8.7 Industrial Application of Le Chatelier's Principle

In industrial chemistry, the goal is to maximize the yield of the desired product while minimizing costs and leftover reactants. Le Chatelier's principle, combined with an understanding of reaction kinetics, is crucial for designing optimal conditions to achieve high, economically viable yields.

8.7.1 Synthesis of Ammonia by Haber's Process

The Haber-Bosch process is a cornerstone of the chemical industry, primarily used for producing ammonia ($\mathrm{N}{\mathrm{H}}_3$), a key component of fertilizers. The reversible reaction is:

${\mathrm{N}}_{2(g)}+3{\mathrm{H}}_{2(g)}\rightleftharpoons 2{\mathrm{NH}}_{3(g)}\Delta H^{\circ }=-92.46\text{kJ}$

This equation reveals two critical pieces of information for applying Le Chatelier's principle:

1. The reaction is exothermic ($\Delta H^{\circ }$ is negative), meaning heat is released.
2. The reaction proceeds with a decrease in the number of moles of gas (4 moles of reactants → 2 moles of product).

Based on these facts, we can determine the optimal conditions to maximize the yield of ammonia.

i. Low Temperature

Principle: Since the forward reaction is exothermic, decreasing the temperature will shift the equilibrium to the right, favoring the production of ammonia ($\mathrm{N}{\mathrm{H}}_3$).

Industrial Practice: While a very low temperature would give the highest yield at equilibrium, the rate of reaction becomes extremely slow. Therefore, a compromise temperature of $400^{\circ }\mathrm{C}$ is used along with a catalyst to achieve a reasonable reaction rate.

ii. High Pressure

Principle: The forward reaction involves a decrease in the number of gas molecules (from 4 to 2). According to Le Chatelier's principle, increasing the pressure will shift the equilibrium to the side with fewer moles to relieve the stress. This favors the formation of ammonia.

Industrial Practice: A high pressure of 200–300 atm is used. This significantly increases the yield of ammonia. The equilibrium mixture under these conditions contains about 35% $\mathrm{N}{\mathrm{H}}_3$ by volume.

iii. Continual Removal of Ammonia

Principle: Removing a product as it is formed will shift the equilibrium to the right to replenish the product.

Industrial Practice: This is a highly effective method. The gaseous mixture is cooled in refrigeration coils to $-33.4^{\circ }\mathrm{C}$, the boiling point of ammonia. At this temperature, $\mathrm{N}{\mathrm{H}}_3$ liquefies and is removed. The unreacted ${\mathrm{N}}_2$ and ${\mathrm{H}}_2$ remain as gases and are recycled back into the reaction chamber. This continuous removal of the product drives the reaction forward, allowing for a practical conversion of nearly 100%.

Catalyst: A catalyst does not affect the position of the equilibrium, but it greatly increases the rate at which equilibrium is reached. In the Haber process, a catalyst of iron (Fe) with promoters $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$ and ${\mathrm{K}}_2\mathrm{O}$ is used. The promoters enhance the activity and stability of the iron catalyst.

8.7.2 Synthesis of Sulphuric Acid by Contact Process

The Contact process is the modern industrial method for producing sulfuric acid (${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$). It involves several steps, with one key equilibrium reaction.

Steps in the Contact Process

1. Formation of Sulphur Dioxide Sulphur is burned in air to produce sulphur dioxide. ${\mathrm{S}}_{(s)}+{\mathrm{O}}_{2(g)}\to {\mathrm{SO}}_{2(g)}$

2. Oxidation of Sulphur Dioxide (The Key Equilibrium Step) Sulphur dioxide is catalytically oxidized to sulphur trioxide. This is the reversible reaction governed by Le Chatelier's principle. $2{\mathrm{SO}}_{2(g)}+{\mathrm{O}}_{2(g)}\rightleftharpoons 2{\mathrm{SO}}_{3(g)}\Delta H^{\circ }=-198\text{kJ}$

3. Conversion to Sulphuric Acid Sulphur trioxide is absorbed in concentrated sulfuric acid to form oleum (${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_7$), which is then diluted with water to produce sulfuric acid. $\mathrm{S}{\mathrm{O}}_3(\mathrm{g})+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{l})\to {\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_7(\mathrm{l})$ ${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_7(\mathrm{l})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{l})$

Optimizing the Yield of Sulphur Trioxide ($\mathrm{S}{\mathrm{O}}_3$)

The key equilibrium reaction has the following characteristics:

• It is exothermic ($\Delta H^{\circ }$ is negative).
• It proceeds with a decrease in the number of moles of gas (3 moles of reactants → 2 moles of product).

a) Low Temperature

Principle: Being an exothermic reaction, a low temperature will favor the forward reaction, increasing the yield of $\mathrm{S}{\mathrm{O}}_3$.

Industrial Practice: Similar to the Haber process, a very low temperature makes the reaction too slow. A compromise temperature of $450–500^{\circ }\mathrm{C}$ is used with a vanadium pentoxide (${\mathrm{V}}_2{\mathrm{O}}_5$) catalyst.

b) High Pressure

Principle: The forward reaction leads to fewer moles of gas (3 → 2). High pressure will shift the equilibrium to the right, increasing the yield of $\mathrm{S}{\mathrm{O}}_3$.

Industrial Practice: High-pressure equipment is expensive to build and maintain. Since the reaction already gives a very high yield (about 98%) at atmospheric pressure, only a slightly elevated pressure of 2 atm is used as a cost-effective compromise.

c) Adding Excess Oxygen

Principle: Increasing the concentration of a reactant (${\mathrm{O}}_2$) will shift the equilibrium to the right to consume the added reactant, thus increasing the yield of $\mathrm{S}{\mathrm{O}}_3$.

Industrial Practice: Excess air (oxygen) is supplied to the reaction chamber. This shifts the equilibrium: $2\mathrm{S}{\mathrm{O}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})\rightleftharpoons 2\mathrm{S}{\mathrm{O}}_3(\mathrm{g})$ to the right, increasing the conversion of $\mathrm{S}{\mathrm{O}}_2$ to $\mathrm{S}{\mathrm{O}}_3$. Additionally, $\mathrm{S}{\mathrm{O}}_3$ is continuously removed by absorption into concentrated ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$, which further drives the forward reaction. This combination of excess ${\mathrm{O}}_2$ and continuous product removal achieves a yield of approximately 98%.

Summary Table: Industrial Conditions