Chapter 18: Chemical Equilibrium

Chapter 18: Chemical Chapter 18: Chemical EquilibriumEquilibrium

1. The Concept of Equilibrium

A. Equilibrium exists when two opposing processes occur at the same rate.

B. Reversible reactions 2NO2 (g) N2O4(g)

1. Products take part in a separate reaction to reform the reactants

2. A chemical reaction in which the products can regenerate the original reactants is a reversible reaction

3. Use two half arrows to indicate a reversible reaction Forward reaction (above arrow) reverse reaction (below

arrow)

1. The Concept of Equilibrium C. Chemical Equilibrium 1. The reaction rate depends on different

factors -The concentrations of the substances in the

reaction: reaction rate is proportional to reactant concentration.

2. When substance enter into the reaction, the concentration of the reactants decreases as the reactant are converted into products.

-Therefore the concentration of the products increase.

The Concept of EquilibriumThe Concept of Equilibrium

3. Chemical equilibrium is the point at which the forward reaction is equal to the rate of the reverse reaction.

- The concentrations of the reactants and products become constant

The Concept of Equilibrium

4. Chemical equilibrium is the state in which the concentration of reactants and products remain constant with time because the rate at which they are being formed is equal to the rate at which they are consumed in the opposite reaction.

5. Square brackets [ ] are used to denote concentration of a substance.

II. The Law of Chemical Equilibrium

A. The Equilibrium Constant 1. The Law of Mass Action

A. Formulated by Guldbert and Waage (1864)

B. Expresses the relative concentrations of reactants and products at equilibrium in

terms of quantity called the equilibrium constant (Keq)

II. The Law of Chemical Equilibrium

C. In the reaction:

aA+ bB cC+ dD

Keq= [C]c [D]d [A]a [B]b

a,b,c,d = coefficients for the substancesA,B,C,D = reactants and products

II. The Law of Chemical Equilibrium

D. This ratio is always a constant value for a given reaction regardless of initial concentrations at a given temperature. 1. This is called The Law of Chemical

Equilibrium. 2. The equilibrium can be reached from either

direction. E. Equilibrium position: Set of equilibrium

concentrations in an experiment

II. The Law of Chemical Equilibrium

F. The equilibrium constant is a measure of the extent to which a reaction proceeds to completion.

II. The Law of Chemical Equilibrium

1. Keq >> 1:The numerator must be larger than the denominator

The equilibrium concentrations of the products must be larger than the reactants

The equilibrium “lies to the right”

Keq <<1 : The numerator must be much smaller than the denominator

The equilibrium concentrations of the products must be smaller than the reactants

The equilibrium “lies to the left”

II. The Law of Chemical Equilibrium

B. Homogeneous and Heterogeneous Equilibria 1. Homogeneous Equilibria: Equilibrium

conditions for reaction in which all the reactants and products are in the same state.

2. Heterogeneous Equilibria: Equilibrium conditions for reactions in which the substances involved are in more than one state.

II. The Law of Chemical Equilibrium

3. Example of heterogeneous equilibria NH4Cl(s) NH3 (g) + HCl(g)

The concentration of a pure liquid or solid is basically constant and unaffected by the temperature. The concentration is density/molar mass.

Therefore the concentration of a pure liquid or solid does NOT change during a reaction

The concentrations of liquids and solids are left out of the equilibrium expression.

So, Keq for our example =

[HCl][NH3]

II. The Law of Chemical Equilibrium

C. The Reaction Quotient 1. Q: Used to determine if a

reaction is at equilibrium. 2. Calculated like Keq except that

it uses the concentrations that exist at the time the measurement is taken not the equilibrium concentrations.

II. The Law of Chemical Equilibrium

3. Example

At 472° C, N2 (g) + 3 H2 (g) 2NH3 (g)

Keq= .105

You measure the concentrations to be

[NH3]=.15M [N2]=.0020M [H2]= .10 M

Find Q and decide if you are at equilibrium

Q= [NH3]2 / [N2] [H2]3

Q= [.15M]2 / [.0020][.10]3 = 1.1 x 104

Not at equilibrium because Q does not equal

Keq

II. The Law of Chemical Equilibrium

3. Which direction will the reaction proceed?Q<Keq : There is too much of the reactants and too little of the products the reaction will shift to the right to make more products.

3. Which Direction will a reaction 3. Which Direction will a reaction proceed? proceed?

Q>Keq : There is too much of the products and too little of the reactants the reaction will shift to the left.

Q=Keq Q=Keq No Shift.

II. The Law of Chemical Equilibrium

4. Try: At 448° C, Keq=50.5 for

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

Find Q and predict how the reaction will shift

[H2] = .150 M [I2 ]=.175M [HI]=.950 M

Q= [HI]2 / [H2] [I2 ] =

[.950 M]2 / [.150 M][.175M]=34.4

Shifts to the right because Q < Keq

III. LeChatelier’s Principle

A. If a change in conditions is imposed on a system at equilibrium, the equilibrium position will shift in the direction that tends to reduce the change in conditions

B. Changes in Concentration 1. If more of a substance is added

to a reaction at equilibrium, the concentration of that substances increases.A. Reaction will return to equilibrium by consuming some of the added substance.

2. If a substance is removed, its concentration decreasesA. The reaction will return to

equilibrium by producing more of the substance that was removed.

3. Only the equilibrium shifts, NOT the equilibrium constant.

III. LeChatelier’s Principle

C. Changes in Pressure 1. If the total pressure of a system

increases, the system will shift to reduce the pressure by proceeding in the direction that produces fewer moleculesA. The reaction changes the

equilibrium position but NOT the equilibrium constant.

D. Effect of Changing Temperature

1. ExampleH2(g)+I2(g) 2HI(g)+ heatKeq=54.5 @ 400° CKeq=45.9 @ 490° CSo raising the temperature

causes the reaction to proceed less completely to the products

III. LeChatelier’s Principle

2. If a forward reaction is exothermic, the reverse reaction is endothermic in a reversible reaction.

3. Changing temperature DOES change the value of the

Equilibrium Constant.

III. LeChatelier’s Principle

E. The Haber Process1. Haber examined the reaction :

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

III. LeChatelier’s Principle

A. This reaction reached equilibrium before producing much ammonia.

B. Haber developed a process and the equipment necessary to reach the pressures and temperatures needed to produce large amounts of ammonia

III. LeChatelier’s Principle

i. Ammonia is continuously removed ii. Pressure is increased to force a reaction to

the right. iii. The forward reaction is exothermic but he

increased the temperature to speed up the reaction even though it drove the reverse reaction. 1. Compensated for this by increasing the pressure. 2.If he had decreased the temperature, the reaction

moved too slowly.

III. LeChatelier’s Principle

C. Haber also developed the use of chlorine as poison gas weapon.

D. Reactions that go to completion 1. Formation of a gas

H2CO3(aq) H2O(g) + CO2(g)

III. LeChatelier’s Principle

2. Formation of a precipitate NaCl(aq) +AgNO3(aq) NaNO3 (aq) + AgCl(s)

3. Formation of a slightly ionized product HCl(aq) + NaOH(aq) NaCl(aq) + H2O

III. LeChatelier’s Principle

G. Common-ion Effect 1. An equilibrium reaction may be driven in

the desired direction by applying LeChatelier’s principle

2. Example: HCl bubbled into NaCl solution NaCl (s) Na+ (aq) + Cl- (aq)Keq = [Na+] [Cl-]

HCl H3O+ +Cl – (aq)

III. LeChatelier’s Principle

3. Common Ion effect: -The addition of an ion common

to two solutes brings about precipitation or reduced ionization.

IV. Equilibria of Acids, Bases and Salts

A. The acid dissociation constant1. When the products and

reactants of a reaction reach

equilibrium, a certain ratio of their concentrations always has the same value

IV. Equilibria of Acids, Bases and Salts

2. For the reaction:

HA (acid) + H2O H3O+ (hydronium ion) + A- (anion)

Keq= [H3O+][A-] / [HA][H2O] Ka = [H3O+][A-] / [HA] Ka= acid dissociation constant

IV. Equilibria of Acids, Bases and Salts

3. The greater the Ka, the further the

action runs to completion. So Ka is the

measure of the strength of an acid 4. For diprotic and triprotic acids, each

dissociation takes place in a separate step.

IV. Equilibria of Acids, Bases and Salts

A. Step 1:H2CO3+ H2O H3O+ +CO3 -

Ka1= [H3O+][HCO3-] / [H2CO3]=4.5x10-7

B. Step 2: HCO3-+H2O H3O+ + CO3

2-

Ka2 = 5.6 x 10-11

IV. Equilibria of Acids, Bases and Salts

B. The Base Dissociation Constant 1. B(base) + H2O OH- + HB+

(cation) 2. Kb: the measure of the strength

of the base Kb= [HB+][OH-] / [B]

IV. Equilibria of Acids, Bases and Salts

C. Calculating Dissociation Constants 1. Example: acetic acid is a weak

monoprotic acid that dissociates into an acetate ion and a hydronium ion in aqueous solution. Calculate Ka for acetic acid if a 1.0 M solution results in an equilibrium hydronium concentration of 0.0042 M

IV. Equilibria of Acids, Bases and Salts

2. You try : Ammonia is a weak base. If the initial concentration of ammonia is 0.150 M and the equilibrium concentration of hydroxide is 0.0016M, calculate Kb for ammonia.

IV. Equilibria of Acids, Bases and Salts

D. Buffers 1. A solution that can resist changes in pH 2. Weak acid and salt of a weak acid 3. Weak base and salt of a weak base E. Ionization Constant of Water

Kw=concentration of

[OH-][H30+] = 1.0 x10-14

IV. Equilibria of Acids, Bases and Salts

F. Hydrolysis of Salts 1. Hydrolysis:

A. reaction between H20 molecules and ions of a dissolved salt.

2. Anion Hydrolysis:

if the acid is a weak acid, its conjugate base (anion) will be strong enough to remove H+ from H20 molecules to form OH- ions.

IV. Equilibria of Acids, Bases and Salts

3. Cation Hydrolysis: if the base is weak its

conjugate acid which is now the cation will be strong enough to donate a H+ ion to a H2O molecule to form H30+ ions

IV. Equilibria of Acids, Bases and Salts

G. Hydrolysis in Acid-Base Reactions 1. Strong acid-Strong base neutral

solution 2. Strong acid-Weak base acidic

solution 3. Strong base -Weak acid basic

solution 4. Weak acid-Weak base acidic, basic,

or neutral solution

V. Solubility equilibrium

A. Solubility product1. Soluble: A substance whose

solubility is greater than 1 gram per 100 grams of H20

2. Insoluble: A substance whose solubility is less than 0.1 grams per 100 grams of H2O.

3. Slightly Soluble: Solubility between 0.1 and 1.0 gram per 100 grams of H20.

4. Saturated Solution: contains the maximum amount of solute possible at a given temperature in equilibrium with an undissolved excess of substances.

V. Solubility equilibrium

5. Example: Ksp= solubility product constant: the product of the molar concentration of its ions in a saturated solution, each raised to the power that is the coefficient of that ion in the balanced chemical equation

V. Solubility equilibrium

6. Calculate the solubility product constant of Lead (II) Chloride which has a solubility of 1.0 g/100 g water at 20 degrees Celsius.

V. Solubility equilibrium

B. Precipitation Calculations 1. If the ion product is greater than

the Ksp, the salt precipitates. 2. Example: Will a precipitate form if

20.0mL of 0.010M Barium Chloride solution is mixed with 20.0mL of 0.0050 M Sodium Sulfate solution?

C. Read about limitations on the use of Ksp on page 620.