ch. 18 chemical equilibrium

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Ch. 18 Chemical Equilibrium


18.1 The Nature of Chemical Equilibrium

• List two everyday processes that can easily be reversed and two that cannot.

• The freezing of water and the melting of ice can be reversed

• The cooking of an egg or the lighting of a match cannot be reversed.

• Reactions that “go to completion” refer to a reaction in which all the reactants are converted to products

Chapter 18


Reversible Reactions

• But theoretically, every reaction can proceed in two directions, forward and reverse.

• Essentially all chemical reactions are considered to be reversible under suitable conditions.

• A chemical reaction in which the products can react to re-form the reactants is called a reversible reaction.

Chapter 18

22HgO( ) 2Hg( ) O ( )s l + g


• Chemical equilibrium is established when the rate of its forward reaction equals the rate of its reverse reaction

• When this happens the amounts of products (concentrations) and reactants remain constant.

• Both reactions continue, but there is no net change in their concentration

Chapter 18


• products of the forward reaction favored, lies to the right

2 2 32SO ( ) O ( ) 2SO ( )g + g g

–

2 3 2 3 3H CO ( ) H O( H O ( ) HCO ( )aq + l) aq + aq

–

2 3 2 3 3H SO ( ) H O( ) H O ( ) HSO ( )aq + l aq + aq

Chapter 18

• Neither reaction is favored

• products of the reverse reaction favored, lies to the left


• After equilibrium is reached, the individual concentrations of A, B, C, and D undergo no further change if conditions remain the same.

• i.e. : the ratio of their concentrations remains constant.

• The equilibrium constant is designated by the letter K.

Chapter 18

A B C Dn m x y

•The expression for K =


• The Equilibrium Constant Expression, K, is the ratio of the product of the concentrations of substances formed at equilibrium to the product of the concentrations of reacting substances. Each concentration is raised to a power equal to the coefficient of that substance in the chemical equation.

• K is dependent on temperature and always determined experimentally

• Only the concentrations of substances that can actually change are included in K.

• So pure solids and liquids are omitted because their concentrations cannot change.

Chapter 18


What would be true if K = 1?•Concentration of products = concentration of reactants

What would be true if K is greater than 1?•Large K values indicate that the products are favored (forward reaction)

What would be true if K is less than 1?•Small K values indicate the reactants are favored (reverse reaction)

Chapter 18


What does knowing the value of Keq help us predict?

• Whether the forward of the reverse reaction is favored• Therefore, tell us if the reactants or products will be in

higher concentrations when equilibrium is established

*Does not tell us how quickly the reaction reaches the equilibrium


Examples of Equilibrium Constant Expressions

• Make sure you always balance an equation before writing an equilibrium constant expression


Examples of Equilibrium Constant Expressions


Sample Problem A

An equilibrium mixture of N2, O2 , and NO gases at

1500 K is determined to consist of 6.4 10–3 mol/L of

N2, 1.7 10–3 mol/L of O2, and 1.1 10–5 mol/L of NO.

What is the equilibrium constant for the system at

this temperature?

Chapter 18


The Equilibrium Expression, continued

Sample Problem A Solution

2N ( ) ) 2NO( )2g + O (g g

Section 1 The Nature of Chemical EquilibriumChapter 18

Given: [N2] = 6.4 10–3 mol/L

[O2] = 1.7 10–3 mol/L

[NO] = 1.1 10–5 mol/L

Solution: The balanced chemical equation is

Unknown: K

The chemical equilibrium expression is


• Imagine children playing on a seesaw.

• Five boys are sitting on one side and five girls on the other, and the seesaw is just balanced.

• Then, one girl gets off, and the system is no longer at equilibrium.

• One way to get the seesaw in balance again is for one of the boys to move toward the girls’ side.

Chapter 18

Section 18.2 Le Châtelier’s principle


Chapter 18

• Le Châtelier’s principle states that if a system at equilibrium is subjected to a stress, the equilibrium is shifted in the direction that tends to relieve the stress.

• This principle is true for all dynamic equilibria, chemical as well as physical.

• Changes in pressure, concentration, and temperature illustrate Le Châtelier’s principle because they alter the equilibrium position and thereby change the relative amounts of reactants and products

Section 2 Shifting Equilibrium


Predicting the Direction of Shift: Pressure

• Change in pressure affects only equilibrium systems in which gases are involved.

• An increase in pressure is an applied stress.

Chapter 18

• The system can reduce the total pressure by reducing the number of molecules.

• Therefore, increase in pressure favors the reaction that produces fewer gas molecules


• Ex: Haber process for the synthesis of ammonia

2 2 3N ( ) 3H ( ) 2NH ( )g + g g

Chapter 18

4 molecules of gas 2 molecules of gas• When pressure is applied, the equilibrium will shift to

the right, and produce more NH3.• A reduction in the total number of molecules

leads to a decrease in pressure.

Note: changes in pressure do not affect the value of the equilibrium constant


Changes in Concentration

• An increase in the concentration of a reactant is a stress on the equilibrium system.

A B C D

Chapter 18

• To relieve the stress, some of the added A reacts with B to form products C and D.

• Increase in concentration of one substance shifts the equilibrium to the opposite side (i.e. opposite side favored)


• Note: changes in concentration have no effect on the value of the equilibrium constant.

Chapter 18

• Such changes have an equal effect on the numerator and the denominator of the chemical equilibrium expression.


Changes in Temperature

• Reversible reactions are exothermic in one direction and endothermic in the other.

• The addition of energy in the form of heat shifts the equilibrium so that energy is absorbed.

• An increase in temperature favors the endothermic reaction

• A decreased in temperature (removal of energy) favors the exothermic reaction

Chapter 18


• A rise in temperature increases the rate of any reaction.

• In an equilibrium system, the rates of the opposing reactions are raised unequally.

**The value of the equilibrium constant for a given system is affected by the temperature.

Chapter 18


• Haber Process: exothermic 2 2 3N ( ) 3H ( ) 2NH ( ) 92 kJg + g g +

Section 2 Shifting EquilibriumChapter 18

• A rise in temperature increases the rate of any reaction, but the rates of the opposing reactions are raised unequally.

• The value of the equilibrium constant for a given system is affected by the temperature.

• A high temperature favors the decomposition of ammonia, the endothermic reaction.

• At low temperatures, the forward reaction is too slow to be commercially useful.


Temperature Changes Affect an Equilibrium System


Reactions That Go to Completion• Some reactions involving compounds formed by the

chemical interaction of ions in solutions appear to go to completion in the sense that the ions are almost completely removed from solution.

Chapter 18


1. Formation of a Gas

Chapter 18

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

• Reaction goes practically to completion because one of the products, CO2, escapes as a gas if the container is open to the air.

3 Examples of Reactions that go to Completion


2. Formation of a Precipitate

aq + aq + aq + aq

aq + aq + s

3

3

Na ( ) Cl ( ) Ag ( ) NO ( )

Na ( ) NO ( ) AgCl( )

Chapter 18

• If chemically equivalent amounts of the two solutes are mixed, almost all of the Ag+ ions and Cl− ions combine and separate from the solution as a precipitate of AgCl.

• AgCl is only very sparingly soluble in water.

• The reaction thus effectively goes to completion because an essentially insoluble product is formed.


3. Formation of a Slightly Ionized Product

• Neutralization reactions between H3O+ ions from aqueous acids and OH− ions from aqueous bases result in the formation of water molecules, which are only slightly ionized.

aq + aq + aq + aq

aq + aq + aq

3

2

H O ( ) Na ( ) Cl ( ) OH ( )

Na ( ) Cl ( ) 2H O( )

Chapter 18

• Hydronium ions and hydroxide ions are almost entirely removed from the solution.

• The reaction effectively runs to completion because the product is only slightly ionized.


Common-Ion Effect

• The phenomenon in which the addition of an ion common to two solutes brings about precipitation or reduced ionization of a reactant

g + l aq + aq2 3HCl( ) H O( ) H O ( ) Cl ( )

s aq + aq NaCl( ) Na ( ) Cl ( )

Chapter 18

• Ex: hydrogen chloride gas is bubbled into a saturated

solution of sodium chloride.


Common-Ion Effect, continued

• As the hydrogen chloride dissolves in sufficient quantity, it increases the concentration of Cl− ions in the solution, which is a stress on the equilibrium system.

• The system can compensate by forming some solid NaCl. The NaCl precipitates out, relieving the stress of added chloride (reactions shifts to the left)

• The new equilibrium has a greater concentration of Cl− ions but a decreased concentration of Na+ ions.

s aq + aq NaCl( ) Na ( ) Cl ( )

Chapter 18


Particle Model for the Common-Ion Effect

Section 2 Shifting EquilibriumChapter 18


• Soluble ionic compounds dissolve in water until they are at equilibrium with their ions

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

Chapter 18

• The solubility product constant, Ksp, of a substance is the product of the molar concentrations of its ions in a saturated solution, each raised to the power that is the coefficient of that ion in the balanced chemical equation.

Solubility Product


Solubility Product, continued

• For dissolution of the ionic compound AaBb (A cation, B anion), the solubility-production expression is

Section 4 Solubility EquilibriumChapter 18

• The equilibrium expression is written without including the solid species (or water)

• The numerical value of Ksp helps us determine how soluble a compound is in water

Ksp = [A]a [B]b

AaBb (s) → aA + bB



• Compounds with large Ksp values are more soluble• Compounds with smaller Ksp values are less soluble


• The solubility of a solid is an equilibrium position that represents the amount of the solid required to form a saturated solution with a specific amount of solvent.

• It has an infinite number of possible values at a given temperature and is dependent on other conditions, such as the presence of a common ion.



• For a saturated solution of CaF2, the equilibrium equation is

2 –

2CaF ( ) Ca ( ) 2F ( )s aq aq


• The expression for the solubility product constant is

• The solubility of CaF2 is 8.6 10−3 g/100 g of water at 25°C. Expressed in moles per liter this concentration becomes 1.1 10−3 mol/L.

Ksp = [Ca] [F]2


Ionization Constant of Acids & Bases• When we write the equiblium constant expression for

the ionization of an acid/base we call them

• Ka :acid ionization constant.

• Kb: base ionization constant.

• The acid ionization constant, Ka , is constant for a specified temperature

• Weak acids, like CH3COOH, remainly largely unionized as will have small Ka values.

• Strong acids, like HNO3, ionize completely and will have larger Ka values.

Section 3 Equilibria of Acids, Bases, and Salts

Chapter 18


Ionization Constant of Water, Kw

• The self-ionization of water is an equilibrium reaction.

l + l aq + aq2 2 3H O( ) H O( ) H O ( ) OH ( )

Section 3 Equilibria of Acids, Bases, and SaltsChapter 18

• Equilibrium is established with a very low concentration of H3O+ and OH− ions.

Kw=[H3O+][OH–] = 1.0 10-14


Determining Ksp for Reactions at Chemical Equilibrium




• CaF2 dissociates to yield twice as many F− ions as Ca2+ ions.

[Ca2+] = 1.1 10−3 mol/L [F− ] = 2.2 10−3 mol/L

spK 2 2[Ca ][F ]


Ksp = 5.3 10-9

• Calculations of Ksp ordinarily should be limited totwo significant figures.


Solubility Product Constants at 25°C




Sample Problem B

Calculate the solubility product constant, Ksp ,for

copper(I) chloride, CuCl, given that the solubility

of this compound at 25°C is 1.08 10–2 g/100. g

H2O.




Sample Problem B Solution

–CuCl( ) Cu ( ) Cl ( )s aq aq

Ksp=[Cu+][Cl–]


Unknown: Ksp

Given: solubility of CuCl = 1.08 10−2 g CuCl/100. g H2O

Solution:

[Cu+] = [Cl–] = solubility in mol/L



Sample Problem B Solution, continued

1.09 10-3 mol/L CuCl


Ksp = (1.09 10-3)(1.09 10-3) =

1.19 10-6

[Cu+] = [Cl–]=1.09 10-3 mol/L


Calculating Solubilities

• The solubility product constant can be used to determine the solubility of a sparingly soluble salt.

2 2–

3 3BaCO ( ) Ba ( ) CO ( )s aq aq


• The molar solubility of BaCO3 is 7.1 10−5 mol/L.

• How many moles of barium carbonate, BaCO3, can be dissolved in 1 L of water at 25°C?


Calculating Solubilities, continued

Sample Problem C

Calculate the solubility of silver bromide, AgBr,

in mol/L, using the Ksp value for this compound.



Calculating Solubilities, continued

Sample Problem C Solution

Given: Ksp = 5.0 10−13

Unknown: solubility of AgBr

Solution:

[Ag+] = [Br−], so let [Ag+] = x and [Br−] = x

s aq) + aqAgBr( ) Ag ( Br ( )



Limitations on the Use of Ksp

• The solubility product principle can be very useful when applied to solutions of sparingly soluble substances.

• It cannot be applied very successfully to solutions of moderately soluble or very soluble substances.

• The positive and negative ions attract each other, and this attraction becomes appreciable when the ions are close together.

• Sometimes it is necessary to consider two equilibria simultaneously.



Equilibrium Calculations



Multiple Choice

1. A chemical reaction is in equilibrium when

A. forward and reverse reactions have ceased.

B. the equilibrium constant equals 1.

C. forward and reverse reaction rates are equal.

D. No reactants remain.

Standardized Test PreparationChapter 18


1. A chemical reaction is in equilibrium when

A. forward and reverse reactions have ceased.

B. the equilibrium constant equals 1.

C. forward and reverse reaction rates are equal.

D. No reactants remain.

Standardized Test Preparation

Multiple Choice

Chapter 18


2. Which change can cause the value of the equilibrium

constant to change?

A. temperature

B. concentration of a reactant

C. concentration of a product

D. None of the above


Multiple Choice

Chapter 18


3. Consider the following reaction:

The equilibrium constant expression for this reaction is

A. C.

B. D.

22C( ) O ( ) 2CO( )s + g g


Multiple Choice

Chapter 18


3. Consider the following reaction:

The equilibrium constant expression for this reaction is

A. C.

B. D.


Multiple Choice

Chapter 18

22C( ) O ( ) 2CO( )s + g g


4. The solubility product of cadmium carbonate, CdCO3, is 1.0 10−12. In a saturated solution of this salt, the concentration of Cd2+(aq) ions is

A. 5.0 . 10−13 mol/L.

B. 1.0 . 10−12 mol/L.

C. 1.0 . 10−6 mol/L.

D. 5.0 . 10−7 mol/L.


Multiple Choice

Chapter 18


4. The solubility product of cadmium carbonate, CdCO3, is 1.0 10−12. In a saturated solution of this salt, the concentration of Cd2+(aq) ions is

A. 5.0 . 10−13 mol/L.

B. 1.0 . 10−12 mol/L.

C. 1.0 . 10−6 mol/L.

D. 5.0 . 10−7 mol/L.

Standardized Test PreparationChapter 18

Multiple Choice


5. Consider the following equation for an equilibrium system:

Which concentration(s) would be included in the denominator of the equilibrium constant expression?

A. Pb(s), CO2(g), and SO2(g)

B. PbS(s), O2(g), and C(s)

C. O2(g), Pb(s), CO2(g), and SO2(g)

D. O2(g)

2 2 22PbS( ) 3O ( ) C( ) 2Pb( ) CO ( ) 2SO ( )s + g + s s + g + g


Multiple Choice

Chapter 18


5. Consider the following equation for an equilibrium system:

Which concentration(s) would be included in the denominator of the equilibrium constant expression?

A. Pb(s), CO2(g), and SO2(g)

B. PbS(s), O2(g), and C(s)

C. O2(g), Pb(s), CO2(g), and SO2(g)

D. O2(g)


Multiple Choice

Chapter 18

2 2 22PbS( ) 3O ( ) C( ) 2Pb( ) CO ( ) 2SO ( )s + g + s s + g + g


6. If an exothermic reaction has reached equilibrium,

then increasing the temperature will

A. favor the forward reaction.

B. favor the reverse reaction.

C. favor both the forward and reverse reactions.

D. have no effect on the equilibrium.


Multiple Choice

Chapter 18


7. Le Châtelier’s principle states that

A. at equilibrium, the forward and reverse reaction

rates are equal.

B. stresses include changes in concentrations, pressure, and temperature.

C. to relieve stress, solids and solvents are omitted from equilibrium constant expressions.

D. chemical equilibria respond to reduce applied stress.


Multiple Choice

Chapter 18

ch. 18 chemical equilibrium

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