Chemical Chemical EquilibriumEquilibrium

In a chemical reaction, the reactions never go in only one direction

In this chapter, we will study:

- the equilibrium concept

- equilibrium constant and its calculation

- activity of ionic species – ion effect

complex formation

- acid-base reactions

a A + b B c C + d D

The Rate ConceptThe Rate Concept

Similarly, in the backward reaction

Rate of backward reaction (Rb ) [C]c[D]d

= kb [C]c[D]d

If A is a solid or liquid molar concentration of A

If A is a gas pressure of A in atmosphere

Solvents are omitted from the equilibrium constant

rate constant: dependent on temperature, pressure and presence of catalyst

The Rate Concept

Rate of a chemical reaction is proportional to the “active masses” of the reacting substances present at any time

Rate of forward reaction (Rf ) [A]a[B]b

= kf [A]a[B]b

a A + b B c C + d D

[C]c[D]d

[A]a[B]b

kf

kb

At equilibrium:

rate of forward reaction = rate of backward reaction

kf [A]a[B]b = kb [C]c[D]d

Rearranging:

= = K

K can be evaluated by measuring the concentrations of A, B, C and D at equilibrium.

The larger the K value the farther to the right is the reaction at equilibrium the more favorable the rate constant for the forward reaction relative to the backward reaction

Manipulating Equilibrium Manipulating Equilibrium ConstantsConstants

Consider the reaction:

HA H+ + A-

K1 =

If the direction of a reaction is reversed, the new value of K is simple the reciprocal of the original value of K

H+ + A- HA

K’1 = = 1/ K1

[H+][A-]

[HA]

[H+][A-]

[HA]

Equilibrium constant for sum of reactions:

K3 = K1K2 = x

=

[H+][A-]

[HA]

[CH+]

[H+][C]

[A-][CH+][HA][C]

[A-][CH+]

If two reactions are added, the new K value is the product of the two individual values:

HA H+ + A- K1

H+ + C CH+ K2

HA + C A- + CH+ K3

Example :

The equilibrium constant at 25oC for the reaction:

H2O H+ + OH- Kw = 1.0 x 10-14

NH3(aq) + H2O NH4+ + OH-

K = 1.8 x 10-5

Find the equilibrium constant for the reaction:

NH4+ NH3 + H+

NH3

The 3rd reaction can be obtained by:

H2O H+ + OH- Kw

NH4+ + OH- NH3 + H2O 1/K

NH4+ H+ + NH3 K3 = Kw /K

= 5.6 x 10-10

NH3

NH3

Equilibrium and Equilibrium and ThermodynamicsThermodynamics

Enthalpy of a reaction : heat absorbed or released by the reaction

Entropy of a reaction : degree of disorder of reactants and products

contribute to the degree to which the reaction is favoured or disfavoured

entropy and enthalpy are related to equilibrium constant

Enthalpy change for a reaction (H) : Hproduct -

Hreactanct

H = + heat is absorbed endothermic

H = - heart is liberated exothermic

K value :

- does not tell us how fast a reaction will proceed toward equilibrium

- tells us the tendency of a reaction to occur and in what direction

EntropyEntropy

S = Sproduct - Sreactanct

S = + products are more disordered than reactants

S = - products are less disordered than reactants

Entropy of a substance, S : is the degree of disorder.

Greater the disorder greater the entropy eg gas is more disorder (has higher entropy) than a liquid which is turn is more disordered than a solid

A system will always tend toward lower energy and increased randomness ie lower enthalpy and higher entropy ie a chemical system is driven towards the formation of products by a negative H or a positive value of S, or both. The combined effects of these is given by the Gibbs free energy, G:

G = H – TS where T = temperature in Kelvins

G is a measure of the energy of the system, and a system spontaneously tends toward lower energy state.

Gibbs Free EnergyGibbs Free Energy

The change in energy of a system at a constant temperature is

G = H -TS

The equation combines the effects of H and S. Hence :G = + reaction is

disfavored

G = - reaction is favored

Go is related to the equilibrium constant of a reaction by:

K = e-G /RT or Go = -RTlnK

where R = gas constant = 8.314 JdegK-

1mol-1

From the equation: a large equilibrium constant results from a large negative free energy

o

Standard enthalpy, Ho, standard entropy, So, and standard free energy, Go represent the thermodynamic quantities at standard state (ie 1 atm, 298K and unit concentration)

Summary:

A chemical reaction is favored by (i) the liberation of heat (H negative), (ii) an increase in disorder (S positive), (iii) Go is negative or, equivalently, if K > 1.

Le Châtelier’s PrincipleLe Châtelier’s Principle

Suppose a system at equilibrium is subjected to a change that disturbs the system (eg the equilibrium concentrations of reactants and products are altered by changing the temperature, the pressure or the concentration of one of the reactants), the effects of such changes can be predicted from Le Châtelier’s Principle which states:

when a change is applied to a system at equilibrium, the equilibrium will shift in a direction that tends to relieve or counteract that change.

Example :

BrO3- + Cr3+ + 4H2O Br - + Cr2O7

2- + 8H+

for which the equilibrium constant is given by

K = =1 x 1011 at 25oC

In a particular equilibrium state of this system, the following concentrations exist:

[Br -][Cr2O72-][H+]8

[BrO3-][Cr3+]2

[H+] = 5.0 M; [Cr2O72-] = 0.10M;

[Cr3+] = 0.003M; [Br -] = 1.0M;

[BrO3-] = 0.043M

Dichromate is added to the solution to increase the concentration of [Cr2O7

2-] to

0.20M. In what direction would the reaction proceed to reach equilibrium?

According to Le Châtelier’s Principle, the reaction should move to the left to partially offset the increase in dichromate. This can be verified by setting up a reaction quotient, Q:

Q = = 2 x 1011 > K

(1.0)(0.20)(5.0)8

(0.043)(0.0030)2

Because Q > K reaction must move to the left to decrease the numerator and increase the denominator until Q = K

Hence to achieve equilibrium and:

If Q < K reaction must proceed to the right

If Q > K reaction must proceed to the left

What if the temperature is changed ?

temperature dependent temperature independent

From:

K = e-G /RT = e-(H - TS )/RT = e-H /RT .e-S /Ro o o o o

If Ho is negative, e-H /RT decreases with increasing temperature in an exothermic reaction, K decreases with increasing temperature

If Ho is positive, e-H /RT increases with increasing temperature in an endothermic reaction, K increases with increasing temperature

o

o

Solubility ProductSolubility Product

When substances have limited solubility and their solubility is exceeded, the ions of the dissolved portion exist in equilibrium with the solid material

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

- the substance will have a definite solubility at a given temperature

- a small very amount of undissociated compound usually exists in equilibrium in the aqueous phase and its concentration is constant

- the overall equilibrium constant for the solubility can be written for the stepwise equilibrium:

Ksp =

=

Since AgCl(s) is the pure solid [AgCl(s)] = 1

[AgCl(aq)]

[AgCl(s)] [AgCl(aq)]

[Ag+] [Cl-]

[Ag+] [Cl-][AgCl(s)]

Hence :

Ksp = [Ag+] [Cl-]

This relationship measures the compound’s solubility. It holds under all equilibrium conditions at the specified temperatures

Example:

What is the solubility of Hg2Cl2, in g/l, if the

solubility product is 1.2 x 10-18 ?

Hg2Cl2(s) Hg22+ + 2Cl-

Ksp = [Hg22+ ][Cl-]2 = 1.2 x 10-18

Let s represent the molar solubility of Hg2Cl2.

Then

[Hg22+ ] = s and [Cl-] = 2s

Thus: (s)(2s)2 = 1.2 x 10-18 s = 6.7 x 10 –7 M

Solubility in g/l = 6.7 x 10 –7 x 472.08g/mol

= 3.162 x 10 –10 g/l

Common Ion EffectCommon Ion Effect

What will be the concentration of Hg22+ if a

2nd source of Cl- was added to the solution (eg 0.030M NaCl) ?

Hg2Cl2(s) Hg22+ + 2Cl-

Initial conc solid 0 0.030

Final conc solid x 2x+ 0.030

Ksp = [Hg22+ ][Cl-]2 = x (2x + 0.030)2 = 1.2 x 10-18

(x)(0.030)2 = 1.2 x 10-18

x = 1.3 x 10-15 M

Without the NaCl, [Hg22+ ] = 6.7 x 10 –7 M

- addition of a product displaces the reaction toward the left

This application of Le Châtelier’s Principle is called the common ion effect : a salt will be less soluble if one of its constituent ions is already present in the solution

Separation by PrecipitationSeparation by Precipitation

Example :

Consider a solution containing lead(II)(Pb2+) and mercury(I)(Hg2

2+) ions, each at

a concentration of 0.010M

PbI2(s) Pb2+ + 2I- Ksp = 7.9 x 10-9

Hg2I2(s) Hg22+ + 2I- Ksp = 1.1 x 10-28

Is it possible to completely separate the Pb2+ and Hg2

2+ by selectively precipitating

the latter with iodide?

Precipitation reactions can sometimes be used to separate ions from each other.

Q = [Pb2+][I-]2 = (0.010)(1.0 x 10-11)2

= 1.0 x 10-24 < Ksp for PbI2

Hg2I2(s) Hg22+ + 2I-

Initial Conc 0 0.010 0

Final Conc solid 1.0 x 10 –6 x

Ksp = [Hg22+ ][I-]2 =(1.0 x 10 –6 )(x)2

= 1.1 x 10-28

x = [I-] = 1.0 x 10-11 M

Will this amount of I - cause Pb2+ to precipitate?

Consider lowering the Hg22+ concentration to

0.010% (ie .010% of 0.010M = 1.0 x 10 –6 M) of its original value without precipitating Pb2+.

Let x be the concentration of I- at equilibrium with 1.0 x 10 –6 M [Hg2

2+]

Complex FormationComplex Formation

If anion X- precipitates metal M+, it is sometimes observed that a high concentration of X- causes solids MX to redissolve. This phenomenon can be attributed to the formation of complex ions, such as MX2

-

In complex ions such as MX2-, X- is known as

the ligand of M+. A ligand is defined as any atom of group of atoms attached to the species of interest. M+ accepts electrons Lewis acid

X- donates electrons Lewis base

Room to accept electrons

Room to donate electrons

Example :

Pb++ + I - [Pb I ]+ adduct

dative or coordinate covalent bond

Effect of Complex Ion Effect of Complex Ion Formation on SolubilityFormation on Solubility

At low I- concentrations, the solubility of lead is governed by precipitation of PbI2 :

Pb2+ + 2I- PbI2(s) Ksp = 7.9 x 10-9

However at high I- concentrations, complex ion formation occurs as the reaction is driven to the right :

Pb2+ + I- PbI+ K1 = 1.0 x 102

Pb2+ + 2I- PbI2(aq) 2 = 1.4 x 103

Pb2+ + 3I- PbI3 – 3 = 8.3 x 103

Pb2+ + 4I- PbI4 – 4 = 3.0 x 104

Example :

Find the concentration of PbI+, PbI2(aq), PbI3–

and PbI42- in a solution saturated with PbI2(s)

and containing dissolved I- with a concentration of 1.0M.

Ksp = 7.9 x 10-9 = [Pb2+][I-]2 = [Pb2+]

[PbI+] = K1 [Pb2+][I-] = (1.0 x 102)(7.9 x 10-9)1

= 7.9 x 10-7 M

[PbI2(aq)] = 2[Pb2 +][I-]2 = (1.4 x 103 )(7.9 x 10-9) = 1.1 x 10-5 M

[PbI3–] = 3[Pb2 +][I-]3 = (8.3 x 103 )(7.9 x 10-9)

= 6.6 x 10-5M

[PbI42-] = 4[Pb2 +][I-]4 = (3.0 x 104 )(7.9 x 10-9)

=2.4 x 10-4M

The total concentration of dissolved lead is:

[Pb]total = [Pb2 +] + [PbI+] +[PbI2(aq)] +

[PbI3–] + [PbI42-]

= 3.2 x 10-4M

% dissolved Pb in PbI42- = (2.4 x 10-4)/(3.2 x 10-

4) x 100

= 75%

Each of the reaction is governed by an equilibrium and an equilibrium constant. The total concentration of dissolved lead is considerably greater than that of

Pb2+ alone

The concentration of Pb2+ that satisfies any one of the equilibria must satisfy all the equilibria.

Protic Acids and BasesProtic Acids and Bases

H+ is known as proton because it is what remains when a hydrogen atom loses its electron

Protic refers to the chemistry that involves the transfer of H+ from one molecule to another

In the Brønsted and Lowry classification:

acid proton donor

eg HCl + H2O H3O+ + Cl-

bases proton acceptors

eg HCl + NH3 NH4+Cl-

Salts an ionic compound

product of an acid-base neutralization reaction

typically strong electrolyte

Conjugate Acids and BasesConjugate Acids and Bases

In the Brønsted and Lowry classification: the products of a reaction between an acid and a base are also classified as acids and bases. What does this mean?

Example :

acetic acid methylamine acetate ion methyl-

ammonium ion

CH3 C

O

O H

+ CH3 NH

H

CH3 C

O

O -+ CH3 N

HH

H

+

acid basebase acid

Acetic acid and the acetate ion are said to be a conjugate acid-base pair

Conjugate acids and bases are related to each other by the gain or loss of one H+

AutoprotolysisAutoprotolysis

Some compounds are able to undergo self-ionization, in which they act as both an acid and a base

Example :

H2O H+ + OH-

2 H2O H3O+ + OH-

2 NH3 NH4+ + NH2

-

What compounds undergo autoprotolysis?Compounds with a reactive H+ protic solvent

Examples of protic solvents:

- CH3 CH2 OH (ethanol)

- CH3 COOH (acetic acid)

Compounds without a reactive H+

aprotic solvent (eg CH3 CH2 O CH2 CH3 , CH3

CN)

pHpH

pH -log[H+]

- value can lie from –2 +16

- pH = -1 means -log[H+] = -1.00 or

[H+] = 10M

very concentrated strong acid

-2 0 2 4 6 8 10 12 14 16

neutral

acidic basic

For water : H2O H+ + OH-

Kw = [H+][OH-] = 1.0 x 10-14 at 25oC

log Kw = log [H+] + log[OH-]

- log Kw = -log [H+] - log[OH-]

14.00 = pH + pOH

Strengths of Acids and Strengths of Acids and BasesBases

Acids and bases are commonly classified as strong or weak, depending on whether they react nearly “completely” or only “partially” to produce H+ or OH-

Example of strong acid: HCl(aq) H+ + Cl-

Example of strong base: KOH(aq) K+ + OH-Weak acids and bases react “partially” to produce H+ or OH-

Weak acids (HA) react with water by donating a proton to H2O ie

HA + H2O H3O+ + A-

or

HA + H2O H+ + A- Ka =

[H+][A-]

[HA]

acid dissociation constant

– value is small for weak acids

Weak bases (B) react with water by abstracting a proton from H2O ie

B + H2O BH+ + OH- Kb =[BH+][OH-]

[B]

Base hydrolysis constant

– value is small for weak bases

Acids and BasesAcids and Bases

HOC-COH H+ + -OC-COH

O O O O

-OC-COH H+ + -OC-CO-

O O O O

Compounds that can donate more than one proton polyprotic acids

Compounds that can accept more than one proton polyprotic bases

P

O

O-

O-O- + 3H2O P

O

OHHO+ 3HO-

HO

P

O

O-

O-O- + H2O P

O

OH+ HO-

O-

O-

P

O

O-O- + H2O P

O

OHOH

HO

+ HO-O-

P

O

O- + H2O P

O

OHOH

HO

+ HO-

HO

HO

Kb1 = 1.4 x 10-2

Kb2 = 1.59 x 10-7

Kb3 = 1.42 x 10-12

P

O

O-

O-O- + 3H2O P

O

OHHO+ 3HO-

Kf = Kb1 .Kb2 .Kb3

HO