Larry Emme Chemeketa Community College

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Reversible ReactionsReversible Reactions

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reversible reaction - a chemical reaction in which the products formed react to produce the original reactants.

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2NO2(g) → N2O4 (g)cooling

N2O4 (g) → 2NO2 (g)heating

The reaction between NO2 and N2O4

is reversible.N2O4 is formed

N2O4 decomposes when heated forming NO2

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2NO2(g) N2O4 (g)→→

reaction to the right (forward)

reaction to the left (reverse)

Ice water Hot water

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Rates of ReactionRates of Reaction

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• The rate of a reaction is variable. It depends on:– concentrations of the reacting species

– reaction temperature

– presence or absence of catalysts

– the nature of the reactants

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Chemical EquilibriumChemical Equilibrium

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equilibrium A dynamic state in which two or more opposing processes are taking place at the same time and at the same rate.

chemical equilibrium The state in which the rate of the forward reaction equals the rate of the reverse reaction in a chemical change.

At equilibrium the concentrations of the products and the reactants are not changing.

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NaCl(s) Na+(aq) + Cl-(aq)→→

A saturated salt solution is in equilibrium with solid salt.

salt crystalsare dissolving

Na+ and Cl- are crystallizing

At equilibrium the rate of salt dissolution equals the rate of salt crystallization.

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Le Chatelier’s PrincipleLe Chatelier’s Principle

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Henri LeChatelier

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In 1888, the French chemist Henri LeChatelier set forth a far-reaching generalization on the behavior of equilibrium systems.

This generalization, known as LeChatelier’s Principle, states

If a stress or strain is applied to a system in equilibrium, the system will respond in such a way as to relieve that stress and restore equilibrium under a new set of conditions.

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Effect of Concentration Effect of Concentration on Equilibriumon Equilibrium

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• For most reactions the rate of reaction increases as reactant concentrations increase.

• The manner in which the rate of reaction changes with concentration must be determined experimentally.

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An equilibrium is disturbed when the concentration of one or more of its components is changed. As a result, the concentration of all species will change and a new equilibrium mixture will be established.

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results in A and B being used faster than they are produced.

Increasing the concentration of B

A + B C + D→→

The system is at equilibrium

results in C and D being produced faster than they are used.

increases the rate of the forward reaction

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The system is again at equilibrium

After enough time has passed, the rates of the forward and reverse reactions become equal.

In the new equilibriumconcentration of A has decreased

concentrations of B, C and D have increased

A + B C + D→→

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Cl2(aq) + 2H2O(l) HOCl(aq) + H3O+(aq) + Cl-(aq)→→

Effect of Concentration Changeson the Chlorine Water Equilibrium

decrease Cl2 concentration

Equilibrium shifts to left

increase H2O concentration

Equilibrium shifts to right

increase HOCl concentration


decrease H3O+ concentration


increase Cl- concentration


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HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2(aq)→→ -2 3 2C H O ( )aq

1 L 0.100 M HC2H3O2

Equilibrium pH = 2.87

Effect of C2H3O2

Concentration Changes on pH

Add 0.100 mol NaC2H3O2

NaC2H3O2(aq) → Na+(aq) + C2H3O2(aq)-2 3 2C H O ( )aq


1 L 0.100 M HC2H3O2


Add 0.200 mol NaC2H3O2

1 L 0.100 M HC2H3O2


-2 3 2C H O

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Effect of Pressure Effect of Pressure on Equilibriumon Equilibrium

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• Changes in pressure significantly affect the reaction rate only when one or more of the reactants or products is a gas and the reaction is run in a closed container.

• The effect of increasing the pressure is to increase the concentrations of any gaseous reactants or products.

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CaCO3(s) CaO(s) + CO2(g)→→increases CO2 concentration


Increase Pressure

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CaCO3(s) CaO(s) + CO2(g)→→decreases CO2 concentration


Decrease Pressure

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In a system composed entirely of gases, a increase in the pressure of the container will cause the reaction and the equilibrium to shift to the side that contains the smallest number of molecules.

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N2(g) + 3H2(g) 2NH3(g)→→

Equilibrium shifts to the right towards fewer molecules.

Increase Pressure

1 mol 3 mol 2 mol

6.02 x 1023

molecules1.81 x 1024 molecules

1.20 x 1024

molecules

2.41 x 1024

molecules

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N2(g) + O2(g) 2NO(g)→→Equilibrium does not shift. The number of molecules is the same on both sides of the equation.

Increase Pressure

1 mol 1 mol 2 mol

6.02 x 1023

molecules6.02 x 1023

molecules1.20 x 1024

molecules

1.20 x 1024

molecules

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Effect of Temperature Effect of Temperature on Equilibriumon Equilibrium

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When the temperature of a system is raised, the rate of reaction increases.

In a reversible reaction, the rates of both the forward and the reverse reactions are increased by an increase in temperature.

The rate of the reaction that absorbs heat is increased to a greater extent, and the equilibrium shifts to favor that reaction. When the process is endothermic, the forward (left to right) reaction is increased. When the process is exothermic, the reverse (right to left) process is increased.

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At room temperature very little CO forms.

C(s) + CO2(g) + heat 2CO(g)→→At 1000oC moles CO2 moles CO

Heat may be treated as a reactant in endothermic reactions.

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Effect of CatalystsEffect of Catalystson Equilibriumon Equilibrium

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A catalyst is a substance that influences the rate of a reaction and can be recovered essentially unchanged at the end of the reaction.

A catalyst does not shift the equilibrium of a reaction. It affects only the speed at which the equilibrium is reached.

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Energy Diagram for an Exothermic Reaction

Activation energy: the minimum energy required for a reaction to occur.A catalyst speeds up a reaction by lowering the activation energy.

A catalyst does not change the energy of a reaction.

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PCl3(l) + S(s) → PSCl3(l)AlCl3

Very little thiophosphoryl chloride is formed in the absence of a catalyst because the reaction is so slow.

In the presence of a catalyst the reaction is complete in a few seconds.

MnO2

Δ 2KClO3(s) → 2KCl + 3O2(l)

The laboratory preparation of oxygen uses manganese dioxide as a catalyst to increase the rate of the reaction.

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Equilibrium ConstantsEquilibrium Constants

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At equilibrium the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products are constant.

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The equilibrium constant (Keq) is a value representing the unchanging concentrations of the reactants and the products in a chemical reaction at equilibrium.

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For the general reaction

aA + bB cC + dD→→c d

q a be

[C] [D]K =

[A] [B]

at a given temperature

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For the reaction

3H2 + N2 2NH3→→

3eq

2

2

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[NH ]K =

[H ] [N ]

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For the reaction

4NH3 + 3O2 2N2 + 6H2O→→

2 6

4 32 2

eq3 2

[N ] [H O]K =

[NH ] [O ]

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The magnitude of an equilibrium constant indicates the extent to which the forward and reverse reactions take place.

H2 + I2 2HI→→

oeq

2 2

2[HI]K = = 54.8 at 425 C

[H ] [I ]

COCl2 CO + Cl2→→

o2q

-4e

2

[CO][Cl ]K = 7.6 x 10= at 400 C

[COCl ]

At equilibrium more product than reactant exists.At equilibrium more reactant than product exists.

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When the molar concentrations of all species in an equilibrium reaction are known, the Keq can be calculated by substituting the concentrations into the equilibrium constant expression.

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Calculate the Keq for the following reaction on concentrations of PCl5 = 0.030 mol/L, PCl3 = 0.97 mol/L and Cl2 = 0.97 mol/L at 300oC.

PCl5(g ) PCI3(g) + Cl2(g)→→

3 2eq

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[PCl ][Cl ]K = =

[PCl ]

(0.97)(0.97) = 31

(0.030)

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Ionization ConstantsIonization Constants

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In addition to Kw, several other ionization constants are used.

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KaKa

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HC2H3O2(aq) H+ + C2H3O2→→ -

2 3 2C H O

When acetic acid ionizes in water, the following equilibrium is established:

Ka is the ionization constant for this equilibrium. + -

2 3 2a

2 3 2

[H ][C H O ]K =

[HC H O ]

Ka is called the acid ionization constant.

Since the concentration of water is large and does not change appreciably, it is omitted from Ka.

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At 25oC, a 0.100 M solution of HC2H3O2 is 1.34% ionized and has an [H+] of 1.34 x 10-3 mol/L. Calculate Ka for acetic acid.

HC2H3O2(aq) H+ + C2H3O2→→ -

2 3 2C H O

Because each molecule of HC2H3O2 that ionizes yields one H+ and one C2H3O2 , the concentrations of the two ions are equal.

-2 3 2C H O

+ - -32 3 2[H ] = [C H O ] = 1.34 x 10 mol/L

The moles of unionized acetic acid per liter are 0.100 mol/L – 0.00134 mol/L = 0.099 mol/L

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+ -2 3 2

a2 3 2

[H ][C H O ]K =

[HC H O ]

-3 -3(1.34x10 )(1.34x10 )

(0.099)

-5 = 1.8x10

+ - -32 3 2[H ] = [C H O ] = 1.34 x 10 mol/L

[HC2H3O2] = 0.099 mol/L

Substitute these concentrations into the equilibrium expression and solve for Ka.

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Because each molecule of HC2H3O2 that ionizes yields one H+ and one C2H3O2 , the concentrations of the two ions are equal.

-2 3 2C H O

What is the [H+] in a 0.50 M HC2H3O2 solution? The ionization constant, Ka, for HC2H3O2 is 1.8 x 10-5.

HC2H3O2(aq) H+ + C2H3O2→→ -

2 3 2C H O

The equilibrium expression and Ka for HC2H3O2 are:

+ -2 3 2

a2 3 2

[H ][C H O ]K =

[HC H O ]

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What is the [H+] in a 0.50 M HC2H3O2 solution? The ionization constant, Ka, for HC2H3O2 is 1.8 x 10-5.

HC2H3O2(aq) H+ + C2H3O2→→ -

2 3 2C H O

The equilibrium expression and Ka for HC2H3O2 are:

+ -2 3 2

a2 3 2

[H ][C H O ]K =

[HC H O ]

Let x = [H+] = -2 3 2[C H O ]+ -

2 3 2[H ] = [C H O ]

[HC2H3O2] at equilibrium is 0.50 – x.

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Substitute these values into Ka for HC2H3O2.

+ -2 3 2

a2 3 2

[H ][C H O ]K =

[HC H O ]

x = [H+] = -2 3 2[C H O ] HC2H3O2 = 0.50 - x

2-5

a

( )( )K = = = 1.8 x 10

0.50 - 0.50 - x x x

x x

Assume x is small compared to 0.50 - x. Then 0.50 – x 0.50

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Substitute these values into Ka for HC2H3O2.

+ -2 3 2

a2 3 2

[H ][C H O ]K =

[HC H O ]

x = [H+] = -2 3 2[C H O ] HC2H3O2 = 0.50 - x

2-5

a

( )( )K = = = 1.8 x 10

0.50 0.50 x x x

2-5 = 1.8 x 10

0.50 x

2 -5 = 0.50 x 1.8 x 10xSolve for x2.

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Take the square root of both sides of the equation. 2 -5 = 0.50 x 1.8 x 10x

-5= 0.90 x 10 -6= 9. 0 x 10

-6 -3 = 9.0 x 10 = 3.0 x 10 mol/Lx+ -3[H ] = 3.0 x 10 mol/L

Making no approximation and using the quadratic equation the answer is 2.99 x 10-3 mol/L, showing that it was justified to assume Y was small compared to 0.5.

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Calculate the percent ionization in a 0.50 M HC2H3O2 solution.

The percent ionization is given by

+ -concentration of [H ] or [A ] x 100= percent ionized

initial concentration of [HA]

The ionization of a weak acid is given by

HA H+ + A- →→

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[H+] was previously calculated as 3.0 x 10-3 mol/L

+ -2 3 2

2 3 2

concentration of [H ] or [C H O ] x 100= percent ionized

initial concentration of [HC H O ]

Calculate the percent ionization in a 0.50 M HC2H3O2 solution.

HC2H3O2(aq) H+ + C2H3O2→→ -

2 3 2C H O

The percent ionization of acetic acid is given by

The ionization of acetic acid is given by

-33.0 x 10 / x 100= 0.60% percent ionized

0.50 mol/Lmol L

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Acid Formula Ka Acid Formula Ka

Acetic HC2H3O2 1.8 x 10-5 Hydrocyanic HCN 4.0 x10-10

Benzoic HC7H5O2 6.3 x 10-5 Hypochlorous HOCl 3.5 x10-8

Carbolic HC6H5O 1.3 x 10-10 Nitrous HNO2 4.5 x10-4

Cyanic HCNO 2.0 x 10-4 Hydrofluoric HF 6.5 x10-4

Formic HCHO2 1.8 x 10-4

Ionization Constants (Ka) of Weak Acids at 25oC

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Ionic Equilibria Ionic Equilibria ProblemsProblems

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Use Ka as a measure of acid strength. Let 1 = Hi and 5 = Lo

Acid Ka order

HC2H3O2 1.8 10–5

H2S 6 10–8

HF 7 10– 4

HC2Cl3O2 2 10–1

HCN7.2 10–

10

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Use Ka as a measure of acid strength. Let 1 = Hi and 5 = Lo

Acid Ka order

HC2H3O2 1.8 10–5 3

H2S 6 10–8 4

HF 7 10– 4 2

HC2Cl3O2 2 10–1 1

HCN7.2 10–

10 5

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Find the pH in 0.1 M HCN if

Ka = 7.2 10–10

Step 1: ionize the acid

+ -

a

[H ][CN ]K =

[HCN]

Step 2: write the Ka expression

+HCN H CN→→

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Step 3: let x = [H+] = [CN–] [HCN] = 0.1 M – x ≈ 0.1 M

Step 4: substitute into Ka expression

10a

[ ][ ]K = 7.2 10

[0.1] x x

Step 5: solve for x:2 11

11 6

= 7.2 10

7.2 10 8.5 10

x

x

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Step 6: solve for pH

6pH - log 8.5 10 5.07

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Find [H+] in 0.1 M H2CO3 if

Ka = 4.0 10–17


+ 2 -23

a2 3

[H ] [CO ]K =

[H CO ]


+ -22 3 3H CO 2H CO→→

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Step 3: let x = [H+] , [CO3–2] = ½ x

[H2CO3] = 0.1 M – ½ x ≈ 0.1 M

Step 4: substitute into Ka expression2 3

172a

[ ] [ ]K = 4.0 10

[0.1] 0.2

xx x


3 18 6

= 8.0 10

8.0 10 2.0 10

x

x

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Find the pH of a 0.001 M solution of bombastic hydroxide (BmOH) if

Kb = 1.6 10–10

Step 1: ionize the base

+ -

b

[Bm ][OH ]K =

[BmOH]

Step 2: write the Kb expression

+BmOH Bm OH→→

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Step 3: let x = [Bm+] = [OH–] [BmOH] = 0.001 M – x ≈ 0.001 M

Step 4: substitute into Kb expression

10b

[ ][ ]K = 1.6 10

[0.001] x x


13 7

= 1.6 10

1.6 10 4.0 10

x

x

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Step 6: solve for pOH

7pOH - log 4.0 10 6.40

Step 7: solve for pH

pH 14 6.40 7.60

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Find the Ka for a 0.01 M HCN solution if the pH = 6.3


+ -

a

[H ][CN ]K =

[HCN]


+HCN H CN→→

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Step 3: use pH to find [H+]

Step 4: substitute into Ka expression7 7

11a

[5.0 10 ][5.0 10 ]K = 2.5 10

[0.01]

[H+] = 10–pH = 10–6.3 = 5.0 10–7

[H+] = [CN–], [HCN] = 0.01- 5.0 10–7 ≈ 0.01

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Solubility Product ConstantSolubility Product Constant

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The solubility product constant, Ksp, is the equilibrium constant of a slightly (sparingly) soluble salt.

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Silver chloride is in equilibrium with its ions in aqueous solution.

AgCl(s) Ag+(aq) + Cl-(aq) →→

+ -

eq

[Ag ][Cl ]K =

[AgCl( )]s

The equilibrium constant is

The amount of solid AgCl does not affect the equilibrium.

The concentration of solid AgCl is a constant.

+ -eq spK x [AgCl( )] = [Ag ][Cl ] = Ks

Rearrange

The product of Keq and [AgCl(s) is a constant.

+ -spK = [Ag ][Cl ]

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Ksp for Sparingly Soluble Solids

To find Ksp:

1. ionize the salt

2. write the Ksp expression

3. determine the M’s of the ions

4. substitute into the Ksp expression

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The solubility of AgCl in water is 1.3 x 10-5

mol/L.

Because each formula unit of AgCl that dissolves yields one Ag+ and one Cl-

, the concentrations of the two ions are equal.

AgCl(s) Ag+(aq) + Cl-(aq) →→Ksp = [Ag+][Cl-]

= (1.3 x 10-5)(1.3 x 10-5)

[Ag+] = [Cl-] = 1.3 x 10-5 mol/L

= 1.7 x 10-10

The Ksp has no denominator.

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Molar Solubility for Sparingly Soluble Solids

1. ionize the salt

2. write the Ksp expression

3. let x equal the amount of salt that

dissolves

4. find concentrations of ions in terms of x

5. substitute in the Ksp expression

6. solve for x

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The Ksp value for lead sulfate is 1.3 x 10-8. Calculate

the solubility of PbSO4 in moles per liter.

The equilibrium equation of PbSO4 is 2+ 2-

4 4PbSO Pb + SO→→

2+ 2-sp 4K = [Pb ][SO ]

The Ksp of PbSO4 is

Because each formula unit of PbSO4 that dissolves yields one Pb2+ and one , the concentrations of the two ions are equal.

2-4SO

2+ 2-4Let = [Pb ] = [SO ]x

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The Ksp value for lead sulfate is 1.3 x 10-8. Calculate

the solubility of PbSO4 in moles per liter.

2+ 2-sp 4K = [Pb ][SO ]

Substitute x into the Ksp equation.

-8spK = ( )( ) = 1.3 x 10x x

2 -8 = 1.3 x 10x

-8 = 1.3 x 10x-4= 1.1 x 10 mol/L

The solubility of PbSO4 is 1.1 x 10-4 mol/L.

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Compound Ksp Compound Ksp

AgCl 1.7 x 10-10 CaF2 3.9 x 10-11

AgBr 5 x 10-13 CuS 9 x 10-45

AgI 8.5 x 10-17 Fe(OH)3 6 x 10-38

AgC2H3O2 2 x 10-3 PbS 7x 10-29

Ag2CrO4 1.9 x 10-12 PbSO4 1.3 x 10-8

BaCrO4 8.5 x 10-11 Mn(OH)2 2.0 x 10-13

BaSO4 1.5 x 10-9

Solubility Product Constants (Ksp) at 25oC

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PracticePractice

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The Ksp value for silver carbonate is 6.2 x 10-12. Calculate the solubility of Ag2CO3 in moles per liter.

+ 2-2 3 3Ag CO 2Ag + CO→→

+ 2 2-sp 3K = [Ag ] [CO ]

If x = amount of salt that dissolves, then [Ag+] = 2x, [CO3

–2] = x

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x = 1.2 x 10-4 mol/L

4x3 = 6.2 x 10-12

x3 = 1.6 x 10-12

3 -12 = 1.6 x 10x

The solubility of silver carbonate is 1.2 x 10-4 mol/L

2spK = [2 ] [ ]x x = 4x3

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The Ksp value for calcium phosphide is 1.7 x 10-40. Calculate the solubility of Ca3P2 in moles per liter.

Solution in lecture!

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Acid-Base Acid-Base Properties of SaltsProperties of Salts

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hydrolysis is the term used for the general reaction in which a water molecule is split.

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→→-2 3 2 2 2

-3C H O (aq) + (l) C H O (aq) OH O+H HH ( )aq

The net ionic equation for the hydrolysis of sodium acetate is

The water molecule splits.

Salts that contain the anion of a weak acid undergo hydrolysis.

The solutionis basic.

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The net ionic equation for the hydrolysis of ammonium chloride is

The water molecule splits.

Salts that contain the cation of a weak base undergo hydrolysis.

The solutionis acidic.

+3

+4 3NH ( ) + ( ) NH ( ) OH O+ ( )H Haq l aq aq→→

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Salts derived from a strong acid and a strong base do not undergo hydrolysis.

+ -Na + Cl + ( ) no re O acH H tionl →

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Type of saltNature of

Aqueous SolutionExamples

Weak base-strong acid Acid NH4Cl, NH4NO3

Strong base-weak acid Basic NaC2H3O2

Weak base-weak acid Depends on the salt NH4C2H3O2, NH4NO2

Strong base-strong acid Neutral NaCl, KBr

Ionic Composition of Salts and the Nature of the Aqueous Solutions They Form

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Buffer Solutions: Buffer Solutions: The Control of pHThe Control of pH

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A buffer solution resists changes in pH when diluted or when small amounts of acid or base are added.

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Sodium acetate when mixed with acetic acid forms a buffer solution.

A weak acid mixed with its conjugate base form a buffer solution.

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+ -2 3 2 2 3 2 H ( ) + C H O ( ) HC H O ( )aq aq aq

If a small amount of HCl is added, the acetate ions of the buffer will react with the H+ of the HCl to form unionized acetic acid.


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- -2 3 2 2 2 3 2OH + HC H O ( ) H O( ) + C H O ( )aq l aq

If a small amount of NaOH is added, the acetic acid molecules of the buffer will react with the OH– of the NaOH to form water.


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A liter of a buffered solution contains 0.1 moles of H2CO3 and 0.05 moles of NaHCO3. Find the pH if Ka for the acid is 1.0 10–7

Solution in lecture!

Larry Emme Chemeketa Community College

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