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CHEMICAL EQUILIBRIUM. CHEMICAL EQUILIBRIUM. the dynamic state in which rates of the forward and reverse reaction s are identical. aA + bB cC + dD. K>1  Forward reaction is favoured. Equilibrium constant a,b,c,d – stoichiometry coefficients

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the dynamic state in which rates of the forward and reverse reactions are identical.

aA + bB cC + dD

K>1 

Forward reaction is


Equilibrium constant

a,b,c,d – stoichiometry coefficients

[A], [B], [C], [D] –concentrations of A, B, C, D

  • in standard state:

  • For solutes – 1M

  • For gases – 1 atm

  • For solids and pure liquids: [X] = 1

  • In these conditions K is dimensionless.

Equilibrium const. for a reverse reaction:K1 = 1/K

cC + dD aA + bB

Equilibrium const. for two reactions added:K3 = K1 x K2

HA H+ + A-

H+ + C CH+

HA + C CH+ + A-


The equilibrium constant is derived from the thermodynamics of a chemical reaction.


ΔH - enthalpy change is the heat absorbed or released during reaction

ΔH> 0

Heat absorbed

Endothermic reaction

ΔH< 0

Heat liberated

Exothermic reaction


KCl(s) K+(aq) + Cl-(aq)

ΔS0 = +76 J/K molat 250C


ΔS – a measure of ‘disorder’ of a substance

Gas  Liquid  Solid

Decrease in disorder

ΔS> 0

Products more disordered than reactants

ΔS< 0

Products less disordered than reactants


Gibbs free energy:ΔG = ΔH - TΔS

ΔG< 0 The reaction is favoured K>1

ΔG> 0 The reaction is not favoured K<1

Free energy and equilibrium:


A + B C + D


When a system at equilibrium is disturbed, the direction in which the system proceeds back to equilibrium is such that the change is partially offset.

If reaction is at equilibrium

and reactants are added

(or products removed),

the reaction goes to the right.

If reaction is at equilibrium

and products added ( or reactants are removed),

the reaction goes to the left.


Q = Reaction quotient

has the same form as equilibrium constant (K), but the solution concentrations do not have to be equilibrium concentrations.

Thus at equilibrium:

Q = K

According to Le Châtelier:

If Q > K  reaction will proceed in the reverse direction

If Q < K reaction will proceed in the forward direction

A + B C + D

At equilibrium:

[A] = 0.002M [B] = 0.025M

[C] = 5.0M [D] = 1.0M

= 1x105 at 250C

Say we add double reactant A, [A] = 0.004M

= 0.5x105 at 250C

Q < K

To reach equilibrium:

Q = K

and the reaction must go to the right

 spontaneous



Independent of T

For endothermic reactions (ΔH > 0):

K increases if T increases.

For exothermic reactions (ΔH < 0):

K decreases if T increases.


K = Ksp (solubility product) when the equilibrium reaction involves a solid salt dissolving to give its constituent ions in solution.

Recall: [Solid] = 1

Saturated solution – in equilibrium with undissolved solid

Thus if an aqueous solution is left in contact with excess solid, the solid will dissolve until Ksp is satisfied.

Thereafter the amount of undissolved solid remains constant.

PbCl2(s) Pb2+(aq) + 2Cl-(aq)


Calculate the mass of PbCl2 that dissolves in 100 ml water. (Ksp = 1.7x10-5 for PbCl2)





m = 0.45 g

PbCl2(s) Pb2+(aq) + 2Cl-(aq)


  • Now add 0.03M NaCl to the PbCl2 solution

  • We added 0.03M Cl-





For this system to be at equilibrium when [Cl-] is added, the [Pb2+] decreases (reverse reaction).

– this is an application of the Le Chatelier’s principle and is called THE COMMON ION EFFECT

The salt will be less soluble if one of its constituent ions is already present in the solution.

The Common Ion Effect - experiment


H+ does not exist on its own in H2O  forms H3O+


In aqueous solution, H3O+ is tightly associated with 3 molecules of H2O through exceptionally strong hydrogen bonds.

One H2O is held by weaker ion-dipole attraction

Can also form H5O2+ cation  H+ shared by 2 water molecules

H3O2- (OH-.H2O) has been observed in solids


Water undergoes self-ionisation autoprotolysis,

since H2O acts as an acid and a base.

H2O + H2O H3O++ OH-

The extent of autoprotolysis is very small.

For H2O: Kw = [H3O+][OH-] = 1.0 x 10-14(at 25oC)


pH ≈ -log[H+]

Approximate definition of pH

pH + pOH = -log(Kw) = 14.00 at 250C

It is generally assumed that the pH range is 0-14. But we can get pH values outside this range.

e.g. pH = -1  [H+] = 10 M

This is attainable in a strong concentrated acid.

aA + bB cC + dD


Equilibrium constant

[A], [B], [C], [D] –concentrations of A, B, C, D

  • BUT in a real solution all charged ions are:

  • surrounded by ions with opposite charge – ionic atmosphere

  • hydrated - surrounded by tightly held water dipoles

Adding an “inert” salt to a sparingly soluble salt increasesthe solubility of the sparingly soluble salt.

“inert” salt = a salt whose ions do not react with the compound of interest



BaSO4 (Ksp = 1.1x10-10) as the sparingly soluble salt and KNO3 as the “inert” salt. In solution:

The cation (Ba2+) is surrounded by anions (SO42-, NO3-)

net positive charge is reduced

The anion (SO42-) is surrounded by cations (Ba2+, K+)

net negative charge is reduced

 attraction between oppositely charged ions is decreased.

The net charge in the ionic atmosphere is less than the charge of the ion at the center.

The ionic atmosphere decrease the attraction between ions.

The greater the ionic strength of a solution, the higher the charge in the ionic atmosphere.

Each ion-plus-atmosphere contains less charge and there is less attraction between any particular cation and anion.

Activity of the ion in a solution depends on its hydrated radius not the size of the bare ion.


A measure of the total concentration of ions in solution. The more highly charged an ion, the more it is counted.

Where ci = concentration of theith species zi= charge for all ions in solution


Find the ionic strength of 0.010 M Na2SO4 solution.

Effect of ionic strength on solubility

Explain all 4 cases


To account for the effect of ionic strength, concentrations are replaced by activities.

Activity coefficient

Activity of C

And general form of equilibrium constant is:

  • Activity coefficient:

  • Measure of deviation of behaviour from ideality (ideal  = 1)

  • Allows for the effect of ionic strength

At low ionic strength:

activity coefficients  1

and K  concentration equilibrium

Thus for the sparingly soluble salt BaSO4, dissolving in the presence of the “inert” salt KNO3:

Ksp = aBa aSO4 = [Ba2+]Ba [SO42-]SO4

If more BaSO4 dissolves in the presence of KNO3,

[Ba2+] and [SO42-] increases and Ba and SO4 decreases


Extended Debye-Hűckel equation relates activity coefficients to ionic strength:

at 250C

 = effective hydrated radius of the ion

Effect of Ionic Strength, Ion charge and Ion Size on the Activity Coefficient

(Over the range of ionic strength from 0 to 0.1M)

As ionic strength increases, the activity coefficient decreases.   1 as   0

  • As the charge of the ion increases, the departure of its activity coefficient from unity increases. Activity corrections are much more important for an ion with a charge of 3 than one with the charge 1.

Activity coefficients for differently charged ions with a constant hydrated radius of 500pm.

3. The smaller the hydrated radius of the ion, the more important activity effects become.

Obtain values for  from the table:

Use interpolation to find values of  for ionic strengths not listed

How to interpolate - SELF STUDY!!

Linear interpolation:

At high ionic strengths:

activity coefficients of most ions increase

Concentrated salt solutions are not the same as dilute aqueous solutions  “different solvents”

H+ in NaClO4 solution of varying ionic strengths


pH ≈ -log[H+]

Approximate definition of pH

The real definition of pH is:


A pH electrode measures activity of H+ and NOT concentration


Chemical equilibrium provides a basis for most techniques in analytical chemistry and application of chemistry to other disciplines such as like biology, geology etc.

The systematic treatment of equilibrium gives us the tool to deal with all types of complicated chemical equilibria.

The systematic procedure is to write as many independent algebraic equations as there are unknowns (species) in the problem. This includes all chemical equilibrium conditions + two balances: charge and mass balances.


The sum of positive charges in solution equals the sum of negative charges.

Charge neutrality

E.g. An aqueous solution of KH2PO4 and KOH contains the following ionic species:

H+, OH-, K+, H2PO4-, HPO42-, PO43-

The charge balance is:

[H+] + [K+] = [OH-] + [H2PO4-] + 2[HPO42-] + 3[PO43-]

The coefficient in front of each species

= the magnitude of the charge on the ion


Conservation of matter.

Quantity of all species in a solution containing a particular atom must equal the amount of that atom delivered to the solution.

E.g. Mass balance for 0.02 M phosphoric acid in water:

0.02 M = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]


Step 1. Write the pertinent reactions.

Step2. Write the charge balance equation.

Step3. Write the mass balance equations.

Step4. Write the equilibrium constant for each chemical reaction.

Step 5. Count the equations and unknowns.

Step 6. Solve for all the unknowns.

E.g.: The ionization of water

H2O H+ + OH- Kw = 1.0x10-14 at 250C

Find the concentrations of H+ and OH- in pure water

Step 1. Pertinent reaction

– only one above.

Step 2. Charge balance:


Step 3. Mass balance:


Step 4. Equilibrium constants

– the only one


Step 5. Count equations and unknowns

– 2 eq. and 2 unknowns

Step 6. Solve.

For pure water the ionic strength approaches 0 and we can write eq.3 as:

Step 1. Pertinent reactionsHg2Cl2 Hg22+ + 2Cl- KspH2O H+ + OH-Kw

E.g.: The solubility of Hg2Cl2

Find the concentration of Hg22+ in a saturated solution of Hg2Cl2

Step 2. Charge balance: (1)

Step 3. Mass balance:


Step 4. Equilibrium constants: (3)


Step 5. Count equations and unknowns – 4 eqs. and 4 unknowns

Step 6. Solve. Using eqn 2we can write eqn 3 as:


Coupled equilibria – the product of one reaction is reactant in the next reaction


The mineral fluorite, CaF2,has a cubic crystal structure and often cleaves to form nearly perfect octahedra.

Find the solubility of CaF2 in water.

CaF2 dissolves:

CaF2(s) Ca2+ + 2F- Ksp = 3.9x10-11

The F- ions reacts with water to give HF:

F- + H2O HF + OH- Kb = 1.5x10-11

For every aqueous solution:

H2O H+ + OH-Kw = 1x10-14

Step 1.Pertinent reaction

Step 2.Charge balance


Step 3.Mass balance

Some fluoride ions react to give HF.



CaF2(s) Ca2+ + 2F- Ksp

F- + H2O HF + OH- Kb

H2O H+ + OH-Kw

Step 4.Equilibrium constants




Step 5.Count equations and unknowns

5 eqs. and 5 unknowns:

Step 6.Solve

To simplify the problem let us solve it for a fixed pH = 3

That means:[H+] = and [OH-] =

Then from eqn 4:

Kb = [HF][OH-]/[F-]

[HF]/[F-] = Kb/[OH-] = 1.5x10-11/1.0x10-11= 1.5

Thus[HF] = 1.5[F-]

Substitute [HF] in the eqn 2:

[F-] + [HF] = 2[Ca2+]

[F-] + 1.5[F-] = 2[Ca2+]

And[F-] = 0.80[Ca2+]

Using this expression in eqn 3:

Ksp = [Ca2+][F-]2 = [Ca2+](0.80[Ca2+])2

Now find:

[F-] = 3.1x10-4 M

[HF] = 4.7x10-4 M

Thus[Ca2+] = (Ksp/0.802)1/3 = 3.9x10-4M

NOTE: To fix the pH of a solution an ionic compound is added. Thus the charge balance equation as written not longer holds.

pH dependence of the conc. of Ca2+, F- and HF in a saturated solution.

Also [OH-] = [H+] + [HF]

No longer holds

SO2(g) + H2O(l)  H2SO3(aq)



Applications of coupled equilibria in the modeling of environmental problems

Found [Ca] in acid rain that has washed off marble stone (largely CaCO3) increases as the [H+] of acid rainincreases.

CaCO3(s) + 2H+(aq) 

Ca2+(aq) + CO2(g) + H2O(l)

Al is usually “locked” into insoluble minerals e.g. kaolinite and bauxite. But due to acid rain, soluble forms of Al are introduced into the environment. (Similarly with other minerals containing Hg, Pb etc.)

Total [Al] as a function of pH in 1000 Norwegian lakes.

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