Question

Describe the manufacture of ${{H}_{2}}S{{O}_{4}}$ by contact process. $$$$

Answer 1

Sulphuric acid (${{H}_{2}}{{S}_{2}}{{O}_{7}}$) is industrially manufactured by the Contact process. ${{H}_{2}}S{{O}_{4}}$is a colorless, dense, syrupy liquid. It is also known as oil of vitriol. The sulphuric acid obtained by the Contact process is free from arsenic impurities and is 96-98% pure.

Complete answer:
Manufacture of sulphuric acid by the Contact process involves the following three steps:
Burning of sulphur or roasting of iron pyrites ($Fe{{S}_{2}}$) to give sulphur dioxide ($S{{O}_{2}}$) Chemical reactions involved:
$\begin{aligned} & {{S}_{8}}+8{{O}_{2}}\to 8S{{O}_{2}} \\\ & 4Fe{{S}_{2}}+11{{O}_{2}}\to 2F{{e}_{2}}{{O}_{3}}+8S{{O}_{2}} \\\ \end{aligned}$
Oxidation of $S{{O}_{2}}$by air to sulphur trioxide ($S{{O}_{2}}$) in the presence of a catalyst vanadium pentoxide (${{V}_{2}}{{O}_{5}}$)
Chemical reaction involved:
$2S{{O}_{2}}(g)+{{O}_{2}}(g)\overset{{{V}_{2}}{{O}_{5}}}{leftrightarrows}2S{{O}_{3}}(g);\text{ }{{\Delta }_{f}}{{H}^{o}}=-196.6\text{ kJ mo}{{\text{l}}^{-1}}$
This is the most important step in the manufacture of${{H}_{2}}S{{O}_{4}}$. The above reaction is reversible. The favorable conditions for the maximum yield of $S{{O}_{3}}$ according to Le Chatelier’s principle are high pressure and low temperature.
High pressure: Since the forward reaction leads to a decrease in volume (or decrease in number of moles), therefore, high pressure will favor the reaction (in accordance with Boyle’s law:$P\propto {}^{1}/{}_{V}$ ).
Low temperature: Since the reaction involved is an exothermic reaction, low temperature will push the reaction forward resulting in higher yield of$S{{O}_{3}}$. However, the rate of reaction also depends on temperature, so the temperature has to be maintained at an optimum value of 720 K so that the reaction rate is not slowed down.

Absorption of $S{{O}_{3}}$ in ${{H}_{2}}S{{O}_{4}}$ to give oleum (${{H}_{2}}{{S}_{2}}{{O}_{7}}$)
$S{{O}_{3}}$ from the catalytic converter (present in the plant used for manufacturing) is absorbed by concentrated ${{H}_{2}}S{{O}_{4}}$to form oleum. Oleum is then diluted with water to get ${{H}_{2}}S{{O}_{4}}$ of desired concentration.
Chemical reactions involved:
$\begin{aligned} & S{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\to {{H}_{2}}{{S}_{2}}{{O}_{7}} \\\ & {{H}_{2}}{{S}_{2}}{{O}_{7}}+{{H}_{2}}O\to 2{{H}_{2}}S{{O}_{4}} \\\ \end{aligned}$

Additional Information: The plant used in the Contact process for manufacturing of ${{H}_{2}}S{{O}_{4}}$has four main parts, namely, sulphur burner where oxidation of sulphur ores is carried out, purifying unit where any dust or arsenic impurities are removed, catalytic converter and absorption tower. Sulphuric acid is used for the manufacture of a number of compounds like fertilizers (e.g. superphosphate, ammonium sulphate, etc), dyes, paints and nitrocellulose products.

Note: Take special caution while writing and balancing the chemical reactions. It is to be noted that$S{{O}_{3}}$ is never directly absorbed in water to produce ${{H}_{2}}S{{O}_{4}}$ but instead first oleum is produced and then dilution of oleum gives desired${{H}_{2}}S{{O}_{4}}$. It is because of the fact that $S{{O}_{3}}$in water forms a dense fog of ${{H}_{2}}S{{O}_{4}}$which is not easily condensed.