Industrial Applications of Le Chatelier's Principle & Haber Process - Notes | Study Chemistry Class 11 - NEET

INDUSTRIAL APPLICATIONS OF LE CHATELIER'S PRINCIPLE

1. HABER PROCESS

In Haber process, the ammonia is synthesized by combining pure nitrogen and hydrogen gases in 1:3 ratio in presence of finely powdered iron catalyst and molybdenum promoter at around 450^oC and at about 250 atm. of pressure.

The le Chatelier's principle helps in choosing these conditions to improve the yields of ammonia as explained as below.

Effect of pressure:

Hence the synthesis of ammonia is favored by increasing the pressure of the system. Industrially, 100 - 250 atm. of pressure is employed. Effect of temperature: Since the forward reaction is exothermic, the increase in temperature favors the backward reaction i.e., the dissociation of ammonia. That means according to le Chatelier's principle, the synthesis of ammonia is favored at lower temperatures. However the reaction will be too slow at lower temperatures (a kinetic restriction). Hence this reaction is carried out at optimal temperatures i.e., at about 450 - 550 oC to overcome the kinetic barrier. Removal of ammonia: The forward reaction can also be favored by removing ammonia from the system from time to time by liquefying it. Catalyst: To increase the speed of the reaction, finely powdered or porous iron is used as catalyst. Its efficiency can be improved by adding molybdenum or oxides of potassium and aluminium.

2. CONTACT PROCESS

In the contact process, sulfuric acid, the king of chemicals, is manufactured on large scale. The major steps involved in the process are:

The crucial step is the oxidation of sulfur dioxide, SO₂ to sulfur trioxide, SO₃. It is a reversible reaction. At normal conditions, the equilibrium lies far to the left and the amount of sulfur trioxide formed is very small. To improve the yield of sulfur trioxide, the reaction is carried out at around 450^oC and 2 atm pressure in presence of V₂O₅ or Pt, which acts as a catalysts.

These conditions are chosen by applying le Chatelier's principle as explained below.

Effect of pressure: In the forward reaction (formation of sulfur trioxide), the number of moles of gaseous components is decreasing.

Hence the forward reaction is favored by increasing the pressure of the system. However, at high pressures, the iron towers used in the contact process are corroded. Hence the process is carried out at optimal pressures like 2 atm.

Effect of temperature: Since the forward reaction is exothermic, at higher temperatures the backward reaction i.e., the dissociation of sulfur dioxide is more favored. However the reaction will be too slow at lower temperatures. Hence this reaction is carried out at optimal temperatures i.e., around 450° C.

Catalyst: To increase the speed of the reaction, V₂O₅ or Pt are used as catalysts

Ex. Under what conditions will the following reactions go in the forward direction?

1. N₂(g) + 3H₂(g)  2NH₃(g) +23 Kcal.

2. 2SO₂(g) + O₂(g)  2SO₃(g)+ 45 Kcal.

3. N₂(g) + O₂(g)  2NO(g) - 43.2 Kcal

4. 2NO(g) + O₂(g)  2NO₂(g)+ 27.8 Kcal

5. C(s) + H₂O(g)  CO₂(g) + H₂(g) + x k cal.

6. PCl₅(g)  PCl₃(g) + Cl₂(g) - X kcal.

7. N₂O₄(g)  2NO₂(g) - 14 kcal

Sol. 1. Low T, High P, excess of N₂ and H₂ .

2. Low T, High P, excess of SO₂ and O₂

3. High T, any P , excess of N₂ and O₂

4. Low T, High P, excess of NO and O₂

5. Low T, Low P, excess of C and H₂O

6. High T, low P, excess of PCl₅

7. High T, Low P, excess of N₂O₄

Simultaneous Equilibrium

A_(s)  B_(g) + C_(g)

D_(s)  B_(g) + E_(g)

 Applicable only when at least one of the product is common in both the reaction.

Ex. The pressure at equilibrium over solid A is 50 atm and over solid D is 68 atm if both solid A and D are heated simultaneously then find

(i) the total pressure over the solids.

(ii) In the above question find the mole ratio of C and B

(iii) mole fraction of C 

Sol. (i)

, 

, 

 ...(i)

 ...(ii)

total pressure = p_B + p_C + p_E = x+ y+ x +y = 2 (x+ y)

Also 

→ 

→ Total pressure = 2 (x + y) = 

(ii) At constant temp and volume P ≈ n

→ pressure Ratio will their mole ratio by eq. (i)/(ii) 

(iii) Mole fraction of C

As we know 

→  →  ,, 

→ mole fraction of C = 

Ex.6 A(s)  At eq., pressure = 18 atm

C(s)  At eq., pressure = 36 atm

Calculation

(i) total pressure at new equilibrium when both the solids are heated simultaneously.

(ii) mole ratio of B and D

(iii) mole fraction of B in the mixture.

Sol. A(s)  H₂S(g) + B(g), Kp₁ = (9)² = 81

                                x + y        x

C(s)  H₂S(g) + D(g), Kp₂ = (18)² = 324

                         x + y        y

total pressure = x + y + x + y = 2 (x + y)

Kp₁ = x(x + y)

Kp₂ = y(x + y)

→  = (x + y)² → x + y = 

→ total pressure = 2 (x + y) = 2  = 

(ii) mole ratio of B & D =  =  = 

(iii) mole fraction of B in mixture =  = 

Heterogeneous reactions: the vapor pressure of solid hydrates

Many common inorganic salts form solids which incorporate water molecules into their crystal structures.

These water molecules are usually held rather loosely and can escape as water vapor. Copper(II) sulfate, for example forms a pentahydrate in which four of the water molecules are coordinated to the Cu2 ion while the fifth is hydrogen-bonded to SO₄^--. This latter water is more tightly bound, so that the pentahydrate loses water in two stages on heating: CuSO₄.5H₂OCuSO₄.H₂OCuSO₄

These dehydration steps are carried out at the temperatures indicated above, but at any

temperature, some moisture can escape from a hydrate. For the complete dehydration of the

pentahydrate we can define an equilibrium constant CuSO₄.5H₂O(s) → CuSO₄(s)  + 5 H₂O(g) K_p = 1.14×10^-10

The vapor pressure of the hydrate (for this reaction) is the partial pressure of water vapor at which the two solids can coexist indefinitely; its value is K_p^1/5 atm. If a hydrate is exposed to air in which the partial pressure of water vapor is less than its vapor pressure, the reaction will proceed to the right and the hydrate will lose moisture. Vapor pressures always increase with temperature, so any of these compounds can be dehydrated by heating. Loss of water usually causes a breakdown in the structure of the crystal; this is commonly seen with sodium sulfate, whose vapor pressure is sufficiently large that it can exceed the partial pressure of water vapor in the air when the relative humidity is low. What one sees is that the well-formed crystals of the decahydrate undergo deterioration into a powdery form, a phenomenon known as efflorescence. When a solid is able to take up moisture from

the air, it is described as hygroscopic. A small number of anhydrous solids that have low vapor pressures not only take up atmospheric moisture on even the driest of days, but will become wet as water molecules are adsorbed onto their surfaces; this is most commonly observed with sodium hydroxide and calcium chloride. With these solids, the concentrated solution that results continues to draw in water from the air so that the entire crystal eventually dissolves into a puddle of its own making; solids exhibiting this behavior are said to be deliquescent.

One of the first hydrates to be investigated in detail was calcium sulfate hemihydrate (CaSO₄×1/2 H₂O) which LeChâtelier showed to be what forms when when the form of CaSO₄ known as plaster of Paris hardens; the elongated crystals of the hydrate bind themselves into a cement-like mass.

Ex. At what relative humidity will copper sulfate pentahydrate lose its waters of hydration when the air temperature is 30°C? What is K_p for this process at this temperature?

Sol. From the table, we see that the vapor pressure of the hydrate is 12.5 torr, which corresponds to a relative humidity of 12.5/31.6 = 0.40 or 40%. This is the humidity that will be maintained if the hydrate is placed in a closed container of dry air.

For this hydrate, Kp = , so the partial pressure of water vapor that will be in equilibrium with the hydrate and the dehydrated solid (remember that both solids must be present to have equilibrium!),expressed in atmospheres, will be (12.5/760)⁵= 1.20 x10^-9.

Relationship Between 

Free energy, G, denotes the self intrinsic electrostatic potential energy of a system. This means that in any molecule if we calcualte the total electrostatic potential energy of all the charges due to all the other charges , we get what is called the free energy of the molecule. It tells about the stability of a molecule with respect to another molecule. Lesser the free energy of a molecule more stable it is.Every reaction proceeds with a decrease in free energy.The free energy change in a process is expressed by ΔG.If it is negative, it means that product have lesser G than reactants, so the reaction goes forward.If it is positive the reaction goes reverse and if it is zero the reaction is at equilibrium.

ΔG is the free energy change at any given concentration of reactants and products . If all the reactants and products are taken at a concentration of 1 mole per liter, the free energy change of the reaction is called ΔG^o ( standard free energy change ).

One must understand that ΔG^o is not the free energy change at equilibrium . It is free energy change when all the reactants and products are at a concentration of 1 mole/L . ΔG^o is related to K (equilibrium constant ) by the relation , ΔG^o= -RT ln K.

K may either be kc or Kp.Accordingly we get . The units of DG^o depends only or R.

T is always is Kelvin, and if R is in Joules, ΔG^o will be in joules, and if R is calories than ΔG^o will be in calories.