equilibrium sooner

Le Châtelier’s Principle

Page 45: Chapter 12 Gaseous Chemical Equilibrium

Catalysts increase the rate of both the forward and reverse reactions.

Page 46: Chapter 12 Gaseous Chemical Equilibrium

Equilibrium is achieved faster, but the equilibrium composition remains unaltered.

Page 47: Chapter 12 Gaseous Chemical Equilibrium

Example

• (e) helium is added and the total pressure increases.

• (f) a catalyst is added

Page 48: Chapter 12 Gaseous Chemical Equilibrium

Chemistry In Action

Life at High Altitudes and Hemoglobin Production

Kc = [HbO2]

[Hb][O2]

Hb (aq) + O2 (aq) HbO2 (aq)

Page 49: Chapter 12 Gaseous Chemical Equilibrium

Chemistry In Action: The Haber Process

N2 (g) + 3H2 (g) 2NH3 (g) H0 = -92.6 kJ/mol

Page 50: Chapter 12 Gaseous Chemical Equilibrium

The Haber Process

The transformation of nitrogen and hydrogen into ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance.

Page 51: Chapter 12 Gaseous Chemical Equilibrium

The Haber Process

If H2 is added to the system, N2 will be consumed and the two reagents will form more NH3.

Page 52: Chapter 12 Gaseous Chemical Equilibrium

The Haber Process

This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid.

Page 53: Chapter 12 Gaseous Chemical Equilibrium

Le Châtelier’s Principle

Change Shift EquilibriumChange Equilibrium

Constant

Concentration yes no

Pressure yes (?) no

Volume yes no

Temperature yes yes

Catalyst no no

Adding an inert gas to increase

the total pressure

no no

Page 54: Chapter 12 Gaseous Chemical Equilibrium

R(g) + heat B(g) + C(g)

• Increase temperature• Increase [C]• Decrease temperature• Add a catalyst• Decrease [C]• Increase pressure by

pumping in helium• Increase pressure by

decreasing the volume

• Increase blue• Increase red• Increase red• Stays purple• Increase blue• Stays purple

(although diluted a bit)

• Increase red


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