equilibrium © 2009, prentice-hall, inc. chapter 14 chemical equilibrium

Equilibrium

© 2009, Prentice-Hall, Inc.

Chapter 14Chemical Equilibrium

Equilibrium


The Concept of Equilibrium

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

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.

Equilibrium


A System at Equilibrium

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

Equilibrium


Depicting Equilibrium

Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow.

N2O4 (g) 2 NO2 (g)

Equilibrium


The Equilibrium Constant

Equilibrium



• Forward reaction:N2O4 (g) 2 NO2 (g)

• Rate Law:Rate = kf [N2O4]

Equilibrium



• Reverse reaction:2 NO2 (g) N2O4 (g)

• Rate Law:Rate = kr [NO2]2

Equilibrium



• Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

• Rewriting this, it becomeskf

kr [NO2]2

[N2O4]=

Equilibrium



The ratio of the rate constants is a constant at that temperature, and the expression becomes

Keq =kf

kr [NO2]2

[N2O4]=

Equilibrium



• Consider the generalized reaction

• The equilibrium expression for this reaction would be

Kc = [C]c[D]d

[A]a[B]b

aA + bB cC + dD

Equilibrium



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

Kp =(PC

c) (PDd)

(PAa) (PB

b)

Equilibrium


Relationship Between Kc and Kp

• From the Ideal Gas Law we know that

• Rearranging it, we get

PV = nRT

P = RTnV

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)

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.

Equilibrium



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

Equilibrium



It doesn’t 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.

Equilibrium


What Does the Value of K Mean?

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

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.

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.

Kc = = 0.212 at 100 C[NO2]2

[N2O4]N2O4 (g)

QuickTime™ and aPhoto - JPEG decompressor

are needed to see this picture. 2 NO2 (g)

Kc = = 4.72 at 100 C[N2O4][NO2]2

N2O4 (g)


are needed to see this picture.2 NO2 (g)

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 100 C[NO2]2

[N2O4]N2O4(g)


are needed to see this picture. 2 NO2(g)

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

[N2O4]22 N2O4(g)


are needed to see this picture. 4 NO2(g)

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.

Equilibrium


Heterogeneous Equilibrium

Equilibrium


The Concentrations of Solids and Liquids Are Essentially Constant

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

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)


are needed to see this picture.

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)



Equilibrium


Writing Equilibrium Constant Expressions

• The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.

• The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.

• The equilibrium constant is a dimensionless quantity.

• In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.

• If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

14.2

Equilibrium


Equilibrium Calculations

Equilibrium


An Equilibrium Problem

A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at 448 C 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 448 C for the reaction taking place, which is

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



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

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


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


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

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)

Equilibrium


Example• Sulfuryl chloride, SO2Cl2, reacts to

produce sulfur dioxide, and chlorine gas.

• 3.509 grams of the compound decomposes in a 1.00 L flask, the temp is raised to 375K, determine the pressure in the flask.

• At equilibrium the total pressure is 1.43 atm, calculate the partial pressure of all the substances.

• Calculate the Kp value for the reaction.

Equilibrium


The Reaction Quotient (Q)

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

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

Equilibrium


If Q = K,

the system is at equilibrium.

Equilibrium


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

equilibrium shifts to the left.

Equilibrium


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

equilibrium shifts to the right.

Equilibrium


The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.

IF

• Qc > Kc system proceeds from right to left to reach equilibrium

• Qc = Kc the system is at equilibrium

• Qc < Kc system proceeds from left to right to reach equilibrium

14.4

Equilibrium


Le Châtelier’s Principle

Equilibrium



“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.”

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.


• Changes in Concentration

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

AddNH3

Equilibrium shifts left to offset stress

14.5

Equilibrium



• 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

left14.5

aA + bB cC + dD

AddAddRemove Remove

Equilibrium



• 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

14.5

Equilibrium



• Changes in Temperature

Change Exothermic Rx

Increase temperature K decreases

Decrease temperature K increases

Endothermic Rx

K increases

K decreases

14.5colder hotter

Equilibrium


uncatalyzed catalyzed

14.5

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


Equilibrium



Change Shift EquilibriumChange Equilibrium

Constant

Concentration yes no

Pressure yes no

Volume yes no

Temperature yes yes

Catalyst no no

14.5

Equilibrium


Catalysts

Equilibrium


Catalysts

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

Equilibrium


Catalysts

When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered.

Equilibrium


Gibbs Free Energy and Chemical Equilibrium

G = G0 + RT lnQ

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

18.6

Equilibrium


G0 = RT lnK

18.6

Equilibrium


Ch 16: Solubility Equilibria

16.6

AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-] Ksp is the solubility product constant

MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp = [Mg2+][F-]2

Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO3

2-]

Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO3

3-]2

Dissolution of an ionic solid in aqueous solution:

Q = Ksp Saturated solution

Q < Ksp Unsaturated solution No precipitate

Q > Ksp Supersaturated solution Precipitate will form

Equilibrium


Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution.

Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.

16.6

Equilibrium


What is the solubility of silver chloride in g/L ?


Ksp = [Ag+][Cl-]Initial (M)

Change (M)

Equilibrium (M)

0.00

+s

0.00

+s

s s

Ksp = s2

s = Ksps = 1.3 x 10-5

[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M

Solubility of AgCl = 1.3 x 10-5 mol AgCl

1 L soln143.35 g AgCl

1 mol AgClx = 1.9 x 10-3 g/L

Ksp = 1.6 x 10-10

16.6

Equilibrium


If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form?

16.6

The ions present in solution are Na+, OH-, Ca2+, Cl-.

Only possible precipitate is Ca(OH)2 (solubility rules).

Is Q > Ksp for Ca(OH)2?

[Ca2+]0 = 0.100 M [OH-]0 = 4.0 x 10-4 M

Ksp = [Ca2+][OH-]2 = 8.0 x 10-6

Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8

Q < Ksp No precipitate will form

Equilibrium


What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br- and Cl- at a concentration of 0.02 M?


Ksp = [Ag+][Cl-]

Ksp = 1.6 x 10-10

16.7

AgBr (s) Ag+ (aq) + Br- (aq) Ksp = 7.7 x 10-13

Ksp = [Ag+][Br-]

[Ag+] = Ksp

[Br-]7.7 x 10-13

0.020= = 3.9 x 10-11 M

[Ag+] = Ksp

[Br-]1.6 x 10-10

0.020= = 8.0 x 10-9 M

3.9 x 10-11 M < [Ag+] < 8.0 x 10-9 M

Equilibrium


The Common Ion Effect and Solubility

The presence of a common ion decreases the solubility of the salt.

What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr?

AgBr (s) Ag+ (aq) + Br- (aq)

Ksp = 7.7 x 10-13

s2 = Ksp

s = 8.8 x 10-7

NaBr (s) Na+ (aq) + Br- (aq)

[Br-] = 0.0010 M

AgBr (s) Ag+ (aq) + Br- (aq)

[Ag+] = s

[Br-] = 0.0010 + s 0.0010

Ksp = 0.0010 x s

s = 7.7 x 10-10

16.8

equilibrium © 2009, prentice-hall, inc. chapter 14 chemical equilibrium

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