Chemical Equilibrium-Applications, Chemistry tutorial

Introduction:

We are familiar that a reversible reaction tends to a state of dynamic equilibrium. At equilibrium there is a fixed relationship among the molar concentrations of the reactants and products. This last statement is the equilibrium law and the fixed relationship is the equilibrium constant for the reaction at the particular temperature. A change of reaction conditions like temperature, pressure and concentration influence a reaction at equilibrium. The effect of such changes in reaction conditions on the equilibrium is illustrated by Le Chatelier's principle.

Most of the industrial methods comprise reversible reactions and optimum conversion of the reactants to products is desired at minimum cost. The conditions of the reactions chamber can greatly influence the position of equilibrium and therefore the economy of the process. The Haber method for the production of ammonia and the contact method for the production of tetraoxo-sulphate (vi) acid are illustrations. In both cases there is a consideration of the best manner of accomplishing a good yield of products at minimum cost.

The equilibrium law has been applied to numerous reactions in solution. Such reactions comprise acid-base dissociation in water, salt hydrolysis, buffer solutions, solubility product and general ion effect.

Applications of the Equilibrium Law to Gas Reactions:

The Haber procedure:

In the Haber method ammonia gas is generated from nitrogen and hydrogen gases. In this large scale production of ammonia, the mixture of one portion of nitrogen and three portion of hydrogen is compressed to very high pressures (200 to 1000 atmospheres) and passed via the catalyst chamber at 450 to 500°C. The catalyst is iron mixed by aluminum oxide (that I, alumina). The alumina promotes the action of the iron as a catalyst.

N2 (g) + 3H2 (g) ↔ 2 NH3 (g), ΔH -ve

                      Catalyst

The industrial conditions for the Haber method are illustrated via Le Chatelier's principle.

a) Very high pressures as the reaction are accompanied via a decrease in volume.

b) A temperature of 450 to 500°C. The reaction is exothermic and a lower temperature would raise the yield of ammonia. The rate of the reaction is though too low at lower temperatures.

c) The iron catalyst is utilized for the equilibrium to be established faster.

Even by the above industrial conditions the yield of ammonia is just 10 to 15 %. The reaction is pushed in the forward direction for more ammonia formation via eliminating the ammonia as it is formed.

Remember that whenever the product concentration is decreased, an equilibrium reaction shifts in the product direction to form more products.

The contact process:

Big scale production of tetraoxosulphate (vi) acid is through the contact process. The major reaction is the catalytic oxidation of sulphur (iv) oxide gas to sulphur (vi) oxide gas. This is a reversible reaction accomplished through evolution of heat (that is, exothermic).

2SO2 (g) + O2 (g) ↔ 2SO3 (g) ΔH -ve

                       Catalyst

The catalyst is vanadium (v) oxide (V2O5).

The mixture of sulphur (iv) oxide and surplus oxygen (air) is first purified and dried. The dry pure mixture of gases is passed via the catalyst chamber at a temperature of around 450°C and at atmospheric pressure.

The industrial conditions are as well illustrated through Le Chatelier's principle.

i) Atmospheric pressure is employed however an increased pressure must enhance the yield. The yield at this low pressure is around 98 percent and a higher pressure effect will be negligible.

ii) A temperature must enhance the yield as the reaction is exothermic. The yield is pretty high at this temperature and a lower temperature will reduce the rate of the reaction.

iii) The catalyst assists to decrease the time of the method.

The two processes are employed to push the reaction in favor of the product.

i) The product SO3 (g) is eliminated as it is made.

ii) Surplus oxygen is employed.

Applications of the Equilibrium Law to Aqueous Equilibria:

Dissociation of weak acids and bases:

The dissociation of a weak acid or base in water is the reversible reaction. The degree of dissociation is deduced in the acid/base dissociation constant (ka or kb). For the acid HA and base MOH in the aqueous solution.

HA (aq) + H2O (l) ↔ H3O+ (aq) + A- (aq)

H+ is known as the hydroxonium ion

For base MOH,

MOH (aq) ↔ M+ (aq) + OH- (aq)

KC = ([H3O+][A-])/([HA][H2O])

KC [H2O] = ([H3O+][A-])/([HA]) = Ka

For base MOH,

[M+][OH-]/[MOH] = Kb

Ka, Kb is the acid and base dissociation constants correspondingly. The acid or base strength is found out by the extent of dissociation. Acids or bases that are strongly ionized will encompass high dissociation constants. Acids/ bases having low dissociation constants are weak acids or bases.

The table represents Ka, Kb values for several acids and bases.

Acid/Base       Ka/Kb values

CH3COOH       1.8 x 10-5

HF                   6.5 x 10-4

HCN                3.1 x 10-8

NH3                1.8 x 10-5

CH3NH2         3.7 x 10-4

Dissociation of water and the pH scale:

Water dissociates slightly to represent:

H2O (l) ↔ H+ (aq) + OH- (aq)

KC = [H+][OH-]/[H2O]

Can you now describe why water conducts electricity?

KC [H2O] = [H+][OH-] = KW

Kw is frequently known as the ion product of water. At 25°C (298K)

Kw = 10-14 and for pure water [H+] = [OH-]

At 25°C, [H] = [OH-] = 10-7

Due to the very small values of [H] and [OH-], an acidity scale known as the pH scale is employed.

pH = - log H+

For pure water pH = - log 10-7

                             = 7

In similar manner we state,

pOH = - log [OH-]

For water pOH = 7

pH = pOH = 7 for water at 25oC

And pH + pOH = 14 at similar temperature.

Whenever an acid dissolves in water the (H+) increases and pH < 7 however if a base dissolves [OH-] increases pOH < 7 or pH > 7

On the pH scale thus, acid solutions encompass pH < 7 whereas alkaline solutions encompass pH > 7 

873_Dissociation of water and the pH scale.jpg

Fig: Dissociation of water and the pH scale

Illustration: Compute the pH of 0.001 HCl solution

HCl is a strong acid

HCl (aq) → H+ (aq) → Cl- (aq)

[H+] = .001 mol dm-3

pH = - log H+

     = - log .001

     = 3

Now compute pH for .005 H2SO4 (aq)

Hydrolysis of salts:

The solution of an acid will turn blue litmus red whereas alkaline solution turns red litmus blue. Several salt solutions are acidic or alkaline to litmus. This is due to the reason of hydrolysis of the salts. For illustration, a solution of sodium ethanoate turns red litmus blue and a solution of ammonium chloride turns blue litmus red. The above observation is illustrated via a reversible reaction taking place between the salts and water that leads to alkaline or acidic behavior. CH3COONa (aq) + H2O ↔ CH3COOH (aq) + NaOH. As sodium hydroxide is a strong base and CH3COOH is a weak acid, the solution is alkali that is, the alkali consists of an upper hand. 

For NH4Cl,

NH4Cl (aq) + H2O (l) ↔ NH4OH (aq) + HCl (aq)

HCl is a strong acid though NH4OH is a weak alkaline (base). The acid consists of the upper hand and the resultant solution is acidic to litmus. The theory of hydrolysis is very useful in illustrating such observations as above and as well in making proper choice of indicator for acid-base titrations.

Buffer Solutions:

Buffer solutions are the solutions of a weak acid or base and one of its salts.

Such solutions hold the capability to resist pH change whenever small amounts of acid or base are added to them. Illustrations of buffers are ethanoic acid and its sodium salt and ammonia solution and ammonium chloride. The first illustration is acid buffer and the second is the alkaline buffer. The capacity of buffer solutions to resist change of pH is illustrated through reversible reactions in the acid-salt mixture.

Take for illustration the buffer having ethanoic acid and its sodium salt. The given reactions take place.

CH3COOH (aq) ↔ CH3COO- (aq) + H+ (aq) Weak acid

CH3COONa (aq) → CH3COO- (aq) + Na+ (aq) Highly dissociated

Whenever a small amount of H+ is added to the buffer solution the weak acid equilibrium is reversed and un-dissociated CH3COOH is made. This will have little effect on the CH3COO- in solution due to CH3COO-from the second reaction. Likewise whenever OH- is added the reaction OH- + H+ → H2O takes place (H+) is reduced and the acid equilibrium shifts in the forward direction. This generates more CH3COO- however not adequate to make much difference as large concentration of CH3COO- from the salt dissociation.

Buffer solutions are significant in reactions which take place only at constant hydrogen ion concentration (pH). They are extremely significant in the body as most of body metabolism takes place only at constant pH.

Common ion effect:

The saturated solution is one in which no more of solute will dissolve in the solvent at that temperature. The saturated solution is in equilibrium having surplus solid in solution.

Take the illustration of a saturated calcium chloride solution

CaCl2 (s) ↔ Ca2+ (aq) + 2Cl- (aq)

If to this solution is added a few sodium chloride crystals (NaCl), the NaCl as well furnishes Cl- in solution.

NaCl (aq) ↔ Na+ (aq) + Cl- (aq)

Cl- is the common ion in this illustration. The extra Cl- ions in solutions will shift the CaCl2 equilibrium in the reverse direction. More solid will come out that is, solubility of CaCl2 reduces:  The common ion effect can be stated as the decrease in solubility caused by the presence of an ion of a salt which is added to the solution of the salt from the external source.

The industrial application of common ion effect is in the soap industry. Soap is generated through treating oil or fat having sodium hydroxide. Fat or oil + NaOH → Soap + Propane 1, 2, 3 triol.

Soap is the sodium salt of the fatty acid. Soap is fairly soluble generating Ne and the anion of the fatty acid. Addition of the sodium chloride (NaCl) to the soap solutions will introduce Ne as common ion. The effect is that the solubility of the soap is decreased and more solid soap precipitates out. The outcome of soap is by this method increased. This is termed as salting out of soap. This is a practical application of common ion effect.

Solubility product:

For sparingly soluble salts, the solubility product is the index of solubility. Salts having high value of solubility product are more soluble than others by low solubility product values.

Take the illustration of a saturated AgCl solution.

AgCl(s) → Ag+ (aq) + Cl- (aq)

KC = [Ag+][Cl-]/[AgCl]

KC [AgCl] = [Ag+][Cl-] = KSP

KSP is the solubility product of the AgCl. The significance of the solubility ion theory lies in its bearing on precipitation from the solution of metal ions.

The solubility product of AgCl at 25°C is 1.1 x 10-10 mol2 dm-6. In any aqueous solution, the product of the molar concentration of Ag+ and Cl- in solution can't surpass this value. If for silver chloride the product of the concentrations is made to go beyond this value. Adding a salt having a common ion or mixing solutions having high ion concentrations of Ag+ and Cl- ions will outcome in precipitation of the salt.

Illustration:

The solubility product of AgCl is 1.1 x 10-10 at 25°C. Describe why precipitation takes place if a drop of 0.01 mol dm-3 NaCl is added to 10.0 cm3 of 0.l0 mol dm-3 AgNO3 solution in a test tube.

Answer:

AgNO3 → Ag+ (aq) + NO3- (aq)

Dissociation is supposed complete.

[Ag+] = .10

[NO3-] = .10

For AgCl to be precipitated:

[Ag+] [Cl-] > 1.1 x 10-10

[Cl-] > (1.1 x 10-10)/0.1 = 1.1 x 10-9

The quantity of Cl- is extremely small that a tiny drop of 0.01 mol dm-3 sodium chloride solution will provide precipitate. We are familiar that this is a test for chlorides.

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